FM1-4 Elastic collisions in one dimension — revision question pack

2 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FM1-4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FM1-4.1 · Direct impact of elastic spheres. Newton's law of restitution. Loss of kinetic energy due to impact.

Explanation

In a direct impact, choose one positive direction and conserve signed momentum along the line of centres. Newton's law of restitution states that speed of separation equals $e$ times speed of approach, with $0\leq e\leq1$ .
In signed form for $A$ behind $B$ , $v_B-v_A=e(u_A-u_B)$ . Solve the momentum and restitution equations simultaneously, then check that the post-impact velocities describe separation.
Spheres may be modelled as particles.
Momentum is conserved for the isolated system, but kinetic energy is generally lost; calculate loss as total kinetic energy before minus total kinetic energy after.
Examiners expect the non-negative root for $e$ and consistent velocity signs.

Worked example

A $3\,\text{kg}$ sphere moving at $5\,\text{m s}^{-1}$ strikes a stationary $2\,\text{kg}$ sphere directly. The coefficient of restitution is $\tfrac12$ . Find both velocities after impact.

1.Momentum: $15=3v_A+2v_B$ .
2.Restitution: $v_B-v_A=\tfrac12(5)=2.5$ .
3.Substitute $v_B=v_A+2.5$ : $15=5v_A+5$ , so $v_A=2$ and $v_B=4.5$ .

Answer: The velocities are $2\,\text{m s}^{-1}$ and $4.5\,\text{m s}^{-1}$ in the original direction.

Common mistakes

Don't write speed of approach minus speed of separation in the restitution equation.
Don't conserve kinetic energy as well as momentum when $e<1$ .
Don't accept a negative algebraic root for $e$ despite $0\leq e\leq1$ .

Exam tip

Write momentum and restitution as two labelled equations before eliminating either final velocity.

Tier 1 · Easy

1.
Sphere $A$ approaches stationary sphere $B$ at $7\,\text{m s}^{-1}$ . After their direct impact, $A$ continues in the same direction at $2\,\text{m s}^{-1}$ . The coefficient of restitution is $0.4$ . Find the velocity of $B$ .
(2)
(Total for Question 1 is 2 marks)
2.
Sphere $A$ has mass $2\,\text{kg}$ and velocity $5\,\text{m s}^{-1}$ . It collides directly with a $3\,\text{kg}$ sphere $B$ whose velocity is $1\,\text{m s}^{-1}$ in the same direction. For this impact $e=1/2$ . Find the speed of each sphere immediately after the impact.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A $2\,\text{kg}$ sphere travelling at $6\,\text{m s}^{-1}$ strikes a stationary $3\,\text{kg}$ sphere directly. The coefficient of restitution is $1/2$ . Find both velocities after impact and the kinetic energy lost.
(5)
(Total for Question 1 is 5 marks)
2.
A $3\,\text{kg}$ sphere $A$ travelling at $4\,\text{m s}^{-1}$ collides directly with a $2\,\text{kg}$ sphere $B$ travelling at $1\,\text{m s}^{-1}$ in the same direction. For this collision take $e=1/2$ . Determine the impulse exerted on $A$ and on $B$ . Calculate the percentage by which their combined kinetic energy decreases.
(6)
(Total for Question 2 is 6 marks)
3.
Sphere $A$ , of mass $4\,\text{kg}$ , moves at $8\,\text{m s}^{-1}$ and strikes a stationary sphere $B$ directly. The coefficient of restitution is $3/8$ . Immediately after the impact, $A$ moves at $1\,\text{m s}^{-1}$ in its original direction. Find the speed of $B$ immediately after the impact and the mass of $B$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A $1\,\text{kg}$ sphere moving at $5\,\text{m s}^{-1}$ collides directly with a stationary $2\,\text{kg}$ sphere. The collision loses $16/3\,\text{J}$ of kinetic energy. Determine the coefficient of restitution and both velocities after impact.
(6)
(Total for Question 1 is 6 marks)
2.
A sphere of mass $km$ moving at speed $u$ collides directly with a stationary sphere of mass $m$ , where $k>0$ . The coefficient of restitution is $1/2$ . Find the set of possible values of $k$ for which the first sphere rebounds while no more than two thirds of the initial kinetic energy is lost.
(7)
(Total for Question 2 is 7 marks)
3.
Sphere $A$ , of mass $2m$ , moves rightwards at speed $u$ . Sphere $B$ , of mass $3m$ , moves leftwards at speed $ku$ , where $k>0$ . They collide directly. Immediately afterwards $A$ is at rest, and the collision has lost $6/7$ of the initial kinetic energy. Determine $k$ , the coefficient of restitution and the velocity of $B$ , verifying that the inferred coefficient lies in $[0,1]$ .
(8)
(Total for Question 3 is 8 marks)
4.
A sphere $A$ of mass $4m$ moves at speed $u$ on a smooth horizontal surface and collides directly with a stationary sphere $B$ of mass $m$ . Immediately after the collision the kinetic energies of the two spheres are equal. Determine the coefficient of restitution, the velocity of each sphere, the impulse on each sphere and the fraction of the initial kinetic energy lost.
(8)
(Total for Question 4 is 8 marks)
5.
Rightwards is positive. Sphere $A$ , of mass $3m$ , moves at velocity $5u$ and collides directly with sphere $B$ , of mass $km$ , moving at velocity $-u$ , where $k>0$ and $u>0$ . The coefficient of restitution is $5/9$ . The impulse on $A$ has magnitude $10mu$ and acts leftwards. Find $k$ and the velocity of each sphere immediately after the collision. State the impulse on each sphere and verify that the spheres separate.
(7)
(Total for Question 5 is 7 marks)

