FM1-5 Elastic collisions in two dimensions — revision question pack

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

Checked against Edexcel 9FM0 section FM1-5. 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

FM1-5.1 · Oblique impact of smooth elastic spheres and a smooth sphere with a fixed surface. Loss of kinetic energy due to impact.

Explanation

For an oblique impact, resolve each velocity into components parallel and perpendicular to the line of centres, or tangential and normal to a fixed surface. Smoothness makes the impulse normal to the contact, so tangential components remain unchanged.
For two equal-radius spheres, apply momentum and Newton's law of restitution only along the line of centres, then recombine the new normal components with the unchanged transverse components.
At a fixed surface, reverse and scale the normal component by $e$ .
Kinetic-energy loss comes only from the changed normal components.
Examiners expect the resolution directions to be shown, signed normal components to be used, and complete final velocity vectors rather than normal components alone.

An oblique approach resolved along and perpendicular to the line of centres.

Worked example

A sphere approaches a fixed smooth wall with velocity $(8\mathbf i+5\mathbf j)\,\text{m s}^{-1}$ , where $\mathbf i$ is normal towards the wall and $\mathbf j$ is tangential. If $e=\tfrac34$ , find the velocity and kinetic-energy fraction remaining after impact.

1.The tangential component remains $5\mathbf j$ .
2.The normal component reverses and becomes $-(\tfrac34)(8)\mathbf i=-6\mathbf i$ .
3.The kinetic-energy fraction is the ratio of speed squares: $\dfrac{6^2+5^2}{8^2+5^2}=\dfrac{61}{89}$ .

Answer: Velocity $(-6\mathbf i+5\mathbf j)\,\text{m s}^{-1}$ ; kinetic-energy fraction remaining $\dfrac{61}{89}$ .

Common mistakes

Don't apply restitution to the whole speed instead of only the normal component.
Don't change the tangential component even though the contact surfaces are smooth.
Don't apply sphere-sphere momentum conservation separately to transverse components affected by no impulse.

Exam tip

Draw the line of centres and resolve every velocity before writing momentum or restitution equations.

Tier 1 · Easy

1.
The unit vector $\mathbf i$ points normally towards a fixed smooth wall and $\mathbf j$ is parallel to it. A sphere arrives with velocity $(3\mathbf i-4\mathbf j)\,\text{m s}^{-1}$ and has coefficient of restitution $1/2$ with the wall. Find its velocity after impact.
(2)
(Total for Question 1 is 2 marks)
2.
A sphere approaches a fixed smooth plane with velocity $(7\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . A unit normal directed towards the plane is $\mathbf n=(5\mathbf i+12\mathbf j)/13$ , and take $\mathbf t=(-12\mathbf i+5\mathbf j)/13$ parallel to the plane. Find the scalar normal and tangential components of the incoming velocity, and state the tangential velocity component immediately after impact.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Two smooth spheres of equal radius and mass $2\,\text{kg}$ collide. At impact, $\mathbf i$ is directed along their line of centres. Their velocities are $(6\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ and $(\mathbf i-3\mathbf j)\,\text{m s}^{-1}$ , and $e=1/2$ . Find both velocities after impact and the impulse on the first sphere.
(5)
(Total for Question 1 is 5 marks)
2.
Two smooth equal-radius spheres, each of mass $1\,\text{kg}$ , collide obliquely. Before impact their velocities are $5\mathbf j\,\text{m s}^{-1}$ and $\mathbf0$ . Afterwards the first sphere has velocity $(-9\mathbf i+13\mathbf j)/5\,\text{m s}^{-1}$ . Find the direction of the line of centres from the first sphere to the second. Hence find the second velocity, $e$ and the kinetic energy lost.
(6)
(Total for Question 2 is 6 marks)
3.
A smooth sphere of mass $2\,\text{kg}$ approaches a fixed smooth plane with velocity $(\mathbf i+7\mathbf j)\,\text{m s}^{-1}$ . A unit normal towards the plane is $\mathbf n=(4\mathbf i+3\mathbf j)/5$ , and $\mathbf t=(-3\mathbf i+4\mathbf j)/5$ is parallel to the plane. The coefficient of restitution is $3/5$ . Find the velocity after impact, the impulse exerted on the sphere and the kinetic energy lost.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
Two smooth spheres have equal radius and each has mass $2\,\text{kg}$ . At collision their line of centres is parallel to $\mathbf i$ , and their velocities are $(5\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ and $(-\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ . Given $e=1/3$ , determine both velocities after impact and the loss of kinetic energy.
(6)
(Total for Question 1 is 6 marks)
2.
Two smooth equal-radius spheres, each of mass $2\,\text{kg}$ , collide obliquely. Sphere $A$ initially has velocity $5\mathbf i\,\text{m s}^{-1}$ and sphere $B$ is at rest. Their coefficient of restitution is $1/2$ . At the instant of impact the line of centres, directed from $A$ to $B$ , makes an angle $\theta$ with $\mathbf i$ , where $30^\circ<\theta<90^\circ$ . After impact, $A$ moves parallel to $5\mathbf i-3\mathbf j$ . Determine $\theta$ , both final velocities and the energy loss, and justify the rejection of any other root of your equation.
(8)
(Total for Question 2 is 8 marks)
3.
Smooth equal-radius spheres $P$ and $Q$ have masses $2\,\text{kg}$ and $3\,\text{kg}$ respectively and collide obliquely. At impact, the unit vector from $P$ to $Q$ along the line of centres is $\mathbf n=(8\mathbf i+15\mathbf j)/17$ , and $\mathbf t=(-15\mathbf i+8\mathbf j)/17$ . Their initial velocities are $(26\mathbf i+121\mathbf j)/17\,\text{m s}^{-1}$ and $(37\mathbf i-39\mathbf j)/17\,\text{m s}^{-1}$ respectively. The collision loses $144/5\,\text{J}$ of kinetic energy. Calculate the coefficient of restitution, the two final velocity vectors and the impulse on $P$ .
(8)
(Total for Question 3 is 8 marks)
4.
Smooth equal-radius spheres $A$ and $B$ have masses $m$ and $2m$ respectively. At impact, resolve along perpendicular unit vectors $\mathbf n$ and $\mathbf t$ , where $\mathbf n$ points from $A$ to $B$ along the line of centres. Before impact the scalar $(\mathbf n,\mathbf t)$ velocity components, in $\text{m s}^{-1}$ , are $(4,1)$ for $A$ and $(-1,3/2)$ for $B$ . Immediately after impact the velocity vectors of the spheres are perpendicular. Find the coefficient of restitution, both final velocity vectors, the impulse on $A$ and the kinetic energy lost.
(8)
(Total for Question 4 is 8 marks)
5.
Smooth equal-radius spheres $A$ and $B$ have masses $m$ and $2m$ respectively. At impact, resolve along perpendicular unit vectors $\mathbf n$ and $\mathbf t$ , where $\mathbf n$ points from $A$ to $B$ along the line of centres. Before impact their velocity components, in $\text{m s}^{-1}$ , are $(6,2)$ for $A$ and $(-2,-1)$ for $B$ . The impulse on $A$ has magnitude $28m/3\,\text{N s}$ . Find the coefficient of restitution, both final velocity vectors, the impulse on each sphere and the kinetic energy lost. Verify that the coefficient is physical.
(8)
(Total for Question 5 is 8 marks)

