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.
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 \(\dfrac{7}{20} \lt \dfrac{2f}{5}\)
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 \(1 > f\).
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 \(e\)
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
Figure 1 represents the plan of part of a smooth horizontal floor, where \(W_1\) and \(W_2\) are two fixed parallel vertical walls. The walls are 3 metres apart.
A particle lies at rest at a point \(O\) on the floor between the two walls, where the point \(O\) is \(d\) metres, \(0 \lt d \leqslant 3\), from \(W_1\)
At time \(t = 0\), the particle is projected from \(O\) towards \(W_1\) with speed \(u\ \text{m s}^{-1}\) in a direction perpendicular to the walls.
The coefficient of restitution between the particle and each wall is \(\dfrac{2}{3}\)
The particle returns to \(O\) at time \(t = T\) seconds, having bounced off each wall once.
(a) Show that \(T = \dfrac{45 - 5d}{4u}\) (6)
The value of \(u\) is fixed, the particle still hits each wall once but the value of \(d\) can now vary.
(b) Find the least possible value of \(T\), giving your answer in terms of \(u\). You must give a reason for your answer. (2)
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 \(= \dfrac{8d}{9}\)
1. A small ball of mass 0.3 kg is released from rest from a point 3.6 m above horizontal ground. The ball falls freely under gravity, hits the ground and rebounds vertically upwards.
In the first impact with the ground, the ball receives an impulse of magnitude 4.2 N s. The ball is modelled as a particle.
(a) Find the speed of the ball immediately after it first hits the ground. (5)
(b) Find the kinetic energy lost by the ball as a result of the impact with the ground. (3)
Mark scheme (a)
Scheme
Marks
AO
Speed just before impact: \(v^2 = u^2 + 2as = 2 \times 9.8 \times 3.6\ (= 70.56)\)
M1
3.4
\(v = 8.4\ (\text{m s}^{-1})\)
A1
1.1b
Use of \(I = mv - mu\): \(4.2 = 0.3\big(w - (-8.4)\big)\)
M1
3.1b
Follow their 8.4
A1ft
1.1b
\(w = 5.6\ (\text{m s}^{-1})\)
A1
1.1b
(5)
Notes
M1: Use the model and suvat or energy to find speed before impact
M1: A complete strategy to find \(w\): Use the model and impulse-momentum equation using given impulse and their speed of impact. Must be using a difference in velocities. Be vigilant for sign fudges that make the original equation incorrect.
A1ft: Correct unsimplified equation using their speed
A1: Correct positive answer
Mark scheme (b)
Scheme
Marks
AO
KE lost \(= \dfrac{1}{2}m\left(v^2 - w^2\right)\)
M1
3.3
\(= \dfrac{0.3}{2}\left(8.4^2 - 5.6^2\right)\) Follow their 8.4 and 5.6
A1ft
1.1b
\(= 5.88\ (\text{J})\)
A1
1.1b
(3)
(8 marks)
Notes
M1: Correct method to find the KE lost in the impact. Need to be using speeds immediately before and immediately after impact.
A1ft: Correct expression for their speeds. Accept subtraction either way round
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: \(\ w - v = \dfrac{3}{4} \times 3u\left(= \dfrac{9}{4}u\right)\)
5. A particle of mass \(m\) kg lies on a smooth horizontal surface. Initially the particle is at rest at a point \(O\) midway between a pair of fixed parallel vertical walls. The walls are 2 m apart. At time \(t = 0\) the particle is projected from \(O\) with speed \(u\) m s\(^{-1}\) in a direction perpendicular to the walls. The coefficient of restitution between the particle and each wall is \(\dfrac{2}{3}\). The magnitude of the impulse on the particle due to the first impact with a wall is \(\lambda mu\) N s.
(a) Find the value of \(\lambda\). (3)
The particle returns to \(O\), having bounced off each wall once, at time \(t = 3\) seconds.
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 \(= 4a - \dfrac{4}{5}a\)
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)
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)