FM1-4.2 · Successive direct impacts of spheres and/or a sphere with a smooth plane surface.

Explanation

Successive impacts must be handled chronologically. Keep one positive direction throughout and use the velocities immediately after one impact as the approach velocities for the next.
At a fixed smooth plane, the sphere's normal velocity reverses and its speed is multiplied by $e$ ; the plane remains stationary. After every calculation, use the ordering and signed velocities to decide which bodies can meet next.
A faster rear sphere may cause another sphere-sphere collision, while increasing velocities from rear to front prevent further impacts.
Spheres are modelled as particles in these direct-impact problems.
Examiners expect a justified impact order and a final no-further-collision argument, not merely a list of algebraic velocity pairs.

Two spheres in line with a fixed wall, a common setup for successive direct impacts.

Worked example

Identical spheres $A$ and $B$ lie in that order before a wall. Initially $A$ moves right at $8\,\text{m s}^{-1}$ and $B$ is stationary. Their mutual coefficient of restitution is $\tfrac12$ ; $B$ rebounds from the wall with coefficient $\tfrac34$ . Find the velocities after the first two impacts and decide whether $A$ and $B$ meet again.

1.For the first equal-mass impact, $v_A+v_B=8$ and $v_B-v_A=4$ , giving $v_A=2$ , $v_B=6$ .
2.At the wall, $B$ reverses with velocity $-(\tfrac34)(6)=-4.5\,\text{m s}^{-1}$ .
3. $A$ moves right while $B$ moves left, so their separation decreases.

Answer: After $A$ - $B$ : $(2,6)\,\text{m s}^{-1}$ ; after the wall: $(2,-4.5)\,\text{m s}^{-1}$ , so they collide again.

Common mistakes

Don't reuse the original velocity of a sphere instead of its velocity after the preceding impact.
Don't multiply the wall-impact speed by $e$ and fail to reverse its direction.
Don't state that another impact occurs without comparing relative positions and velocities.

Exam tip

After each impact, draw the new signed velocity arrows before deciding which impact happens next.

