Three small balls, \(A\), \(B\) and \(C\), have masses \(2m\), \(3m\) and \(4m\) respectively.
The balls are initially at rest in a straight line on a smooth horizontal surface, with \(B\) between \(A\) and \(C\), as shown in Figure 2.
Ball \(A\) is projected towards \(B\) with speed \(5u\) and \(A\) and \(B\) collide directly. Immediately after the collision, the speed of \(B\) is \(3u\)
The balls are modelled as uniform spheres with equal radii.
(a) Find, in terms of \(u\), the speed of \(A\) immediately after the collision with \(B\). (3)
(b) Find the coefficient of restitution between \(A\) and \(B\). (3)
After the collision between \(A\) and \(B\), ball \(C\) is projected towards \(B\) with speed \(u\) Balls \(B\) and \(C\) collide directly. The coefficient of restitution between \(B\) and \(C\) is \(f\)
Given that there is a second direct collision between \(A\) and \(B\)
(c) find the complete range of possible values of \(f\) (7)
M1: Use of CLM for \(A\) and \(B\), to form an equation in \(v\) and \(u\) (and \(m\)). Dimensionally correct with correct mass and velocity pairings. Condone sign errors. May use two impulse-momentum equations: \(-2m(v - 5u) = 3m(3u - 0)\)
A1: Correct unsimplified equation.
A1: Correct only, must be positive.
Mark scheme (b)
Scheme
Marks
AO
Use of NEL
M1
3.4
\(3u - v = e \times 5u\) OR \(3u + v = e \times 5u\)
A1
1.1b
\(\Rightarrow e = \dfrac{1}{2}\)
A1
1.1b
(3)
Notes
M1: Correct use of Impact Law for \(A\) and \(B\) to form an equation in \(u\) (and \(v\)). Allow consistent omission of \(u\). Dimensionally correct, condone sign errors on the velocities. M0 if separation and approach are on the wrong sides.
\(y - x = f \times (3u - -u)\) OR \(y + x = f \times (3u - -u)\)
A1
1.1b
E.g. \(\begin{cases} 3x + 4y = 5u \\ y - x = 4uf \end{cases} \;\Rightarrow\; x = \dfrac{u}{7}(5 - 16f)\) o.e. OR E.g. \(\begin{cases} -3x + 4y = 5u \\ y + x = 4uf \end{cases} \;\Rightarrow\; x = \dfrac{u}{7}(16f - 5)\) o.e.
M1
1.1b
\(v \gt x \;\Rightarrow\; \dfrac{1}{2}u \gt \dfrac{u}{7}(5 - 16f)\) OR \(v \gt -x \;\Rightarrow\; \dfrac{1}{2}u \gt -\dfrac{u}{7}(16f - 5)\)
dM1
3.1b
\(\dfrac{3}{32} \lt f \leqslant 1\)
A1
2.2a
(7)
(13 marks)
Notes
M1: Use CLM, dimensionally correct, with correct mass and velocity pairings. Must have all non-zero velocities. Condone sign errors. Follow their directions for the unknown velocities of \(B\) and \(C\) after impact. (Ignore the diagram if it benefits the candidate.)
A1: Correct unsimplified equation.
M1: Correct use of NEL for \(B\) and \(C\), dimensionally correct. Condone sign errors on velocity. Must have all non-zero velocities. Condone use of another letter for \(f\). M0 if separation and approach are on the wrong sides.
A1: Correct unsimplified equation, the directions of \(B\) and \(C\) after impact must be consistent with their CLM equation.
M1: Solve simultaneous equations with two unknowns to find an expression for the velocity of \(B\) after the second collision in terms of \(f\) and \(u\). Condone use of another letter for \(f\). Must use CLM and NEL to reach \(x = \ldots\) or a multiple of \(x = \ldots\)
dM1: Dependent on 3 previous M’s. Complete method to find the values of \(f\) for a second collision. Must consider \(A\) and \(B\) moving in the same direction and the speed of \(A\) > speed of \(B\).
A1: \(0.094 \lt f \leqslant 1\) (0.094 or better) Both ends required with correct inequality signs and must use \(f\) as coefficient of restitution.
A particle \(A\) of mass \(m\) is moving on a smooth horizontal plane when it collides directly with a fixed vertical wall which is perpendicular to the direction of motion of \(A\). Immediately after the collision, the speed of \(A\) is \(4u\), directly away from the wall, as shown in Figure 1.
The coefficient of restitution between \(A\) and the wall is \(e\).
Given that the magnitude of the impulse exerted on \(A\) by the wall in the collision is \(9mu\),
(a) find the value of \(e\). (5)
Figure 2
At the instant when \(A\) rebounds from the wall, a particle \(B\) of mass \(2m\) is projected along the plane towards \(A\), as shown in Figure 2.
The particles are moving in opposite directions along the same straight line when they collide directly. Immediately before the collision, the speed of A is \(4u\) and the speed of \(B\) is \(u\). Immediately after the collision, the total kinetic energy of the two particles is \(mu^2\)
(b) Determine, showing your method clearly, whether there are any further collisions between \(A\) and the wall. (7)
Mark scheme (a)
Scheme
Marks
AO
Use of impulse-momentum
M1
3.4
\(9mu = m(4u - (-v))\) OR \(9mu = m(4u - v)\)
A1
1.1b
\(e = \dfrac{4u}{v}\) oe OR \(e = \dfrac{4u}{-v}\) oe
A1
1.1b
Use of NEL
M1
3.4
\(e = \dfrac{4}{5}\) oe
A1
1.1b
(5)
Notes
N.B. In part (a), the first two (M1A1) marks are for use of impulse-momentum to produce an equation in (\(m\)), \(u\) and \(v\). The second two marks (A1M1) are for use of NEL to produce an equation in \(e\), \(u\) and \(v\).
M1: Correct no. of terms, condone sign errors. Allow consistent missing \(m\)’s. N.B. Must be using \(4u\).
A1: Correct impulse-momentum equation.
A1: Correct NEL equation. (Must be consistent with their other equation but \(v\) does not need to be substituted.
M1: Condone sign errors but M0 if ratio of speeds is inverted.
A1: cao
Mark scheme (b)
Scheme
Marks
AO
Use of CLM
M1
3.1a
\(4mu - 2mu = mv_A + 2mv_B\)
A1
1.1b
Use of sum of kinetic energies to obtain equation.
Solve for \(v_A\) (must be 0 or a multiple of \(u\)) N.B. Allow slips with signs, brackets etc but must come from a substitution method of solution of their equations to give a quadratic. Allow if they only have one solution. N.B. They may use NEL as well to find \(v_A\left[= \dfrac{2u}{3}(1-5e)\right]\) and \(v_B\left[= \dfrac{u}{3}(2+5e)\right]\) in terms of \(e\) then sub. these into the energy equation to give a quadratic in \(e\), then find \(e\) (\(= \dfrac{1}{5}\) or \(-\dfrac{1}{5}\), which is rejected) which is then used to find \(v_A\).
DM1
1.1b
\(v_A = 0\) or \(\dfrac{4}{3}u\) Need both unless they’ve used NEL and rejected \(\dfrac{4}{3}u\) as it comes from \(e = -\dfrac{1}{5}\). N.B. \(\dfrac{4}{3}u\) is impossible but candidates do not need to show that it is impossible, since it would imply no further collision of \(A\) with the wall.
A1
1.1b
No further collision of \(A\) with the wall, with a correct justification in both cases if they have the two possible values for \(v_A\).
A1
2.4
(7)
(12 marks)
Notes
M1: Correct no. of terms, condone sign errors. Allow consistent missing \(m\)’s. Treat an incorrect mass as an A error.
A1: Correct unsimplified equation. The \(4mu\) and \(2mu\) terms must have opposite signs but the signs on the other terms could be + or \(-\).
M1: Correct no. of terms. Must be ADDING the kinetic energies. Allow consistent missing \(m\)’s. Treat an incorrect mass as an A error.
A1: Correct unsimplified equation, \(v_A\) and \(v_B\) do not need to be substituted.
DM1: Solve for speed of \(A\), dependent on previous two M marks.
2. A particle \(Q\) of mass \(3m\) is at rest on a smooth horizontal plane. A particle \(P\) of mass \(m\) is moving along the plane when it collides directly with \(Q\).
The speed of \(P\) immediately before the collision is \(u\).
The direction of motion of \(P\) is reversed by the collision.
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(P\) immediately after the collision is \(\dfrac{u(3e-1)}{4}\) (6)
(b) State the full range of possible values of \(e\). (1)
Given that \(e = \dfrac{1}{2}\)
(c) find, in terms of \(m\) and \(u\), the magnitude of the impulse exerted by \(P\) on \(Q\) in the collision. (3)
Mark scheme (a)
Scheme
Marks
AO
\[\begin{array}{ccc} u \rightarrow & \qquad & \rightarrow 0 \\ (P)\ m & & 3m\ (Q) \\ v \leftarrow & & \rightarrow w \end{array}\]
Use of CLM
M1
3.1a
\(-mv + 3mw = mu\)
A1
1.1b
Use of NEL
M1
3.4
\(v + w = eu\)
A1
1.1b
Solve for \(v\):
M1
1.1b
\(\dfrac{u(3e-1)}{4}\)*
A1*
2.2a
(6)
Notes
N.B. When checking for consistency between their equations, mark the CLM equation FIRST.
M1: Use of CLM, with correct no. of terms, condone sign errors and consistent missing \(m\)’s
A1: Correct unsimplified equation. Allow \(v\) replaced by \(-v\)
M1: Use of NEL with \(e\) on the correct side of the equation, condone sign errors.
A1: Correct unsimplified equation consistent with CLM equation.
M1: Solve for \(v\) (must be dimensionally correct but allow slips in algebra)
A1*: Given answer correctly obtained, with no errors seen. Allow \(\dfrac{u}{4}(3e-1)\) or \(\dfrac{1}{4}u(3e-1)\) or \(\dfrac{u}{4}(-1+3e)\) or \(\dfrac{1}{4}u(-1+3e)\) or \(\dfrac{u(-1+3e)}{4}\) If they have \(v\) in the initial direction of \(P\) and obtain \(v = \dfrac{u(1-3e)}{4}\), we need to see a clear explanation of why the signs are changed.
Mark scheme (b)
Scheme
Marks
AO
\(1 \geqslant e \gt \dfrac{1}{3}\)
B1
2.2a
(1)
Notes
B1: cao
Mark scheme (c)
Scheme
Marks
AO
Use of impulse-momentum for \(P\) or \(Q\)
M1
3.1a
\(P\): \(\pm m(v+u)\) OR \(Q\): \(\pm 3mw\)
A1
1.1b
\(\dfrac{9mu}{8}\) or \(1\dfrac{1}{8}mu\)
A1
1.1b
(3)
(10 marks)
Notes
M1: Condone sign errors but must have correct terms (M0 if \(m\) omitted). M0 if \(m\) is used with \(w\) or \(3m\) is used with \(v\).
A1: \(\pm m(v+u)\) or \(\pm 3mw\). N.B. \(v\) and \(w\) do not need to be substituted.
A1: Accept \(1.1mu\) or better. N.B. Must be of form \(kmu\) and must be positive.
1. A particle \(A\) has mass \(4m\) and a particle \(B\) has mass \(3m\). The particles are moving along the same straight line on a smooth horizontal table. The particles are moving in opposite directions towards each other when they collide directly.
As a result of the collision, the direction of motion of each particle is reversed.
Immediately before the collision, the speed of \(A\) is \(u\) and the speed of \(B\) is \(ku\), where \(k\) is a constant.
Immediately after the collision, the speed of \(A\) is \(2v\) and the speed of \(B\) is \(3v\).
