A smooth plane is inclined at an angle \(\theta\) to horizontal ground, where \(\sin\theta = \dfrac{1}{3}\)
The points \(A\), \(B\) and \(C\) lie on a line of greatest slope of the plane, with \(B\) between \(A\) and \(C\), with \(A\) above \(B\) and \(AC = 3a\), as shown in Figure 3.
A light elastic string of natural length \(2a\) has one end attached to the fixed point \(A\). The other end of the string is attached to a parcel \(P\) of mass \(m\).
The modulus of elasticity of the string is \(\dfrac{8}{3}mg\)
The parcel rests in equilibrium on the plane at the point \(B\).
The parcel is modelled as a particle.
(a) Show that \(AB = \dfrac{9}{4}a\) (4)
The parcel is now held at \(C\) and released from rest.
(b) Find the elastic potential energy lost by the string when \(P\) moves from \(C\) to \(B\). (3)
(c) Use the principle of conservation of mechanical energy to find the speed of \(P\) at the instant it passes \(B\). (3)
Mark scheme (a)
Scheme
Marks
AO
Correct use of Hooke’s law \(T = \dfrac{\frac{8}{3}mge}{2a}\) or \(T = \dfrac{\frac{8}{3}mg(AB - 2a)}{2a}\)
\(\Rightarrow e = \dfrac{2a}{8}\left(= \dfrac{a}{4}\right)\), \(AB = \dfrac{9}{4}a\) * OR \(\Rightarrow AB - 2a = \dfrac{2a}{8}\left(= \dfrac{a}{4}\right)\), \(AB = \dfrac{9}{4}a\) *
A1*
2.2a
(4)
Notes
M1: Correct use of Hooke’s law with \(\dfrac{8mg}{3}\) and \(2a\) substituted.
M1: Resolve parallel to the slope with all required terms and no extras. Dimensionally correct. Weight must be resolved but condone sin/cos confusion. \(T\) does not need to be replaced.
A1: Correct unsimplified equilibrium equation, \(T\) does not need to be replaced.
A1*: A complete and correct method using Hooke’s law with the parallel equilibrium equation to obtain the given answer. There must be at least one line of working between the initial equations and the given answer. Must see ‘\(AB = \ldots\)’ Accept fractions \(\dfrac{9a}{4}\) or \(\dfrac{9}{4}a\).
M1: Correct method for difference in EPE at \(B\) and \(C\), allow either way round. Dimensionally correct and of the correct structure. No need to substitute \(\lambda\). For M mark, condone denominator of \(2a\). Must use extensions \((3a - 2a)\) and \(\left(\dfrac{9a}{4} - 2a\right).\)
A1: Correct unsimplified expression for change in EPE. Allow \(\pm\)
A1: Correct answer, o.e. ISW. Accept \(0.625mga\) and \(0.63mga\). Must be positive but allow a negative expression to change to a positive expression without justification.
M1: Use of conservation of energy principle from \(C\) to \(B\) to form an equation with all terms of the correct structure and dimensionally correct. Condone sign errors. Condone sin/cos confusion on vertical height. For GPE, \(mg\left(3a - \dfrac{9a}{4}\right)\sin\theta\) o.e for example \(mg\dfrac{3a}{4}\sin\theta,\ \ mg\dfrac{3a}{4}\left(\dfrac{1}{3}\right),\ \ mg\dfrac{a}{4}\) For EPE change, may use their answer from (b) if dimensionally correct (of the form \(kma\) where \(k\) is a constant) or start again with 2 EPE terms: \(\dfrac{\frac{8mg}{3}(3a - 2a)^2}{2(2a)}\) and \(\dfrac{\frac{8mg}{3}\left(\frac{9}{4}a - 2a\right)^2}{2(2a)}\). For M mark, condone denominator of \(2a\). (Corrected from the printed mark scheme: the second EPE term is printed with \(\left(3a - \frac{9}{4}a\right)^2\); the extension at \(B\) is \(\frac{9}{4}a - 2a\).)
A1ft: Correct unsimplified equation. Terms must be correct when the equation is formed but follow their answer to (b) for EPE if used.
A1: Correct answer in terms of \(\sqrt{ag}\) ISW. Accept eg \(\dfrac{1}{2}\sqrt{3ag}\), \(0.87\sqrt{ag}\) or better. N.B. If the final mark in (b) is A0 due to substituting \(g = 9.8\ \text{m s}^{-2}\), do not penalise again for the same reason here.
A rough straight ramp is fixed to horizontal ground. The ramp is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{14}\). The points \(A\) and \(B\) are on a line of greatest slope of the ramp with \(AB = 18\) m and \(B\) above \(A\), as shown in Figure 1.
A package of mass 0.5 kg is projected up the ramp from \(A\) with speed \(8\ \text{m s}^{-1}\) and comes to instantaneous rest at \(B\).
The work done against friction as the package moves from \(A\) to \(B\) is \(W\) joules.
The package is modelled as a particle and air resistance is ignored.
(a) Use the work-energy principle to show that \(W = 9.7\) (3)
The coefficient of friction between the package and the ramp is \(\mu\)
M1: Form work-energy equation in terms of \(W\) (and \(g\) and \(\theta\)) only. All required terms present and no extras. All terms dimensionally correct (of correct structure). Condone \(\pm\) sign errors on terms and sin/cos confusion on vertical height. M0 if a term is missing or for incorrect trig use eg \(18\tan\theta,\ \dfrac{18}{\sin\theta},\ \dfrac{18}{\cos\theta}\). M0 for use of suvat
A1: Correct unsimplified equation, no need to replace trig.
A1*: Obtain given answer from complete and correct working. Must see a line of working between the initial equation and the given answer. Condone missing \(W\) during working but must see ‘\(W = \ldots\)’ for the final mark.
Mark scheme (b)
Scheme
Marks
AO
Use of \(F = \mu R = \mu \times 0.5g\cos\theta\)
M1
3.1b
Complete method to form a dimensionally correct equation in \(\mu\) (and \(\theta\)) using
Either work done \(= F \times 18\)
Or suvat and N2L
M1
3.4
Either \(9.7 = \mu \times 0.5g\cos\theta \times 18\)
Or relevant suvat to find \(a\) \(\left(= -\dfrac{16}{9}\right)\) and N2L to form \(-\mu \times 0.5g\cos\theta - 0.5g\sin\theta = 0.5\left(-\dfrac{16}{9}\right)\)
A1
1.1b
\(\mu = 0.11 \quad (0.110)\)
A1
1.1b
(4)
(7 marks)
Notes
M1: Correct use of \(F = \mu R\) and \(R = 0.5g\cos\theta\) to form an expression for Friction. Dimensionally correct. Condone sin/cos confusion. Missing \(g\) is an accuracy error not a method error.
M1: Complete method to form a dimensionally correct equation in \(\mu\) (\(\theta\) and \(g\)). Trig does not need to be replaced for M mark. M0 for \(W = \mu R\). May use 9.7 and work done \(= F \times 18\) to form a dimensionally correct equation in \(\mu\). May see relevant suvat to find acceleration, followed by N2L to form dimensionally correct equation \(\mu\). N2L must contain all relevant terms and no extras. Condone sin/cos confusion on the weight components.
A1: Correct unsimplified equation in \(\mu\) (and \(g\)) with trig replaced correctly.
A1: 2 sf or 3 sf only. A0 for use of \(g = 9.81\ \text{m s}^{-2}\)
(Corrected from the printed mark scheme: the N2L equation in the second method is printed as \(-\mu \times 0.5g\cos\theta \times 18 - 0.5g\sin\theta = 0.5\left(-\dfrac{16}{9}\right)\); the friction force has no factor of 18 in an equation of motion.)
3. A plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \dfrac{4}{3}\). A small block of mass \(m\) is held at a point \(A\) on the plane and released from rest.
Initially the block is modelled as a particle, air resistance is modelled as being negligible and the plane is modelled as being smooth.
(a) Using the model and the principle of conservation of mechanical energy, find the distance of the block from \(A\) at the instant when the speed of the block is \(7\ \text{m s}^{-1}\) (3)
In a refined model, the block is again modelled as a particle and air resistance is modelled as being negligible, but the plane is modelled as being rough with the coefficient of friction between the block and the plane being \(\dfrac{2}{3}\)
Using the refined model,
(b) find, in terms of \(m\) and \(g\), the magnitude of the frictional force acting on the block as it slides down the plane, (3)
(c) find, using the work-energy principle, the distance of the block from \(A\) at the instant when the speed of the block is \(7\ \text{m s}^{-1}\) (4)
Mark scheme (a)
Scheme
Marks
AO
Use of conservation of energy principle: \(\dfrac{1}{2}m \times 7^2 = mgh\)
M1
3.4
\(\dfrac{1}{2}m \times 7^2 = mgd\sin\alpha\)
A1
1.1b
\(d = 3.1\) or 3.13 (m)
A1
1.1b
(3)
Notes
N.B. Only penalise overaccuracy in answer to (a) or fractional answers to (a) and (c) ONCE for the question.
Consistent use of \(\sin\alpha = \dfrac{3}{5}\) and \(\cos\alpha = \dfrac{4}{5}\) could be treated as a MR of \(\tan\alpha = \dfrac{4}{3}\) for \(\tan\alpha = \dfrac{3}{4}\), if there is no evidence to the contrary e.g. a correct triangle, and leads to (a) \(d = 4.17\) or 4.2 (m) (b) \(\dfrac{8mg}{15}\) (c) \(x = 37.5\) or 38 (m). It can score MAX (a) M1A1A0 (b) M1M1A0 (c) B1M1A1A1
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, allow consistent missing \(m\)’s. N.B. \(h\) does not need to be substituted.
A1: A correct equation in (\(m\)), \(d\) and \(\alpha\), seen or implied. N.B. Could be e.g. \(2.5 = d\sin\alpha\) if they find \(h\) first.
A1: Either answer. 25/8 is A0 as is \(245/8g\)
Mark scheme (b)
Scheme
Marks
AO
Resolve perpendicular to the plane: \(R = mg\cos\alpha\)
M1
3.1b
Use of \(F = \dfrac{2}{3}R\)
M1
1.2
\(F = \dfrac{2}{5}mg\) or \(\dfrac{6}{15}mg\) or \(0.4mg\) (must be in terms of \(m\) and \(g\))
A1
1.1b
(3)
Notes
N.B. Only penalise overaccuracy in answer to (a) or fractional answers to (a) and (c) ONCE for the question.
Consistent use of \(\sin\alpha = \dfrac{3}{5}\) and \(\cos\alpha = \dfrac{4}{5}\) could be treated as a MR of \(\tan\alpha = \dfrac{4}{3}\) for \(\tan\alpha = \dfrac{3}{4}\), if there is no evidence to the contrary e.g. a correct triangle, and leads to (a) \(d = 4.17\) or 4.2 (m) (b) \(\dfrac{8mg}{15}\) (c) \(x = 37.5\) or 38 (m). It can score MAX (a) M1A1A0 (b) M1M1A0 (c) B1M1A1A1
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion.
M1: Use of \(F = \dfrac{2}{3}R\)
A1: cao
Mark scheme (c)
Scheme
Marks
AO
WD against friction \(= Fx\) or \(F\left(\dfrac{h}{\sin\alpha}\right)\)
B1
3.4
Use of work-energy principle: \(mgh - \dfrac{1}{2}m \times 7^2 = \dfrac{2}{5}mgx\)
N.B. Only penalise overaccuracy in answer to (a) or fractional answers to (a) and (c) ONCE for the question.
Consistent use of \(\sin\alpha = \dfrac{3}{5}\) and \(\cos\alpha = \dfrac{4}{5}\) could be treated as a MR of \(\tan\alpha = \dfrac{4}{3}\) for \(\tan\alpha = \dfrac{3}{4}\), if there is no evidence to the contrary e.g. a correct triangle, and leads to (a) \(d = 4.17\) or 4.2 (m) (b) \(\dfrac{8mg}{15}\) (c) \(x = 37.5\) or 38 (m). It can score MAX (a) M1A1A0 (b) M1M1A0 (c) B1M1A1A1
B1: Seen or implied, \(F\) does not need to be substituted but must be \(F\) not \(R\).
M1: Correct no. of terms, dimensionally correct, condone sin/cos confusion and sign errors, allow consistent missing \(m\)’s. N.B. Allow if they clearly make a slip and use \(Rd\) instead of \(Fd\) for WD against friction. N.B. M0 if they use \(h\) from part (a) or any other numerical value for \(h\).
A1: Correct equation in (\(m\)), \(x\) and \(\alpha\), seen or implied. N.B. Could be e.g. \(5 = x\sin\alpha\) if they find \(h\) first.
Figure 1 shows part of the end elevation of a building which sits on horizontal ground. The side of the building is vertical and has height \(h\).
A small stone of mass \(m\) is at rest on the roof of the building at the point \(A\). The stone slides from rest down a line of greatest slope of the roof and reaches the edge \(B\) of the roof with speed \(\sqrt{2gh}\)
The stone then moves under gravity before hitting the ground with speed \(W\).
In a model of the motion of the stone from \(\boldsymbol{B}\) to the ground
the stone is modelled as a particle
air resistance is ignored
Using the principle of conservation of mechanical energy and the model,
(a) find \(W\) in terms of \(g\) and \(h\). (4)
In a model of the motion of the stone from \(\boldsymbol{A}\) to \(\boldsymbol{B}\)
the stone is modelled as a particle of mass \(m\)
air resistance is ignored
the roof of the building is modelled as a rough plane inclined to the horizontal at an angle \(\theta\), where \(\tan\theta = \dfrac{3}{4}\)
the coefficient of friction between the stone and the roof is \(\dfrac{1}{3}\)
\(AB = d\)
Using this model,
(b) find, in terms of \(m\) and \(g\), the magnitude of the frictional force acting on the stone as it slides down the roof, (3)
(c) use the work–energy principle to find \(d\) in terms of \(h\). (5)
Mark scheme (a)
Scheme
Marks
AO
Use the principle of conservation of mechanical energy and model
2. A rough plane is inclined to the horizontal at an angle \(\theta\), where \(\tan\theta = \dfrac{3}{4}\)
A particle \(P\) of mass \(m\) is at rest at a point on the plane.
The particle is projected up the plane with speed \(\sqrt{2ag}\)
The particle moves up a line of greatest slope of the plane and comes to instantaneous rest after moving a distance \(d\).
The coefficient of friction between \(P\) and the plane is \(\dfrac{1}{7}\)
(a) Show that the magnitude of the frictional force acting on \(P\) as it moves up the plane is \(\dfrac{4mg}{35}\) (3)
Air resistance is assumed to be negligible.
Using the work-energy principle,
(b) find \(d\) in terms of \(a\). (4)
Mark scheme (a)
Scheme
Marks
AO
Resolve perpendicular to the plane and use \(F = \mu R\)
M1
3.1a
\(\tfrac{1}{7}mg\cos\theta\)
A1
1.1b
\(\dfrac{1}{7}mg \times \dfrac{4}{5} = \dfrac{4mg}{35}\) * or \(\dfrac{4}{35}mg\) *
A1*
1.1b
(3)
Notes
M1: Resolve perpendicular to the plane to find an expression for \(R\) and use \(\mu R\). Condone sin/cos confusion on weight component. All required terms present and no extras. Dimensionally correct.