Tier 1 · Easy

1.
A sphere moving normally towards a fixed smooth wall at $5\,\text{m s}^{-1}$ has coefficient of restitution $0.7$ with the wall. State its velocity immediately after impact, taking motion towards the wall as positive.
(2)
(Total for Question 1 is 2 marks)
2.
A sphere moves normally between two fixed parallel smooth walls. Its speed before the first impact is $16\,\text{m s}^{-1}$ , and its coefficient of restitution with each wall is $3/4$ . Write down its speed immediately after the $n$ th impact and hence find its speed immediately after the fourth impact.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Identical spheres $A$ and $B$ lie on a line, with $B$ between $A$ and a wall. Sphere $A$ moves towards stationary $B$ at $6\,\text{m s}^{-1}$ . Their coefficient of restitution is $1/2$ , and $B$ 's coefficient with the wall is $2/3$ . Find the velocities just after the first sphere-sphere impact and explain why $A$ and $B$ collide again after $B$ rebounds from the wall.
(5)
(Total for Question 1 is 5 marks)
2.
Two identical smooth spheres $A$ and $B$ move along the positive $x$ -direction of a straight line on a smooth horizontal surface. $A$ is at $x=0$ with speed $8\,\text{m s}^{-1}$ and $B$ is at $x=12\,\text{m}$ with speed $4\,\text{m s}^{-1}$ . For their impact the coefficient of restitution is $\tfrac12$ . Find the time and position of the first impact and the velocity of each sphere immediately afterwards. State, with a reason, whether the spheres collide again.
(5)
(Total for Question 2 is 5 marks)
3.
Balls $A$ and $B$ fall along separate vertical lines towards the same fixed smooth horizontal plane. At time $t=0$ , each is released from rest. Ball $A$ is $4.9\,\text{m}$ above the plane and has coefficient of restitution $1/2$ ; ball $B$ is $441/40\,\text{m}$ above the plane and has coefficient of restitution $3/5$ . Taking $g=9.8\,\text{m s}^{-2}$ , determine the first three impacts with the plane in chronological order, giving the time and the speed immediately after each one.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Three identical spheres $A$ , $B$ and $C$ are arranged in that order on a straight line. Initially $A$ moves towards the other two at $8\,\text{m s}^{-1}$ while $B$ and $C$ are at rest. Every impact has coefficient of restitution $1/2$ . Find the velocities after all impacts have occurred, and justify that no further collision follows.
(7)
(Total for Question 1 is 7 marks)
2.
A ball of mass $3m$ is dropped from rest at a height $2h$ above a fixed smooth horizontal plane. At every impact with the plane the coefficient of restitution is $1/2$ . Find the speed immediately after the $n$ th impact and the greatest height reached after that impact. Find also the total kinetic energy lost in the first $n$ impacts and hence the total kinetic energy lost over all impacts.
(8)
(Total for Question 2 is 8 marks)
3.
A ball is projected vertically upwards at $14\,\text{m s}^{-1}$ from a fixed smooth horizontal floor towards a parallel smooth ceiling $5\,\text{m}$ above it. The coefficients of restitution at the ceiling and floor are $1/2$ and $3/4$ respectively. Taking $g=9.8\,\text{m s}^{-2}$ , find the time and velocity immediately after each of the first two impacts. Determine whether the ball reaches the ceiling again after the second impact.
(8)
(Total for Question 3 is 8 marks)
4.
Two identical spheres $A$ and $B$ move on a smooth horizontal line towards a fixed wall at $x=12\,\text{m}$ . Initially $A$ is at $x=0$ with velocity $9\,\text{m s}^{-1}$ and $B$ is at $x=d\,\text{m}$ with velocity $3\,\text{m s}^{-1}$ , where $0<d<12$ . Find the set of possible values of $d$ for which $A$ collides with $B$ before $B$ reaches the wall. For the rest of the question take $d=6$ , the sphere-sphere coefficient of restitution to be $3/4$ , and the wall coefficient for $B$ to be $1/3$ . Find the velocities after each subsequent impact and show that no further impact occurs after $B$ 's second collision with the wall.
(8)
(Total for Question 4 is 8 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FM1-4.1 · Direct impact of elastic spheres. Newton's law of restitution. Loss of kinetic energy due to impact.