The magnitude of the impulse received by \(A\) in the collision is \(20mv\).
(a) Find \(u\) in terms of \(v\) only. (3)
(b) Find the exact value of \(k\). (3)
Mark scheme (a)
Scheme
Marks
AO
Impulse-momentum equation for \(A\) or other complete method to form an equation in \(u\) and \(v\) only (and \(m\))
M1
3.4
For \(A\): \(20mv = 4m\big(2v - (-u)\big)\)
A1
1.1b
\(u = 3v\)
A1
1.1b
(3)
Notes
Working in parts (a) and (b) may be marked together.
M1: Use of \(I = m(v-u)\) on \(A\) or equivalent complete method to form an equation in terms of \(u\) and \(v\) (and \(m\)) only (not \(k\)). Must consider change in momenta but condone in the wrong order or velocity sign errors. Might see \((2v+u)\) rather than \((2v - -u)\). Dimensionally correct equation with correct mass and velocities pairings. If CLM equation is formed in (a), then an impulse-equation must also be used to eliminate \(k\) for this mark. Allow consistent omission of \(m\).
A1: Correct unsimplified equation in \(u\) and \(v\) (and \(m\))
A1: Correct only, ISW after \(u = 3v\) A0 for only \(v = \dfrac{u}{3}\)
Mark scheme (b)
Scheme
Marks
AO
Impulse momentum equation for \(B\) or use of CLM to form an equation in \(k\) (and \(u\), \(v\), \(m\))
M1
3.4
Either For \(B\): \(20mv = 3m\big(3v - (-ku)\big)\) Or CLM: \(4m(u) - 3m(ku) = 3m(3v) - 4m(2v)\)
A1ft
1.1b
\(k = \dfrac{11}{9}\)
A1
1.1b
(3)
(6 marks)
Notes
Working in parts (a) and (b) may be marked together.
M1: Correct use of impulse-momentum equation or CLM to form an equation in \(k\), \(u\), \(v\) (and \(m\)). All terms required, allow consistent omission of \(m\). Dimensionally correct.
A1ft: Correct unsimplified equation in \(k\), \(u\) and \(v\). No need to replace \(u\) and \(v\) (follow through their positive \(u\) if substituted)
A1: Correct exact value. Accept \(1\dfrac{2}{9}\) or recurring decimal \(1.\dot{2}\) with correct notation.
4. A particle \(A\) of mass \(2m\) is moving in a straight line with speed \(3u\) on a smooth horizontal plane. Particle \(A\) collides directly with a particle \(B\) of mass \(m\) which is at rest on the plane.
The coefficient of restitution between \(A\) and \(B\) is \(e\), where \(e \gt 0\)
(a) Show that the speed of \(B\) immediately after the collision is \(2u(1 + e)\). (6)
After the collision, \(B\) hits a smooth fixed vertical wall which is perpendicular to the direction of motion of \(B\).
(b) Show that there will be a second collision between \(A\) and \(B\). (3)
The coefficient of restitution between \(B\) and the wall is \(\dfrac{1}{2}\)
Find, in simplified form, in terms of \(m\), \(u\) and \(e\),
(c) the magnitude of the impulse received by \(B\) in its collision with the wall, (3)
(d) the loss in kinetic energy of \(B\) due to its collision with the wall. (3)
M1: Use of CLM, all terms required, dimensionally correct (mass \(\times\) velocity in each term). Mass and velocity paired correctly. Condone sign errors on velocities. Condone consistent extra \(g\) and/or consistent missing \(m\) (in every term).
A1: Correct unsimplified equation.
M1: Correct use of Impact Law, dimensionally correct, condone sign errors on velocity. M0 if separation and approach are on the wrong sides.
A1: Correct unsimplified equation, the direction of \(A\) must be consistent with their CLM.
M1: Use their correctly formed equations to solve for \(v_B\)
A1*: Given answer correctly obtained and exactly as printed. Working should include an equation in \(v_B\) only before reaching the given answer.
Mark scheme (b)
Scheme
Marks
AO
Solve for \(v_A\)
M1
3.1a
\(v_A = u(2 - e)\) OR \(v_A = u(e - 2)\)
A1
1.1b
Complete and correct explanation: \(0 \leqslant e \leqslant 1 \;\Rightarrow\; v_A \gt 0\) \(\Rightarrow A\) continues to move towards the wall \(\Rightarrow A\) will collide again with \(B\). OR Complete and correct explanation: \(0 \leqslant e \leqslant 1 \;\Rightarrow\; v_A \lt 0\) \(\Rightarrow A\) continues to move towards the wall \(\Rightarrow A\) will collide again with \(B\).
A1
2.4
(3)
Notes
M1: Use given answer in (a) to solve for \(v_A\). If seen in (a) it must be used in (b) to earn this mark.
A1: Correct expression seen for velocity of \(A\) after impact.
A1*: Correct and complete explanation with no incorrect statements. Must include all of:
\(0 \lt e \leqslant 1\) or \(0 \leqslant e \leqslant 1\) or \(0 \leqslant e \lt 1\) or ‘for all \(e\)’
\(v_A \gt 0\) or \(v_A \lt 0\) (must be correct for their \(v_A\))
Must refer to the wall.
\(A\) continues to move with unchanged direction or towards the wall (eg do not accept descriptions for direction of travel as ‘to the right’ or similar)
Conclude second collision between \(A\) and \(B\)
Note: A0* for explanations that rely on comparing speed of \(v_B\) and \(v_A\)
Mark scheme (c)
Scheme
Marks
AO
Rebound speed or velocity of \(B = \pm\dfrac{1}{2} \times 2u(1 + e)\)
B1
3.4
\(\pm m\left[-u(1 + e) - 2u(1 + e)\right]\)
M1
3.1a
\(3(1 + e)mu\)
A1
1.1b
(3)
Notes
B1: Correct unsimplified expression for rebound speed or velocity of \(B\), may be seen on diagram or elsewhere in working. Accept positive or negative.
M1: Attempt to find difference in momenta, dimensionally correct (mass \(\times\) velocity). Must use given answer from (a) for Vb. Condone use of \(e\) or \(e'\) in place of ½
A1: Correct answer, must be positive. Accept any equivalent form that is either factorised or reduced to 2 terms eg \(3(1 + e)mu\), \(3mu + 3emu\)
M1: Attempt at KE loss of \(B\): clear attempt at a difference in KE of \(B\) before and after impact with the wall. Must use velocity of \(B\) from (a). Condone use of \(e\) or \(e'\) in place of ½. Dimensionally correct, condone subtraction either way round.
A1: Correct unsimplified expression for KE loss of \(B\)
A1: Any equivalent factorised form eg \(\dfrac{3mu^2(1 + e)^2}{2}\), \(\dfrac{3mu^2(1 + 2e + e^2)}{2}\), oe
A particle \(P\) of mass \(m\) and a particle \(Q\) of mass \(4m\) are at rest on a smooth horizontal plane, as shown in Figure 2.
Particle \(P\) is projected with speed \(u\) along the plane towards \(Q\) and the particles collide.
The coefficient of restitution between the particles is \(e\), where \(e \gt \dfrac{1}{4}\)
As a result of the collision, the direction of motion of \(P\) is reversed and \(P\) has speed \(\dfrac{u}{5}(4e-1)\).
(a) Find, in terms of \(u\) and \(e\), the speed of \(Q\) after the collision. (3)
After the collision, \(P\) goes on to hit a vertical wall which is fixed at right angles to the direction of motion of \(P\).
The coefficient of restitution between \(P\) and the wall is \(f\), where \(f \gt 0\)
Given that \(e = \dfrac{3}{4}\)
(b) find, in terms of \(m\), \(u\) and \(f\), the kinetic energy lost by \(P\) as a result of its impact with the wall. Give your answer in its simplest form. (4)
After its impact with the wall, \(P\) goes on to collide with \(Q\) again.
(c) Find the complete range of possible values of \(f\). (4)
M1: CLM: Correct no. of terms, condone consistent extra \(g\)’s, sign errors, cancelled \(m\)’s OR NEL: \(e\) on the correct side but condone sign errors
A1: Correct equation
A1: Cao. Accept any equivalent two term expression.
Mark scheme (b)
Scheme
Marks
AO
\(v_P = \pm\dfrac{fu}{5}(4e-1)\)
B1
3.3
\(= \pm\dfrac{2fu}{5}\)
B1
1.1b
KE Loss \(= \dfrac{1}{2}m\left(\dfrac{2u}{5}\right)^2 - \dfrac{1}{2}m\left(\dfrac{2fu}{5}\right)^2\)
M1
3.1a
\(= \dfrac{2mu^2}{25}(1 - f^2)\)
A1
1.1b
(4)
Notes
B1: Seen or implied.
B1: Seen or implied.
M1: Allow negative of this and without \(e\) being substituted. N.B. Allow anything of the form: \(\pm\left(\dfrac{1}{2}m\left(v_P\right)^2 - \dfrac{1}{2}m\left(fv_P\right)^2\right)\), provided that \(v_P\) has come from an attempt to put \(e = \tfrac{3}{4}\) in the given expression
A1: Cao. Accept any equivalent two term expression, isw
Mark scheme (c)
Scheme
Marks
AO
\(v_Q = \dfrac{7u}{20}\)
M1
1.1b
\(\dfrac{7u}{20} \lt \dfrac{2fu}{5}\)
M1
2.1
\(\dfrac{7}{8} \lt f \leqslant 1\)
A1 B1
1.1b 1.2
(4)
(11 marks)
Notes
M1: For attempt to put \(e = \dfrac{3}{4}\) in their \(v_Q\) expression to give a multiple of \(u\), seen or implied at some stage.
M1: Correct inequality for their speeds (which could involve \(e\)), provided it’s dimensionally correct Not available if \(Q\) is moving towards the wall.
A1: \(\dfrac{7}{8} \lt f\) oe
B1: For \(f \leqslant 1\)
N.B. All the marks are available if they go straight to \(\dfrac{7}{20} \lt \dfrac{2f}{5}\)
1. A particle \(A\) has mass \(2m\) and a particle \(B\) has mass \(3m\). The particles are moving in opposite directions along the same straight line and collide directly.
Immediately before the collision, the speed of \(A\) is \(2u\) and the speed of \(B\) is \(u\). Immediately after the collision, the speed of \(A\) is \(0.5u\) and the speed of \(B\) is \(w\).
Given that the direction of motion of each particle is reversed by the collision,
(a) find \(w\) in terms of \(u\) (3)
(b) find the coefficient of restitution between the particles, (3)
(c) find, in terms of \(m\) and \(u\), the magnitude of the impulse received by \(A\) in the collision. (3)
Mark scheme (a)
Scheme
Marks
AO
Use of CLM:
M1
3.1a
\(2m \times 2u - 3mu = -2m \times 0.5u + 3mw\)
OR: \((I =)\ \ 2m(0.5u - -2u) = 3m(w - -u)\)
A1
1.1b
\(w = \dfrac{2u}{3}\) accept \(0.67u\) or better
A1
1.1b
(3)
Notes
M1: Use of CLM, with correct no. of terms, condone consistent extra \(g\)’s, sign errors, cancelled \(m\)’s, to produce an equation in (\(m\)), \(u\) and \(w\) only. OR: Use of two imp-mom equations, with \(I\) then eliminated, to produce an equation in (\(m\)), \(u\) and \(w\) only, condone consistent extra \(g\)’s, sign errors, cancelled \(m\)’s. N.B. Allow the use of another letter for \(w\) in the working.
A1: Correct equation in (\(m\)), \(u\) and their \(w\) only.