A1: Correct unsimplified expression for friction. Allow with \(\cos\theta\) or \(\tfrac{4}{5}\)
A1*:Given answer correctly obtained. Working out must include both \(\tfrac{1}{7}\) and \(\tfrac{4}{5}\) in the same line before reaching the given answer.
M1: Use of work-energy principle with correct number of terms: 1 work, 1 KE, 1 GPE. Condone \(\pm\) sign errors on terms. All required terms present and no extras. Resolve only when required and condone cos/sin confusion for the method mark. Must use given answer from (a) in work term. M0 if Friction is not multiplied by distance (dimensionally incorrect) M0 if a term is missing. M0 for incorrect use of trig eg \(d\tan\theta,\ \dfrac{d}{\sin\theta},\ \dfrac{d}{\cos\theta}\ \ldots\)
A1: Correct equation with at most one error
A1: Correct equation
A1: Correct answer for \(d\). Any equivalent fraction or decimal multiple of \(a\)
4. A light elastic string has natural length \(2a\) and modulus of elasticity \(4mg\).
One end of the elastic string is attached to a fixed point \(O\). A particle \(P\) of mass \(m\) is attached to the other end of the elastic string.
The particle \(P\) hangs freely in equilibrium at the point \(E\), which is vertically below \(O\)
(a) Find the length \(OE\). (4)
Particle \(P\) is now pulled vertically downwards to the point \(A\), where \(OA = 4a\), and released from rest. The resistance to the motion of \(P\) is a constant force of magnitude \(\dfrac{1}{4}mg\).
(b) Find, in terms of \(a\) and \(g\), the speed of \(P\) after it has moved a distance \(a\). (7)
Particle \(P\) is now held at \(O\)
Particle \(P\) is released from rest and reaches its maximum speed at the point \(B\).
The resistance to the motion of \(P\) is again a constant force of magnitude \(\dfrac{1}{4}mg\).
(c) Find the distance \(OB\). (4)
Mark scheme (a)
Scheme
Marks
AO
\(T = \dfrac{4mge}{2a}\)
B1
3.3
\(T = mg\)
M1
3.1a
\(e = \dfrac{1}{2}a\)
A1
1.1b
\(OE = \dfrac{5a}{2}\)
A1
1.1b
(4)
Notes
B1: Hooke’s Law seen with \(4mg\) and \(2a\) substituted.
M1: Resolving vertically. Correct number of terms.
A1: cao for extension.
A1: cao for \(OE\). Note that if the extension is \((OE - 2a)\) in their equation, \(OE\) can be found directly and both A’s can be earned together.
Mark scheme (b)
Scheme
Marks
AO
GPE term, \(\pm\, mga\)
B1
3.4
Work done against resistance, \(\pm\, \dfrac{1}{4}mga\)
B1: Work term seen \(\dfrac{mga}{4}\), ignore sign. Allow B1 for the case where WD = \(\dfrac{5mga}{4}\). This is a special case where the work done against resistance is included within the term. \(\dfrac{5mga}{4}\) = WD against resistance + WD against weight.
M1: Use of EPE formula. Accept EPE in the form \(\dfrac{\lambda x^2}{ka}\)
A1: Difference between two correct EPE terms seen, unsimplified.
M1: Work-energy equation is formed with all relevant terms and no extras.: KE, GPE, 2EPE, WD. Condone sign errors. M0: For work-energy equation with WD = \(\dfrac{5mga}{4}\) and a GPE term. This is because weight is considered twice and so the equation contains an extra term.
A1: Correct unsimplified equation
A1: Correct answer in terms of \(a\) and \(g\), do not allow 9.8 for \(g\) \(\quad v = \sqrt{\dfrac{7ag}{2}}\), \(\ v = \dfrac{1}{2}\sqrt{14ag}\)
Mark scheme (c)
Scheme
Marks
AO
\(mg - T - \dfrac{1}{4}mg = 0\)
M1
3.1a
\(mg - \dfrac{4mgx}{2a} - \dfrac{1}{4}mg = 0\)
A1
1.1b
\(x = \dfrac{3a}{8}\)
A1
1.1b
\(OB = \dfrac{19a}{8}\) oe
A1
1.1b
(4)
(15 marks)
Notes
M1: Vertical equilibrium equation or equation of motion with \(a = 0\). Condone sign errors. Correct no. of terms - all 3 forces must be included although \(\left(mg \pm \dfrac{mg}{4}\right)\) may already be simplified. Hooke’s Law does not need to be substituted but M0 if the equilibrium position from (a) is used.
A1: Correct equation in one unknown.
A1: cao
A1: cao Note that if the extension is \((OB - 2a)\) in their equation, \(OB\) can be found directly and both A’s can be earned together.
4(c) Alt 1: Using differentiation with a Work - energy equation from the point of release
M1: Forming work-energy equation with the usual rules: all relevant terms to be included and of the correct form and no extra terms. \(\dfrac{1}{2}mv^2 = mgh - \dfrac{4mg(h - 2a)^2}{2(2a)} - \dfrac{mgh}{4}\)
A1: Correct equation for \(v^2\) and \(h\) (may use a different letter)
A1: Correct equation after differentiating \(v^2\) or \(v\) with respect to \(h\) and setting it equal to zero. \(\dfrac{\mathrm{d}}{\mathrm{dh}}\left(v^2\right) = 0 \quad \rightarrow \quad \dfrac{3g}{2} = \dfrac{4g(h - 2a)}{a}\) oe
3. A stone of mass 0.5 kg is projected vertically upwards with a speed \(U\ \text{m s}^{-1}\) from a point \(A\). The point \(A\) is 2.5 m above horizontal ground.
The speed of the stone as it hits the ground is \(25\ \text{m s}^{-1}\)
The motion of the stone from the instant it is projected from \(A\) until the instant it hits the ground is modelled as that of a particle moving freely under gravity.
(a) Use the model and the principle of conservation of mechanical energy to find the value of \(U\). (4)
In reality, the stone will be subject to air resistance as it moves from \(A\) to the ground.
(b) State how this would affect your answer to part (a). (1)
The ground is soft and the stone sinks a vertical distance \(d\) cm into the ground. The resistive force exerted on the stone by the ground is modelled as a constant force of magnitude 2000 N and the stone is modelled as a particle.
(c) Use the model and the work-energy principle to find the value of \(d\), giving your answer to 3 significant figures. (5)
Mark scheme (a)
Scheme
Marks
AO
Attempt at use of conservation of energy principle
M1: Correct no. of terms (two KE and one PE), dimensionally correct equation, condone sign errors \(m\) does not need to be substituted and allow cancelled \(m\)’s. N.B. M0 if clearly using \(v^2 = u^2 + 2as\) for the whole motion. M0 if they use \(0.5g \times (2.5 + 0.01d)\) since extra term
A1: Correct equation in \(U\) only with at most one error
7. A spring of natural length \(a\) has one end attached to a fixed point \(A\). The other end of the spring is attached to a package \(P\) of mass \(m\). The package \(P\) is held at rest at the point \(B\), which is vertically below \(A\) such that \(AB = 3a\). After being released from rest at \(B\), the package \(P\) first comes to instantaneous rest at \(A\). Air resistance is modelled as being negligible.
By modelling the spring as being light and modelling \(P\) as a particle,
(a) show that the modulus of elasticity of the spring is \(2mg\) (5)
(b)
(i) Show that \(P\) attains its maximum speed when the extension of the spring is \(\dfrac{1}{2}a\)
(ii) Use the principle of conservation of mechanical energy to find the maximum speed, giving your answer in terms of \(a\) and \(g\).
(6)
In reality, the spring is not light.
(c) State one way in which this would affect your energy equation in part (b). (1)
Mark scheme (a)
Scheme
Marks
AO
EPE at \(A = \dfrac{\lambda a^2}{2a}\) or EPE at \(B = \dfrac{\lambda(2a)^2}{2a}\)
M1: Correct method for EPE seen or implied Need something of the form \(\dfrac{1}{2}kx^2\) where \(k = \dfrac{\lambda}{a}\) Must be using the formula for EPE correctly at least once
M1: Require all terms. Dimensionally correct. Condone their EPE. Condone sign errors
A1 A1: Unsimplified equation with at most one error. A repeated error in EPE formula is one error Correct unsimplified equation.
A1*: Obtain given answer from correct working
Mark scheme (b)
Scheme
Marks
AO
Extension at equilibrium:
M1
2.1
\(\dfrac{2mgx}{a} = mg \Rightarrow x = \dfrac{a}{2}\) *
M1: Use correct method for tension to find the extension at equilibrium. Need to see the formula for tension used. Allow verification with an appropriate conclusion If they use SHM they must use \(F = ma\) to prove that \(P\) is moving with SHM, otherwise 0/2.
A1*: Correct answer from correct work Allow verification with an appropriate conclusion
M1: Use given \(x\) to form work-energy equation. Need all terms, and dimensionally correct. Condone sign errors. Accept with values of \(\lambda\) and \(x\) not substituted
A1 A1: Unsimplified equation with at most one error. Need given \(\lambda\) and given \(x\) substituted at some point. A repeated error in the formula for EPE is one error. Correct unsimplified equation with given \(\lambda\) and given \(x\) substituted at some point
A1: Use correct method for tension to find the extension at equilibrium. Any equivalent form. \(2.1\sqrt{ag}\) or better
Alternative for the first M1A1
Scheme
Marks
AO
Use the work-energy equation to obtain \(\dfrac{\mathrm{d}V^2}{\mathrm{d}x}\) and set the derivative equal to zero
Alt: M1: Or an equivalent method for finding the turning point of a quadratic
Alt: A1*: Correct answer from correct work
Mark scheme (c)
Scheme
Marks
AO
e.g. for B1 Need to include the GPE of the spring The extension of the spring at equilibrium will be different The spring will have KE You would need to include the KE of the spring in the energy equation You would need to include the GPE of the spring in the energy equation The GPE of the system changes It would take work to raise the spring so the package would have less KE If the spring has mass then GPE of the spring would need to be included
B1
3.5b
(1)
(12 marks)
Notes
B1: Any valid response. B0 if answer includes an additional incorrect factor. Must be specific e.g. not just “the GPE changes”, but the GPE of the system changes is OK. Must relate to an effect on the energy equation E.g. for B0 The extension changes \(AB\) will increase The tension/energy/GPE/work done etc would increase The KE/GPE/EPE/acceleration/extension/velocity changes The mass of the spring would drag down and the EPE would change The EPE/KE/GPE etc would be variable There would be tension in the spring as well It has weight The velocity would decrease as energy is converted
Two blocks, \(A\) and \(B\), of masses 2 kg and 4 kg respectively are attached to the ends of a light inextensible string.
Initially \(A\) is held on a fixed rough plane. The plane is inclined to horizontal ground at an angle \(\theta\), where \(\tan\theta = \dfrac{3}{4}\)
The string passes over a small smooth light pulley \(P\) that is fixed at the top of the plane. The part of the string from \(A\) to \(P\) is parallel to a line of greatest slope of the plane.
Block \(A\) is held on the plane with the distance \(AP\) greater than 3 m. Block \(B\) hangs freely below \(P\) at a distance of 3 m above the ground, as shown in Figure 4.
The coefficient of friction between \(A\) and the plane is \(\mu\)
Block \(A\) is released from rest with the string taut.
By modelling the blocks as particles,
(a) find the potential energy lost by the whole system as a result of \(B\) falling 3 m. (3)
Given that the speed of \(B\) at the instant it hits the ground is \(4.5\ \text{m s}^{-1}\) and ignoring air resistance,
(b) use the work-energy principle to find the value of \(\mu\) (6)
After \(B\) hits the ground, \(A\) continues to move up the plane but does not reach the pulley in the subsequent motion. Block \(A\) comes to instantaneous rest after moving a total distance of \((3 + d)\) m from its point of release.
Ignoring air resistance,
(c) use the work-energy principle to find the value of \(d\) (4)
Work done against friction \(= 3 \times F_{\max}\ (= 47.04\mu)\)
B1ft
3.4
Work-energy equation: their GPE lost = their KE gained + their WD against friction
M1
3.4
\(82.32 = 60.75 + 47.04\mu\)
A1
1.1b
\(\mu = 0.459\ (0.46)\)
A1
1.1b
(6)
Notes
B1: Gain in KE for the system (not just for one block)
B1: Correct unsimplified expression for \(F_{\max}\) seen or implied
B1ft: Correct expression for work done: follow their \(F_{\max}\). This is dependent on them having found an expression for \(F_{\max}\)
M1: Complete method using work-energy to form an equation in \(\mu\). Require all terms (needs to consider the KE and GPE of both blocks). Dimensionally correct. Condone sign errors.
A1: Correct unsimplified equation in \(\mu\)
A1: 3 sf or 2 sf only
NB: It is possible to find the value of \(\mu\) by finding the tension in the string and forming a work-energy equation for particle \(B\), but in this case the first B1 is for KE of \(B\) and correct tension (25.7(N)) B1 for \(F_{\max}\) B1ft is for work done by the tension in the string and against friction M1 for \(3 \times 25.7 = 20.25 + 35.28 + 3 \times 15.68\mu\) O.E.
M1: All terms required. Dimensionally correct. Condone sign errors and sin / cos confusion. If the equation uses \(d + 3\) in place of \(d\) in the PE term it is correct if it also includes a term for the initial PE. If the equation uses \(d + 3\) in place of \(d\) in the term for work done then it scores M0.
A1 A1: Unsimplified equation in \(d\) and \(\mu\) with at most one error Correct unsimplified equation in \(d\) and \(\mu\) The ft is on their \(\mu\) if they have substituted a value.
\(v = 24\ (\text{m s}^{-1})\) (23.99895831… = 24 to 2SF if \(g = 9.81\))
A1
1.1b
(4)
Notes
N.B. If consistent use of a specific value of \(m\), allow all the marks but deduct the final A mark in each part but allow full marks if \(m\)’s have been cancelled or don’t appear.
B1: Seen anywhere
M1: Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion M0 for non-energy methods. Allow max M1A0A0 if 25/6 not resolved or not resolved correctly in PE term
A1: Correct equation in \(m\), \(g\), \(v\) and \(\alpha\)
\(v = 23.2\) or 23 \((\text{m s}^{-1})\) (23.16700… = 23.2 or 23 to 3SF or 2SF if \(g = 9.81\))
A1
1.1b
(8)
(12 marks)
Notes
N.B. If consistent use of a specific value of \(m\), allow all the marks but deduct the final A mark in each part but allow full marks if \(m\)’s have been cancelled or don’t appear.
M1: Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion
A1: Correct equation
B1: Seen anywhere
B1: Seen anywhere
M1: Correct no. of terms, dimensionally correct, condone sign errors and sin/cos confusion M0 for non work-energy methods Allow max M1A1A0A0 if 25/6 not resolved or not resolved correctly in PE term
A1: Equation in \(m\), \(g\), \(v\) and \(\alpha\) with at most one error N.B. If KE terms reversed, only penalise ONCE.