FM1-5.2 · Successive oblique impacts of a sphere with smooth plane surfaces.

Explanation

At each smooth plane, retain the tangential velocity component and reverse the normal component with its magnitude multiplied by that plane's coefficient of restitution. The complete post-impact vector becomes the pre-impact vector for the next plane, and different surfaces may have different coefficients.
Geometry and direction determine the order of impacts, so the updated vector must actually point towards the stated next surface.
In a right-angled corner, successive impacts can alter perpendicular components independently.
Speed and kinetic energy usually decrease because normal components are scaled by numbers in $[0,1]$ , while a component parallel to a plane is unchanged at that impact.
Examiners expect one labelled vector after each collision and a direction check.

Worked example

A sphere in a right-angled corner has velocity $(12\mathbf i+5\mathbf j)\,\text{m s}^{-1}$ . It strikes the wall normal to $\mathbf i$ with $e=\tfrac12$ , then the wall normal to $\mathbf j$ with $e=\tfrac35$ . Find its final velocity and the fraction of initial kinetic energy remaining.

1.After the first wall, the velocity is $-6\mathbf i+5\mathbf j$ .
2.After the second wall, the velocity is $-6\mathbf i-3\mathbf j$ .
3.The energy fraction is the ratio of speed squares: $\dfrac{36+9}{144+25}=\dfrac{45}{169}$ .

Answer: Final velocity $(-6\mathbf i-3\mathbf j)\,\text{m s}^{-1}$ ; fraction remaining $\dfrac{45}{169}$ .

Common mistakes

Don't start the second impact from the original velocity rather than the first post-impact velocity.
Don't use the first surface's coefficient of restitution for both plane impacts.
Don't reverse both velocity components at a smooth plane instead of only its normal component.

Exam tip

Write a short vector sequence, one line per plane, so that each unchanged tangential component is visible.