A1: Cao. Must see \(w = ku\)
Mark scheme (b)
Scheme
Marks
AO
Use of NEL
M1
3.4
\(e = \dfrac{0.5u + \dfrac{2u}{3}}{2u + u}\)
A1ft
1.1b
\(\dfrac{7}{18}\) accept 0.39 or better
A1
1.1b
(3)
Notes
M1: Use of NEL, must be the right way up with the ‘correct’ terms but condone sign errors. Allow without the \(u\)’s
A1ft: Correct unsimplified expression, ft on their \(w\)
A1: cao
Mark scheme (c)
Scheme
Marks
AO
Use of impulse-momentum principle for \(A\) or \(B\)
M1
3.1a
\(\pm 2m(0.5u - -2u)\) OR \(\pm 3m\left(\dfrac{2u}{3} - -u\right)\)
A1
1.1b
\(5mu\)
A1
1.1b
(3)
(9 marks)
Notes
M1: Correct structure but condone sign errors (M0 if \(g\) included or \(m\)’s missing) Can score this mark if they use \(m\) for the mass with a ‘correct’ pair of velocities.
Three particles, \(P\), \(Q\) and \(R\), lie at rest on a smooth horizontal plane. The particles are in a straight line with \(Q\) between \(P\) and \(R\), as shown in Figure 1.
Particle \(P\) is projected towards \(Q\) with speed \(u\). At the same time, \(R\) is projected with speed \(\dfrac{1}{2}u\) away from \(Q\), in the direction \(QR\).
Particle \(P\) has mass \(m\) and particle \(Q\) has mass \(2m\).
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision between \(P\) and \(Q\) is \(\dfrac{u(1+e)}{3}\) (6)
It is given that \(e \gt \dfrac{1}{2}\)
(b) Determine whether there is a collision between \(Q\) and \(R\). (2)
(c) Determine the direction of motion of \(P\) immediately after the collision between \(P\) and \(Q\). (2)
(d) Find, in terms of \(m\), \(u\) and \(e\), the total kinetic energy lost in the collision between \(P\) and \(Q\), simplifying your answer. (3)
(e) Explain how using \(e = 1\) could be used to check your answer to part (d). (1)
Mark scheme (a)
Scheme
Marks
AO
\[\begin{array}{cc} u \rightarrow & \leftarrow 0 \\ (P)\ m & (Q)\,2m \\ \longleftarrow & \longrightarrow \\ v_P & v_Q \end{array}\]
M1: Correct no. of terms with correct pairings, condone sign errors, allow consistently cancelled \(m\)’s or extra \(g\)’s. M0 if they assume the same speeds
A1: Correct equation
M1: Correct no. of terms, condone sign errors. M0 if \(e\) on the wrong side of the equation
A1: Correct equation consistent with their CLM equation
Hence collision between \(Q\) and \(R\) will occur
A1
1.1b
(2)
Notes
M1: Use of \(e \gt \dfrac{1}{2}\) in answer to (a). Allow M1 for: if \(e = \tfrac{1}{2}\), \(v_Q = \dfrac{1}{2}u\) hence ( if \(e \gt \dfrac{1}{2}\) ), \(v_Q \gt \dfrac{1}{2}u\),
A1: cso
Mark scheme (c)
Scheme
Marks
AO
\(v_P = \dfrac{u(2e-1)}{3}\)
M1
3.1a
\(e \gt \dfrac{1}{2} \ \Rightarrow v_P \gt 0\) so \(P\) moves in the opposite direction to its original direction oe e.g. direction (motion) of \(P\) is reversed, opposite direction to \(Q\)
A0 for any of: direction changes, \(P\) moves away from \(Q\), goes left, \(P\) will travel in the negative direction, \(P\) moves backwards
M1: Correct no. of terms, dimensionally correct. Allow if \(m\) and \(2m\) are swapped or \(m\) used for mass of \(Q\) but otherwise correct. Allow an expression for the KE gain.
A1: Correct unsimplified expression
A1: Any correct simplified two term expression, not necessarily factorised.
Mark scheme (e)
Scheme
Marks
AO
The answer for part (d) should equal 0 when \(e = 1\).
B1
2.4
(1)
(14 marks)
Notes
B1: Correct explanation. This can be scored after an incorrect answer to (d). Need to say that when \(e = 1\), KE change = 0 Not enough to say KE Before = KE After
3. A particle \(P\) of mass \(2m\) is moving in a straight line with speed \(3u\) on a smooth horizontal plane. It collides directly with a particle \(Q\) of mass \(m\) that is moving on the plane with speed \(2u\) in the opposite direction to \(P\).
The coefficient of restitution between \(P\) and \(Q\) is \(e\), where \(e > \dfrac{4}{5}\)
(a) Show that the speed of \(Q\) immediately after the collision is \(\dfrac{(4 + 10e)u}{3}\) (6)
After the collision \(Q\) hits a smooth fixed vertical wall that is perpendicular to the direction of motion of \(Q\). The coefficient of restitution between \(Q\) and the wall is \(f\).
(b) Find, in terms of \(e\), the set of values of \(f\) for which there will be a second collision between \(P\) and \(Q\). (4)
M1: Attempt to solve for \(v_P\). If \(v_P\) is found in (a) it must be used in (b) to score this mark. Note that if \(P\) is assumed to reverse direction in (a) then \(v_P = \dfrac{(5e - 4)u}{3}\) oe
B1: Correct expression seen for speed or velocity of \(Q\) after rebound \(\pm\dfrac{f(4 + 10e)u}{3}\). This may appear on a diagram.
M1: Correct unsimplified inequality seen. The inequality must be correct, accepts cancelled \(u\)’s and/or 3’s
A1: Correct inequality, do not ISW. Allow \(1 \geqslant f\) to be omitted but do not allow the strict inequality \(1 > f\).
1. Two particles, \(P\) and \(Q\), of masses \(3m\) and \(2m\) respectively, are moving on a smooth horizontal plane. They are moving in opposite directions along the same straight line when they collide directly.
Immediately before the collision, \(P\) is moving with speed \(2u\).
The magnitude of the impulse exerted on \(P\) by \(Q\) in the collision is \(\dfrac{9mu}{2}\)
(a) Find the speed of \(P\) immediately after the collision. (3)
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
Given that the speed of \(Q\) immediately before the collision is \(u\),
(b) find the value of \(e\). (5)
Mark scheme (a)
Scheme
Marks
AO
\[\begin{array}{cccc} & 2u \rightarrow & \leftarrow u & \\ \dfrac{9mu}{2} \longleftarrow & (P)\ 3m & (Q)\,2m & \longrightarrow \dfrac{9mu}{2} \\ & \rightarrow v & \rightarrow y & \end{array}\]
Use of Impulse-momentum principle for \(P\)
M1
3.1a
\(-\dfrac{9mu}{2} = 3m(v - 2u)\)
A1
1.1b
\(v = \dfrac{u}{2}\)
A1
1.1b
(3)
Notes
M1: Use of Impulse-momentum principle for \(P\), condone sign errors but M0 if dimensionally incorrect e.g. if \(m\) missing or \(g\) included.
A1: Correct unsimplified equation (may have \(-v\))
A1: cao (must be positive)
Mark scheme (b)
Scheme
Marks
AO
Use of Impulse-momentum principle for \(Q\) or CLM
M1: Use of Impulse-momentum principle for \(Q\), must be using \(2m\), condone sign errors but M0 if dimensionally incorrect e.g. \(g\) included Or use of CLM, condone sign errors and consistent omission of \(m\)’s or consistent extra \(g\)’s, with their \(v\). Must be using correct masses on all four terms.
A1: Correct unsimplified equation
M1: Use of NEL with their \(v\) and their \(y\), condone sign errors but M0 if ratio for \(e\) is inverted
A1: Correct unsimplified equation in \(u\) and \(e\) only
5. Two particles, \(P\) and \(Q\), are moving in opposite directions along the same straight line on a smooth horizontal surface when they collide directly. The mass of \(P\) is \(3m\) and the mass of \(Q\) is \(4m\). Immediately before the collision the speed of \(P\) is \(2u\) and the speed of \(Q\) is \(u\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision is \(\dfrac{u}{7}(9e + 2)\) (6)
After the collision with \(P\), particle \(Q\) collides directly with a fixed vertical wall and rebounds. The wall is perpendicular to the direction of motion of \(Q\).
The coefficient of restitution between \(Q\) and the wall is \(\dfrac{1}{2}\)
(b) Find the complete range of possible values of \(e\) for which there is a second collision between \(P\) and \(Q\). (4)
M1: Use of CLM. Need all terms. Must be dimensionally correct. Condone sign errors. Accept consistent cancelling of \(m\)
A1: Correct unsimplified equation for CLM. They can have \(v\) in either direction
M1: Correct use of the impact law (used the right way round) Condone sign errors in finding speed of approach and speed of separation.
A1: Correct unsimplified equation. Signs consistent with equation for CLM.
M1: Complete method to find \(w\) e.g. by forming simultaneous equations using CLM and Impact Law and solving. This requires both of the preceding M marks
A1*: Obtain given answer from correct working. Accept with \(2 + 9e\) in place of \(9e + 2\) Check that the answer does follow from the working.
4. A particle \(P\) of mass \(2m\) kg is moving with speed \(2u\ \text{m s}^{-1}\) on a smooth horizontal plane. Particle \(P\) collides with a particle \(Q\) of mass \(3m\) kg which is at rest on the plane. The coefficient of restitution between \(P\) and \(Q\) is \(e\). Immediately after the collision the speed of \(Q\) is \(v\ \text{m s}^{-1}\)
(a) Show that \(v = \dfrac{4u(1+e)}{5}\) (6)
(b) Show that \(\dfrac{4u}{5} \leqslant v \leqslant \dfrac{8u}{5}\) (2)
Given that the direction of motion of \(P\) is reversed by the collision,
(c) find, in terms of \(u\) and \(e\), the speed of \(P\) immediately after the collision. (2)
After the collision, \(Q\) hits a wall, that is fixed at right angles to the direction of motion of \(Q\), and rebounds.
The coefficient of restitution between \(Q\) and the wall is \(\dfrac{1}{6}\)
Given that \(P\) and \(Q\) collide again,
(d) find the full range of possible values of \(e\). (5)
Mark scheme (a)
Scheme
Marks
AO
\[\begin{array}{cc} 2u \rightarrow & 0 \\ P\ (2m) & Q(3m) \\ w \leftarrow & \rightarrow v \end{array}\]
Use of CLM
M1
3.4
\(2m \times 2u = -2mw + 3mv\)
A1
1.1b
Use of NEL
M1
3.4
\(2ue = w + v\)
A1
1.1b
Solve for \(v\)
D M1
1.1b
\(v = \dfrac{4u(1+e)}{5}\)*
A1*
2.2a
(6)
Notes
M1: Correct no. of terms, condone sign errors, allow consistently cancelled \(m\)’s or extra \(g\)’s or common factors throughout
A1: Correct equation; they may have \(w\) instead of \(-w\)
M1: Correct no. of terms, condone sign errors. M0 if \(e\) on the wrong side of the equation
A1: Correct equation; they may have \(w\) instead of \(-w\)
DM1: Solve for \(v\), dependent on previous two marks
A1*: Correct answer correctly obtained
Mark scheme (b)
Scheme
Marks
AO
Since \(0 \leqslant e \leqslant 1\), \(\dfrac{4u(1+0)}{5} \leqslant v \leqslant \dfrac{4u(1+1)}{5}\)
M1
3.1a
i.e. \(\dfrac{4u}{5} \leqslant v \leqslant \dfrac{8u}{5}\)*
A1*
2.2a
(2)
Notes
M1: Use of \(0 \leqslant e \leqslant 1\) in the given answer; allow use of \(e = 0\) and \(e = 1\) to obtain the min and max expressions M1A0 for ‘verification’.