A1: Correct equation in \(m\), \(g\), \(v\) and \(\alpha\)
A light elastic spring has natural length \(3l\) and modulus of elasticity \(3mg\).
One end of the spring is attached to a fixed point \(X\) on a rough inclined plane.
The other end of the spring is attached to a package \(P\) of mass \(m\).
The plane is inclined to the horizontal at an angle \(\alpha\) where \(\tan\alpha = \dfrac{3}{4}\)
The package is initially held at the point \(Y\) on the plane, where \(XY = l\). The point \(Y\) is higher than \(X\) and \(XY\) is a line of greatest slope of the plane, as shown in Figure 2.
The package is released from rest at \(Y\) and moves up the plane.
The coefficient of friction between \(P\) and the plane is \(\dfrac{1}{3}\)
By modelling \(P\) as a particle,
(a) show that the acceleration of \(P\) at the instant when \(P\) is released from rest is \(\dfrac{17}{15}g\) (5)
(b) find, in terms of \(g\) and \(l\), the speed of \(P\) at the instant when the spring first reaches its natural length of \(3l\). (6)
Mark scheme (a)
Scheme
Marks
AO
Thrust in the spring \(= \dfrac{3mg\,2l}{3l}\ \ (= 2mg)\)
6. A light elastic string with natural length \(l\) and modulus of elasticity \(kmg\) has one end attached to a fixed point \(A\) on a rough inclined plane. The other end of the string is attached to a package of mass \(m\).
The plane is inclined at an angle \(\theta\) to the horizontal, where \(\tan\theta = \dfrac{5}{12}\)
The package is initially held at \(A\). The package is then projected with speed \(\sqrt{6gl}\) up a line of greatest slope of the plane and first comes to rest at the point \(B\), where \(AB = 3l\).
The coefficient of friction between the package and the plane is \(\dfrac{1}{4}\)
By modelling the package as a particle,
(a) show that \(k = \dfrac{15}{26}\) (6)
(b) find the acceleration of the package at the instant it starts to move back down the plane from the point \(B\). (5)
Mark scheme (a)
Scheme
Marks
AO
Work done against friction \(= 3l \times \mu mg\cos\theta\ \ \left(= \dfrac{9mgl}{13}\right)\) Gain in EPE \(= \dfrac{kmg \times 4l^2}{2l}\ \ (= 2kmgl)\) Gain in GPE \(= mg \times 3l\sin\theta\ \ \left(= \dfrac{15mgl}{13}\right)\)
4. A small ball, of mass \(m\), is thrown vertically upwards with speed \(\sqrt{8gH}\) from a point \(O\) on a smooth horizontal floor. The ball moves towards a smooth horizontal ceiling that is a vertical distance \(H\) above \(O\). The coefficient of restitution between the ball and the ceiling is \(\dfrac{1}{2}\)
In a model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to air resistance of constant magnitude \(\dfrac{1}{2}mg\).
Using this model,
(a) use the work-energy principle to find, in terms of \(g\) and \(H\), the speed of the ball immediately before it strikes the ceiling, (5)
(b) find, in terms of \(g\) and \(H\), the speed of the ball immediately before it strikes the floor at \(O\) for the first time. (5)
In a simplified model of the motion of the ball, it is assumed that the ball, as it moves up or down, is subject to no air resistance.
Using this simplified model,
(c) explain, without any detailed calculation, why the speed of the ball, immediately before it strikes the floor at \(O\) for the first time, would still be less than \(\sqrt{8gH}\) (1)
A1ft: Correct equation with at most one error ft on their answer to (a)
M1A1ft is available to a candidate who has not scored the first M1
A1: Correct equation (no ft)
A1: Correct answer (any equivalent but must be in terms of \(g\) and \(H\))
Mark scheme (c)
Scheme
Marks
AO
Since \(e \lt 1\), ball loses energy in its collision with the ceiling.
B1
2.4
(1)
(11 marks)
Notes
B1: Clear explanation
Need to identify that the loss of KE occurs in the impact with the ceiling. Do not insist on seeing \(e \lt 1\) or equivalent. If they include incorrect additional statements then B0
M1: Use conservation of energy with EPE \(= \dfrac{\lambda x^2}{2a}\). (Condone EPE \(= \dfrac{\lambda x^2}{a}\) here). All three terms required. Must be dimensionally correct. Condone sign errors.
M1: Maximum speed at equilibrium seen or implied, and correct method to find \(e\)
A1: Correct \(e\)
Alternative: form energy equation for movement through a height of \(h\) and differentiate \(v^2\) wrt \(h\) to find \(h\) for max \(v\) M1 \(h = \dfrac{5a}{4}\) A1
M1: Form energy equation for movement from \(A\) to equilibrium position. Need all 4 terms. Correct form for EPE. Dimensionally correct. Condone sign errors. Allow in \(a\), \(g\) and \(e\) (with \(e\) defined)
A1ft A1ft: Unsimplified equation in their \(e\) with at most one error Correct unsimplified equation (using their \(e\)) for \(v\)
A1: Any equivalent form. Accept \(1.44\sqrt{ag}\) or \(1.4\sqrt{ag}\)
SHM is not on this specification, but you might see some candidates using it. See below for SHM alternative for parts (b) and (c)
SHM alternative for parts (b) and (c)
Scheme
Marks
AO
At equilibrium, \(\dfrac{4mge}{3a} = mg\), \(e = \dfrac{3a}{4}\) Equation of motion: \(mg - \dfrac{4mg}{3a}(e + x) = m\ddot{x}\), so \(\ddot{x} = -\dfrac{4g}{3a}x\) Hence SHM They need to start by showing that they have SHM in order to justify using the standard results. No marks scored for this at this stage.
(b) Use of \(x = \dfrac{5a}{4}\) and their \(\omega^2\) Substitute to find acceleration
M1
\(\ddot{x} = -\dfrac{4g}{3a} \times \dfrac{5a}{4} = -\dfrac{5g}{3}\), \(|\ddot{x}| = \dfrac{5g}{3}\) Correct only ISW. Condone \(1.7g\) or better
A1
(2)
(c) \(\dfrac{4mge}{3a} = mg\),
M1
\(e = \dfrac{3a}{4}\) This work now scores the two marks provided it is used in part (c)
A1
Use of \(v_{\max} = \omega a\) Correct method to find max \(v\)
M1
\(v_{\max} = \sqrt{\dfrac{4g}{3a}} \times \dfrac{5a}{4}\) Follow their \(e\) and \(\omega\)
A2ft
\(v_{\max} = \dfrac{5}{2}\sqrt{\dfrac{ga}{3}}\) Any equivalent form. Accept \(1.44\sqrt{ag}\) or \(1.4\sqrt{ag}\)
4. A car of mass 600 kg pulls a trailer of mass 150 kg along a straight horizontal road. The trailer is connected to the car by a light inextensible towbar, which is parallel to the direction of motion of the car. The resistance to the motion of the trailer is modelled as a constant force of magnitude 200 N. At the instant when the speed of the car is \(v\ \text{m s}^{-1}\), the resistance to the motion of the car is modelled as a force of magnitude \((200 + \lambda v)\) N, where \(\lambda\) is a constant.
When the engine of the car is working at a constant rate of 15 kW, the car is moving at a constant speed of \(25\ \text{m s}^{-1}\)
(a) Show that \(\lambda = 8\) (4)
Later on, the car is pulling the trailer up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{15}\)
The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 200 N at all times. At the instant when the speed of the car is \(v\ \text{m s}^{-1}\), the resistance to the motion of the car from non-gravitational forces is modelled as a force of magnitude \((200 + 8v)\) N.
The engine of the car is again working at a constant rate of 15 kW.
When \(v = 10\), the towbar breaks. The trailer comes to instantaneous rest after moving a distance \(d\) metres up the road from the point where the towbar broke.
(b) Find the acceleration of the car immediately after the towbar breaks. (4)
(c) Use the work-energy principle to find the value of \(d\). (4)
Mark scheme (a)
Scheme
Marks
AO
Use of \(P = Fv\): \(F = \dfrac{15000}{25}\ (= 600)\)
B1
3.3
Equation of motion:
M1
3.4
\(F - (200 + 200 + 25\lambda) = 0\)
A1
1.1b
\(\lambda = 8\) *
A1*
2.2a
(4)
Notes
B1: 600 or equivalent
M1: Use the model to form the equation of motion If they start with two separate equations each one must be correct.
M1: Use the model to form the equation of motion for the car (with \(v = 10\) used). All terms required. Dimensionally correct. Condone sign error and sin/cos confusion
A1 A1: Unsimplified equation with at most one error. Correct unsimplified equation
3. A particle, \(P\), of mass \(m\) kg is projected with speed \(5\ \text{m s}^{-1}\) down a line of greatest slope of a rough plane. The plane is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{3}{5}\)
The total resistance to the motion of \(P\) is a force of magnitude \(\dfrac{1}{5}mg\)
Use the work-energy principle to find the speed of \(P\) at the instant when it has moved a distance 8 m down the plane from the point of projection. (7)
Mark scheme
Scheme
Marks
AO
Work done \(= \dfrac{1}{5}mg \times 8 \quad (15.68m)\)
Figure 1 shows a ramp inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{2}{7}\)
A parcel of mass 4 kg is projected, with speed \(5\ \text{m s}^{-1}\), from a point \(A\) on the ramp. The parcel moves up a line of greatest slope of the ramp and first comes to instantaneous rest at the point \(B\), where \(AB = 2.5\) m. The parcel is modelled as a particle.
The total resistance to the motion of the parcel from non-gravitational forces is modelled as a constant force of magnitude \(R\) newtons.
(a) Use the work-energy principle to show that \(R = 8.8\) (4)
After coming to instantaneous rest at \(B\), the parcel slides back down the ramp. The total resistance to the motion of the particle is modelled as a constant force of magnitude 8.8 N.
(b) Find the speed of the parcel at the instant it returns to \(A\). (3)
(c) Suggest two improvements that could be made to the model. (2)
Mark scheme (a)
Scheme
Marks
AO
Work-energy equation: KE lost = PE gained + Work Done
M1: A complete method to obtain \(R\). The question requires the use of work-energy. Need to consider all three terms with no duplication. Condone sign error and sin/cos confusion.
A1: Unsimplified equation with at most one error
A1: Correct unsimplified
A1*: Correct answer with sufficient working shown to justify given answer
Mark scheme (b)
Scheme
Marks
AO
Work-energy equation: KE after = initial KE – 2 (Work Done)
\(\Rightarrow 2v^2 = 6,\ v = 1.7\ (\text{m s}^{-1})\)
A1
1.1b
(3)
Notes
M1: Work-energy equation considering \(A \rightarrow A\) or \(B \rightarrow A\). Requires all relevant terms with no duplication. Condone sign errors and sin/cos confusion
A1: Correct unsimplified equation
A1: Accept 1.7 or 1.73 (answer depends on use of g). Not \(\sqrt{3}\)
Alternative (b)
Scheme
Marks
AO
Work-energy equation: KE at \(A\) = PE lost – Work Done
\(\Rightarrow 2v^2 = 6,\ v = 1.7\ (\text{m s}^{-1})\)
A1
(3)
(corrected from the printed mark scheme: “KE at \(B\)”, but the kinetic energy found is at \(A\); the parcel is at rest at \(B\))
Alternative (b)
Scheme
Marks
AO
Equation of motion and suvat: \(4g\sin\theta - 8.8 = 4a \quad (a = 0.6)\)
M1
\(v^2 = 2 \times a \times 2.5\)
A1
\(v = 1.7\ (\text{m s}^{-1})\)
A1
(3)
M1: Complete method to find \(v\) or \(v^2\).
A1: Correct unsimplified expression for \(v\) or \(v^2\).
A1: Accept 1.7 or 1.73 (answer depends on use of g)
Mark scheme (c)
Scheme
Marks
AO
A valid improvement
B1
3.5c
A second valid, distinct, improvement
B1
3.5c
(2)
(9 marks)
Notes
B1: it has assumed a constant resistance - have variable resistance -have air resistance proportional to speed ……
B1: Do not model the parcel as a particle - so can consider the possibility that the parcel rotates as it moves up/down the slope - consider the dimensions of the parcel
The comments need to relate to the 2 modelling assumptions in the question. Air resistance and friction are already included in "non-gravitational forces".
4. One end of a light elastic string, of modulus of elasticity \(2mg\) and natural length \(l\), is fixed to a point \(O\) on a rough plane. The plane is inclined at angle \(\alpha\) to the horizontal, where \(\sin\alpha = \dfrac{3}{5}\). The other end of the string is attached to a particle \(P\) of mass \(m\) which is held at rest on the plane at the point \(O\). The coefficient of friction between \(P\) and the plane is \(\dfrac{1}{4}\). The particle is released from rest and slides down the plane, coming to instantaneous rest at the point \(A\), where \(OA = kl\).
Given that \(k > 1\), find, to 3 significant figures, the value of \(k\). (7)
Mark scheme
Scheme
Marks
Elastic energy \(= \dfrac{1}{2} \times 2mg\dfrac{x^2}{l}\)
B1
Work done by friction = \((l + x)\mu mg\cos\alpha\)
B1
Energy from release: \((l + x)\mu mg\cos\alpha + \dfrac{1}{2} \times 2mg\dfrac{x^2}{l} = (l + x)mg\sin\alpha\)
M1 Attempt a work-energy equation. Must have 3 terms: work done by friction, elastic energy, GPE. EPE term must be of the form \(= k\lambda\dfrac{x^2}{l}\) \(k = 1, 2\) or \(\dfrac{1}{2}\) Work done term must be of the form distance \(\times\ \mu mg\cos\) or \(\sin\alpha\)
A1ft Correct equation, ft their EPE and work terms
dM1 Solve their 3 term quadratic to obtain a value for the extension as a multiple of \(l\). Award if correct answer follows a correct quadratic. If the quadratic is incorrect award only if working shown (ie general formula shown explicitly and used or by implication through substitution correct for their equation, pos root only needed)
A1 Correct extension decimal or exact
A1 Complete by adding 1 to the numerical multiple of \(l\) Must be 3 significant figures.
(Corrected from the printed mark scheme: the fifth line is printed as \(l^2 + 4lx + 5x^2 = 3l^2 + 3lx\).)