Tier 1 · Easy

Mark scheme for FM1-4.1 Tier 1 · Easy
Question	Scheme	Marks
1	$4.8\,\text{m s}^{-1}$ in $A$ 's original direction	2
(2 marks)		2
Notes
Newton's law gives $v_B-v_A=e(u_A-u_B)$ . Hence $v_B-2=0.4(7-0)=2.8$ , so $v_B=4.8\,\text{m s}^{-1}$ .
2	Conservation of momentum gives $13=2v_A+3v_B$ Restitution gives $v_B-v_A=2$ Solving gives $v_A=7/5\,\text{m s}^{-1}$ and $v_B=17/5\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
Taking the direction of motion as positive, momentum gives $13=2v_A+3v_B$ and restitution gives $v_B-v_A=\tfrac12(5-1)=2$ . Solving gives $v_A=7/5$ and $v_B=17/5\,\text{m s}^{-1}$ ; both are positive, so these are also the requested speeds.

Tier 2 · Standard

Mark scheme for FM1-4.1 Tier 2 · Standard
Question	Scheme	Marks
1	Velocities $0.6\,\text{m s}^{-1}$ and $3.6\,\text{m s}^{-1}$ in the original direction Kinetic energy lost $16.2\,\text{J}$	5
(5 marks)		5
Notes
Momentum gives $12=2v_A+3v_B$ . Restitution gives $v_B-v_A=(1/2)(6)=3$ . Solving yields $v_A=0.6$ and $v_B=3.6$ . Initial kinetic energy is $\tfrac12(2)(6^2)=36\,\text{J}$ ; final kinetic energy is $\tfrac12(2)(0.6^2)+\tfrac12(3)(3.6^2)=19.8\,\text{J}$ . The loss is $36-19.8=16.2\,\text{J}$ .
2	Conservation of momentum gives $14=3v_A+2v_B$ Restitution gives $v_B-v_A=\tfrac12(4-1)=3/2$ Solving gives $v_A=11/5\,\text{m s}^{-1}$ and $v_B=37/10\,\text{m s}^{-1}$ The impulse on $A$ is $3(11/5-4)=-27/5\,\text{N s}$ The impulse on $B$ is $2(37/10-1)=27/5\,\text{N s}$ The kinetic energy falls from $25\,\text{J}$ to $419/20\,\text{J}$ , a loss of $81/5\%$	6
(6 marks)		6
Notes
Taking the common initial direction as positive, conservation of momentum gives $3(4)+2(1)=3v_A+2v_B$ , while restitution gives $v_B-v_A=\tfrac12(4-1)=3/2$ . Hence $v_A=11/5$ and $v_B=37/10\,\text{m s}^{-1}$ . The impulses are $3(11/5-4)=-27/5\,\text{N s}$ on $A$ and $2(37/10-1)=27/5\,\text{N s}$ on $B$ . The initial kinetic energy is $25\,\text{J}$ and the final kinetic energy is $419/20\,\text{J}$ , so the loss is $81/20\,\text{J}$ , which is $(81/20)/25\times100=81/5\%$ .
3	Restitution gives $v_B-1=\tfrac38(8-0)$ Hence the speed of $B$ is $v_B=4\,\text{m s}^{-1}$ If the mass of $B$ is $M\,\text{kg}$ , momentum gives $4(8)=4(1)+4M$ Therefore $M=7\,\text{kg}$	4
(4 marks)		4
Notes
Taking $A$ 's original direction as positive, Newton's law of restitution gives $v_B-1=\tfrac38(8-0)$ , so $v_B=4\,\text{m s}^{-1}$ . If $B$ has mass $M\,\text{kg}$ , conservation of momentum gives $4(8)=4(1)+4M$ , and hence $M=7\,\text{kg}$ . The positive post-impact velocities and $v_B>v_A$ are consistent with the spheres separating.