A1*: Correct answer correctly obtained (including use of max and min)
Mark scheme (c)
Scheme
Marks
AO
Solve for \(w\)
M1
1.1b
\(w = \dfrac{2u(3e-2)}{5}\) oe \((\text{m s}^{-1})\) or \(\left|\dfrac{2u(2-3e)}{5}\right|\) oe
A1
1.1b
(2)
Notes
M1: Solve for their \(w\)
A1: cao
Mark scheme (d)
Scheme
Marks
AO
Speed of \(Q\) after hitting the wall \(= \dfrac{1}{6}v\ \ (\text{m s}^{-1})\)
M1
3.4
For a further collision between \(P\) and \(Q\), \(\dfrac{1}{6}v \gt w\)
M1
3.1a
Substitute for \(v\) and \(w\) and solve for \(e\)
M1
1.1b
\(e \lt \dfrac{7}{8}\)
A1
1.1b
\(\dfrac{2}{3} \lt e \lt \dfrac{7}{8}\)
A1
1.1b
(5)
(15 marks)
Notes
M1: Speed so must see a positive quantity M0 if \(\dfrac{1}{6}\) is on the wrong side of the equation
M1: Correct inequality for their \(w\) (allow even if their \(w\) is dimensionally incorrect)
M1: Independent M mark but must have an inequality in \(v\) and \(w\): Substitute for \(v\), using given answer, and \(w\) and solve for \(e\)
2. Two particles, \(A\) and \(B\), have masses \(m\) and \(3m\) respectively. The particles are moving in opposite directions along the same straight line on a smooth horizontal plane when they collide directly.
Immediately before they collide, \(A\) is moving with speed \(2u\) and \(B\) is moving with speed \(u\).
The direction of motion of each particle is reversed by the collision.
In the collision, the magnitude of the impulse exerted on \(A\) by \(B\) is \(\dfrac{9mu}{2}\)
(a) Find the value of the coefficient of restitution between \(A\) and \(B\). (7)
(b) Hence, write down the total loss in kinetic energy due to the collision, giving a reason for your answer. (1)
Mark scheme (a)
Scheme
Marks
AO
\[\begin{array}{cccc} & 2u \rightarrow & \leftarrow u & \\ \dfrac{9mu}{2} \longleftarrow & m & 3m & \longrightarrow \dfrac{9mu}{2} \\ & v \leftarrow & \rightarrow w & \end{array}\]
Use of Impulse-momentum principle for \(A\) or \(B\)
N.B.Ignore diagrams if it helps the candidate. Equations need to be consistent, where appropriate, to earn A marks.
M1: Use of Impulse-momentum principle for \(A\) or \(B\), condone sign errors but M0 if dimensionally incorrect e.g. if \(m\) missing
A1: Correct unsimplified equation
M1: Use of Impulse-momentum principle for other particle or CLM, condone sign errors but M0 if dimensionally incorrect e.g. if \(m\) missing from impulse For CLM, allow consistent missing \(m\)’s or extra \(g\)’s.
A1: Correct unsimplified equation
A1: Cao for both. Allow one or both negative if correct for their symbols.
M1: Use of NEL to obtain \(e = \ldots\), condone sign errors in numerator but must be terms in \(u\) only AND must be \((2u + u)\) in denominator. M0 if inverted
A1: cso
ALTERNATIVE
Scheme
Marks
AO
NEL is written down before \(v\) and \(w\) are found: \(v + w = 3ue\)
3rd M1
Use of Impulse-momentum principle for \(A\) or \(B\)
Use of Impulse-momentum principle for \(B\) or \(A\) or CLM
2nd M1
\(\dfrac{9mu}{2} = 3m(w - -u)\) or \(\dfrac{9mu}{2} = m(v - -2u)\) or \(2mu - 3mu = -mv + 3mw\)
2nd A1
An equation (not an identity) in \(u\) and \(e\) only is produced
3rd A1
\(e = 1\)
A1cso
Mark scheme (b)
Scheme
Marks
AO
Perfectly elastic (or the coefficient of restitution is 1) so no loss in kinetic energy. Allow a direct evaluation of the KE loss i.e. \(\dfrac{1}{2}m(2u)^2 + \dfrac{1}{2} \times 3mu^2 - \left(\dfrac{1}{2}m\left(\dfrac{5u}{2}\right)^2 + \dfrac{1}{2} \times 3m\left(\dfrac{u}{2}\right)^2\right) = 0\) B0 if incorrect extras
DB1
2.4
(1)
(8 marks)
Notes
N.B.Ignore diagrams if it helps the candidate. Equations need to be consistent, where appropriate, to earn A marks.
DB1: Dependent on \(e = 1\) correctly obtained in (a) A correct statement e.g. zero, 0 etc and a correct reason
1. A particle \(A\) of mass \(3m\) and a particle \(B\) of mass \(m\) are moving along the same straight line on a smooth horizontal surface. The particles are moving in opposite directions towards each other when they collide directly.
Immediately before the collision, the speed of \(A\) is \(ku\) and the speed of \(B\) is \(u\). Immediately after the collision, the speed of \(A\) is \(v\) and the speed of \(B\) is \(2v\).
The magnitude of the impulse received by \(B\) in the collision is \(\dfrac{3}{2}mu\).
(a) Find \(v\) in terms of \(u\) only. (3)
(b) Find the two possible values of \(k\). (5)
Mark scheme (a)
Scheme
Marks
AO
Note that if they start with their 2v to the left this creates an impossible situation (the particles need to pass through each other). The maximum score is M1M1M1.
Impulse received by \(B\):
M1
3.4
\(\dfrac{3}{2}mu = m\big(2v - (-u)\big)\)
A1
1.1b
\(v = \dfrac{u}{4}\)
A1
1.1b
(3)
Notes
M1: Form impulse-momentum equation for \(B\) (or \(A\)). May be expressed as either \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) or \(\mathbf{I} + m\mathbf{u} = m\mathbf{v}\). Dimensionally correct. Must be considering difference in velocities Must have a correct combination of mass and velocity: pairing velocity of one with the mass of the other scores M0 Allow for subtraction the wrong way round or impulse in the wrong direction. Assuming that you have not seen an incorrect formula stated, allow for \(2v + u\) without overt evidence of subtraction. Allow if the common factor of \(m\) is not seen
A1: Correct unsimplified equation for \(B\) (or \(A\)). Allow without \(m\)
A1: Correct answer only
Mark scheme (b)
Scheme
Marks
AO
Use of CLM or Impulse-momentum for one option for \(A\):
M1
3.4
\(3kmu - mu = 2mv + 3mv\left(= \dfrac{5mu}{4}\right)\) or \(3m(v - ku) = -\dfrac{3mu}{2}\) \(\left(3mu\left(\dfrac{1}{4} + \dfrac{1}{2}\right) = 3mku\right)\)
A1ft
1.1b
\(k = \dfrac{3}{4}\)
A1
1.1b
Form a second equation in \(k\) \(\left(3mku - mu = 2mv - 3mv\left(= -\dfrac{mu}{4}\right)\ \text{or}\ 3m(v + ku) = \dfrac{3mu}{2}\right)\)
M1
3.1a
\(k = \dfrac{1}{4}\)
A1
1.1b
(5)
(8 marks)
Notes
M1: Correct method to form an equation in \(k\). Must be dimensionally correct Condone sign errors in CLM. Allows marks for CLM equation here if seen in (a) and used correctly to find \(k\) here. Rules for impulse-momentum as above. M1 is available if they have not reversed the direction of the impulse. An equation which allows for the change in direction by using \(\mathbf{u} - \mathbf{v}\) can score full marks. Could be working with either option for the direction of motion of \(A\)
A1ft: Correct unsimplified equation in \(u\), \(v\) or their \(v\)
A1: One correct solution Be aware that a sign error in the impulse-momentum equation for \(A\) can lead to a fortuitous answer. A fortuitous answer scores A0 (FYI the incorrect answers are \(\tfrac{-7}{4}\) and \(\tfrac{1}{4}\))
M1: Correct method to form a second equation in \(k\) (reversing the direction of motion of \(A\))
2. Two particles, \(A\) and \(B\), are moving in opposite directions along the same straight line on a smooth horizontal surface when they collide directly.
Particle \(A\) has mass \(5m\) and particle \(B\) has mass \(3m\).
The coefficient of restitution between \(A\) and \(B\) is \(e\), where \(e \gt 0\)
Immediately after the collision the speed of \(A\) is \(v\) and the speed of \(B\) is \(2v\).
Given that \(A\) and \(B\) are moving in the same direction after the collision,
(a) find the set of possible values of \(e\). (8)
Given also that the kinetic energy of \(A\) immediately after the collision is 16% of the kinetic energy of \(A\) immediately before the collision,
(b) find
(i) the value of \(e\),
(ii) the magnitude of the impulse received by \(A\) in the collision, giving your answer in terms of \(m\) and \(v\).
3. Two particles, \(A\) and \(B\), have masses \(3m\) and \(4m\) respectively. The particles are moving in the same direction along the same straight line on a smooth horizontal surface when they collide directly. Immediately before the collision the speed of \(A\) is \(2u\) and the speed of \(B\) is \(u\).
The coefficient of restitution between \(A\) and \(B\) is \(e\).
(a) Show that the direction of motion of each of the particles is unchanged by the collision. (8)
After the collision with \(A\), particle \(B\) collides directly with a third particle, \(C\), of mass \(2m\), which is at rest on the surface.
The coefficient of restitution between \(B\) and \(C\) is also \(e\).
(b) Show that there will be a second collision between \(A\) and \(B\). (6)
3. Three particles \(A\), \(B\) and \(C\) are at rest on a smooth horizontal plane. The particles lie along a straight line with \(B\) between \(A\) and \(C\).
Particle \(B\) has mass \(4m\) and particle \(C\) has mass \(km\), where \(k\) is a positive constant. Particle \(B\) is projected with speed \(u\) along the plane towards \(C\) and they collide directly.
The coefficient of restitution between \(B\) and \(C\) is \(\dfrac{1}{4}\)
(a) Find the range of values of \(k\) for which there would be no further collisions. (8)
The magnitude of the impulse on \(B\) in the collision between \(B\) and \(C\) is \(3mu\)
1. Two particles \(P\) and \(Q\) have masses \(m\) and \(4m\) respectively. The particles are at rest on a smooth horizontal plane. Particle \(P\) is given a horizontal impulse, of magnitude \(I\), in the direction \(PQ\). Particle \(P\) then collides directly with \(Q\). Immediately after this collision, \(P\) is at rest and \(Q\) has speed \(w\). The coefficient of restitution between the particles is \(e\).
(a) Find \(I\) in terms of \(m\) and \(w\). (2)
(b) Show that \(e = \dfrac{1}{4}\) (1)
(c) Find, in terms of \(m\) and \(w\), the total kinetic energy lost in the collision between \(P\) and \(Q\). (2)
Mark scheme (a)
Scheme
Marks
AO
Use of CLM: \(m \times \dfrac{I}{m} = 4mw\)
M1
3.1a
\(I = 4mw\)
A1
1.1b
(2)
Notes
M1: Correct no. of terms, condone extra \(g\) s, sign errors (must be equation in \(I\), \(m\) and \(w\) only)
A1: Correct equation
Answer not given, so a correct answer with no clear error seen will score M1A1 An answer that relies on an impulse-momentum equation using \(4m\) will score M0
Mark scheme (b)
Scheme
Marks
AO
\(e = \dfrac{w}{4w} = \dfrac{1}{4}\) *
B1*
3.4
(1)
Notes
B1*: Use of NLR to obtain given answer
Mark scheme (c)
Scheme
Marks
AO
KE Loss \(= \dfrac{1}{2}m(4w)^2 - \dfrac{1}{2}4mw^2\)
5. A particle \(P\) of mass \(3m\) and a particle \(Q\) of mass \(2m\) are moving along the same straight line on a smooth horizontal plane. The particles are moving in opposite directions towards each other and collide directly.