1. A small bead is threaded on a smooth straight horizontal wire. The wire is modelled as a line with vector equation \(\mathbf{r} = (2 + \lambda)\mathbf{i} + (2\lambda - 1)\mathbf{j}\), where the unit of length is the metre. The bead is moved a distance of \(\sqrt{80}\) m along the wire by a force \(\mathbf{F} = (4\mathbf{i} - 3\mathbf{j})\) N. Find the magnitude of the work done by \(\mathbf{F}\). (5)
Mark scheme
Scheme
Marks
ALTERNATIVE 1
Direction of line is \((\mathbf{i} + 2\mathbf{j})\)
First M1 for attempt to find the angle (or cos thereof) between their \(\mathbf{d}\) (must be a multiple of \((\mathbf{i} + 2\mathbf{j})\)) and \((4\mathbf{i} - 3\mathbf{j})\)
First A1 for \(\cos\theta = \dfrac{-2}{5\sqrt{5}}\) oe
Second M1 for W.D. \(= \left|(4\mathbf{i} - 3\mathbf{j})\right|\left|\cos\theta\right| \times \sqrt{80}\)
Second A1 for 8 (J)
ALTERNATIVE 3
Direction of line is \((\mathbf{i} + 2\mathbf{j})\)
First M1 for attempt to find a unit vector in the direction of their direction vector
First A1 for \(\hat{\mathbf{d}} = \dfrac{(\mathbf{i} + 2\mathbf{j})}{\sqrt{5}}\) oe
Second M1 for W.D. \(= \left|(4\mathbf{i} - 3\mathbf{j})\cdot\hat{\mathbf{d}}\right| \times \sqrt{80}\) (\(\hat{\mathbf{d}}\) must be a multiple of \((\mathbf{i} + 2\mathbf{j})\))
A uniform rod \(AB\) has mass \(m\) and length \(4a\). The end \(A\) of the rod is freely hinged to a fixed point. One end of a light elastic string, of natural length \(a\) and modulus \(\dfrac{1}{4}mg\), is attached to the end \(B\) of the rod. The other end of the string is attached to a small light smooth ring \(R\). The ring can move freely on a smooth horizontal wire which is fixed at a height \(a\) above \(A\), and in a vertical plane through \(A\). The angle between the rod and the horizontal is \(\theta\), where \(0 \lt \theta \lt \dfrac{\pi}{2}\), as shown in Figure 1. Given that the elastic string is vertical,
(a) show that the potential energy of the system is \[2mga(\sin^2\theta - \sin\theta) + \text{constant}\] (4)
(b) Show that when \(\theta = \dfrac{\pi}{6}\) the rod is in stable equilibrium. (7)
Mark scheme (a)
Scheme
Marks
For the rod: GPE \(= -2mga\sin\theta\) Must be working from a fixed point
B1
Extension in the string \(= 4a\sin\theta\)
B1
GPE in the string \(= \dfrac{\frac{1}{4}mgx^2}{2a}\)
M1
Total \(\dfrac{mg}{8a}\times(4a\sin\theta)^2 - 2mga\sin\theta = 2mga\left(\sin^2\theta - \sin\theta\right) +\) constant Given Answer
Second derivative: \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2amg\left(2\cos^2\theta - 2\sin^2\theta + \sin\theta\right)\)
M1A1
Substitute \(\theta = \dfrac{\pi}{6}\) in both:
M1
\(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 2mga\left(2\times\dfrac{1}{2}\cos\theta - \cos\theta\right) = 0\) hence equilibrium cso Allow from working to find solutions for \(\theta\)
Figure 2 shows four uniform rods, each of mass \(m\) and length \(2a\). The rods are freely hinged at their ends to form a rhombus \(ABCD\). Point \(A\) is attached to a fixed point on a ceiling and the rhombus hangs freely with \(C\) vertically below \(A\). A light elastic spring of natural length \(2a\) and modulus of elasticity \(7mg\) connects the points \(A\) and \(C\). A particle of mass \(3m\) is attached to point \(C\).
(a) Show that, when \(AD\) is at an angle \(\theta\) to the downward vertical, the potential energy \(V\) of the system is given by \[V = 28mga\cos^2\theta - 48mga\cos\theta + \text{constant}\] (5)
Given that \(\theta \gt 0\)
(b) find the value of \(\theta\) for which the system is in equilibrium, (4)
(c) determine the stability of this position of equilibrium. (4)
Mark scheme (a)
Scheme
Marks
Relative to the fixed point A, PE of mass at C \(= -3mg\times 4a\cos\theta\)
B1
PE of rods \(= -2mg\times a\cos\theta - 2mg\times 3a\cos\theta\)
6. The ends of a light elastic string, of natural length 0.4 m and modulus of elasticity \(\lambda\) newtons, are attached to two fixed points \(A\) and \(B\) which are 0.6 m apart on a smooth horizontal table. The tension in the string is 8 N.
(a) Show that \(\lambda = 16\) (3)
A particle \(P\) is attached to the midpoint of the string. The particle \(P\) is now pulled horizontally in a direction perpendicular to \(AB\) to a point 0.4 m from the midpoint of \(AB\). The particle is held at rest by a horizontal force of magnitude \(F\) newtons acting in a direction perpendicular to \(AB\), as shown in Figure 5 below.
Figure 5
(b) Find the value of \(F\). (4)
The particle is released from rest. Given that the mass of \(P\) is 0.3 kg,
(c) find the speed of \(P\) as it crosses the line \(AB\). (6)
Mark scheme (a)
Scheme
Marks
\(8 = \dfrac{\lambda \times 0.20}{0.40}\)
M1A1
\(\lambda = 16\) *
A1cso
(3)
Notes
M1 Attempt Hooke's Law using the whole string or a half string.
A1 Correct equation.
A1cso Correct given value of \(\lambda\) obtained with no errors seen.
M1 Use Hooke's Law with the new longer length for the string or half string. \(\lambda\) must be 16, but length need not be correct but use of 0.2 for extension of full string or 0.1 for extension of half string scores M0.
A1 Obtain \(T = 24\)
M1 Resolve parallel to \(F\) or in another direction which gives an equation connecting \(T\) and \(F\).
2. A truck of mass 900 kg is towing a trailer of mass 150 kg up an inclined straight road with constant speed 15 m s\(^{-1}\). The trailer is attached to the truck by a light inextensible towbar which is parallel to the road. The road is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{9}\). The resistance to motion of the truck from non-gravitational forces has constant magnitude 200 N and the resistance to motion of the trailer from non-gravitational forces has constant magnitude 50 N.
(a) Find the rate at which the engine of the truck is working. (5)
When the truck and trailer are moving up the road at 15 m s\(^{-1}\) the towbar breaks, and the trailer is no longer attached to the truck. The rate at which the engine of the truck is working is unchanged. The resistance to motion of the truck from non-gravitational forces and the resistance to motion of the trailer from non-gravitational forces are still forces of constant magnitudes 200 N and 50 N respectively.
(b) Find the acceleration of the truck at the instant after the towbar breaks. (3)
(c) Use the work-energy principle to find out how much further up the road the trailer travels before coming to instantaneous rest. (4)
Mark scheme (a)
Scheme
Marks
Constant speed \(\Rightarrow\) no acceleration. Driving force \(= 200 + 50 + 900g\sin\theta + 150g\sin\theta\)
M1
Or \(D - T - 200 - 900g\sin\theta = 0\) and \(T - 50 - 150g\sin\theta = 0\)
M1 Equation of motion for the truck at instant after the towbar breaks. All terms required & dimensionally correct. Allow for an equation to find acceleration down the slope
A1ft Correct for their driving force \(\left(1393\dfrac{1}{3}\right)\).
A1 Accept 0.24, not \(\dfrac{32}{135}\) must be +ve
1. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal, x-y plane.]
A bead \(P\) of mass 0.08 kg is threaded on a smooth straight horizontal wire which lies along the line with equation \(y = 2x - 1\). The unit of length on both axes is the metre. Initially the bead is at rest at the point \((a, b)\). A force \((6\mathbf{i} - 2\mathbf{j})\) N acts on \(P\) and moves it along the wire so that \(P\) passes through the point \((5, 9)\) with speed 10 m s\(^{-1}\).
Find the value of \(a\) and the value of \(b\). (7)
Figure 3 shows a uniform rod \(AB\), of length \(2l\) and mass \(4m\). A particle of mass \(2m\) is attached to the rod at \(B\). The rod can turn freely in a vertical plane about a fixed smooth horizontal axis through \(A\). One end of a light elastic spring, of natural length \(2l\) and modulus of elasticity \(kmg\), where \(k \gt 4\), is attached to the rod at \(B\). The other end of the spring is attached to a fixed point \(C\) which is vertically above \(A\), where \(AC = 2l\). The angle \(BAC\) is \(2\theta\), where \(\dfrac{\pi}{6} \lt \theta \leqslant \dfrac{\pi}{2}\)
(a) Show that the potential energy of the system is \[4mgl\{(k - 4)\sin^2\theta - k\sin\theta\} + \text{constant}\] (6)
Given that there is a position of equilibrium with \(\theta \neq \dfrac{\pi}{2}\)
(b) show that \(k \gt 8\) (6)
Given that \(k = 10\)
(c) determine the stability of this position of equilibrium. (4)
3. One end of a light elastic string, of natural length 1.5 m and modulus of elasticity 14.7 N, is attached to a fixed point \(O\) on a ceiling. A particle \(P\) of mass 0.6 kg is attached to the free end of the string. The particle is held at \(O\) and released from rest. The particle comes to instantaneous rest for the first time at the point \(A\).
Find
(a) the distance \(OA\), (6)
(b) the magnitude of the instantaneous acceleration of \(P\) at \(A\). (3)
2. A car of mass 800 kg is moving on a straight road which is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{20}\). The resistance to the motion of the car from non-gravitational forces is modelled as a constant force of magnitude \(R\) newtons. When the car is moving up the road at a constant speed of 12.5 m s\(^{-1}\), the engine of the car is working at a constant rate of \(3P\) watts. When the car is moving down the road at a constant speed of 12.5 m s\(^{-1}\), the engine of the car is working at a constant rate of \(P\) watts.
(a) Find
(i) the value of \(P\),
(ii) the value of \(R\). (6)
When the car is moving up the road at 12.5 m s\(^{-1}\) the engine is switched off and the car comes to rest, without braking, in a distance \(d\) metres. The resistance to the motion of the car from non-gravitational forces is still modelled as a constant force of magnitude \(R\) newtons.
(b) Use the work-energy principle to find the value of \(d\). (4)
1. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.]
A bead \(P\) of mass 0.4 kg is threaded on a smooth straight horizontal wire. The wire lies along the line with vector equation \(\mathbf{r} = (\mathbf{i} + 2\mathbf{j}) + \lambda(-2\mathbf{i} + 3\mathbf{j})\). The bead is initially at rest at the point \(A\) with position vector \((-\mathbf{i} + 5\mathbf{j})\) m. A constant horizontal force \((0.5\mathbf{i} + \mathbf{j})\) N acts on \(P\) and moves it along the wire to the point \(B\). At \(B\) the speed of \(P\) is 5 m s\(^{-1}\).
A smooth wire, with ends \(A\) and \(B\), is in the shape of a semicircle of radius \(r\). The line \(AB\) is horizontal and the midpoint of \(AB\) is \(O\). The wire is fixed in a vertical plane. A small ring \(R\) of mass \(2m\) is threaded on the wire and is attached to two light inextensible strings. One string passes through a small smooth ring fixed at \(A\) and is attached to a particle of mass \(\sqrt{6}m\). The other string passes through a small smooth ring fixed at \(B\) and is attached to a second particle of mass \(\sqrt{6}m\). The particles hang freely under gravity, as shown in Figure 3. The angle between the radius \(OR\) and the downward vertical is \(2\theta\), where \(-\dfrac{\pi}{4} < \theta < \dfrac{\pi}{4}\)
(a) Show that the potential energy of the system is \[2mgr\left(2\sqrt{3}\cos\theta - \cos 2\theta\right) + \text{constant}\] (6)
(b) Find the values of \(\theta\) for which the system is in equilibrium. (4)
(c) Determine the stability of the position of equilibrium for which \(\theta > 0\) (3)
Mark scheme (a)
Scheme
Marks
GPE of the ring: \(-2mgr\cos 2\theta\)
B1
GPE of suspended particles: \(-\sqrt{6}mg(L_1 - a) - \sqrt{6}mg(L_2 - b)\)
M1 Second derivative - needs to be the full expression.
M1 Substitute \(\theta = \dfrac{\pi}{6}\)
A1 No errors seen
(Corrected from the printed mark scheme: the bracket is printed as \(8\left(\frac{3}{4} - \frac{1}{2}\right)\); \(\sin^2\frac{\pi}{6} = \frac{1}{4}\), which gives the printed \(-2mgr\).)
A particle \(P\) of mass 10 kg is projected from a point \(A\) up a line of greatest slope \(AB\) of a fixed rough plane. The plane is inclined at angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{5}{12}\) and \(AB = 6.5\) m, as shown in Figure 2. The coefficient of friction between \(P\) and the plane is \(\mu\). The work done against friction as \(P\) moves from \(A\) to \(B\) is 245 J.
(a) Find the value of \(\mu\). (5)
The particle is projected from \(A\) with speed 11.5 m s\(^{-1}\). By using the work-energy principle,
(b) find the speed of the particle as it passes through \(B\). (4)
Mark scheme (a)
Scheme
Marks
Max friction \(= \mu \times 10g\cos\alpha\)
B1
Work done against friction \(= 6.5 \times 10g\mu\cos\alpha\ (= 245)\)
1. A particle \(P\) moves from the point \(A\), with position vector \((2\mathbf{i} + 4\mathbf{j} + a\mathbf{k})\) m, where \(a\) is a positive constant, to the point \(B\), with position vector \((-\mathbf{i} + a\mathbf{j} - \mathbf{k})\) m, under the action of a constant force \(\mathbf{F} = (2\mathbf{i} + a\mathbf{j} - 3\mathbf{k})\) N. The work done by \(\mathbf{F}\), as it moves the particle \(P\) from \(A\) to \(B\), is 3 J. Find the value of \(a\). (6)
1. A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 1.2 m and modulus of elasticity \(\lambda\) newtons. The other end of the spring is attached to a fixed point \(A\) on a ceiling. The particle is hanging freely in equilibrium at a distance 1.5 m vertically below \(A\).
(a) Find the value of \(\lambda\). (3)
The particle is now raised to the point \(B\), where \(B\) is vertically below \(A\) and \(AB = 0.8\) m. The spring remains straight. The particle is released from rest and first comes to instantaneous rest at the point \(C\).
(b) Find the distance \(AC\). (4)
Mark scheme (a)
Scheme
Marks
\(0.5g = T = \dfrac{\lambda \times 0.3}{1.2}\)
M1A1
\(\lambda = 2g = 19.6\)
A1
(3)
Notes
M1 Use Hooke's law to obtain the tension and equate to the weight
A1 Correct equation
A1 Solve to get \(\lambda = 19.6\) Accept 20 or \(2g\)
A uniform rod \(AB\), of length \(2l\) and mass \(12m\), has its end \(A\) smoothly hinged to a fixed point. One end of a light inextensible string is attached to the other end \(B\) of the rod. The string passes over a small smooth pulley which is fixed at the point \(C\), where \(AC\) is horizontal and \(AC = 2l\). A particle of mass \(m\) is attached to the other end of the string and the particle hangs vertically below \(C\).
The angle \(BAC\) is \(\theta\), where \(0 < \theta < \dfrac{\pi}{2}\), as shown in Figure 1.