Tier 3 · Hard

Mark scheme for FM1-4.1 Tier 3 · Hard
Question	Scheme	Marks
1	$e=3/5$ The $1\,\text{kg}$ sphere has velocity $-1/3\,\text{m s}^{-1}$ and the $2\,\text{kg}$ sphere has velocity $8/3\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
Conservation of momentum gives $5=v_A+2v_B$ , and Newton's law of restitution gives $v_B-v_A=5e$ . Solving simultaneously, $v_A=(5-10e)/3$ and $v_B=(5+5e)/3$ . The kinetic energy loss is $\tfrac12(1)(5^2)-\tfrac12(1)v_A^2-\tfrac12(2)v_B^2$ ; setting this equal to $16/3$ and simplifying gives $1-e^2=16/25$ , so $e^2=9/25$ . Since $0\le e\le1$ , $e=3/5$ . Substituting back gives $v_A=-1/3$ and $v_B=8/3\,\text{m s}^{-1}$ .
2	Momentum and restitution give $ku=kv_A+v_B$ and $v_B-v_A=u/2$ Solving gives $v_A=u(k-1/2)/(k+1)$ Solving also gives $v_B=3ku/[2(k+1)]$ The first sphere rebounds exactly when $k<1/2$ The kinetic energy lost is $3kmu^2/[8(k+1)]$ The fraction of the initial kinetic energy lost is $3/[4(k+1)]$ Requiring this to be at most $2/3$ gives $k\ge1/8$ , hence $1/8\le k<1/2$	7
(7 marks)		7
Notes
Let the velocities after impact be $v_A$ and $v_B$ . Momentum and restitution give $ku=kv_A+v_B$ and $v_B-v_A=u/2$ . Solving, $v_A=u(k-1/2)/(k+1)$ and $v_B=3ku/[2(k+1)]$ , so $A$ rebounds when $k<1/2$ . The kinetic energies before and after are $\tfrac12kmu^2$ and $\tfrac12kmv_A^2+\tfrac12mv_B^2$ respectively. Substituting the derived velocities and subtracting gives a loss of $3kmu^2/[8(k+1)]$ , which is the fraction $3/[4(k+1)]$ of the initial energy. Requiring this fraction to be at most $2/3$ gives $k\ge1/8$ , so the required interval is $1/8\le k<1/2$ .
3	Conservation of momentum gives $(2-3k)u=3v_B$ Hence $v_B=(2-3k)u/3$ Separation with $B$ moving rightwards requires $0<k<2/3$ Restitution gives $v_B=e(1+k)u$ Therefore $e=(2-3k)/[3(1+k)]$ Retaining $1/7$ of the initial kinetic energy gives $27k^2-42k+11=0$ The roots are $k=1/3$ and $k=11/9$ ; the second violates $k<2/3$ Thus $k=1/3$ , $e=1/4$ and $v_B=u/3$ rightwards, with $e\in[0,1]$	8
(8 marks)		8
Notes
Take rightwards as positive. Momentum gives $(2-3k)mu=3mv_B$ , so $v_B=(2-3k)u/3$ . Since $A$ is at rest, separation requires $v_B>0$ , hence $k<2/3$ . Restitution gives $v_B=e(u+ku)$ , so $e=(2-3k)/[3(1+k)]$ . The initial kinetic energy is $mu^2+(3/2)mk^2u^2$ and the final kinetic energy is $(3/2)mv_B^2$ . Setting the latter equal to one seventh of the former simplifies to $27k^2-42k+11=0$ , with roots $1/3$ and $11/9$ . Only $k=1/3$ meets the separation condition, giving $e=1/4$ and $v_B=u/3$ .
4	Momentum and restitution give $4u=4v_A+v_B$ and $v_B-v_A=eu$ Solving gives $v_A=(4-e)u/5$ and $v_B=4(1+e)u/5$ Equal kinetic energies give $4v_A^2=v_B^2$ Since separation requires $v_B>v_A$ , this gives $2(4-e)=4(1+e)$ and hence $e=2/3$ The velocities are $v_A=2u/3$ and $v_B=4u/3$ in the original direction The impulse on $A$ is $4m(2u/3-u)=-4mu/3$ The impulse on $B$ is $m(4u/3)=4mu/3$ The final kinetic energy is $16mu^2/9$ , so the fraction lost from the initial $2mu^2$ is $1/9$	8
(8 marks)		8
Notes
Solving momentum and restitution gives $v_A=(4-e)u/5$ and $v_B=4(1+e)u/5$ . Equal post-impact kinetic energies require $4v_A^2=v_B^2$ . The separating velocity ordering selects $2v_A=v_B$ , which yields $e=2/3$ . Substitution gives the two velocities and equal opposite impulses of magnitude $4mu/3$ . The final energy is twice either equal contribution, namely $16mu^2/9$ , compared with $2mu^2$ initially, so the lost fraction is $1/9$ .
5	For $A$ , $3m(v_A-5u)=-10mu$ Hence $v_A=5u/3$ rightwards Restitution gives $v_B-v_A=(5/9)[5u-(-u)]=10u/3$ Hence $v_B=5u$ rightwards For $B$ , $km[5u-(-u)]=10mu$ , so $k=5/3$ The impulses are $-10mu$ on $A$ and $+10mu$ on $B$ $v_B-v_A=10u/3>0$ , so the spheres separate	7
(7 marks)		7
Notes
The stated impulse changes $A$ 's velocity from $5u$ to $5u-10u/3=5u/3$ . Restitution makes the separation speed $(5/9)(6u)=10u/3$ , so $B$ has velocity $5u$ after the collision. Its momentum change is $km[5u-(-u)]$ , which equals the opposite impulse $10mu$ , giving $k=5/3$ . The positive separation speed confirms that the spheres separate.