Immediately before the collision the speed of \(P\) is \(u\) and the speed of \(Q\) is \(2u\).
Immediately after the collision \(P\) and \(Q\) are moving in opposite directions.
The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Find the range of possible values of \(e\), justifying your answer. (8)
Given that \(Q\) loses 75% of its kinetic energy as a result of the collision,
(b) find the value of \(e\). (3)
Mark scheme (a)
Scheme
Marks
AO
Use of CLM
M1
3.1a
\(3mu - 4mu = 2mw - 3mv\) \((-u = -3v + 2w)\)
A1
1.1b
Use of impact law
M1
3.4
\(w + v = 3ue\)
A1
1.1b
Correct strategy to form equation in \(w\) and find critical value of \(e \in (0, 1)\) \(\big(5w = u(9e - 1)\big)\)
M1
3.1a
\(w \gt 0: e \gt \dfrac{1}{9}\)
A1
1.1b
Complete strategy to justify the range of values of \(e\) \(\big(5v = u(1 + 6e)\big)\) \(v \gt 0\): true for all \(e\)
M1
3.1a
Therefore \(\dfrac{1}{9} \lt e \leqslant 1\)
A1
2.2a
(8)
Notes
M1: Use of CLM. All terms required. Must be dimensionally correct. Condone sign errors
A1: Correct unsimplified equation
M1: Use of impact law. Must be dimensionally correct and used correctly. Condone sign errors
A1: Correct unsimplified equation Signs consistent with CLM equation
M1: Correct overall strategy to find the critical value of \(e\) in \((0, 1)\) in \(e\) eg by using CLM and impact law to form equation or inequality in \(w\) and solve for \(e\).
A1: One inequality for \(e\) correct Condone \(e \geqslant \dfrac{1}{9}\)
M1: Correct strategy to find the range of possible value of \(e\). i.e find second speed and form second inequality
4. Three particles, \(P\), \(Q\) and \(R\), are at rest on a smooth horizontal plane. The particles lie along a straight line with \(Q\) between \(P\) and \(R\). The particles \(Q\) and \(R\) have masses \(m\) and \(km\) respectively, where \(k\) is a constant.
Particle \(Q\) is projected towards \(R\) with speed \(u\) and the particles collide directly.
The coefficient of restitution between each pair of particles is \(e\).
(a) Find, in terms of \(e\), the range of values of \(k\) for which there is a second collision. (9)
Given that the mass of \(P\) is \(km\) and that there is a second collision,
(b) write down, in terms of \(u\), \(k\) and \(e\), the speed of \(Q\) after this second collision. (1)
Mark scheme (a)
Scheme
Marks
AO
Use of conservation of momentum
M1
3.1a
\(mu = -mv_Q + kmv_R\)
A1
1.1b
Use of NLR
M1
3.4
\(eu = v_Q + v_R\)
A1
1.1b
Using correct strategy to solve problem by finding \(v_Q\)
M1
3.1a
\(v_Q = \dfrac{u(ke-1)}{k+1}\) or \(v_Q = \dfrac{v_R(ke-1)}{1+e}\)
A1
1.1b
For second collision, \(v_Q \gt 0\)
M1
3.1a
\(\dfrac{u(ke-1)}{k+1} \gt 0\)
M1
1.1b
\(k \gt \dfrac{1}{e}\)
A1
1.1b
(9)
Notes
M1: Correct no. of terms and dimensionally correct but condone sign errors
2. Two particles, \(A\) and \(B\), of masses \(2m\) and \(3m\) respectively, are moving on a smooth horizontal plane. The particles are moving in opposite directions towards each other along the same straight line when they collide directly. Immediately before the collision the speed of \(A\) is \(2u\) and the speed of \(B\) is \(u\). In the collision the impulse of \(A\) on \(B\) has magnitude \(5mu\).
(a) Find the coefficient of restitution between \(A\) and \(B\). (9)
(b) Find the total loss in kinetic energy due to the collision. (4)
Could find \(v_A\) before \(v_B\): M1A1A1 for first velocity, M1A1A1 for second M1A1A1 for \(e\) found correctly
Candidates are approaching this in many different ways. They need - two of momentum impulse equation for each particle and CLM - impact law M1A1 for each correct equation (in the order seen) Of the remaining 3 A marks, A1 for a correct expression for \(v_A\) or \(v_B\) A1 for a correct expression in \(e\) A1 for the correct answer
M1: Correct no. of terms and must be a difference. Must be dimensionally correct at the point when they state their expression for the loss (change) in KE
A1ft: Unsimplified expression in \(u\) with at most 1 error, ft on their speeds from (a)
A1ft: Correct unsimplified expression in \(u\). (These first 3 marks can be scored for a correct loss or gain in KE), ft on their speeds from (a)
4. A particle \(P\) of mass \(3m\) is moving in a straight line on a smooth horizontal floor. A particle \(Q\) of mass \(5m\) is moving in the opposite direction to \(P\) along the same straight line.
The particles collide directly.
Immediately before the collision, the speed of \(P\) is \(2u\) and the speed of \(Q\) is \(u\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision is \(\dfrac{u}{8}(9e + 1)\) (6)
(b) Find the range of values of \(e\) for which the direction of motion of \(P\) is not changed as a result of the collision. (2)
When \(P\) and \(Q\) collide they are at a distance \(d\) from a smooth fixed vertical wall, which is perpendicular to their direction of motion. After the collision with \(P\), particle \(Q\) collides directly with the wall and rebounds so that there is a second collision between \(P\) and \(Q\). This second collision takes place at a distance \(x\) from the wall.
Given that \(e = \dfrac{1}{18}\) and the coefficient of restitution between \(Q\) and the wall is \(\dfrac{1}{3}\)
(c) find \(x\) in terms of \(d\). (6)
Mark scheme (a)
Scheme
Marks
AO
Complete strategy to find speed of \(Q\)
M1
3.1b
Use of CLM
M1
3.1a
\(6mu - 5mu\ (= mu) = 3mv + 5mw\)
A1
1.1b
Use of impact law
M1
3.1a
\(w - v = 3ue\)
A1
1.1b
\(\left.\begin{array}{l} 3v + 5w = u \\ 3w - 3v = 9ue \end{array}\right\} \Rightarrow 8w = u + 9ue, \quad w = \dfrac{u}{8}(9e + 1)\) *
A1*
2.1
(6)
Notes
M1: Complete strategy e.g. use of CLM, impact law and solution of simultaneous equations.
M1: CLM equation. Requires all terms and dimensionally correct. Condone sign errors.
A1: Correct unsimplified equation
M1: Impact law. Condone sign error. Must be used the right way round.
A1: Correct unsimplified equation Signs consistent with CLM equation.
A1*: Obtain given answer from correct working
Mark scheme (b)
Scheme
Marks
AO
\(v = w - 3ue = \dfrac{u}{8}(1 - 15e)\) and \(v \gt 0\)
M1
3.1b
\(\Rightarrow (0 \leqslant)\ e \lt \dfrac{1}{15}\)
A1
1.1b
(2)
Notes
M1: Find speed of \(P\) and form correct inequality consistent with their directions.
A1: Correct solution. Need not mention the lower limit.
Mark scheme (c)
Scheme
Marks
AO
Complete strategy to find time for \(Q\) to get to second collision
M1
3.1a
Speed of \(Q\) after impact with wall \(= \dfrac{u}{16}\)
B1
1.1b
Time for Q: \(\dfrac{16d}{3u} + \dfrac{16x}{u}\) follow their \(\dfrac{u}{16}\) and \(\dfrac{16d}{3u}\)
A1ft
1.1b
Complete strategy to find time for \(P\) to get to second collision \(= \dfrac{48(d - x)}{u}\)
B1ft
1.1b
Use both at the same place at the same
M1
2.1
\(x = \dfrac{128d}{192} = \dfrac{2d}{3}\)
A1
1.1b
(6)
(14 marks)
Notes
M1: Complete strategy e.g. find time to wall and back again
B1: Correct use of impact law
A1ft: Correct unsimplified equation using \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\) and following their \(\dfrac{u}{16}\) and \(\dfrac{16d}{3u}\)
B1ft: Correct use of \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\) Follow their \(\dfrac{u}{48}\)
M1: find \(x\) by putting both particles in the same place at the same time. Must be valid expressions for the times.
A1: Correct answer or exact equivalent
Alternative (c)
Scheme
Marks
AO
Complete strategy to find position of second collision
M1
3.1a
Speed of \(Q\) after impact with wall \(= \dfrac{u}{16}\)
B1
1.1b
Distance apart when Q strikes the wall \(= \dfrac{8d}{9}\)
5. A particle \(A\) of mass \(3m\) is moving in a straight line with speed \(2u\) on a smooth horizontal floor. Particle \(A\) collides directly with another particle \(B\) of mass \(2m\) which is moving along the same straight line with speed \(u\) but in the opposite direction to \(A\). The coefficient of restitution between \(A\) and \(B\) is \(\dfrac{1}{3}\).
(a)
(i) Show that the speed of \(B\) immediately after the collision is \(\dfrac{7}{5}u\)
(ii) Find the speed of \(A\) immediately after the collision. (7)
After the collision, \(B\) hits a smooth vertical wall which is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the wall is \(\dfrac{1}{2}\). The first collision between \(A\) and \(B\) occurred at a distance \(x\) from the wall. The particles collide again at a distance \(y\) from the wall.
DM1 Dependent on previous M1. Form equation in \(x\) and \(y\)
A1 Or equivalent. 0.45x or better
Alt 2
Speed of B after collision with wall: \(\dfrac{1}{2} \times \dfrac{7}{5}u = \left(\dfrac{7}{10}u\right)\)
B1
\(x -\) distance moved by \(A = x - \dfrac{2}{5}u \times \dfrac{5x}{7u} = \dfrac{5}{7}x\)
B1
Gap closing at \(\ \dfrac{7}{10}u + \dfrac{2}{5}u = \dfrac{11}{10}u\)
Time to collision: \(\ \left(\dfrac{5}{7}x\right) \div \left(\dfrac{11}{10}u\right) = \dfrac{50x}{77u}\)
M1A1
Distance moved by \(B\): \(\ y = \dfrac{7}{10}u \times \dfrac{50x}{77u} = \dfrac{5}{11}x\)
DM1 A1
B1 Accept +/-
B1 Distance apart when \(B\) hits the wall
M1A1 Use of \(\dfrac{9x}{7}\) for \(\dfrac{5x}{7}\) is M0
DM1 A1 Dependent on previous M1. Or equivalent. 0.45x or better
Alt 3
Speed of B after collision with wall: \(\dfrac{1}{2} \times \dfrac{7}{5}u = \left(\dfrac{7}{10}u\right)\)
B1
\(x -\) distance moved by \(A = x - \dfrac{2}{5}u \times \dfrac{5x}{7u} = \dfrac{5}{7}x\)
B1
Ratio of speeds 4:7
M1A1
Distance moved by \(B\): \(\ y = \dfrac{7}{11} \times \dfrac{5x}{7} = \dfrac{5}{11}x\)
DM1 A1
B1 Accept +/-
B1 Distance apart when \(B\) hits the wall
DM1 A1 Dependent on previous M1. Use of \(\dfrac{9x}{7}\) for \(\dfrac{5x}{7}\) is M0. Or equivalent. 0.45x or better
Alt 4
Speed of B after collision with wall: \(\dfrac{1}{2} \times \dfrac{7}{5}u\left(= \dfrac{7}{10}u\right)\)
B1
\(x -\) distance moved by \(A = x - \dfrac{2}{5}u \times \dfrac{5x}{7u} = \dfrac{5}{7}x\)
B1
Equate times for each particle to cover the residual distance. \(\dfrac{5}{2u}\left(\dfrac{5x}{7} - y\right) = \dfrac{10}{7u} \times y,\ \ \dfrac{1}{2}\left(\dfrac{5x}{7} - y\right) = \dfrac{11}{7}y\)
M1A1
Distance moved by \(B\): \(\ y = \dfrac{5}{11}x\)
DM1 A1
B1 Accept +/-
B1 Distance apart when \(B\) hits the wall
M1A1 Use of \(\dfrac{9x}{7}\) for \(\dfrac{5x}{7}\) is M0
DM1 A1 Dependent on previous M1. Or equivalent. 0.45x or better
7. Two particles \(A\) and \(B\), of masses \(3m\) and \(4m\) respectively, lie at rest on a smooth horizontal surface. Particle \(B\) lies between \(A\) and a smooth vertical wall which is perpendicular to the line joining \(A\) and \(B\). Particle \(B\) is projected with speed \(5u\) in a direction perpendicular to the wall and collides with the wall. The coefficient of restitution between \(B\) and the wall is \(\dfrac{3}{5}\).