(a) Show that the potential energy of the system is \[4mgl\left(\sin\frac{\theta}{2} - 3\sin\theta\right) + \text{constant}\] (4)
(b) Find the value of \(\theta\) when the system is in equilibrium and determine the stability of this equilibrium position. (10)
A particle \(P\) of mass 2 kg is released from rest at a point \(A\) on a rough inclined plane and slides down a line of greatest slope. The plane is inclined at 30\(^\circ\) to the horizontal. The point \(B\) is 5 m from \(A\) on the line of greatest slope through \(A\), as shown in Figure 3.
(a) Find the potential energy lost by \(P\) as it moves from \(A\) to \(B\). (2)
The speed of \(P\) as it reaches \(B\) is 4 m s\(^{-1}\).
(b)
(i) Use the work-energy principle to find the magnitude of the constant frictional force acting on \(P\) as it moves from \(A\) to \(B\).
(ii) Find the coefficient of friction between \(P\) and the plane. (7)
The particle \(P\) is now placed at \(A\) and projected down the plane towards \(B\) with speed 3 m s\(^{-1}\). Given that the frictional force remains constant,
(c) find the speed of \(P\) as it reaches \(B\). (4)
Mark scheme (a)
Scheme
Marks
PE lost \(= mgh = 2 \times 9.8 \times 5\sin 30 = 49\) J
3. One end \(A\) of a light elastic string \(AB\), of modulus of elasticity \(mg\) and natural length \(a\), is fixed to a point on a rough plane inclined at an angle \(\theta\) to the horizontal. The other end \(B\) of the string is attached to a particle of mass \(m\) which is held at rest on the plane. The string \(AB\) lies along a line of greatest slope of the plane, with \(B\) lower than \(A\) and \(AB = a\). The coefficient of friction between the particle and the plane is \(\mu\), where \(\mu < \tan\theta\). The particle is released from rest.
(a) Show that when the particle comes to rest it has moved a distance \(2a(\sin\theta - \mu\cos\theta)\) down the plane. (6)
(b) Given that there is no further motion, show that \(\mu \geqslant \dfrac{1}{3}\tan\theta\). (5)
1. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.]
A bead \(P\) of mass 0.2 kg is threaded on a smooth straight horizontal wire. The bead is at rest at the point \(A\) with position vector \((4\mathbf{i} - \mathbf{j})\) m. A force \((0.2\mathbf{i} + 0.3\mathbf{j})\) N acts on \(P\) and moves it to the point \(B\) with position vector \((13\mathbf{i} + 5\mathbf{j})\) m.
8. The points \(A\) and \(B\) are 10 m apart on a line of greatest slope of a fixed rough inclined plane, with \(A\) above \(B\). The plane is inclined at 25\(^\circ\) to the horizontal. A particle \(P\) of mass 5 kg is released from rest at \(A\) and slides down the slope. As \(P\) passes \(B\), it is moving with speed 7 m s\(^{-1}\).
(a) Find, using the work-energy principle, the work done against friction as \(P\) moves from \(A\) to \(B\). (4)
(b) Find the coefficient of friction between the particle and the plane. (5)
Mark scheme (a)
Scheme
Marks
Work done against friction = Loss in GPE– Gain in KE \(= 5 \times 9.8 \times 10\sin 25 - \dfrac{1}{2} \times 5 \times 7^2 = 84.58\ldots\)
M1 A2
\(= 85\) (J) (84.6)
A1
(4)
Notes
M1 Must be using Work-energy principle Needs to consider (work done) KE & GPE and no other terms. Condone sign errors. Watch out for incorrect solutions including both change in GPE and the work done against the weight – this is a method error.
A2 -1 each error
A1 Max 3 s.f. Must be +ve. Accept as \(10F = 84.6\) or equiv.
A bead \(B\) of mass \(m\) is threaded on a smooth circular wire of radius \(r\), which is fixed in a vertical plane. The centre of the circle is \(O\), and the highest point of the circle is \(A\). A light elastic string of natural length \(r\) and modulus of elasticity \(kmg\) has one end attached to the bead and the other end attached to \(A\). The angle between the string and the downward vertical is \(\theta\), and the extension in the string is \(x\), as shown in Figure 2.
Given that the string is taut,
(a) show that the potential energy of the system is \[2mgr\{(k - 1)\cos^2\theta - k\cos\theta\} + \text{constant}\] (6)
Given also that \(k = 3\),
(b) find the positions of equilibrium and determine their stability. (9)
Mark scheme (a)
Scheme
Marks
Measuring GPE from A, GPE \(= -mg\cos\theta(r + x)\)
B1
EPE \(= \dfrac{kmgx^2}{2r}\)
B1
From the isosceles triangle, \(\cos\theta = \dfrac{x + r}{2r}\)
One end of a light elastic string, of natural length \(l\) and modulus of elasticity \(3mg\), is fixed to a point \(A\) on a fixed plane inclined at an angle \(\alpha\) to the horizontal, where \(\sin\alpha = \dfrac{3}{5}\)
A small ball of mass \(2m\) is attached to the free end of the string. The ball is held at a point \(C\) on the plane, where \(C\) is below \(A\) and \(AC = l\) as shown in Figure 3. The string is parallel to a line of greatest slope of the plane. The ball is released from rest. In an initial model the plane is assumed to be smooth.
(a) Find the distance that the ball moves before first coming to instantaneous rest. (5)
In a refined model the plane is assumed to be rough. The coefficient of friction between the ball and the plane is \(\mu\). The ball first comes to instantaneous rest after moving a distance \(\dfrac{2}{5}l\).
(b) Find the value of \(\mu\). (6)
Mark scheme (a)
Scheme
Marks
\(\dfrac{3mgx^2}{2l} = 2mgx\sin\alpha\)
M1A1 B1(A1 on e-pen)
\(3x^2 = 4xl \times \dfrac{3}{5}\)
\(5x^2 = 4xl\)
\(x = \dfrac{4}{5}l\)
DM1A1
(5)
Notes
M1 for an energy equation with an EPE term of the form \(\dfrac{kmgx^2}{l}\) and a GPE term. If a KE term is included it must become 0 later.
A1 for a correct EPE term
B1 for a correct GPE term. This can be in terms of the distance moved down the plane or the vertical distance fallen
M1 dep for solving their equation to obtain the distance moved or using the vertical distance and obtaining the distance moved along the plane.
A1 for \(x = \dfrac{4}{5}l\) oe eg \(x = \dfrac{12}{15}l\)
If \(m\) used instead of \(2m\), assuming correct otherwise: (a) M1A1B0M1A0 (so 2 penalties for mis-read)
B1 for resolving perpendicular to the plane to obtain \(R = 2mg\cos\alpha\). May only be seen in an equation.
M1 for an work-energy equation with an EPE term of the form \(\dfrac{kmgx^2}{l}\), a GPE term and the work done against friction. The work term must include a distance along the plane.
A1 for EPE and GPE terms correct and work subtracted from the GPE
B1 ft for the work term ft their \(R\)
M1 dep for solving to obtain a value for \(\mu\)
A1 cso for \(\mu = \dfrac{3}{8}\) oe inc 0.375 but not 0.38
If \(m\) used instead of \(2m\), assuming correct otherwise:
(b) B1 \(R = mg\cos\alpha\)
M1, A1 Equation, with EPE correct and \(mg \times \dfrac{2}{5}l \times \dfrac{3}{5}\)
1. A small bead is threaded on a smooth, straight horizontal wire which passes through the point \(A(-3, 1)\) and the point \(B(2, 5)\) in the \(x\)-\(y\) plane. The bead moves under the action of a horizontal force \(\mathbf{F}\) of magnitude 8.5 N whose line of action is parallel to the line with equation \(15x - 8y + 4 = 0\). The unit on both the \(x\) and \(y\) axes has length one metre. Find the work done by \(\mathbf{F}\) as it moves the bead from \(A\) to \(B\). (8)
Work done \(= \tfrac{1}{2}(8\mathbf{i} + 15\mathbf{j}) \cdot (5\mathbf{i} + 4\mathbf{j})\)
M1
\(= 50\) (J)
A1
(8 marks)
Notes
First M1 for \(\lambda(\pm 8\mathbf{i} \pm 15\mathbf{j})\)
First A1 for correct expression
Second M1 for \(\lambda^2(8^2 + 15^2) = 8.5^2\) from previous incorrect vector
Second A1 for \(\lambda = \pm\tfrac{1}{2}\)
Third A1 for correct \(\mathbf{F}\)
B1 for correct \(\mathbf{AB}\)
Third M1 for their \(\mathbf{F}\cdot\mathbf{AB}\)
Fourth A1 for 50 (J). (−50 is A0)
(Corrected from the printed mark scheme: the work done line is printed as \(\tfrac{1}{2}(8\mathbf{i} + 15\mathbf{j})\) without the \(\cdot(5\mathbf{i} + 4\mathbf{j})\).)
A uniform rod \(AB\) has mass \(4m\) and length \(4l\). The rod can turn freely in a vertical plane about a fixed smooth horizontal axis through \(A\). A particle of mass \(km\), where \(k < 7\), is attached to the rod at \(B\). One end of a light elastic string, of natural length \(l\) and modulus of elasticity \(4mg\), is attached to the point \(D\) of the rod, where \(AD = 3l\). The other end of the string is attached to a fixed point \(E\) which is vertically above \(A\), where \(AE = 3l\), as shown in Figure 2. The angle between the rod and the upward vertical is \(2\theta\), where \(\arcsin\left(\dfrac{1}{6}\right) < \theta \leqslant \dfrac{\pi}{2}\).
(a) Show that, while the string is stretched, the potential energy of the system is \[8mgl\{(7 - k)\sin^2\theta - 3\sin\theta\} + \text{constant}\] (6)
There is a position of equilibrium with \(\theta \leqslant \dfrac{\pi}{6}\).
(b) Show that \(k \leqslant 4\) (5)
Given that \(k = 4\),
(c) show that this position of equilibrium is stable. (5)
Mark scheme (a)
Scheme
Marks
Length of string \(= 2 \times 3l\sin\theta\)
B1
Extension \(= 6l\sin\theta - l\)
E.P.E. \(= \dfrac{4mg}{2l}(6l\sin\theta - l)^2\)
G.P.E. of rod \(= 4mg \times 2l\cos 2\theta\)
G.P.E. of mass at \(B\) \(= kmg \times 4l\cos 2\theta\)
3. A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 2 m and modulus of elasticity 20 N. The other end of the spring is attached to a fixed point \(A\). The particle \(P\) is held at rest at the point \(B\), which is 1 m vertically below \(A\), and then released.
(a) Find the acceleration of \(P\) immediately after it is released from rest. (4)
The particle comes to instantaneous rest for the first time at the point \(C\).
(b) Find the distance \(BC\). (6)
Mark scheme (a)
Scheme
Marks
Weight + thrust = mass x accn.
M1
\(0.5 \times g + \dfrac{20 \times 1}{2} = 0.5a\)
B1(thrust) A1ft
\(a = g + 20 = 29.8 \approx 30\) (m s\(^{-2}\))
A1
(4)
Mark scheme (b)
Scheme
Marks
Change in GPE \(= mg(x + 1)\)
B1
EPE at B \(= \dfrac{20 \times 1^2}{2 \times 2}\) or EPE at C \(= \dfrac{20 \times x^2}{2 \times 2}\)
2. A ball of mass 0.2 kg is projected vertically upwards from a point \(O\) with speed 20 m s\(^{-1}\). The non-gravitational resistance acting on the ball is modelled as a force of constant magnitude 1.24 N and the ball is modelled as a particle. Find, using the work-energy principle, the speed of the ball when it first reaches the point which is 8 m vertically above \(O\). (6)
Mark scheme
NB This question tells candidates to use work-energy - suvat approach scores 0/6
A small smooth peg \(P\) is fixed at a distance \(d\) from a fixed smooth vertical wire. A particle of mass \(3m\) is attached to one end of a light inextensible string which passes over \(P\). The particle hangs vertically below \(P\). The other end of the string is attached to a small ring \(R\) of mass \(m\), which is threaded on the wire, as shown in Figure 3.
(a) Show that when \(R\) is at a distance \(x\) below the level of \(P\) the potential energy of the system is \[3mg\sqrt{(x^2 + d^2)} - mgx + \text{constant}\] (4)
(b) Hence find \(x\), in terms of \(d\), when the system is in equilibrium. (3)
(c) Determine the stability of the position of equilibrium. (3)
Mark scheme (a)
Scheme
Marks
PE of ring \(= -mgx\)
B1
PE of particle \(= -3mg\left(L - \sqrt{x^2 + d^2}\right)\)
M1 A1
\(\Rightarrow V = 3mg\sqrt{x^2 + d^2} - mgx\) + constant. AG
4. A particle \(P\) of mass 2 kg is attached to one end of a light elastic string of natural length 1.2 m. The other end of the string is attached to a fixed point \(O\) on a rough horizontal plane. The coefficient of friction between \(P\) and the plane is \(\dfrac{2}{5}\). The particle is held at rest at a point \(B\) on the plane, where \(OB = 1.5\) m. When \(P\) is at \(B\), the tension in the string is 20 N. The particle is released from rest.
(a) Find the speed of \(P\) when \(OP = 1.2\) m. (7)
M1 for attempting Hooke's Law, formula must be correct, either explicitly or by correct substitution.
A1 for \(20 = \dfrac{\lambda \times 0.3}{1.2}\)
A1 for obtaining \(\lambda = 80\)
B1 for the initial EPE \(\dfrac{"\lambda" \times 0.3^2}{2.4}\ \ (= 3\text{ J})\) their value for \(\lambda\) allowed. May only be seen in the eqaution.
M1 for a work-energy equation with one EPE term, one KE term and work done against friction (Award if second EPE/KE terms included provided these become 0). The EPE must be dimensionally correct, but need not be fully correct (eg denominator 1.2 instead of 2.4)
A1ft for a completely correct equation follow through their EPE
A1 cao for \(v = 0.80\) or 0.805 must be 2 or 3 sf
NB: This is damped harmonic motion (due to friction) so all SHM attempts lose the last 4 marks.
Mark scheme (b)
Scheme
Marks
Comes to rest \(0.4 \times 2g \times y = 3\)
M1
\(y = \dfrac{3}{0.4 \times 2 \times 9.8} = 0.38\) or 0.383 m
A1
(2)
(9 marks)
Notes
M1 for any complete method leading to a value for either \(BC\). If the distance travelled after the string becomes slack is found the work must be completed by adding 0.3 Their EPE found in (a) used in energy methods.
MS method is energy from \(B\) to \(C\) ie work done against friction = loss of EPE.
OR Energy from point where the string becomes slack to \(C\) ie work done against friction = loss of KE and completed for the required distance
OR NL2 to obtain the acceleration \(\left(-\dfrac{2g}{5}\right)\) while the string is slack and \(v^2 = u^2 + 2as\) to find the distance and completed for the required distance
A1cso for \(BC = 0.38\) or 0.383 (m) must be 2 or 3 sf
2. A particle \(P\) of mass 3 kg moves from point \(A\) to point \(B\) up a line of greatest slope of a fixed rough plane. The plane is inclined at 20\(^\circ\) to the horizontal. The coefficient of friction between \(P\) and the plane is 0.4
Given that \(AB = 15\) m and that the speed of \(P\) at \(A\) is 20 m s\(^{-1}\), find
(a) the work done against friction as \(P\) moves from \(A\) to \(B\), (3)
(b) the speed of \(P\) at \(B\). (4)
Mark scheme (a)
Scheme
Marks
Work done \(= 15\mu R = 15 \times 0.4 \times 3g\cos 20^\circ\)
M1 M1
\(= 18g\cos 20 = 166\) (J)
A1
(3)
Notes
M1 \(F_{\max} = \mu \times 3g\cos 20\ (11.05)\). \(R\) must be resolved but condone trig confusion.