FM1-4.2 · Successive direct impacts of spheres and/or a sphere with a smooth plane surface.

Tier 1 · Easy

Mark scheme for FM1-4.2 Tier 1 · Easy
Question	Scheme	Marks
1	$-3.5\,\text{m s}^{-1}$	2
(2 marks)		2
Notes
The sphere separates from the fixed wall at $e$ times its approach speed. Its speed becomes $0.7(5)=3.5\,\text{m s}^{-1}$ and the direction reverses, so the velocity is $-3.5\,\text{m s}^{-1}$ .
2	Each impact multiplies the speed by $3/4$ The speed after the $n$ th impact is $16(3/4)^n\,\text{m s}^{-1}$ After the fourth impact the speed is $16(3/4)^4=81/16\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
Each wall impact multiplies the speed by $3/4$ , so the speeds form a geometric sequence. After the $n$ th impact the speed is $16(3/4)^n\,\text{m s}^{-1}$ . At $n=4$ this is $16(3/4)^4=81/16\,\text{m s}^{-1}$ .

Tier 2 · Standard

Mark scheme for FM1-4.2 Tier 2 · Standard
Question	Scheme	Marks
1	After the first impact, $v_A=1.5\,\text{m s}^{-1}$ and $v_B=4.5\,\text{m s}^{-1}$ towards the wall $B$ rebounds at $3\,\text{m s}^{-1}$ towards $A$ , so they approach one another	5
(5 marks)		5
Notes
For equal masses, momentum gives $v_A+v_B=6$ and restitution gives $v_B-v_A=(1/2)(6)=3$ . Hence $v_A=1.5$ and $v_B=4.5\,\text{m s}^{-1}$ towards the wall. At the wall, $B$ reverses with speed $(2/3)(4.5)=3\,\text{m s}^{-1}$ . It then moves towards $A$ , while $A$ is still moving towards the wall, so their separation decreases and a second collision occurs.
2	The gap closes at $8-4=4\,\text{m s}^{-1}$ , so the first impact occurs after $12/4=3\,\text{s}$ The impact position is $x=0+8(3)=24\,\text{m}$ Momentum and restitution give $v_A+v_B=12$ and $v_B-v_A=\tfrac12(8-4)=2$ Thus $v_A=5\,\text{m s}^{-1}$ and $v_B=7\,\text{m s}^{-1}$ The spheres do not collide again, because $v_B>v_A$ so $B$ moves away from $A$	5
(5 marks)		5
Notes
The initial gap closes at $8-4=4\,\text{m s}^{-1}$ , so the first impact occurs after $3\,\text{s}$ at $x=24\,\text{m}$ . For identical masses, momentum and restitution give $v_A+v_B=12$ and $v_B-v_A=\tfrac12(8-4)=2$ , hence $v_A=5$ and $v_B=7\,\text{m s}^{-1}$ . After the impact the separation $v_B-v_A=2\,\text{m s}^{-1}$ is positive, so $B$ moves away from $A$ and no second collision occurs.
3	Ball $A$ first reaches the plane at $t=1\,\text{s}$ with speed $9.8\,\text{m s}^{-1}$ It rebounds at $4.9\,\text{m s}^{-1}$ and returns to the plane $1\,\text{s}$ later, at $t=2\,\text{s}$ Ball $B$ first reaches the plane at $t=3/2\,\text{s}$ with speed $14.7\,\text{m s}^{-1}$ It rebounds at $8.82\,\text{m s}^{-1}$ and its next flight lasts $1.8\,\text{s}$ , so its second impact is at $t=3.3\,\text{s}$ Therefore the first three impacts are $A$ at $t=1$ , $B$ at $t=3/2$ , then $A$ at $t=2\,\text{s}$ Their respective post-impact speeds are $4.9$ , $8.82$ and $2.45\,\text{m s}^{-1}$	6
(6 marks)		6
Notes
For a fall from rest, $h=gt^2/2$ . Ball $A$ therefore hits at $t=1$ with incoming speed $9.8$ and rebounds at $4.9\,\text{m s}^{-1}$ . Its up-and-down flight lasts $2(4.9)/9.8=1\,\text{s}$ , so its second impact is at $t=2$ . Ball $B$ first hits at $t=3/2$ with incoming speed $14.7$ and rebounds at $8.82\,\text{m s}^{-1}$ . Its next flight lasts $2(8.82)/9.8=1.8\,\text{s}$ , so its second impact is later, at $t=3.3$ . The third event is therefore $A$ 's second impact, after which its speed is $(1/2)(4.9)=2.45\,\text{m s}^{-1}$ .