(a) Find the magnitude of the impulse received by \(B\) in the collision with the wall. (3)
After the collision with the wall, \(B\) rebounds from the wall and collides directly with \(A\). The coefficient of restitution between \(A\) and \(B\) is \(e\).
(b) Show that, immediately after they collide, \(A\) and \(B\) are both moving in the same direction. (7)
The kinetic energy of \(B\) immediately after it collides with \(A\) is one quarter of the kinetic energy of \(B\) immediately before it collides with \(A\).
(c) Find the value of \(e\). (4)
Mark scheme (a)
Scheme
Marks
Impact with wall: \(\ v = \dfrac{3}{5} \times 5u = 3u\)
B1
Impulse \(\pm 4m\left(3u - (-5u)\right)\)
M1
Magnitude \(32mu\) (Ns)
A1
(3)
Notes
B1 or \(-3u\)
M1 M0 if clearly using \(mv + mu\), otherwise bod
Mark scheme (b)
Scheme
Marks
CLM: \(\ 3mx + 4mw = 4m \times 3u\)
M1 A1ft
Impact: \(\ x - w = e \times 3u\)
M1 A1ft
\(3m(w + 3eu) + 4mw = 7mw + 9emu = 12mu\)
\(7w = u(12 - 9e)\)
DM1
Use of \(e \leqslant 1\) in their \(w\): \(7w \geqslant 3u\)
M1
Hence \(w > 0\) and \(A\) and \(B\) are moving in the same direction
A1
(7)
Notes
M1 Need all 4 terms. Condone sign errors. Use of 5u is M0
A1ft follow their 3u
M1 Used the right way round. Use of 5u is M0
A1ft follow their 3u signs consistent with CLM equation
DM1 Solve for \(w\) or \(kw\). Dependent on two preceding M marks
M1 Condone use of \(\lt\)
A1 Complete argument leading to *given answer*
Mark scheme (c)
Scheme
Marks
KE of \(B\) before collision \(= \dfrac{1}{2} \times 4m \times (3u)^2\ \left(= 18mu^2\right)\)
7. Two particles \(A\) and \(B\), of mass \(2m\) and \(3m\) respectively, are initially at rest on a smooth horizontal surface. Particle \(A\) is projected with speed \(3u\) towards \(B\). Particle \(A\) collides directly with particle \(B\). The coefficient of restitution between \(A\) and \(B\) is \(\dfrac{3}{4}\)
(a) Find
(i) the speed of \(A\) immediately after the collision,
(ii) the speed of \(B\) immediately after the collision. (7)
After the collision \(B\) hits a fixed smooth vertical wall and rebounds. The wall is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the wall is \(e\). The magnitude of the impulse received by \(B\) when it hits the wall is \(\dfrac{27}{4}mu\).
(b) Find the value of \(e\). (3)
(c) Determine whether there is a further collision between \(A\) and \(B\) after \(B\) rebounds from the wall. (2)
Mark scheme (a)
Scheme
Marks
CLM: \(\ 6mu = 2mv + 3mw\)
M1
\((6u = 2v + 3w)\)
A1
Impact: \(\ w - v = \dfrac{3}{4} \times 3u\left(= \dfrac{9}{4}u\right)\)
8. Three identical particles \(P\), \(Q\) and \(R\), each of mass \(m\), lie in a straight line on a smooth horizontal plane with \(Q\) between \(P\) and \(R\). Particles \(P\) and \(Q\) are projected directly towards each other with speeds \(4u\) and \(2u\) respectively, and at the same time particle \(R\) is projected along the line away from \(Q\) with speed \(3u\). The coefficient of restitution between each pair of particles is \(e\). After the collision between \(P\) and \(Q\) there is a collision between \(Q\) and \(R\).
(a) Show that \(e > \dfrac{2}{3}\) (7)
It is given that \(e = \dfrac{3}{4}\)
(b) Show that there will not be a further collision between \(P\) and \(Q\). (6)
Mark scheme (a)
Scheme
Marks
\(4mu - 2mu = mv + mw\)
M1 A1
\(w - v = 6eu\)
M1 A1
\(v + w = 2u\) \(w - v = 6eu\ \ \ \ \ 2w = 2u + 6eu,\ \ (w = u + 3eu)\)
DM1
For \(Q\) and \(R\) to collide require \(w > 3u\),
M1
\(u + 3eu > 3u,\ \ \ \ \ e > \dfrac{2}{3}\)
A1
(7)
Notes
M1 Equation for CLM. Requires all 4 terms. Condone sign errors. Condone \(m\) missing throughout.
M1 Impact law. \(e\) must be used correctly. Condone sign errors
A1 Signs should be consistent with equation for CLM.
DM1 Solve for \(w\). Dependent on the two previous M marks.
M1 Use inequality to compare their \(w\) with \(3u\).
A1 Reach *Given answer* with no errors seen.
8a alt
Collision between \(Q\) and \(R \Rightarrow w > 3u\)
M1
Magnitude of impulse on \(Q > 5mu\) Magnitude of impulse on \(P > 5mu\)
A1
\(\Rightarrow v < -u\)
M1
\(e = \dfrac{w - v}{4u + 2u}\)
M1
Speed of separation after collision \(> u + 3u\)
M1
\(e > \dfrac{4u}{4u + 2u}\)
A1
\(e > \dfrac{2}{3}\)
A1
M1 Impact law
A1 Reach given inequality with no errors seen.
Mark scheme (b)
Scheme
Marks
\(w = u + 3eu = \dfrac{13}{4}u,\ \ v = -\dfrac{5}{4}u\)
7. A particle \(P\) of mass \(2m\) is moving in a straight line with speed \(3u\) on a smooth horizontal table. A second particle \(Q\) of mass \(3m\) is moving in the opposite direction to \(P\) along the same straight line with speed \(u\). The particle \(P\) collides directly with \(Q\). The direction of motion of \(P\) is reversed by the collision. The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision is \(\dfrac{u}{5}(8e + 3)\) (6)
(b) Find the range of possible values of \(e\). (4)
The total kinetic energy of the particles before the collision is \(T\). The total kinetic energy of the particles after the collision is \(kT\). Given that \(e = \dfrac{1}{2}\)
(c) find the value of \(k\). (4)
Mark scheme (a)
Scheme
Marks
\(6mu - 3mu = 3my - 2mx\)
M1
\(3y - 2x = 3u\)
A1
\(4ue = x + y\)
M1 A1
\(y = \dfrac{u}{5}(8e + 3)\ \ \ \ **\)
DM1 A1
(6)
Notes
M1 CLM Needs all the terms. Condone sign errors
A1 Correct equation
M1 Impact law. Must be used the right way round.
A1 Correct equation. Signs with \(x\), \(y\) must be consistent with the CLM equation.
DM1 Dependent on the two preceding M marks
A1 Obtain the given result correctly
This is a given result – the candidate needs to show sufficient working to support the answer.
5. Two particles \(P\) and \(Q\), of masses \(2m\) and \(m\) respectively, are on a smooth horizontal table. Particle \(Q\) is at rest and particle \(P\) collides directly with it when moving with speed \(u\). After the collision the total kinetic energy of the two particles is \(\dfrac{3}{4}mu^2\). Find
(a) the speed of \(Q\) immediately after the collision, (10)
(b) the coefficient of restitution between the particles. (3)
7. Three particles \(P\), \(Q\) and \(R\) lie at rest in a straight line on a smooth horizontal table with \(Q\) between \(P\) and \(R\). The particles \(P\), \(Q\) and \(R\) have masses \(2m\), \(3m\) and \(4m\) respectively. Particle \(P\) is projected towards \(Q\) with speed \(u\) and collides directly with it. The coefficient of restitution between each pair of particles is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision with \(P\) is \(\dfrac{2}{5}(1 + e)u\). (6)
After the collision between \(P\) and \(Q\) there is a direct collision between \(Q\) and \(R\).
Given that \(e = \dfrac{3}{4}\), find
(b)
(i) the speed of \(Q\) after this collision,
(ii) the speed of \(R\) after this collision. (6)
Immediately after the collision between \(Q\) and \(R\), the rate of increase of the distance between \(P\) and \(R\) is \(V\).
7. A particle \(A\) of mass \(m\) is moving with speed \(u\) on a smooth horizontal floor when it collides directly with another particle \(B\), of mass \(3m\), which is at rest on the floor. The coefficient of restitution between the particles is \(e\). The direction of motion of \(A\) is reversed by the collision.
(a) Find, in terms of \(e\) and \(u\),
(i) the speed of \(A\) immediately after the collision,
(ii) the speed of \(B\) immediately after the collision. (7)
After being struck by \(A\) the particle \(B\) collides directly with another particle \(C\), of mass \(4m\), which is at rest on the floor. The coefficient of restitution between \(B\) and \(C\) is \(2e\). Given that the direction of motion of \(B\) is reversed by this collision,
(b) find the range of possible values of \(e\), (6)
(c) determine whether there will be a second collision between \(A\) and \(B\). (3)
Mark scheme (a)
If the signs on their diagram and in their working are inconsistent, ignore the diagram. Penalise inconsistency between the two equations in the second accuracy mark.
Scheme
Marks
\(mu = -mv + 3mw\)
M1
\(u = -v + 3w\)
A1
\(eu = w + v\)
M1 A1
\(w = \dfrac{u}{4}(1 + e)\)
DM1 A1
\(v = -w + eu = \dfrac{u}{4}(3e - 1)\)
A1
(7)
Notes
M1 CLM. Allow for \(v\) in either direction. Needs all 3 terms. Condone sign errors.
A1 \(v\) in either direction. Ignore diagram if equations "correct" but inconsistent with diagram.
M1 Impact law. Must be the right way round, but condone sign errors
A1 Correct equation. Signs consistent with CLM equn.
DM1 Solve for \(v\) or \(w\).
A1 One correct
A1 Both correct. \(1 - 3e \rightarrow\) A0 for \(v\)
Mark scheme (b)
If the signs on their diagram and in their working are inconsistent, ignore the diagram. Penalise inconsistency between the two equations in the B mark.