Tier 1 · Easy

1.
A sphere approaches a vertical smooth wall with velocity $(6\mathbf i+8\mathbf j)\,\text{m s}^{-1}$ , where $\mathbf i$ is normal to the wall. The coefficient of restitution is $1/2$ . Find its velocity and speed immediately after impact.
(3)
(Total for Question 1 is 3 marks)
2.
A ball moves between two parallel vertical smooth walls. Its velocity before hitting the right-hand wall is $(10\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ . The coefficients of restitution at the right-hand and left-hand walls are $2/5$ and $3/4$ respectively. Find the velocity immediately after the subsequent impact with the left-hand wall.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A sphere moves inside a right-angled corner with velocity $(8\mathbf i+6\mathbf j)\,\text{m s}^{-1}$ . It strikes first the wall normal to $\mathbf i$ and then the wall normal to $\mathbf j$ . Its coefficient of restitution with each wall is $1/2$ . Find its velocity after the second impact and the fraction of its initial kinetic energy that remains.
(5)
(Total for Question 1 is 5 marks)
2.
A ball of mass $2\,\text{kg}$ moves on a smooth horizontal plane inside the rectangle $0\le x\le9\,\text{m}$ , $0\le y\le7\,\text{m}$ , whose walls are smooth. It starts from $(3,2)$ with velocity $(6\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ , strikes the wall $x=9$ and then the wall $y=7$ . The restitution coefficients at these walls are $e_1$ and $e_2$ respectively. Immediately after the second impact its velocity is $(-3\mathbf i-\tfrac32\mathbf j)\,\text{m s}^{-1}$ . Find $e_1$ , $e_2$ and the magnitude of the impulse at each wall.
(6)
(Total for Question 2 is 6 marks)
3.
A ball moves on a smooth horizontal plane in the strip between the parallel smooth walls $y=2x$ and $y=2x+15$ . It starts at $(0,5)$ with velocity $(3\sqrt5\mathbf i+\sqrt5\mathbf j)\,\text{m s}^{-1}$ . At the lower and upper walls, the coefficients of restitution are $1/2$ and $3/4$ respectively. Find the time and point of each of the first two impacts, the velocity after each impact and the fraction of the initial kinetic energy remaining after the second impact.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
A sphere travels inside a right-angled corner with velocity $(10\mathbf i+6\mathbf j)\,\text{m s}^{-1}$ . It hits the wall normal to $\mathbf i$ with coefficient of restitution $e$ , then the wall normal to $\mathbf j$ with coefficient $1/2$ . After both impacts its direction is $30^\circ$ below the negative $\mathbf i$ direction. Find $e$ and the fraction of its initial kinetic energy lost.
(6)
(Total for Question 1 is 6 marks)
2.
A small ball moves on a smooth horizontal plane inside $0<x<10\,\text{m}$ , $0<y<6\,\text{m}$ , whose walls are smooth. From $(2,1)$ its velocity is $(4\mathbf i+3\mathbf j)\,\text{m s}^{-1}$ . The top-wall and right-wall restitution coefficients are $p$ and $q$ , where $0<p,q\le1$ . Show that the ball hits the top wall first and then the right-hand wall, then show that the bottom wall is hit before the left wall exactly when $p>12q/(15+2q)$ . Hence, for $p=2/3$ and $q=3/4$ , find the kinetic-energy fraction remaining after the first two impacts.
(8)
(Total for Question 2 is 8 marks)
3.
A ball of mass $2\,\text{kg}$ moves on a smooth horizontal plane from $(2,3)$ with velocity $(4\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . It first strikes the smooth wall $x=10$ , where the coefficient of restitution is $1/2$ . It then strikes the smooth wall $-x+3y=10$ , whose coefficient of restitution is $q$ . Immediately after the second impact the ball moves parallel to $-3\mathbf i-2\mathbf j$ . Verify the stated impact order by position and time, then find $q$ , the final velocity and the impulse at the second wall.
(8)
(Total for Question 3 is 8 marks)
4.
A ball of mass $2\,\text{kg}$ moves on a smooth horizontal surface with velocity $(12\mathbf i+3\mathbf j)\,\text{m s}^{-1}$ . It first strikes a fixed smooth vertical plane whose unit normal towards the plane is $\mathbf i$ and whose coefficient of restitution is $1/2$ . It next strikes a fixed smooth plane whose unit normal towards the plane is $(-3\mathbf i+4\mathbf j)/5$ and whose coefficient of restitution is $2/3$ . Find the velocity and impulse after each impact, the kinetic energy lost at each impact, and the final speed. Verify that the velocity after the first impact points towards the second plane.
(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.

FM1-5.1 · Oblique impact of smooth elastic spheres and a smooth sphere with a fixed surface. Loss of kinetic energy due to impact.

Tier 1 · Easy

Mark scheme for FM1-5.1 Tier 1 · Easy
Question	Scheme	Marks
1	$(-\tfrac32\mathbf i-4\mathbf j)\,\text{m s}^{-1}$	2
(2 marks)		2
Notes
The tangential component $-4\mathbf j$ is unchanged. The normal component $3\mathbf i$ reverses and is multiplied by $e=1/2$ , becoming $-(3/2)\mathbf i$ . Therefore the new velocity is $(-\tfrac32\mathbf i-4\mathbf j)\,\text{m s}^{-1}$ .
2	The normal scalar component is $47/13\,\text{m s}^{-1}$ The tangential scalar component is $-79/13\,\text{m s}^{-1}$ The tangential velocity remains $(948\mathbf i-395\mathbf j)/169\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
Taking scalar products, $(7\mathbf i+\mathbf j)\boldsymbol\cdot\mathbf n=47/13$ and $(7\mathbf i+\mathbf j)\boldsymbol\cdot\mathbf t=-79/13$ . Thus the incoming velocity is $(47/13)\mathbf n-(79/13)\mathbf t$ . Smooth contact produces no tangential impulse, so the tangential velocity $-(79/13)\mathbf t=(948\mathbf i-395\mathbf j)/169\,\text{m s}^{-1}$ is unchanged after impact.