Tier 3 · Hard

Mark scheme for FM1-4.2 Tier 3 · Hard
Question	Scheme	Marks
1	Final velocities $v_A=13/8\,\text{m s}^{-1}$ , $v_B=15/8\,\text{m s}^{-1}$ and $v_C=9/2\,\text{m s}^{-1}$ , all in $A$ 's original direction No further collision occurs because the velocities increase from the rear sphere to the front sphere	7
(7 marks)		7
Notes
For equal masses, each impact conserves the sum of the two velocities and makes their separation speed half their approach speed. The first $A$ - $B$ impact gives $(v_A,v_B)=(2,6)$ . The following $B$ - $C$ impact gives $(v_B,v_C)=(3/2,9/2)$ . Since $A$ at speed $2$ is behind $B$ at speed $3/2$ , they collide again. Solving $v_A+v_B=2+3/2=7/2$ and $v_B-v_A=(1/2)(2-3/2)=1/4$ gives $v_A=13/8$ and $v_B=15/8$ . Now $13/8<15/8<9/2$ , so no rear sphere can catch the one ahead.
2	The speed just before the first impact is $\sqrt{2g(2h)}=2\sqrt{gh}$ Each impact multiplies the speed by $1/2$ The speed immediately after the $n$ th impact is $2^{1-n}\sqrt{gh}$ The greatest subsequent height is $[2^{1-n}\sqrt{gh}]^2/(2g)=2h/4^n$ Just before the $k$ th impact the speed is $2^{2-k}\sqrt{gh}$ The kinetic energy lost at the $k$ th impact is $9mgh/[2(4^{k-1})]$ Summing the geometric series, the loss in the first $n$ impacts is $6mgh(1-4^{-n})$ Letting $n\to\infty$ , the total kinetic energy lost over all impacts is $6mgh$	8
(8 marks)		8
Notes
Energy during the first fall gives the pre-impact speed $\sqrt{2g(2h)}=2\sqrt{gh}$ . Each impact reverses the velocity and multiplies its magnitude by $e=1/2$ , while the intervening rise and fall restore the same speed at the plane. Hence the speed immediately after the $n$ th impact is $(1/2)^n2\sqrt{gh}=2^{1-n}\sqrt{gh}$ , and the rise height is its speed squared divided by $2g$ , namely $2h/4^n$ . Just before impact $k$ , the speed is $2^{2-k}\sqrt{gh}$ . The impact removes the fraction $1-e^2=3/4$ of the incoming kinetic energy, so its loss is $(3/4)(1/2)(3m)(2^{2-k}\sqrt{gh})^2=9mgh/[2(4^{k-1})]$ . Summing from $k=1$ to $n$ gives $6mgh(1-4^{-n})$ ; its limit is $6mgh$ , equal to the initial gravitational potential energy.