2. A particle \(P\) of mass \(3m\) is moving with speed \(2u\) in a straight line on a smooth horizontal plane. The particle \(P\) collides directly with a particle \(Q\) of mass \(4m\) moving on the plane with speed \(u\) in the opposite direction to \(P\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Find the speed of \(Q\) immediately after the collision. (6)
Given that the direction of motion of \(P\) is reversed by the collision,
(b) find the range of possible values of \(e\). (5)
Mark scheme (a)
Scheme
Marks
\(3m.2u - 4mu = 3mv_1 + 4mv_2\)
M1 A1
\(e(2u + u) = -v_1 + v_2\)
M1 A1
\(\dfrac{u(2 + 9e)}{7} = v_2\)
DM1 A1
(6)
Notes
M1 CLM. Need all terms. Condone sign slips.
A1 Correct but check their directions for \(v_1\) & \(v_2\).
M1 Impact law. Must be used the right way round, but condone sign slips.
A1 Directions of \(v_1\) & \(v_2\) must be consistent between the two equations. (Ignore the diagram if necessary)
DM1 Eliminate \(v_1\) to produce an equation in \(v_2\) only. Dependent on both previous M marks – must be using both equations.
A1 DO NOT accept the negative. The question asks for speed.
Mark scheme (b)
Scheme
Marks
\(v_1 = \dfrac{2u(1 - 6e)}{7}\)
M1 A1
\(v_1 < 0 \Rightarrow e > \dfrac{1}{6}\)
DM1 A1
\(1 \geqslant e > \dfrac{1}{6}\)
B1
(5)
(11 marks)
Notes
M1 Use the work from (a) or restart to find \(v_1\) or \(\lambda v_1\) for a constant \(\lambda\). If using work from (a) this mark is dependent on the first 2 M marks.
A1 a.e.f. Correct for their direction. Allow for \(\lambda v_1\)
DM1 An appropriate inequality for their \(v_1\) (seen or implied) – requires previous M1 scored. Work on \(v_1 = 0\) scores M0 until the inequality is formed.
A1 Accept \(\dfrac{2}{12}\). Answer must follow from correct work for \(v_1\)
6. Three identical particles, \(A\), \(B\) and \(C\), lie at rest in a straight line on a smooth horizontal table with \(B\) between \(A\) and \(C\). The mass of each particle is \(m\). Particle \(A\) is projected towards \(B\) with speed \(u\) and collides directly with \(B\). The coefficient of restitution between each pair of particles is \(\frac{2}{3}\).
(a) Find, in terms of \(u\),
(i) the speed of \(A\) after this collision,
(ii) the speed of \(B\) after this collision. (7)
(b) Show that the kinetic energy lost in this collision is \(\dfrac{5}{36}mu^2\) (4)
After the collision between \(A\) and \(B\), particle \(B\) collides directly with \(C\).
(c) Find, in terms of \(u\), the speed of \(C\) immediately after this collision between \(B\) and \(C\). (4)
2. A particle \(P\) of mass \(m\) is moving in a straight line on a smooth horizontal surface with speed \(4u\). The particle \(P\) collides directly with a particle \(Q\) of mass \(3m\) which is at rest on the surface. The coefficient of restitution between \(P\) and \(Q\) is \(e\). The direction of motion of \(P\) is reversed by the collision.
8. A particle \(P\) of mass \(m\) kg is moving with speed 6 m s\(^{-1}\) in a straight line on a smooth horizontal floor. The particle strikes a fixed smooth vertical wall at right angles and rebounds. The kinetic energy lost in the impact is 64 J. The coefficient of restitution between \(P\) and the wall is \(\frac{1}{3}\).
(a) Show that \(m = 4\). (6)
After rebounding from the wall, \(P\) collides directly with a particle \(Q\) which is moving towards \(P\) with speed 3 m s\(^{-1}\). The mass of \(Q\) is 2 kg and the coefficient of restitution between \(P\) and \(Q\) is \(\frac{1}{3}\).
(b) Show that there will be a second collision between \(P\) and the wall. (7)
Mark scheme (a)
Scheme
Marks
KE lost: \(\ \dfrac{1}{2} \times m \times 36 - \dfrac{1}{2} \times m \times v^2 = 64\)
M1A1
Restitution: \(\ v = 1/3 \times 6 = 2\)
M1A1
Substitute and solve for m: \(\ \dfrac{1}{2} \times m \times 36 - \dfrac{1}{2} \times m \times 4 = 64 = 16m\)
DM1
\(m = 4\) answer given
A1
(6)
Mark scheme (b)
Scheme
Marks
Conservation of momentum: \(\ 6 - 8 = 4w - 2v\) their "2"
M1A1ft
Restitution: \(\ v + w = \frac{1}{3}(2 + 3)\) their "2"
8. A small ball \(A\) of mass \(3m\) is moving with speed \(u\) in a straight line on a smooth horizontal table. The ball collides directly with another small ball \(B\) of mass \(m\) moving with speed \(u\) towards \(A\) along the same straight line. The coefficient of restitution between \(A\) and \(B\) is \(\frac{1}{2}\). The balls have the same radius and can be modelled as particles.
(a) Find
(i) the speed of \(A\) immediately after the collision,
(ii) the speed of \(B\) immediately after the collision. (7)
After the collision \(B\) hits a smooth vertical wall which is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the wall is \(\frac{2}{5}\).
(b) Find the speed of \(B\) immediately after hitting the wall. (2)
The first collision between \(A\) and \(B\) occurred at a distance \(4a\) from the wall. The balls collide again \(T\) seconds after the first collision.
(c) Show that \(T = \dfrac{112a}{15u}\). (6)
Mark scheme (a)
Scheme
Marks
(i) Con. of Mom: \(\ 3mu - mu = 3mv + mw\)
\(2u = 3v + w\qquad (1)\)
M1# A1
N.L.R: \(\ \frac{1}{2}(u + u) = w - v\)
M1# A1
\(u = w - v\qquad (2)\)
(1) − (2) \(\quad u = 4v\)
DM1#
\(v = \frac{1}{4}u\)
A1
(ii) In (2) \(\quad u = w - \frac{1}{4}u\)
\(w = \frac{5}{4}u\)
A1
(7)
Mark scheme (b)
Scheme
Marks
\(B\) to wall: N.L.R: \(\ \frac{5}{4}u \times \frac{2}{5} = V\)
M1
\(V = \frac{1}{2}u\)
A1ft
(2)
Mark scheme (c)
Scheme
Marks
\(B\) to wall: \(\quad\) time \(= 4a \div \dfrac{5}{4}u = \dfrac{16a}{5u}\)
B1ft
Dist. Travelled by \(A = \dfrac{1}{4}u \times \dfrac{16a}{5u} = \dfrac{4}{5}a\)
B1ft
In \(t\) secs, \(A\) travels \(\dfrac{1}{4}ut\), \(B\) travels \(\dfrac{1}{2}ut\)
Collide when speed of approach \(= \dfrac{1}{2}ut + \dfrac{1}{4}ut\), distance to cover \(= 4a - \dfrac{4}{5}a\)
2. Two particles, \(P\), of mass \(2m\), and \(Q\), of mass \(m\), are moving along the same straight line on a smooth horizontal plane. They are moving in opposite directions towards each other and collide. Immediately before the collision the speed of \(P\) is \(2u\) and the speed of \(Q\) is \(u\). The coefficient of restitution between the particles is \(e\), where \(e < 1\). Find, in terms of \(u\) and \(e\),
(i) the speed of \(P\) immediately after the collision,
(ii) the speed of \(Q\) immediately after the collision. (7)
8. Particles \(A\), \(B\) and \(C\) of masses \(4m\), \(3m\) and \(m\) respectively, lie at rest in a straight line on a smooth horizontal plane with \(B\) between \(A\) and \(C\). Particles \(A\) and \(B\) are projected towards each other with speeds \(u\) m s\(^{-1}\) and \(v\) m s\(^{-1}\) respectively, and collide directly.
As a result of the collision, \(A\) is brought to rest and \(B\) rebounds with speed \(kv\) m s\(^{-1}\). The coefficient of restitution between \(A\) and \(B\) is \(\dfrac{3}{4}\).
(a) Show that \(u = 3v\). (6)
(b) Find the value of \(k\). (2)
Immediately after the collision between \(A\) and \(B\), particle \(C\) is projected with speed \(2v\) m s\(^{-1}\) towards \(B\) so that \(B\) and \(C\) collide directly.
(c) Show that there is no further collision between \(A\) and \(B\). (4)
7. A particle \(P\) of mass \(3m\) is moving in a straight line with speed \(2u\) on a smooth horizontal table. It collides directly with another particle \(Q\) of mass \(2m\) which is moving with speed \(u\) in the opposite direction to \(P\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) immediately after the collision is \(\tfrac{1}{5}(9e + 4)u\). (5)
The speed of \(P\) immediately after the collision is \(\tfrac{1}{2}u\).
(b) Show that \(e = \tfrac{1}{4}\). (4)
The collision between \(P\) and \(Q\) takes place at the point \(A\). After the collision \(Q\) hits a smooth fixed vertical wall which is at right-angles to the direction of motion of \(Q\). The distance from \(A\) to the wall is \(d\).
(c) Show that \(P\) is a distance \(\tfrac{3}{5}d\) from the wall at the instant when \(Q\) hits the wall. (4)
Particle \(Q\) rebounds from the wall and moves so as to collide directly with particle \(P\) at the point \(B\). Given that the coefficient of restitution between \(Q\) and the wall is \(\tfrac{1}{5}\),
(d) find, in terms of \(d\), the distance of the point \(B\) from the wall. (4)
2. A particle \(A\) of mass \(4m\) is moving with speed \(3u\) in a straight line on a smooth horizontal table. The particle \(A\) collides directly with a particle \(B\) of mass \(3m\) moving with speed \(2u\) in the same direction as \(A\). The coefficient of restitution between \(A\) and \(B\) is \(e\). Immediately after the collision the speed of \(B\) is \(4eu\).
(a) Show that \(e = \dfrac{3}{4}\). (5)
(b) Find the total kinetic energy lost in the collision. (4)
Mark scheme (a)
Scheme
Marks
LM \(12mu + 6mu = 4mx + 12meu\)
B1
NEL \(4eu - x = eu\)
M1 A1
Eliminating \(x\) to obtain equation in \(e\)
DM1
Leading to \(e = \dfrac{3}{4}\) * cso
A1
(5)
Mark scheme (b)
Scheme
Marks
\(x = 3eu\) or \(\dfrac{9}{4}u\) or \(4.5u - 3eu\) seen or implied in (b)
B1
Loss in KE \(= \dfrac{1}{2}4m(3u)^2 + \dfrac{1}{2}3m(2u)^2 - \dfrac{1}{2}4m\left(\dfrac{9}{4}u\right)^2 - \dfrac{1}{2}3m(3u)^2\) ft their \(x\)
7. A particle \(P\) of mass \(2m\) is moving with speed \(2u\) in a straight line on a smooth horizontal plane. A particle \(Q\) of mass \(3m\) is moving with speed \(u\) in the same direction as \(P\). The particles collide directly. The coefficient of restitution between \(P\) and \(Q\) is \(\tfrac{1}{2}\).
(a) Show that the speed of \(Q\) immediately after the collision is \(\tfrac{8}{5}u\). (5)
(b) Find the total kinetic energy lost in the collision. (5)
After the collision between \(P\) and \(Q\), the particle \(Q\) collides directly with a particle \(R\) of mass \(m\) which is at rest on the plane. The coefficient of restitution between \(Q\) and \(R\) is \(e\).