Tier 2 · Standard

Mark scheme for FM1-5.1 Tier 2 · Standard
Question	Scheme	Marks
1	First sphere: $(\tfrac94\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ Second sphere: $(\tfrac{19}{4}\mathbf i-3\mathbf j)\,\text{m s}^{-1}$ Impulse on the first sphere: $-\tfrac{15}{2}\mathbf i\,\text{N s}$	5
(5 marks)		5
Notes
The $\mathbf j$ components stay unchanged. Along $\mathbf i$ , equal masses give $v_{1i}+v_{2i}=7$ and $v_{2i}-v_{1i}=(1/2)(6-1)=5/2$ . Solving gives $v_{1i}=9/4$ and $v_{2i}=19/4$ . Thus the stated velocity vectors follow. The impulse on the first sphere is $2[(9/4)-6]\mathbf i=-15\mathbf i/2\,\text{N s}$ .
2	The first sphere's change in velocity is $(-9\mathbf i-12\mathbf j)/5\,\text{m s}^{-1}$ Therefore the line-of-centres unit vector from the first sphere to the second is $(3\mathbf i+4\mathbf j)/5$ Vector momentum gives the second velocity as $(9\mathbf i+12\mathbf j)/5\,\text{m s}^{-1}$ The normal speed of approach is $(5\mathbf j)\boldsymbol\cdot(3\mathbf i+4\mathbf j)/5=4\,\text{m s}^{-1}$ The normal speed of separation is $2\,\text{m s}^{-1}$ , so $e=1/2$ The kinetic energies are $25/2\,\text{J}$ and $19/2\,\text{J}$ , so the loss is $3\,\text{J}$	6
(6 marks)		6
Notes
The first sphere's velocity change is $(-9\mathbf i-12\mathbf j)/5$ , so the line-of-centres unit vector from it to the second sphere is $\mathbf n=(3\mathbf i+4\mathbf j)/5$ . Vector momentum gives the second velocity as $(9\mathbf i+12\mathbf j)/5\,\text{m s}^{-1}$ . The normal speed of approach is $(5\mathbf j)\boldsymbol\cdot\mathbf n=4$ . The normal speed of separation is $[(18\mathbf i-\mathbf j)/5]\boldsymbol\cdot\mathbf n=2$ , so $e=1/2$ . The initial and final kinetic energies are $25/2\,\text{J}$ and $19/2\,\text{J}$ . Hence the kinetic energy lost is $3\,\text{J}$ .
3	The incoming scalar components are $\mathbf u\boldsymbol\cdot\mathbf n=5$ and $\mathbf u\boldsymbol\cdot\mathbf t=5$ After impact the scalar components are $-3$ along $\mathbf n$ and $5$ along $\mathbf t$ The final velocity is $-3\mathbf n+5\mathbf t=(-27\mathbf i+11\mathbf j)/5\,\text{m s}^{-1}$ The impulse is $2[(-27\mathbf i+11\mathbf j)/5-(\mathbf i+7\mathbf j)]=(-64\mathbf i-48\mathbf j)/5\,\text{N s}$ This is $-16\mathbf n\,\text{N s}$ , so it is parallel to the plane normal and has magnitude $16\,\text{N s}$ The kinetic energy lost is $\tfrac12(2)[(5^2+5^2)-((-3)^2+5^2)]=16\,\text{J}$	6
(6 marks)		6
Notes
Resolving along the orthonormal basis gives incoming components $5$ along $\mathbf n$ and $5$ along $\mathbf t$ . Smoothness leaves the tangential component unchanged, while restitution changes the normal component to $-(3/5)(5)=-3$ . Hence the final velocity is $-3\mathbf n+5\mathbf t=(-27\mathbf i+11\mathbf j)/5$ . Multiplying the velocity change by the mass gives impulse $(-64\mathbf i-48\mathbf j)/5=-16\mathbf n\,\text{N s}$ , which has magnitude $16\,\text{N s}$ . The kinetic energy falls from $50\,\text{J}$ to $34\,\text{J}$ , so the loss is $16\,\text{J}$ .