3	Since $14^2>2(9.8)(5)$ , the first impact is with the ceiling The speed just before the ceiling impact is $\sqrt{14^2-2(9.8)(5)}=7\sqrt2\,\text{m s}^{-1}$ The ceiling is hit at $t=5(2-\sqrt2)/7\,\text{s}$ Immediately afterwards the velocity is $-7\sqrt2/2\,\text{m s}^{-1}$ , taking upwards as positive The speed just before the floor impact is $\sqrt{(7\sqrt2/2)^2+2(9.8)(5)}=7\sqrt{10}/2\,\text{m s}^{-1}$ The floor is hit a further $5(\sqrt{10}-\sqrt2)/14\,\text{s}$ later, at $t=5(4+\sqrt{10}-3\sqrt2)/14\,\text{s}$ Immediately after the floor impact the velocity is $21\sqrt{10}/8\,\text{m s}^{-1}$ upwards The subsequent rise is $225/64\,\text{m}<5\,\text{m}$ , so the ceiling is not reached again	8
(8 marks)		8
Notes
The ceiling is reachable because the projected kinetic head $14^2/(2g)=10\,\text{m}$ exceeds $5\,\text{m}$ . Energy gives ceiling approach speed $7\sqrt2$ . Solving $5=14t-4.9t^2$ for the earlier root gives $t=5(2-\sqrt2)/7$ . The ceiling impact produces downward speed $7\sqrt2/2$ . Over the $5\,\text{m}$ fall, energy gives floor approach speed $7\sqrt{10}/2$ ; the additional time is $5(\sqrt{10}-\sqrt2)/14$ , so the absolute floor-impact time is $5(4+\sqrt{10}-3\sqrt2)/14$ . The floor rebound speed is $21\sqrt{10}/8$ . Its rise is $v^2/(2g)=225/64\,\text{m}$ , which is below the ceiling height.
4	The catch time is $d/6$ and the wall time for $B$ is $(12-d)/3$ $d/6<(12-d)/3$ gives $0<d<8$ For $d=6$ , the first $A$ - $B$ collision occurs at $t=1\,\text{s}$ and $x=9\,\text{m}$ Momentum and restitution give velocities $15/4$ and $33/4\,\text{m s}^{-1}$ after this collision $B$ reaches the wall next and rebounds with velocity $-11/4\,\text{m s}^{-1}$ The spheres then approach, so a second $A$ - $B$ collision occurs That collision gives $v_A=-31/16\,\text{m s}^{-1}$ and $v_B=47/16\,\text{m s}^{-1}$ $B$ then rebounds from the wall at $-47/48\,\text{m s}^{-1}$ ; since $A$ is ahead moving left faster at $31/16\,\text{m s}^{-1}$ , no further impact occurs	8
(8 marks)		8
Notes
Compare the catch time $d/(9-3)$ with $B$ 's wall time. For $d=6$ , the first impact has pre-impact velocities $9$ and $3$ ; equal-mass momentum and restitution give $15/4$ and $33/4$ . The leading sphere reaches the wall and returns at $-(1/3)(33/4)=-11/4$ , so it meets $A$ again. The second equal-mass collision uses sum $15/4-11/4=1$ and separation speed $(3/4)(15/4+11/4)=39/8$ , yielding $-31/16$ and $47/16$ . Sphere $B$ reaches the wall once more and rebounds at $-47/48$ . Sphere $A$ is ahead and has the greater leftward speed, so the gap then increases.