(c) Calculate the range of values of \(e\) for which there will be a second collision between \(P\) and \(Q\). (7)
Mark scheme (a)
Scheme
Marks
LM \(4mu + 3mu = 2mx + 3my\)
M1 A1
NEL \(y - x = \tfrac{1}{2}u\)
B1
Solving to \(y = \tfrac{8}{5}u\) * cso
M1 A1
(5)
Mark scheme (b)
Scheme
Marks
\(x = \tfrac{11}{10}u\) or equivalent
B1
Energy loss \(\ \tfrac{1}{2} \times 2m\left((2u)^2 - \left(\tfrac{11}{10}u\right)^2\right) + \tfrac{1}{2} \times 3m\left(u^2 - \left(\tfrac{8}{5}u\right)^2\right)\)
M1 A(2,1,0)
\(= \tfrac{9}{20}mu^2\)
A1
(5)
Mark scheme (c)
Scheme
Marks
LM \(\tfrac{24}{5}mu = 3ms + mt\)
M1 A1
NEL \(t - s = \tfrac{8}{5}eu\)
B1
Solving to \(s = \tfrac{2}{5}u(3 - e)\)
M1 A1
For a further collision \(\tfrac{11}{10}u > \tfrac{2}{5}u(3 - e)\)
7. Two small spheres \(P\) and \(Q\) of equal radius have masses \(m\) and \(5m\) respectively. They lie on a smooth horizontal table. Sphere \(P\) is moving with speed \(u\) when it collides directly with sphere \(Q\) which is at rest. The coefficient of restitution between the spheres is \(e\), where \(e > \dfrac{1}{5}\).
(a)
(i) Show that the speed of \(P\) immediately after the collision is \(\dfrac{u}{6}(5e - 1)\).
(ii) Find an expression for the speed of \(Q\) immediately after the collision, giving your answer in the form \(\lambda u\), where \(\lambda\) is in terms of \(e\). (6)
Three small spheres \(A\), \(B\) and \(C\) of equal radius lie at rest in a straight line on a smooth horizontal table, with \(B\) between \(A\) and \(C\). The spheres \(A\) and \(C\) each have mass \(5m\), and the mass of \(B\) is \(m\). Sphere \(B\) is projected towards \(C\) with speed \(u\). The coefficient of restitution between each pair of spheres is \(\dfrac{4}{5}\).
(b) Show that, after \(B\) and \(C\) have collided, there is a collision between \(B\) and \(A\). (3)
(c) Determine whether, after \(B\) and \(A\) have collided, there is a further collision between \(B\) and \(C\). (4)
Mark scheme (a)
Scheme
Marks
CLM: \(mv + 5mw = mu\)
B1
NLI: \(w - v = eu\)
B1
Solve \(v\): \(\ v = \tfrac{1}{6}(1 - 5e)u\), so speed \(= \dfrac{1}{6}(5e - 1)u\) (NB – answer given on paper)
M1* A1
Solve \(w\): \(\ w = \tfrac{1}{6}(1 + e)u\)
M1* A1
* The M’s are dependent on having equations (not necessarily correct) for CLM and NLI
(6)
Notes
B1 Conservation of momentum – signs consistent with their diagram/between the two equations
B1 Impact equation
M1 Attempt to eliminate w
A1 correct expression for v. Q asks for speed so final answer must be verified positive with reference to \(e > 1/5\).
Answer given so watch out for fudges.
M1 Attempt to eliminate v
A1 correct expression for w
Mark scheme (b)
Scheme
Marks
After \(B\) hits \(C\), velocity of \(B\) = “\(v\)” \(= \tfrac{1}{6}\left(1 - 5 \cdot \tfrac{4}{5}\right)u = -\tfrac{1}{2}u\)
M1 A1
velocity \(< 0 \Rightarrow\) change of direction \(\Rightarrow B\) hits \(A\)
A1 CSO
(3)
Notes
M1 Substitute for e in speed or velocity of P to obtain \(\boldsymbol{v}\) in terms of \(\boldsymbol{u}\). Alternatively, can obtain v in terms of w
A1 (+/-) u/2 \(\left(v = -\dfrac{5w}{3}\right)\)
A1 CSO Justify direction (and correct conclusion)
Mark scheme (c)
Scheme
Marks
velocity of \(C\) after \(= \tfrac{3}{10}u\)
B1
When \(B\) hits \(A\), “\(u\)” \(= \tfrac{1}{2}u\), so velocity of \(B\) after \(= -\tfrac{1}{2}\left(-\tfrac{1}{2}u\right) = \tfrac{1}{4}u\)
B1
Travelling in the same direction but \(\tfrac{1}{4} < \tfrac{3}{10} \Rightarrow\) no second collision
M1 A1 CSO
(4)
(13 marks)
Notes
B1 speed of C = value of w \(= (\pm)\dfrac{3u}{10}\) (Must be referred to in (c) to score the B1.)
B1 speed of B after second collision \((\pm)\dfrac{1}{4}u\) or \((\pm)\dfrac{5}{6}w\)
M1 Comparing their speed of B after 2nd collision with their speed of C after first collision.
4. A particle \(P\) of mass \(m\) is moving in a straight line on a smooth horizontal table. Another particle \(Q\) of mass \(km\) is at rest on the table. The particle \(P\) collides directly with \(Q\). The direction of motion of \(P\) is reversed by the collision. After the collision, the speed of \(P\) is \(v\) and the speed of \(Q\) is \(3v\). The coefficient of restitution between \(P\) and \(Q\) is \(\tfrac{1}{2}\).
(a) Find, in terms of \(v\) only, the speed of \(P\) before the collision. (3)
(b) Find the value of \(k\). (3)
After being struck by \(P\), the particle \(Q\) collides directly with a particle \(R\) of mass \(11m\) which is at rest on the table. After this second collision, \(Q\) and \(R\) have the same speed and are moving in opposite directions. Show that
(c) the coefficient of restitution between \(Q\) and \(R\) is \(\tfrac{3}{4}\), (4)
(d) there will be a further collision between \(P\) and \(Q\). (2)
Mark scheme (a)
Scheme
Marks
NEL \(3v - (-v) = eu\)
M1 A1
\(u = 8v\)
A1
(3)
Mark scheme (b)
Scheme
Marks
LM \(8mv = -mv + 3kmv\) ft their \(u\)
M1 A1ft
\((m \times (u) = -mv + 3kmv)\)
\(k = 3\)
A1
(3)
Mark scheme (c)
Scheme
Marks
LM \(9mv = -3my + 11my\) ft their \(k\)
M1 A1ft
NEL \(2y = e \times 3v\)
M1
\(y = \dfrac{9}{8}v \Rightarrow e = \dfrac{3}{4}\ \ *\) cso
A1
(4)
Mark scheme (d)
Scheme
Marks
\(y = \tfrac{9}{8}v > v\ \ \Rightarrow\ \ \) further collision between \(P\) and \(Q\)
M1 A1
(2)
(12 marks)
Notes
A1 is cso – watch out for incorrect statements re. velocity
8. Two particles \(A\) and \(B\) move on a smooth horizontal table. The mass of \(A\) is \(m\), and the mass of \(B\) is \(4m\). Initially \(A\) is moving with speed \(u\) when it collides directly with \(B\), which is at rest on the table. As a result of the collision, the direction of motion of \(A\) is reversed. The coefficient of restitution between the particles is \(e\).
(a) Find expressions for the speed of \(A\) and the speed of \(B\) immediately after the collision. (7)
In the subsequent motion, \(B\) strikes a smooth vertical wall and rebounds. The wall is perpendicular to the direction of motion of \(B\). The coefficient of restitution between \(B\) and the wall is \(\tfrac{4}{5}\). Given that there is a second collision between \(A\) and \(B\),
(b) show that \(\tfrac{1}{4} < e < \tfrac{9}{16}\). (5)
Given that \(e = \tfrac{1}{2}\),
(c) find the total kinetic energy lost in the first collision between \(A\) and \(B\). (3)
Mark scheme (a)
Scheme
Marks
\(mu = 4mw - mv\)
M1 A1
\(eu = w + v\)
M1 A1
\(\Rightarrow\ w = \left(\dfrac{1 + e}{5}\right)u,\ \ v = \left(\dfrac{4e - 1}{5}\right)u\) indep
4. A particle \(A\) of mass \(2m\) is moving with speed \(3u\) in a straight line on a smooth horizontal table. The particle collides directly with a particle \(B\) of mass \(m\) moving with speed \(2u\) in the opposite direction to \(A\). Immediately after the collision the speed of \(B\) is \(\tfrac{8}{3}u\) and the direction of motion of \(B\) is reversed.
(a) Calculate the coefficient of restitution between \(A\) and \(B\). (6)
(b) Show that the kinetic energy lost in the collision is \(7mu^2\). (3)
After the collision \(B\) strikes a fixed vertical wall that is perpendicular to the direction of motion of \(B\). The magnitude of the impulse of the wall on \(B\) is \(\tfrac{14}{3}mu\).
(c) Calculate the coefficient of restitution between \(B\) and the wall. (4)
5. Two small spheres \(A\) and \(B\) have mass \(3m\) and \(2m\) respectively. They are moving towards each other in opposite directions on a smooth horizontal plane, both with speed \(2u\), when they collide directly. As a result of the collision, the direction of motion of \(B\) is reversed and its speed is unchanged.
(a) Find the coefficient of restitution between the spheres. (7)
Subsequently, \(B\) collides directly with another small sphere \(C\) of mass \(5m\) which is at rest. The coefficient of restitution between \(B\) and \(C\) is \(\tfrac{3}{5}\).
(b) Show that, after \(B\) collides with \(C\), there will be no further collisions between the spheres. (7)
Mark scheme (a)
Scheme
Marks
CLM: \(6mu - 4mu = 3mv + 4mu\)
M1 A1
\(\Rightarrow v = -\tfrac{2}{3}u\)
A1
NLI: \(2u - v = e.4u\)
M1 A1
\(\Rightarrow 4eu = \tfrac{8}{3}u \Rightarrow e = \tfrac{2}{3}\).
M1 A1
(7)
Mark scheme (b)
Scheme
Marks
\(5my + 2mx = 4mu\)
M1 A1
\(y - x = \tfrac{3}{5}.2u = \tfrac{6}{5}u\)
A1
Solve: \(x = -\tfrac{2}{7}u\)
M1 A1
\(\tfrac{2}{7}u < \tfrac{2}{3}u\) so \(B\) does not overtake \(A\)
6. A particle \(P\) of mass \(3m\) is moving with speed \(2u\) in a straight line on a smooth horizontal table. The particle \(P\) collides with a particle \(Q\) of mass \(2m\) moving with speed \(u\) in the opposite direction to \(P\). The coefficient of restitution between \(P\) and \(Q\) is \(e\).
(a) Show that the speed of \(Q\) after the collision is \(\tfrac{1}{5}u(9e + 4)\). (5)
As a result of the collision, the direction of motion of \(P\) is reversed.
(b) Find the range of possible values of \(e\). (5)
Given that the magnitude of the impulse of \(P\) on \(Q\) is \(\tfrac{32}{5}mu\),
(c) find the value of \(e\). (4)
Mark scheme (a)
Scheme
Marks
LM \(6mu - 2mu = 3mx + 2my\)
M1 A1
NEL \(y - x = 3eu\)
B1
Solving to \(y = \tfrac{1}{5}u(9e + 4)\ \ *\) cso
M1 A1
(5)
Mark scheme (b)
Scheme
Marks
Solving to \(x = \tfrac{2}{5}u(2 - 3e)\) oe
M1 A1
\(x < 0\ \Rightarrow\ e > \tfrac{2}{3}\)
M1 A1
\(\tfrac{2}{3} < e \leqslant 1\) ft their \(e\) for glb