Tier 3 · Hard

Mark scheme for FM1-5.1 Tier 3 · Hard
Question	Scheme	Marks
1	Velocities $(\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ and $(3\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ Kinetic energy lost $16\,\text{J}$	6
(6 marks)		6
Notes
The $\mathbf j$ components remain $4$ and $2$ . Along $\mathbf i$ , momentum for equal masses gives $v_{1i}+v_{2i}=5+(-1)=4$ , while restitution gives $v_{2i}-v_{1i}=(1/3)(5-(-1))=2$ . Hence $v_{1i}=1$ and $v_{2i}=3$ . Since each mass is $2\,\text{kg}$ , the initial kinetic energy is $5^2+4^2+(-1)^2+2^2=46\,\text{J}$ and the final kinetic energy is $1^2+4^2+3^2+2^2=30\,\text{J}$ . The loss is $16\,\text{J}$ .
2	With $\mathbf n=(\cos\theta,\sin\theta)$ , normal momentum and restitution give $\mathbf v_A=5\mathbf i-(15/4)\cos\theta\,\mathbf n$ The direction of $A$ gives $5\sin\theta\cos\theta=4-3\cos^2\theta$ Writing $t=\tan\theta$ gives $4t^2-5t+1=0$ , so $t=1$ or $t=1/4$ The root $t=1/4$ gives $\theta\approx14^\circ$ and is rejected by $30^\circ<\theta<90^\circ$ Hence $\theta=45^\circ$ $\mathbf v_A=(25\mathbf i-15\mathbf j)/8\,\text{m s}^{-1}$ Momentum gives $\mathbf v_B=15(\mathbf i+\mathbf j)/8\,\text{m s}^{-1}$ The kinetic energy falls from $25\,\text{J}$ to $325/16\,\text{J}$ , so the loss is $75/16\,\text{J}$	8
(8 marks)		8
Notes
Let $\mathbf n=(\cos\theta,\sin\theta)$ . For equal masses with $B$ initially at rest, normal momentum and restitution give $\mathbf v_A=5\mathbf i-(15/4)\cos\theta\,\mathbf n$ . Its stated direction gives $v_{Ay}/v_{Ax}=-3/5$ , so $5\sin\theta\cos\theta=4-3\cos^2\theta$ . With $t=\tan\theta$ , this becomes $4t^2-5t+1=0$ , giving $t=1$ or $t=1/4$ . The other root gives $\tan\theta=1/4$ , so $\theta\approx14^\circ$ , which lies outside $30^\circ<\theta<90^\circ$ and is rejected. Hence $\theta=45^\circ$ . Substitution gives $\mathbf v_A=(25\mathbf i-15\mathbf j)/8$ and momentum gives $\mathbf v_B=15(\mathbf i+\mathbf j)/8$ . The initial energy is $25\,\text{J}$ and the final energy is $425/32+225/32=325/16\,\text{J}$ , so the loss is $75/16\,\text{J}$ . The given $e=1/2$ lies in $[0,1]$ .
3	The initial $(\mathbf n,\mathbf t)$ scalar components are $(7,2)$ for $P$ and $(-1,-3)$ for $Q$ If the final normal components are $a,b$ , then $2a+3b=11$ and $b-a=8e$ Solving gives $a=(11-24e)/5$ and $b=(11+16e)/5$ The total kinetic energy before impact is $\tfrac12(2)(7^2+2^2)+\tfrac12(3)((-1)^2+(-3)^2)=68\,\text{J}$ The total kinetic energy after impact is $a^2+4+\tfrac32(b^2+9)=(148+192e^2)/5\,\text{J}$ Hence $68-(148+192e^2)/5=144/5$ , so $e=1/2$ The final velocities are $(-158\mathbf i/85+13\mathbf j/17)\,\text{m s}^{-1}$ for $P$ and $(377\mathbf i/85+33\mathbf j/17)\,\text{m s}^{-1}$ for $Q$ The impulse on $P$ is $-(72/5)\mathbf n=(-576\mathbf i/85-216\mathbf j/17)\,\text{N s}$	8
(8 marks)		8
Notes
Resolving the initial velocities gives $(7,2)$ for $P$ and $(-1,-3)$ for $Q$ along $(\mathbf n,\mathbf t)$ . If the final normal components are $a,b$ , normal momentum and restitution give $2a+3b=11$ and $b-a=8e$ , so $a=(11-24e)/5$ and $b=(11+16e)/5$ . The kinetic energy before impact is $68\,\text{J}$ . The kinetic energy afterwards, retaining the unchanged tangential components, is $a^2+4+\tfrac32(b^2+9)=(148+192e^2)/5\,\text{J}$ . Therefore the stated loss gives $68-(148+192e^2)/5=144/5$ , so $e=1/2$ . Then $a=-1/5$ and $b=19/5$ ; recombining gives the stated velocities. The impulse on $P$ is $2(-1/5-7)\mathbf n=-(72/5)\mathbf n=(-576\mathbf i/85-216\mathbf j/17)\,\text{N s}$ .
4	If the final normal components are $a,b$ , then $a+2b=2$ and $b-a=5e$ Hence $a=(2-10e)/3$ and $b=(2+5e)/3$ The tangential components remain $1$ and $3/2$ Perpendicular final velocities give $ab+3/2=0$ This gives $100e^2+20e-35=0$ , so the physical root is $e=1/2$ The final velocities are $-\mathbf n+\mathbf t$ for $A$ and $(3\mathbf n+3\mathbf t)/2$ for $B$ , in $\text{m s}^{-1}$ The impulse on $A$ is $m(-1-4)\mathbf n=-5m\mathbf n\,\text{N s}$ The kinetic energy falls from $47m/4\,\text{J}$ to $11m/2\,\text{J}$ , so the loss is $25m/4\,\text{J}$	8
(8 marks)		8
Notes
Normal momentum and restitution give $a+2b=2$ and $b-a=5e$ . Smoothness preserves the tangential components. The perpendicularity condition is therefore $ab+(1)(3/2)=0$ , leading to roots $e=1/2$ and $e=-7/10$ ; only $1/2$ is physical. Substitution gives normal components $-1$ and $3/2$ , hence the stated vectors. The normal momentum change of $A$ is $-5m\mathbf n$ . Direct calculation of the two total kinetic energies gives the loss $25m/4$ .
5	The smooth-sphere impulse on $A$ is $-28m\mathbf n/3\,\text{N s}$ The final normal component of $A$ is $6-28/3=-10/3\,\text{m s}^{-1}$ The equal opposite impulse gives $B$ final normal component $-2+(28/3)/2=8/3\,\text{m s}^{-1}$ Restitution gives $e=[8/3-(-10/3)]/[6-(-2)]=3/4$ $0<3/4<1$ , so the coefficient is physical The final velocities are $-10\mathbf n/3+2\mathbf t$ for $A$ and $8\mathbf n/3-\mathbf t$ for $B$ , in $\text{m s}^{-1}$ The impulses are $-28m\mathbf n/3$ on $A$ and $28m\mathbf n/3$ on $B$ , in $\text{N s}$ The kinetic energy falls from $25m\,\text{J}$ to $47m/3\,\text{J}$ , so the loss is $28m/3\,\text{J}$	8
(8 marks)		8
Notes
For smooth spheres the impulse acts along the line of centres, opposing $A$ 's approach, while both tangential components are unchanged. The stated impulse changes $A$ 's normal component from $6$ to $-10/3$ and $B$ 's from $-2$ to $8/3$ . The separation-to-approach ratio is therefore $6/8=3/4$ . Recombining the unchanged tangential components gives the stated velocities, and the impulses are equal and opposite. Direct calculation gives initial kinetic energy $25m\,\text{J}$ and final kinetic energy $47m/3\,\text{J}$ .