M1 \(15 \times\) their \(F_{\max}\). Independent M. \(15 \times F_{\max} + \ldots\) is M0
A1 or 170 (J)
Mark scheme (b)
Scheme
Marks
Energy: WD against \(F\) + GPE + final KE = initial KE
M1A2ft Must include all four correct terms (including resolving). Condone sign errors and trig confusion. Any sign errors in the KE terms count as a single error. Follow their WD. A2ft: -1ee Follow their WD
A1 or 14
Or 2b
\(3a = -\ 0.4 \times 3g\cos 20 + 3g\sin 20\) and use of \(v^2 = u^2 + 2as\)
7. A particle \(P\) of mass 1.5 kg is attached to the mid-point of a light elastic string of natural length 0.30 m and modulus of elasticity \(\lambda\) newtons. The ends of the string are attached to two fixed points \(A\) and \(B\), where \(AB\) is horizontal and \(AB = 0.48\) m. Initially \(P\) is held at rest at the mid-point, \(M\), of the line \(AB\) and the tension in the string is 240 N.
(a) Show that \(\lambda = 400\) (3)
The particle is now held at rest at the point \(C\), where \(C\) is 0.07 m vertically below \(M\). The particle is released from rest at \(C\).
(b) Find the magnitude of the initial acceleration of \(P\). (6)
(c) Find the speed of \(P\) as it passes through \(M\). (6)
5. The point \(A\) lies on a rough plane inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{24}{25}\). A particle \(P\) is projected from \(A\), up a line of greatest slope of the plane, with speed \(U\) m s\(^{-1}\). The mass of \(P\) is 2 kg and the coefficient of friction between \(P\) and the plane is \(\dfrac{5}{12}\). The particle comes to instantaneous rest at the point \(B\) on the plane, where \(AB = 1.5\) m. It then moves back down the plane to \(A\).
(a) Find the work done against friction as \(P\) moves from \(A\) to \(B\). (4)
(b) Use the work-energy principle to find the value of \(U\). (4)
(c) Find the speed of \(P\) when it returns to \(A\). (3)
Mark scheme (a)
Scheme
Marks
\(N = 2g\cos\theta = \dfrac{14}{25}g\)
M1
\(F = \mu N = \dfrac{5}{12} \times \dfrac{14}{25}g = \dfrac{7g}{30}\)
6. A car of mass 1200 kg pulls a trailer of mass 400 kg up a straight road which is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{1}{14}\). The trailer is attached to the car by a light inextensible towbar which is parallel to the road. The car’s engine works at a constant rate of 60 kW. The non-gravitational resistances to motion are constant and of magnitude 1000 N on the car and 200 N on the trailer.
At a given instant, the car is moving at 10 m s\(^{-1}\). Find
(a) the acceleration of the car at this instant, (5)
(b) the tension in the towbar at this instant. (4)
The towbar breaks when the car is moving at 12 m s\(^{-1}\).
(c) Find, using the work-energy principle, the further distance that the trailer travels before coming instantaneously to rest. (5)
M1 Use of \(F = ma\) parallel to the slope for the car
A1 ft At most one error (their \(a\))
A1 ft All correct (their \(a\))
A1 only
OR (b) (following OR (a))
\(-800a = 2T + 800g\sin\alpha + 800 - 6000\)
M1A1A1
\(2T = 5200 - 800g\sin\alpha - 800 \times 2.3\)
\(T = 1400\)
A1
M1A1A1 Subtract and / or substitute to eliminate \(a\)
Mark scheme (c)
Scheme
Marks
\(200d = \dfrac{1}{2}400.12^2 - 400gd\sin\alpha\)
M1 A1 A1
\(d = 60\) (m)
DM1 A1
(5)
(14 marks)
Notes
M1 Use of work-energy. Must have all three terms. Do not accept duplication of terms, but condone sign errors. Equation in only one unknown, but could be vertical distance.
A1 At most one error in the equation
A1 All correct in one unknown
DM1 Solve for \(d\) – dependent on M for work-energy equation.
A1 only
For vertical distance \(\left(= \dfrac{60}{14} = 4.29\right)\) allow 3/5
A uniform rod \(AB\), of length \(4a\) and weight \(W\), is free to rotate in a vertical plane about a fixed smooth horizontal axis which passes through the point \(C\) of the rod, where \(AC = 3a\). One end of a light inextensible string of length \(L\), where \(L \gt 10a\), is attached to the end \(A\) of the rod and passes over a small smooth fixed peg at \(P\) and another small smooth fixed peg at \(Q\). The point \(Q\) lies in the same vertical plane as \(P\), \(A\) and \(B\). The point \(P\) is at a distance \(3a\) vertically above \(C\) and \(PQ\) is horizontal with \(PQ = 4a\). A particle of weight \(\dfrac{1}{2}W\) is attached to the other end of the string and hangs vertically below \(Q\). The rod is inclined at an angle \(2\theta\) to the vertical, where \(-\pi \lt 2\theta \lt \pi\), as shown in Figure 1.
(a) Show that the potential energy of the system is\[Wa(3\cos\theta - \cos 2\theta) + \text{constant}\] (4)
(b) Find the positions of equilibrium and determine their stability. (8)
3. A cyclist and her cycle have a combined mass of 75 kg. The cyclist is cycling up a straight road inclined at 5\(^\circ\) to the horizontal. The resistance to the motion of the cyclist from non-gravitational forces is modelled as a constant force of magnitude 20 N. At the instant when the cyclist has a speed of 12 m s\(^{-1}\), she is decelerating at 0.2 m s\(^{-2}\).
(a) Find the rate at which the cyclist is working at this instant. (5)
When the cyclist passes the point \(A\) her speed is 8 m s\(^{-1}\). At \(A\) she stops working but does not apply the brakes. She comes to rest at the point \(B\). The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 20 N.
(b) Use the work-energy principle to find the distance \(AB\). (5)
Mark scheme (a)
Scheme
Marks
Driving force = F
M1
Resolving parallel to the plane: \(\ F - 20 - 75g\sin 5 = -75 \times 0.2 = -15\)
A2 – 1ee
\(F = 5 + 75g\sin 5^\circ\)
\(P = Fv\quad \therefore\) working at \(\ 12 \times \left(5 + 75g\sin 5^\circ\right) = 828.7\ldots\)
DM1
\(\approx 830\) W
A1
(5)
Mark scheme (b)
Scheme
Marks
Loss in KE = gain in GPE + work done against resistance
M1
\(\dfrac{1}{2} \times 75 \times 64 = 75 \times 9.8 \times \sin 5^\circ d + 20d = d \times 84.059\ldots\)
1. A particle of mass 0.8 kg is attached to one end of a light elastic string of natural length 0.6 m. The other end of the string is attached to a fixed point \(A\). The particle is released from rest at \(A\) and comes to instantaneous rest 1.1 m below \(A\).
Figure 3 shows a framework \(ABC\), consisting of two uniform rods rigidly joined together at \(B\) so that \(\angle ABC = 90^\circ\). The rod \(AB\) has length \(2a\) and mass \(4m\), and the rod \(BC\) has length \(a\) and mass \(2m\). The framework is smoothly hinged at \(A\) to a fixed point, so that the framework can rotate in a fixed vertical plane. One end of a light elastic string, of natural length \(2a\) and modulus of elasticity \(3mg\), is attached to \(A\). The string passes through a small smooth ring \(R\) fixed at a distance \(2a\) from \(A\), on the same horizontal level as \(A\) and in the same vertical plane as the framework. The other end of the string is attached to \(B\).
The angle \(ARB\) is \(\theta\), where \(0 \lt \theta \lt \dfrac{\pi}{2}\).
(a) Show that the potential energy \(V\) of the system is given by\[V = 8amg\sin 2\theta + 5amg\cos 2\theta + \text{constant}\] (7)
(b) Find the value of \(\theta\) for which the system is in equilibrium. (4)
(c) Determine the stability of this position of equilibrium. (3)
5. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(l\) and modulus of elasticity \(3mg\). The other end of the string is attached to a fixed point \(O\) on a rough horizontal table. The particle lies at rest at the point \(A\) on the table, where \(OA = \dfrac{7}{6}l\). The coefficient of friction between \(P\) and the table is \(\mu\).
(a) Show that \(\mu \geqslant \dfrac{1}{2}\). (4)
The particle is now moved along the table to the point \(B\), where \(OB = \dfrac{3}{2}l\), and released from rest. Given that \(\mu = \dfrac{1}{2}\), find
(b) the speed of \(P\) at the instant when the string becomes slack, (5)
(c) the total distance moved by \(P\) before it comes to rest again. (3)
A particle \(P\) of mass 0.5 kg is projected from a point \(A\) up a line of greatest slope \(AB\) of a fixed plane. The plane is inclined at 30\(^\circ\) to the horizontal and \(AB = 2\) m with \(B\) above \(A\), as shown in Figure 2. The particle \(P\) passes through \(B\) with speed 5 m s\(^{-1}\). The plane is smooth from \(A\) to \(B\).
(a) Find the speed of projection. (4)
The particle \(P\) comes to instantaneous rest at the point \(C\) on the plane, where \(C\) is above \(B\) and \(BC = 1.5\) m. From \(B\) to \(C\) the plane is rough and the coefficient of friction between \(P\) and the plane is \(\mu\).
1. A particle moves from the point \(A\) with position vector \((3\mathbf{i} - \mathbf{j} + 3\mathbf{k})\) m to the point \(B\) with position vector \((\mathbf{i} - 2\mathbf{j} - 4\mathbf{k})\) m under the action of the force \((2\mathbf{i} - 3\mathbf{j} - \mathbf{k})\) N. Find the work done by the force. (4)
A small ball of mass \(3m\) is attached to the ends of two light elastic strings \(AP\) and \(BP\), each of natural length \(l\) and modulus of elasticity \(kmg\). The ends \(A\) and \(B\) of the strings are attached to fixed points on the same horizontal level, with \(AB = 2l\). The mid-point of \(AB\) is \(C\). The ball hangs in equilibrium at a distance \(\tfrac{3}{4}l\) vertically below \(C\) as shown in Figure 4.
(a) Show that \(k = 10\) (7)
The ball is now pulled vertically downwards until it is at a distance \(\tfrac{12}{5}l\) below \(C\). The ball is released from rest.
(b) Find the speed of the ball as it reaches \(C\). (6)
A box of mass 30 kg is held at rest at point \(A\) on a rough inclined plane. The plane is inclined at 20\(^\circ\) to the horizontal. Point \(B\) is 50 m from \(A\) up a line of greatest slope of the plane, as shown in Figure 1. The box is dragged from \(A\) to \(B\) by a force acting parallel to \(AB\) and then held at rest at \(B\). The coefficient of friction between the box and the plane is \(\dfrac{1}{4}\). Friction is the only non-gravitational resistive force acting on the box. Modelling the box as a particle,
(a) find the work done in dragging the box from \(A\) to \(B\). (6)
The box is released from rest at the point \(B\) and slides down the slope. Using the work-energy principle, or otherwise,
(b) find the speed of the box as it reaches \(A\). (5)
Mark scheme (a)
Scheme
Marks
Work done against friction \(= 50 \times \mu\text{R}\)
The end \(A\) of a uniform rod \(AB\), of length \(2a\) and mass \(4m\), is smoothly hinged to a fixed point. The end \(B\) is attached to one end of a light inextensible string which passes over a small smooth pulley, fixed at the same level as \(A\). The distance from \(A\) to the pulley is \(4a\). The other end of the string carries a particle of mass \(m\) which hangs freely, vertically below the pulley, with the string taut. The angle between the rod and the downward vertical is \(\theta\), where \(0 \lt \theta \lt \frac{\pi}{2}\), as shown in Figure 1.
(a) Show that the potential energy of the system is\[2mga(\sqrt{(5 - 4\sin\theta)} - 2\cos\theta) + \text{constant}.\] (5)
(b) Hence, or otherwise, show that any value of \(\theta\) which corresponds to a position of equilibrium of the system satisfies the equation\[4\sin^3\theta - 6\sin^2\theta + 1 = 0.\] (5)
(c) Given that \(\theta = \frac{\pi}{6}\) corresponds to a position of equilibrium, determine its stability. (5)
A particle of mass 0.5 kg is attached to one end of a light elastic spring of natural length 0.9 m and modulus of elasticity \(\lambda\) newtons. The other end of the spring is attached to a fixed point \(O\) on a rough plane which is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{3}{5}\). The coefficient of friction between the particle and the plane is 0.15. The particle is held on the plane at a point which is 1.5 m down the line of greatest slope from \(O\), as shown in Figure 2. The particle is released from rest and first comes to rest again after moving 0.7 m up the plane.
2. A particle \(P\) of mass 0.6 kg is released from rest and slides down a line of greatest slope of a rough plane. The plane is inclined at 30\(^\circ\) to the horizontal. When \(P\) has moved 12 m, its speed is 4 m s\(^{-1}\). Given that friction is the only non-gravitational resistive force acting on \(P\), find
(a) the work done against friction as the speed of \(P\) increases from 0 m s\(^{-1}\) to 4 m s\(^{-1}\), (4)
(b) the coefficient of friction between the particle and the plane. (4)
7. A light elastic string has natural length \(a\) and modulus of elasticity \(\dfrac{3}{2}mg\). A particle \(P\) of mass \(m\) is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). The particle is released from rest at \(A\) and falls vertically. When \(P\) has fallen a distance \(a + x\), where \(x > 0\), the speed of \(P\) is \(v\).
(a) Show that \(v^2 = 2g(a + x) - \dfrac{3gx^2}{2a}\). (4)
(b) Find the greatest speed attained by \(P\) as it falls. (4)
After release, \(P\) next comes to instantaneous rest at a point \(D\).
(c) Find the magnitude of the acceleration of \(P\) at \(D\). (6)
3. A particle of mass 0.5 kg is projected vertically upwards from ground level with a speed of 20 m s\(^{-1}\). It comes to instantaneous rest at a height of 10 m above the ground. As the particle moves it is subject to air resistance of constant magnitude \(R\) newtons. Using the work-energy principle, or otherwise, find the value of \(R\). (6)
A particle \(P\) of mass 2 kg is projected up a rough plane with initial speed 14 m s\(^{-1}\), from a point \(X\) on the plane, as shown in Figure 4. The particle moves up the plane along the line of greatest slope through \(X\) and comes to instantaneous rest at the point \(Y\). The plane is inclined at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{7}{24}\). The coefficient of friction between the particle and the plane is \(\dfrac{1}{8}\).