FM1-5.2 · Successive oblique impacts of a sphere with smooth plane surfaces.

Tier 1 · Easy

Mark scheme for FM1-5.2 Tier 1 · Easy
Question	Scheme	Marks
1	Velocity $(-3\mathbf i+8\mathbf j)\,\text{m s}^{-1}$ Speed $\sqrt{73}\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
The $\mathbf j$ component is tangential and remains $8$ . The normal component reverses and becomes $-(1/2)(6)=-3$ . Hence the velocity is $-3\mathbf i+8\mathbf j$ , with speed $\sqrt{(-3)^2+8^2}=\sqrt{73}\,\text{m s}^{-1}$ .
2	After the right-hand impact, the normal component is $-4\mathbf i\,\text{m s}^{-1}$ The tangential component remains $4\mathbf j\,\text{m s}^{-1}$ at both impacts After the left-hand impact, the velocity is $(3\mathbf i+4\mathbf j)\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
At the right-hand wall, the normal component $10\mathbf i$ reverses and is multiplied by $2/5$ , becoming $-4\mathbf i$ , while the tangential component remains $4\mathbf j$ . At the left-hand wall, $-4\mathbf i$ reverses and is multiplied by $3/4$ , becoming $3\mathbf i$ ; the tangential component is again unchanged. The velocity after the second impact is therefore $(3\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ .

Tier 2 · Standard

Mark scheme for FM1-5.2 Tier 2 · Standard
Question	Scheme	Marks
1	$(-4\mathbf i-3\mathbf j)\,\text{m s}^{-1}$ Fraction remaining $1/4$	5
(5 marks)		5
Notes
The first impact changes $8\mathbf i$ to $-4\mathbf i$ and leaves $6\mathbf j$ , giving $-4\mathbf i+6\mathbf j$ . The second leaves $-4\mathbf i$ and changes $6\mathbf j$ to $-3\mathbf j$ . The final speed squared is $4^2+3^2=25$ , compared with initial speed squared $8^2+6^2=100$ . For unchanged mass, the kinetic energy fraction is $25/100=1/4$ .
2	After the first impact the velocity is $(-6e_1\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ After the second impact it is $(-6e_1\mathbf i-2e_2\mathbf j)\,\text{m s}^{-1}$ Comparing $\mathbf i$ -components gives $e_1=1/2$ Comparing $\mathbf j$ -components gives $e_2=3/4$ The first-wall impulse has magnitude $2\lvert-3-6\rvert=18\,\text{N s}$ The second-wall impulse has magnitude $2\lvert-3/2-2\rvert=7\,\text{N s}$	6
(6 marks)		6
Notes
At the wall $x=9$ , only the normal $\mathbf i$ -component changes, so the velocity becomes $-6e_1\mathbf i+2\mathbf j$ . At the wall $y=7$ , only the normal $\mathbf j$ -component changes, giving $-6e_1\mathbf i-2e_2\mathbf j$ . Comparing this with the given vector yields $e_1=1/2$ and $e_2=3/4$ . These values are consistent with the stated order: the first impact is at $(9,4)$ after $1\,\text{s}$ ; the ball then reaches $y=7$ after a further $3/2\,\text{s}$ at $x=9/2$ , before it can reach $x=0$ . The velocity changes at the two impacts are $-9\mathbf i$ and $-(7/2)\mathbf j\,\text{m s}^{-1}$ . Multiplying their magnitudes by the mass $2\,\text{kg}$ gives impulses $18\,\text{N s}$ and $7\,\text{N s}$ .
3	With $\mathbf n=(-2\mathbf i+\mathbf j)/\sqrt5$ and $\mathbf t=(\mathbf i+2\mathbf j)/\sqrt5$ , the initial scalar components are $-5$ and $5$ The value of $y-2x$ falls from $5$ to $0$ at rate $5\sqrt5$ , so the first impact is at time $1/\sqrt5\,\text{s}$ and point $(3,6)$ on $y=2x$ After the first impact the scalar components are $5/2$ along $\mathbf n$ and $5$ along $\mathbf t$ , giving velocity $(5\sqrt5/2)\mathbf j\,\text{m s}^{-1}$ The value of $y-2x$ then rises from $0$ to $15$ in $6/\sqrt5\,\text{s}$ , so the second impact is at total time $7/\sqrt5\,\text{s}$ and point $(3,21)$ After the second impact the scalar components are $-15/8$ along $\mathbf n$ and $5$ along $\mathbf t$ , giving velocity $(7\sqrt5\mathbf i/4+13\sqrt5\mathbf j/8)\,\text{m s}^{-1}$ The kinetic-energy fraction remaining is $[(-15/8)^2+5^2]/[(-5)^2+5^2]=73/128$	6
(6 marks)		6
Notes
Take $\mathbf n=(-2\mathbf i+\mathbf j)/\sqrt5$ towards increasing $y-2x$ and $\mathbf t=(\mathbf i+2\mathbf j)/\sqrt5$ . The initial components are $(-5,5)$ . Since $y-2x$ decreases from $5$ at rate $5\sqrt5$ , the lower wall is hit after $1/\sqrt5\,\text{s}$ at $(3,6)$ . Restitution changes the normal component to $5/2$ and leaves the tangential component $5$ , giving velocity $(5\sqrt5/2)\mathbf j$ . Now $y-2x$ increases at $5\sqrt5/2$ , so the upper wall is reached after a further $6/\sqrt5\,\text{s}$ , at total time $7/\sqrt5\,\text{s}$ and point $(3,21)$ . The second impact changes the normal component to $-(3/4)(5/2)=-15/8$ and again leaves the tangential component unchanged. Recombining gives $7\sqrt5\mathbf i/4+13\sqrt5\mathbf j/8$ . The speed-squared ratio is $[225/64+25]/50=73/128$ .

Tier 3 · Hard

Mark scheme for FM1-5.2 Tier 3 · Hard
Question	Scheme	Marks
1	$e=3\sqrt3/10$ Fraction lost $25/34$	6
(6 marks)		6
Notes
After the first impact the velocity is $-10e\mathbf i+6\mathbf j$ ; after the second it is $-10e\mathbf i-3\mathbf j$ . The final direction gives $3/(10e)=\tan30^\circ=1/\sqrt3$ , so $e=3\sqrt3/10$ . The final speed squared is $100e^2+9=27+9=36$ , whereas the initial speed squared is $10^2+6^2=136$ . Thus the fraction of kinetic energy lost is $(136-36)/136=25/34$ .
2	Initially the times to the top and right walls are $5/3\,\text{s}$ and $2\,\text{s}$ , so the top wall is hit first The top-wall impact point is $(26/3,6)$ and the new velocity is $(4\mathbf i-3p\mathbf j)\,\text{m s}^{-1}$ The time from there to the right wall is $1/3\,\text{s}$ , less than the competing bottom-wall time $2/p\ge2\,\text{s}$ The right-wall impact point is $(10,6-p)$ and the new velocity is $(-4q\mathbf i-3p\mathbf j)\,\text{m s}^{-1}$ The subsequent times to the bottom and left walls are $(6-p)/(3p)$ and $5/(2q)$ The bottom wall is next exactly when $2q(6-p)<15p$ , equivalent to $p>12q/(15+2q)$ For $p=2/3$ and $q=3/4$ , the boundary is $6/11$ , so the bottom wall is next The kinetic-energy fraction after two impacts is $(16q^2+9p^2)/25=13/25$	8
(8 marks)		8
Notes
Initially the top-wall time is $(6-1)/3=5/3\,\text{s}$ and the right-wall time is $(10-2)/4=2\,\text{s}$ , so the top is first. Its impact point has $x=26/3$ and the velocity becomes $(4,-3p)$ . The right wall is $4/3\,\text{m}$ away, giving time $1/3\,\text{s}$ ; the competing bottom-wall time is $6/(3p)=2/p\ge2\,\text{s}$ , so the right wall is second. The point is then $(10,6-p)$ and the velocity is $(-4q,-3p)$ . The times to the bottom and left walls are $(6-p)/(3p)$ and $5/(2q)$ . Bottom comes first exactly when $2q(6-p)<15p$ , or $p>12q/(15+2q)$ . For $q=3/4$ the boundary is $6/11$ , and $p=2/3$ exceeds it. The speed-squared ratio after the two impacts is $(16q^2+9p^2)/25=(9+4)/25=13/25$ .
3	Before the first impact, $-x+3y$ decreases from $7$ at $1\,\text{m s}^{-1}$ , so that wall is not approached, while $x=10$ is reached at $t=2\,\text{s}$ The first impact point is $(10,5)$ After the first impact the velocity is $(-2\mathbf i+\mathbf j)\,\text{m s}^{-1}$ Now $-x+3y$ increases from $5$ at $5\,\text{m s}^{-1}$ , so the second wall is reached $1\,\text{s}$ later at $(8,6)$ After the second impact the velocity is $[(q-3)\mathbf i-(1+3q)\mathbf j]/2\,\text{m s}^{-1}$ Parallelism to $-3\mathbf i-2\mathbf j$ gives $3(1+3q)=2(3-q)$ , hence $q=3/11$ Since $0<3/11<1$ , the final velocity is $(-15\mathbf i/11-10\mathbf j/11)\,\text{m s}^{-1}$ The second-wall impulse is $(14\mathbf i/11-42\mathbf j/11)\,\text{N s}$ , of magnitude $14\sqrt{10}/11\,\text{N s}$	8
(8 marks)		8
Notes
Initially $x=10$ is reached after $2\,\text{s}$ at $(10,5)$ . During this interval $-x+3y$ has rate $-4+3=-1$ , so its value moves away from $10$ . The first impact changes the velocity to $(-2,1)$ . The second wall expression then has rate $2+3=5$ and rises from $5$ to $10$ in one second, giving impact point $(8,6)$ . A unit normal towards the second wall is $(-\mathbf i+3\mathbf j)/\sqrt{10}$ ; applying restitution gives $[(q-3)\mathbf i-(1+3q)\mathbf j]/2$ . The stated direction yields $q=3/11$ , within the physical range. Substitution gives the final vector. Multiplying its change from $(-2,1)$ by the mass gives the stated impulse.
4	After the first impact the velocity is $(-6\mathbf i+3\mathbf j)\,\text{m s}^{-1}$ The first impulse is $2[(-6,3)-(12,3)]=-36\mathbf i\,\text{N s}$ The first kinetic-energy loss is $\tfrac12(2)[12^2-(-6)^2]=108\,\text{J}$ With $\mathbf n=(-3\mathbf i+4\mathbf j)/5$ and $\mathbf t=(-4\mathbf i-3\mathbf j)/5$ , the post-first-impact components are $6$ and $3$ The positive normal component $6$ confirms motion towards the second plane After the second impact the components are $-4$ along $\mathbf n$ and $3$ along $\mathbf t$ , giving velocity $-5\mathbf j\,\text{m s}^{-1}$ The second impulse is $2(-4-6)\mathbf n=-20\mathbf n=(12\mathbf i-16\mathbf j)\,\text{N s}$ The second kinetic-energy loss is $20\,\text{J}$ and the final speed is $5\,\text{m s}^{-1}$	8
(8 marks)		8
Notes
The first smooth-plane impact reverses and halves only the $\mathbf i$ component, giving $(-6,3)$ , impulse $-36\mathbf i$ and energy loss $108\,\text{J}$ . Resolving this velocity along the second plane's normal and tangent gives components $(6,3)$ , so the ball is indeed approaching that plane. The second impact changes these to $(-4,3)$ , which recombine to $-5\mathbf j$ . The normal component changes by $-10$ , so the impulse is $-20\mathbf n$ . The speed-squared reduction at the second impact is $36-16=20$ , and with mass $2$ this is also a $20\,\text{J}$ loss.