(a) Use the work-energy principle to show that \(XY = 25\) m. (7)
After reaching \(Y\), the particle \(P\) slides back down the plane.
(b) Find the speed of \(P\) as it passes through \(X\). (4)
Mark scheme (a)
Scheme
Marks
KE at \(X = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 2 \times 14^2\)
B1
GPE at \(Y =\) \(mgd\sin\alpha\left(= 2 \times g \times d \times \dfrac{7}{25}\right)\)
A light inextensible string of length \(2a\) has one end attached to a fixed point \(A\). The other end of the string is attached to a particle \(P\) of mass \(m\). A second light inextensible string of length \(L\), where \(L \gt \frac{12a}{5}\), has one of its ends attached to \(P\) and passes over a small smooth peg fixed at a point \(B\). The line \(AB\) is horizontal and \(AB = 2a\). The other end of the second string is attached to a particle of mass \(\frac{7}{20}m\), which hangs vertically below \(B\), as shown in Figure 2.
(a) Show that the potential energy of the system, when the angle \(PAB = 2\theta\), is\[\tfrac{1}{5}mga(7\sin\theta - 10\sin 2\theta) + \text{constant}.\] (4)
(b) Show that there is only one value of \(\cos\theta\) for which the system is in equilibrium and find this value. (8)
(c) Determine the stability of the position of equilibrium. (4)
1. At time \(t = 0\), a particle \(P\) of mass 3 kg is at rest at the point \(A\) with position vector \((\mathbf{j} - 3\mathbf{k})\) m. Two constant forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) then act on the particle \(P\) and it passes through the point \(B\) with position vector \((8\mathbf{i} - 3\mathbf{j} + 5\mathbf{k})\) m.
Given that \(\mathbf{F}_1 = (4\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})\) N and \(\mathbf{F}_2 = (8\mathbf{i} - 4\mathbf{j} + 7\mathbf{k})\) N and that \(\mathbf{F}_1\) and \(\mathbf{F}_2\) are the only two forces acting on \(P\), find the velocity of \(P\) as it passes through \(B\), giving your answer as a vector. (7)
One end \(A\) of a light elastic string, of natural length \(a\) and modulus of elasticity \(6mg\), is fixed at a point on a smooth plane inclined at 30\(^\circ\) to the horizontal. A small ball \(B\) of mass \(m\) is attached to the other end of the string. Initially \(B\) is held at rest with the string lying along a line of greatest slope of the plane, with \(B\) below \(A\) and \(AB = a\). The ball is released and comes to instantaneous rest at a point \(C\) on the plane, as shown in Figure 2.
Find
(a) the length \(AC\), (5)
(b) the greatest speed attained by \(B\) as it moves from its initial position to \(C\). (7)
Mark scheme (a)
Scheme
Marks
Let \(x\) be the distance from the initial position of \(B\) to \(C\) GPE lost = EPE gained
\(mgx\sin 30^\circ = \dfrac{6mgx^2}{2a}\)
M1 A1=A1
Leading to \(x = \dfrac{a}{6}\)
M1
\(AC = \dfrac{7a}{6}\)
A1
(5)
Mark scheme (b)
Scheme
Marks
The greatest speed is attained when the acceleration of \(B\) is zero, that is where the forces on \(B\) are equal.
3. A block of mass 10 kg is pulled along a straight horizontal road by a constant horizontal force of magnitude 70 N in the direction of the road. The block moves in a straight line passing through two points \(A\) and \(B\) on the road, where \(AB = 50\) m. The block is modelled as a particle and the road is modelled as a rough plane. The coefficient of friction between the block and the road is \(\tfrac{4}{7}\).
(a) Calculate the work done against friction in moving the block from \(A\) to \(B\). (4)
The block passes through \(A\) with a speed of 2 m s\(^{-1}\).
(b) Find the speed of the block at \(B\). (4)
Mark scheme (a)
Scheme
Marks
\(R(\updownarrow): R = 10g\)
B1
\(F = \mu R \ \Rightarrow\ F = \dfrac{4}{7}(10g) = 56\)
B1
\(\therefore\) WD against friction \(= \dfrac{4}{7}(10g)(50)\)
A uniform rod \(AB\), of length \(2a\) and mass \(kM\) where \(k\) is a constant, is free to rotate in a vertical plane about the fixed point \(A\). One end of a light inextensible string of length \(6a\) is attached to the end \(B\) of the rod and passes over a small smooth pulley which is fixed at the point \(P\). The line \(AP\) is horizontal and of length \(2a\). The other end of the string is attached to a particle of mass \(M\) which hangs vertically below the point \(P\), as shown in Figure 3. The angle \(PAB\) is \(2\theta\), where \(0^\circ \leqslant \theta \leqslant 180^\circ\).
(a) Show that the potential energy of the system is\[Mga(4\sin\theta - k\sin 2\theta) + \text{constant}.\] (5)
The system has a position of equilibrium when \(\cos\theta = \frac{3}{4}\).
(b) Find the value of \(k\). (5)
(c) Hence find the value of \(\cos\theta\) at the other position of equilibrium. (3)
(d) Determine the stability of each of the two positions of equilibrium. (5)
A package of mass 3.5 kg is sliding down a ramp. The package is modelled as a particle and the ramp as a rough plane inclined at an angle of 20\(^\circ\) to the horizontal. The package slides down a line of greatest slope of the plane from a point \(A\) to a point \(B\), where \(AB = 14\) m. At \(A\) the package has speed 12 m s\(^{-1}\) and at \(B\) the package has speed 8 m s\(^{-1}\), as shown in Figure 1. Find
(a) the total energy lost by the package in travelling from \(A\) to \(B\), (5)
(b) the coefficient of friction between the package and the ramp. (5)
Mark scheme (a)
Scheme
Marks
\(\Delta\text{KE} = \dfrac{1}{2} \times 3.5(12^2 - 8^2)\ \ (= 140)\) or KE at A, B correct separately
B1
\(\Delta\text{PE} = 3.5 \times 9.8 \times 14\sin 20^\circ\ \ (\approx 164.238)\) or PE at A, B correct separately
M1 A1
\(\Delta E = \Delta\text{KE} + \Delta\text{PE} \approx 304,\ \ 300\)
DM1 A1
(5)
Mark scheme (b)
Scheme
Marks
Using Work-Energy
\(F_r = \mu \times 3.5g\cos 20^\circ\)
M1 A1
\(304.238\ldots = F_r \times 14\) ft their (a), \(F_r\)
1. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors.]
A small bead of mass 0.5 kg is threaded on a smooth horizontal wire. The bead is initially at rest at the point with position vector \((\mathbf{i} - 6\mathbf{j})\) m. A constant horizontal force \(\mathbf{P}\) N then acts on the bead causing it to move along the wire. The bead passes through the point with position vector \((7\mathbf{i} - 14\mathbf{j})\) m with speed \(2\sqrt{7}\) m s\(^{-1}\).
Given that \(\mathbf{P}\) is parallel to \((6\mathbf{i} + \mathbf{j})\), find \(\mathbf{P}\).
A light elastic spring, of natural length \(L\) and modulus of elasticity \(\lambda\), has a particle \(P\) of mass \(m\) attached to one end. The other end of the spring is fixed to a point \(O\) on the closed end of a fixed smooth hollow tube of length \(L\).
The tube is placed horizontally and \(P\) is held inside the tube with \(OP = \tfrac{1}{2}L\), as shown in Figure 1. The particle \(P\) is released and passes through the open end of the tube with speed \(\sqrt{(2gL)}\).
(a) Show that \(\lambda = 8mg\). (4)
The tube is now fixed vertically and \(P\) is held inside the tube with \(OP = \tfrac{1}{2}L\) and \(P\) above \(O\). The particle \(P\) is released and passes through the open top of the tube with speed \(u\).
4. A particle \(P\) of mass \(m\) lies on a smooth plane inclined at an angle 30\(^\circ\) to the horizontal. The particle is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(2mg\). The other end of the string is attached to a fixed point \(O\) on the plane. The particle \(P\) is in equilibrium at the point \(A\) on the plane and the extension of the string is \(\tfrac{1}{4}a\). The particle \(P\) is now projected from \(A\) down a line of greatest slope of the plane with speed \(V\). It comes to instantaneous rest after moving a distance \(\tfrac{1}{2}a\).
By using the principle of conservation of energy,
(a) find \(V\) in terms of \(a\) and \(g\), (6)
(b) find, in terms of \(a\) and \(g\), the speed of \(P\) when the string first becomes slack. (4)
Mark scheme (a)
Scheme
Marks
Energy equation with at least three terms, including K.E term
In parts (a) and (b) A marks need to have the correct signs
In part (b) for M1 need one KE term in energy equation of at least 3 terms with distance \(\dfrac{3a}{4}\) to indicate first method, and two KE terms in energy equation of at least 4 terms with distance \(\dfrac{a}{4}\) to indicate second method.
Alternative (using point of projection and point where string becomes slack):
3. A car of mass 1000 kg is moving at a constant speed of 16 m s\(^{-1}\) up a straight road inclined at an angle \(\theta\) to the horizontal. The rate of working of the engine of the car is 20 kW and the resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 550 N.
(a) Show that \(\sin\theta = \dfrac{1}{14}\). (5)
When the car is travelling up the road at 16 m s\(^{-1}\), the engine is switched off. The car comes to rest, without braking, having moved a distance \(y\) metres from the point where the engine was switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 550 N.
(b) Find the value of \(y\). (4)
Mark scheme (a)
Scheme
Marks
\(20\,000 = 16F\ \ (F = 1250)\)
M1 A1
\(\nearrow\) \(F = 550 + 1000 \times 9.8\sin\theta\) ft their \(F\)
1. A parcel of mass 2.5 kg is moving in a straight line on a smooth horizontal floor. Initially the parcel is moving with speed 8 m s\(^{-1}\). The parcel is brought to rest in a distance of 20 m by a constant horizontal force of magnitude \(R\) newtons. Modelling the parcel as a particle, find
(a) the kinetic energy lost by the parcel in coming to rest, (2)
(b) the value of \(R\). (3)
Mark scheme (a)
Scheme
Marks
KE lost is \(\dfrac{1}{2} \times 2.5 \times 8^2 = 80\) (J)
M1 A1
(2)
Mark scheme (b)
Scheme
Marks
Work energy \(80 = R \times 20\) ft their (a)
M1 A1 ft
\(R = 4\)
A1
(3)
(5 marks)
Alternative to (b)
\(0^2 = 8^2 - 2 \times a \times 20 \ \Rightarrow\ a = (-)1.6\)
A light elastic string, of natural length \(3l\) and modulus of elasticity \(\lambda\), has its ends attached to two points \(A\) and \(B\), where \(AB = 3l\) and \(AB\) is horizontal. A particle \(P\) of mass \(m\) is attached to the mid-point of the string. Given that \(P\) rests in equilibrium at a distance \(2l\) below \(AB\), as shown in Figure 1,
(a) show that \(\lambda = \dfrac{15mg}{16}\). (9)
The particle is pulled vertically downwards from its equilibrium position until the total length of the elastic string is \(7.8l\). The particle is released from rest.
(b) Show that \(P\) comes to instantaneous rest on the line \(AB\). (6)
Two particles \(A\) and \(B\), of mass \(m\) and \(2m\) respectively, are attached to the ends of a light inextensible string. The particle \(A\) lies on a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{3}{4}\). The string passes over a small light smooth pulley \(P\) fixed at the top of the plane. The particle \(B\) hangs freely below \(P\), as shown in Figure 2. The particles are released from rest with the string taut and the section of the string from \(A\) to \(P\) parallel to a line of greatest slope of the plane. The coefficient of friction between \(A\) and the plane is \(\dfrac{5}{8}\). When each particle has moved a distance \(h\), \(B\) has not reached the ground and \(A\) has not reached \(P\).
(a) Find an expression for the potential energy lost by the system when each particle has moved a distance \(h\). (2)
When each particle has moved a distance \(h\), they are moving with speed \(v\). Using the work-energy principle,
(b) find an expression for \(v^2\), giving your answer in the form \(kgh\), where \(k\) is a number. (5)
Mark scheme (a)
Scheme
Marks
PE lost \(= 2mgh - mgh\sin\alpha\ \ (= 7mgh/5)\)
M1 A1
(2)
Notes
M1 Two term expression for PE lost. Condone sign errors and sin/cos confusion, but must be vertical distance moved for A
A1 Both terms correct, \(\sin\alpha\) correct, but need not be simplified. Allow \(13.72mh\). Unambiguous statement.
B1 Normal reaction between A and the plane. Allow when seen in (b) provided it is clearly the normal reaction. Must use \(\cos\alpha\) but need not be substituted.
M1 (NB QUESTION SPECIFIES WORK & ENERGY) substitute into equation of the form PE lost = Work done against friction plus KE gained. Condone sign errors. They must include KE of both particles.
A1A1 All three elements correct (including signs)
A1A0 Two elements correct, but follow their GPE and \(\mu \times\) their \(R \times h\).
A1 \(V^2\) correct (NB \(kgh\) specified in the Q)
A framework consists of two uniform rods \(AB\) and \(BC\), each of mass \(m\) and length \(2a\), joined at \(B\). The mid-points of the rods are joined by a light rod of length \(a\sqrt{2}\), so that angle \(ABC\) is a right angle. The framework is free to rotate in a vertical plane about a fixed smooth horizontal axis. This axis passes through the point \(A\) and is perpendicular to the plane of the framework. The angle between the rod \(AB\) and the downward vertical is denoted by \(\theta\), as shown in Fig. 1.
(a) Show that the potential energy of the framework is \[-mga(3\cos\theta + \sin\theta) + \text{constant}.\] (4)
(b) Find the value of \(\theta\) when the framework is in equilibrium, with \(B\) below the level of \(A\). (4)
(c) Determine the stability of this position of equilibrium. (4)
1. A bead of mass 0.5 kg is threaded on a smooth straight wire. The only forces acting on the bead are a constant force \((4\mathbf{i} + 7\mathbf{j} + 2\mathbf{k})\) N and the normal reaction of the wire. The bead starts from rest at the point \(A\) with position vector \((\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})\) m and moves to the point \(B\) with position vector \((4\mathbf{i} + 3\mathbf{j} - 2\mathbf{k})\) m.
3. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(3.6mg\). The other end of the string is fixed at a point \(O\) on a rough horizontal table. The particle is projected along the surface of the table from \(O\) with speed \(\sqrt{(2ag)}\). At its furthest point from \(O\), the particle is at the point \(A\), where \(OA = \tfrac{4}{3}a\).
(a) Find, in terms of \(m\), \(g\) and \(a\), the elastic energy stored in the string when \(P\) is at \(A\). (3)
(b) Using the work-energy principle, or otherwise, find the coefficient of friction between \(P\) and the table. (6)
↓ marks a mark that depends on the M mark above it (an arrow in the scheme).
1st M1: allow for attempt to find work done by frictional force (i.e. not just finding friction). 2nd M1: “relevant” terms, i.e. energy or work terms! A1 f.t. on their work done by friction
1. A particle of mass 0.8 kg is moving in a straight line on a rough horizontal plane. The speed of the particle is reduced from 15 m s\(^{-1}\) to 10 m s\(^{-1}\) as the particle moves 20 m. Assuming that the only resistance to motion is the friction between the particle and the plane, find
(a) the work done by friction in reducing the speed of the particle from 15 m s\(^{-1}\) to 10 m s\(^{-1}\), (2)
(b) the coefficient of friction between the particle and the plane. (4)
7. A particle \(P\) has mass 4 kg. It is projected from a point \(A\) up a line of greatest slope of a rough plane inclined at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \tfrac{3}{4}\). The coefficient of friction between \(P\) and the plane is \(\tfrac{2}{7}\). The particle comes to rest instantaneously at the point \(B\) on the plane, where \(AB = 2.5\) m. It then moves back down the plane to \(A\).
(a) Find the work done by friction as \(P\) moves from \(A\) to \(B\). (4)
(b) Using the work-energy principle, find the speed with which \(P\) is projected from \(A\). (4)
(c) Find the speed of \(P\) when it returns to \(A\). (4)
5. Two light elastic strings each have natural length 0.75 m and modulus of elasticity 49 N. A particle \(P\) of mass 2 kg is attached to one end of each string. The other ends of the strings are attached to fixed points \(A\) and \(B\), where \(AB\) is horizontal and \(AB = 1.5\) m.
Figure 2
The particle is held at the mid-point of \(AB\). The particle is released from rest, as shown in Figure 2.
(a) Find the speed of \(P\) when it has fallen a distance of 1 m. (6)
Given instead that \(P\) hangs in equilibrium vertically below the mid-point of \(AB\), with \(\angle APB = 2\alpha\),
(b) show that \(\tan\alpha + 5\sin\alpha = 5\). (6)
Mark scheme (a)
Scheme
Marks
\(AP = \sqrt{\left(0.75^2 + 1^2\right)} = 1.25\)
M1 A1
Conservation of energy
\(\dfrac{1}{2} \times 2 \times v^2 + 2 \times \dfrac{49 \times 0.5^2}{2 \times 0.75} = 2g \times 1\) −1 for each incorrect term
M1 A2 (1, 0)
Leading to \(v \approx 1.8\) (m s\(^{-1}\)) accept 1.81
A uniform rod \(PQ\) has mass \(m\) and length \(2l\). A small smooth light ring is fixed to the end \(P\) of the rod. This ring is threaded on to a fixed horizontal smooth straight wire. A second small smooth light ring \(R\) is threaded on to the wire and is attached by a light elastic string, of natural length \(l\) and modulus of elasticity \(kmg\), to the end \(Q\) of the rod, where \(k\) is a constant.
(a) Show that, when the rod \(PQ\) makes an angle \(\theta\) with the vertical, where \(0 \lt \theta \leqslant \dfrac{\pi}{3}\), and \(Q\) is vertically below \(R\), as shown in Figure 1, the potential energy of the system is \[mgl\left[2k\cos^2\theta - (2k + 1)\cos\theta\right] + \text{constant}.\] (7)
Given that there is a position of equilibrium with \(\theta \gt 0\),
(b) show that \(k \gt \tfrac{1}{2}\). (5)
Mark scheme (a)
Scheme
Marks
PE of rod \(= -mgl\cos\theta\)
B1
EPE of string \(= \dfrac{kmg}{2l}(2l\cos\theta - l)^2\)
M1 A1
Total PE of system, \(\ V = -mgl\cos\theta + \dfrac{kmgl}{2}(2\cos\theta - 1)^2 + c\)
2. A particle of mass 0.5 kg is at rest at the point with position vector \((2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k})\) m. The particle is then acted upon by two constant forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\). These are the only two forces acting on the particle. Subsequently, the particle passes through the point with position vector \((4\mathbf{i} + 5\mathbf{j} - 5\mathbf{k})\) m with speed 12 m s\(^{-1}\). Given that \(\mathbf{F}_1 = (\mathbf{i} + 2\mathbf{j} - \mathbf{k})\) N, find \(\mathbf{F}_2\).
A smooth wire with ends \(A\) and \(B\) is in the shape of a semi-circle of radius \(a\). The mid-point of \(AB\) is \(O\). The wire is fixed in a vertical plane and hangs below \(AB\) which is horizontal. A small ring \(R\), of mass \(m\sqrt{2}\), is threaded on the wire and is attached to two light inextensible strings. The other end of each string is attached to a particle of mass \(\dfrac{3m}{2}\). The particles hang vertically under gravity, as shown in Figure 1.
(a) Show that, when the radius \(OR\) makes an angle \(2\theta\) with the vertical, the potential energy, \(V\), of the system is given by \[V = \sqrt{2}mga(3\cos\theta - \cos 2\theta) + \text{constant}.\] (7)
(b) Find the values of \(\theta\) for which the system is in equilibrium. (6)
(c) Determine the stability of the position of equilibrium for which \(\theta \gt 0\). (4)
Mark scheme (a)
Scheme
Marks
PE of R \(= -\sqrt{2}mga\cos 2\theta\ \ (+c)\qquad (1)\)
B1
PE of LH mass \(= -\dfrac{3}{2}mg(2a - 2a\sin(45 + \theta))\ \ (+c)\qquad (2)\)
M1 A1
PE of RH mass \(= -\dfrac{3}{2}mg(2a - 2a\sin(45 - \theta))\ \ (+c)\qquad (3)\)
A1
\(V = (1) + (2) + (3)\qquad\) (in terms of \(\theta\) etc.)
2. A small smooth sphere \(S\) of mass \(m\) is attached to one end of a light inextensible string of length \(2a\). The other end of the string is attached to a fixed point \(A\) which is at a distance \(a\sqrt{3}\) from a smooth vertical wall. The sphere \(S\) hangs at rest in equilibrium. It is then projected horizontally towards the wall with a speed \(\sqrt{\left(\dfrac{37ga}{5}\right)}\).
(a) Show that \(S\) strikes the wall with speed \(\sqrt{\left(\dfrac{27ga}{5}\right)}\). (4)
Given that the loss in kinetic energy due to the impact with the wall is \(\dfrac{3mga}{5}\),
(b) find the coefficient of restitution between \(S\) and the wall. (7)
where the unit of length is the metre. The bead is moved from a point \(A\) on the wire through a distance of 6 m along the wire to a point \(B\) by a force \(\mathbf{F} = (7\mathbf{i} + 4\mathbf{j} - 2\mathbf{k})\) N.
Find the magnitude of the work done by \(\mathbf{F}\) in moving the bead from \(A\) to \(B\).
1. A brick of mass 3 kg slides in a straight line on a horizontal floor. The brick is modelled as a particle and the floor as a rough plane. The initial speed of the brick is 8 m s\(^{-1}\). The brick is brought to rest after moving 12 m by the constant frictional force between the brick and the floor.
(a) Calculate the kinetic energy lost by the brick in coming to rest, stating the units of your answer. (2)
(b) Calculate the coefficient of friction between the brick and the floor. (4)
7. At a demolition site, bricks slide down a straight chute into a container. The chute is rough and is inclined at an angle of 30\(^\circ\) to the horizontal. The distance travelled down the chute by each brick is 8 m. A brick of mass 3 kg is released from rest at the top of the chute. When it reaches the bottom of the chute, its speed is 5 m s\(^{-1}\).
(a) Find the potential energy lost by the brick in moving down the chute. (2)
(b) By using the work-energy principle, or otherwise, find the constant frictional force acting on the brick as it moves down the chute. (5)
(c) Hence find the coefficient of friction between the brick and the chute. (3)
Another brick of mass 3 kg slides down the chute. This brick is given an initial speed of 2 m s\(^{-1}\) at the top of the chute.
(d) Find the speed of this brick when it reaches the bottom of the chute. (5)
Mark scheme (a)
Scheme
Marks
PE lost \(= 3 \times g \times 8\sin 30 = 3 \times g \times 8 \times 0.5 = 117.6\) J \(\approx\) 118J or 120J
5. A non-uniform rod \(BC\) has mass \(m\) and length \(3l\). The centre of mass of the rod is at distance \(l\) from \(B\). The rod can turn freely about a fixed smooth horizontal axis through \(B\). One end of a light elastic string, of natural length \(l\) and modulus of elasticity \(\dfrac{mg}{6}\), is attached to \(C\). The other end of the string is attached to a point \(P\) which is at a height \(3l\) vertically above \(B\).
(a) Show that, while the string is stretched, the potential energy of the system is \[mgl(\cos^2\theta - \cos\theta) + \text{constant},\] where \(\theta\) is the angle between the string and the downward vertical and \(-\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}\). (6)
(b) Find the values of \(\theta\) for which the system is in equilibrium with the string stretched. (6)
3. A light elastic string has natural length \(2l\) and modulus of elasticity \(4mg\). One end of the string is attached to a fixed point \(A\) and the other end to a fixed point \(B\), where \(A\) and \(B\) lie on a smooth horizontal table, with \(AB = 4l\). A particle \(P\) of mass \(m\) is attached to the mid-point of the string.
The particle is released from rest at the point of the line \(AB\) which is \(\dfrac{5l}{3}\) from \(B\). The speed of \(P\) at the mid-point of \(AB\) is \(V\).
(a) Find \(V\) in terms of \(g\) and \(L\). (7)
(b) Explain why \(V\) is the maximum speed of \(P\). (2)
Mark scheme (a)
Scheme
Marks
Elastic energy when \(P\) is at \(X\): \(E = \dfrac{4mg\left(\frac{2}{3}l\right)^2}{2l} + \dfrac{4mg\left(\frac{4}{3}l\right)^2}{2l}\ \ \left(= \dfrac{40mgl}{9}\right)\)
1. Two constant forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) are the only forces acting on a particle. \(\mathbf{F}_1\) has magnitude 9 N and acts in the direction of \(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\). \(\mathbf{F}_2\) has magnitude 18 N and acts in the direction of \(\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}\).
Find the total work done by the two forces in moving the particle from the point with position vector \((\mathbf{i} + \mathbf{j} + \mathbf{k})\) m to the point with position vector \((3\mathbf{i} + 2\mathbf{j} - \mathbf{k})\) m.
A smooth wire \(PMQ\) is in the shape of a semicircle with centre \(O\) and radius \(a\). The wire is fixed in a vertical plane with \(PQ\) horizontal and the mid-point \(M\) of the wire vertically below \(O\). A smooth bead \(B\) of mass \(m\) is threaded on the wire and is attached to one end of a light elastic string. The string has modulus of elasticity \(4mg\) and natural length \(\tfrac{5}{4}a\). The other end of the string is attached to a fixed point \(F\) which is a distance \(a\) vertically above \(O\), as shown in Fig. 1.
(a) Show that, when \(\angle BFO = \theta\), the potential energy of the system is \[\tfrac{1}{10}mga(8\cos\theta - 5)^2 - 2mga\cos^2\theta + \text{constant}.\] (6)
(b) Hence find the values of \(\theta\) for which the system is in equilibrium. (6)
(c) Determine the nature of the equilibrium at each of these positions. (5)
Mark scheme (a)
Scheme
Take \(O\) as zero p.e.
Mechanical potential energy \((mgh) = -mga\cos 2\theta\)
Total p.e. \(= -mga\left(2\cos^2\theta - 1\right) + \dfrac{8mg}{5a}\left(\dfrac{8a\cos\theta - 5a}{4}\right)^2\)
\(= -2mga\cos^2\theta + mga + \dfrac{mga}{10}(8\cos\theta - 5)^2\)
\(= \dfrac{mga}{10}(8\cos\theta - 5)^2 - 2mga\cos^2\theta + c \qquad\) (change of constant with referral of p.e. to any other zero position.)
Notes
The published mark scheme for this paper is a set of worked answers: no mark allocation is printed.
(Corrected from the printed mark scheme: the extension in the elastic potential energy is printed as \(2a\cos 2\theta - \tfrac{5}{4}a\); the string length is \(FB = 2a\cos\theta\), as used in the next line.)
Mark scheme (b)
Scheme
Equilibrium when p.e. is max/min so \(\dfrac{\mathrm{d}E}{\mathrm{d}\theta} = 0\)
When \(\theta = 0\), \(\ \dfrac{\mathrm{d}^2E}{\mathrm{d}\theta^2} = mga\left(8 - \tfrac{44}{5}\right) = -\tfrac{4}{5}mga\) which is \(\lt 0\) so max \(E\) so unstable.
5. A car of mass 1000 kg is towing a trailer of mass 1500 kg along a straight horizontal road. The tow-bar joining the car to the trailer is modelled as a light rod parallel to the road. The total resistance to motion of the car is modelled as having constant magnitude 750 N. The total resistance to motion of the trailer is modelled as of magnitude \(R\) newtons, where \(R\) is a constant. When the engine of the car is working at a rate of 50 kW, the car and the trailer travel at a constant speed of 25 m s\(^{-1}\).
(a) Show that \(R = 1250\). (3)
When travelling at 25 m s\(^{-1}\) the driver of the car disengages the engine and applies the brakes. The brakes provide a constant braking force of magnitude 1500 N to the car. The resisting forces of magnitude 750 N and 1250 N are assumed to remain unchanged. Calculate
(b) the deceleration of the car while braking, (3)
(c) the thrust in the tow-bar while braking, (2)
(d) the work done, in kJ, by the braking force in bringing the car and the trailer to rest. (4)
(e) Suggest how the modelling assumption that the resistances to motion are constant could be refined to be more realistic. (1)
Mark scheme (a)
Scheme
Marks
\(50\,000 = F \times 25\ \ (F = 2000)\) or equivalent
M1
\(\rightarrow\) \(F = R + 750\)
M1
\(R = 1250\ \ *\) cso
A1
(3)
Mark scheme (b)
Scheme
Marks
N2L \(1500 + 2000 = 2500a\) ignore sign of \(a\)
M1 A1
\(a = 1.4\ \ (\text{m s}^{-2})\) cao
A1
(3)
Mark scheme (c)
Scheme
Marks
Trailer: \(T + R = 1500 \times 1.4\) or Car: \(T - 1500 - 750 = 1000 \times -1.4\)
A small package \(P\) is modelled as a particle of mass 0.6 kg. The package slides down a rough plane from a point \(S\) to a point \(T\), where \(ST = 12\) m. The plane is inclined at an angle of 30\(^\circ\) to the horizontal and \(ST\) is a line of greatest slope of the plane, as shown in Figure 3. The speed of \(P\) at \(S\) is 10 m s\(^{-1}\) and the speed of \(P\) at \(T\) is 9 m s\(^{-1}\). Calculate
(a) the total loss of energy of \(P\) in moving from \(S\) to \(T\), (4)
(b) the coefficient of friction between \(P\) and the plane. (5)
Mark scheme (a)
Scheme
Marks
KE lost is \(\tfrac{1}{2} \times 0.6 \times (10^2 - 9^2)\ \ (= 5.7\ \text{J})\)
B1
PE lost is \(0.6 \times 9.8 \times 12\sin 30^\circ\ (= 35.28\ \text{J})\)
