M2.5 Collisions and restitution
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This is an M2 (A2) sub-topic. The one-dimensional basics (\(p=mv\), coalescing particles, jerks in strings) are on the M1 momentum and impulse page; oblique impacts are not part of M2 and belong to Further Mechanics in IAL Further Mathematics.
What this page covers
7 points
- Momentum as a vector and the impulse–momentum principle \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\) with \(\mathbf{i}\) and \(\mathbf{j}\) components
- Conservation of linear momentum, and the equal and opposite impulses in a collision
- Newton's law of restitution for a direct impact between two particles, with \(0\le e\le 1\)
- The kinetic energy lost in an impact
- A particle striking a fixed smooth plane surface (a wall or the ground) at right angles
- Successive impacts: up to three particles in a line, or two particles and a wall
- Deciding whether any further collision happens
In one minute
7 points
- One number, the coefficient of restitution \(e\), together with conservation of momentum, gives two equations for the two unknown velocities after any direct impact.
- Impulse = change in momentum: \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\). It is a vector; work in \(\mathbf{i}\) and \(\mathbf{j}\) components. Unit N s.
- In a collision the particles exert equal and opposite impulses on each other, so total momentum is conserved.
- Newton's law of restitution: speed of separation = \(e\) × speed of approach, with \(0\le e\le 1\).
- \(e=1\): perfectly elastic, no kinetic energy lost. \(e=0\): the particles coalesce. Otherwise some kinetic energy is always lost.
- Wall or ground (fixed, smooth, hit at right angles): rebound speed = \(e\) × impact speed.
- Successive impacts: another collision happens only if a particle behind is moving faster (in the positive direction) than the one in front of it.
Contents
12 sections
- How do momentum and impulse work as vectors?
- What is conserved when two particles collide?
- What is Newton's law of restitution?
- How much kinetic energy is lost in an impact?
- What happens when a particle hits a fixed smooth surface?
- How do you decide whether there are further collisions?
- Exam tips for M2.5
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do momentum and impulse work as vectors?
M1 taught you that momentum is conserved. M2 adds two things: momentum in two dimensions, and the missing ingredient for collisions, which is how bouncy they are.
The momentum of a particle of mass \(m\) moving with velocity \(\mathbf{v}\) is \(m\mathbf{v}\). It has the same direction as the velocity, and its unit is kg m s⁻¹, which is the same as N s. An impulse \(\mathbf{I}\) acting on the particle changes its momentum. For a constant force \(\mathbf{F}\) acting for time \(t\), \(\mathbf{I}=\mathbf{F}t\); for a sudden blow you are usually given \(\mathbf{I}\) directly.
\[\mathbf{I}=m\mathbf{v}-m\mathbf{u}\]Here \(\mathbf{u}\) and \(\mathbf{v}\) are the velocities just before and just after. This is not in the formulae booklet: learn it. Use it on one particle at a time. Because these are vectors, the equation holds component by component: the \(\mathbf{i}\) parts balance and the \(\mathbf{j}\) parts balance separately. Drawn head to tail, \(m\mathbf{u}+\mathbf{I}=m\mathbf{v}\) forms a triangle, which is a useful check on directions. In one dimension it becomes \(I=mv-mu\) with signs showing direction.
A particle of mass 0.4 kg is moving with velocity \((5\mathbf{i}-2\mathbf{j})\) m s⁻¹ on a smooth horizontal surface. It receives an impulse of \((-3\mathbf{i}+4\mathbf{j})\) N s. (a) Find the velocity of the particle immediately after the impulse. (b) Find its speed immediately after the impulse.
(a) Momentum before: \(m\mathbf{u}=0.4(5\mathbf{i}-2\mathbf{j})=2\mathbf{i}-0.8\mathbf{j}\). Since \(m\mathbf{v}=m\mathbf{u}+\mathbf{I}\): \(0.4\mathbf{v}=(2-3)\mathbf{i}+(-0.8+4)\mathbf{j}=-\mathbf{i}+3.2\mathbf{j}\), so \(\mathbf{v}=(-2.5\mathbf{i}+8\mathbf{j})\) m s⁻¹
(b) \(|\mathbf{v}|=\sqrt{2.5^2+8^2}=\sqrt{70.25}=\) 8.38 m s⁻¹
Adding the impulse to the velocity: \(5\mathbf{i}-2\mathbf{j}+(-3\mathbf{i}+4\mathbf{j})\). The impulse changes momentum, so divide it by the mass first (or add it to \(m\mathbf{u}\)).
A ball of mass 0.25 kg, modelled as a particle, is moving with velocity \((6\mathbf{i}+8\mathbf{j})\) m s⁻¹ when it is kicked. Immediately after the kick its velocity is \((-9\mathbf{i}+12\mathbf{j})\) m s⁻¹. Find (a) the impulse of the kick, (b) the magnitude of this impulse, (c) the angle through which the direction of motion of the ball is turned, (d) the kinetic energy gained by the ball.
(a) \(\mathbf{I}=m(\mathbf{v}-\mathbf{u})=0.25(-15\mathbf{i}+4\mathbf{j})=\) \((-3.75\mathbf{i}+\mathbf{j})\) N s
(b) \(|\mathbf{I}|=\sqrt{3.75^2+1^2}=\) 3.88 N s
(c) The speeds are \(|\mathbf{u}|=10\) and \(|\mathbf{v}|=15\). By the scalar product, \(\cos\theta=\dfrac{6(-9)+8(12)}{10\times 15}=\dfrac{42}{150}=0.28\), so \(\theta=73.7°\). (Without the scalar product: the direction before is \(\tan^{-1}(8/6)=53.1°\) above \(\mathbf{i}\); after, it is \(180°-\tan^{-1}(12/9)=126.9°\); the difference is 73.7°.)
(d) Kinetic energy is a scalar, so use the speeds: \(\tfrac12(0.25)(15^2-10^2)=15.625\) J \(\approx\) 15.6 J
What is conserved when two particles collide?
During the impact, A pushes on B and B pushes on A. By Newton's third law these forces are equal and opposite and act for the same time, so the impulses on A and B are equal and opposite. Adding the two impulse–momentum equations, the impulses cancel: total momentum before = total momentum after, provided no external impulse acts along the line of motion.
\[m_1u_1+m_2u_2=m_1v_1+m_2v_2\]This is not in the formulae booklet: learn it. It is also true in vector form, \(m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2\), for particles moving in a plane.
Choose a positive direction and draw a before-and-after diagram. Give every unknown velocity an arrow in the positive direction; a negative answer then simply means it moves the other way. Spheres are assumed smooth and of equal radius so that, in a direct impact, every force acts along the line of centres.
Two small blocks A and B, of masses 0.6 kg and 0.9 kg, slide in the same direction along a smooth straight groove. A moves at 5 m s⁻¹ and B at 2 m s⁻¹, so A catches up with B. Immediately after the collision, B moves at 4 m s⁻¹ in the same direction as before. (a) Find the velocity of A immediately after the collision. (b) Find the impulse exerted on B, and deduce the impulse exerted on A.
(a) Positive in the direction of motion: \(0.6(5)+0.9(2)=0.6v_A+0.9(4)\), so \(4.8=0.6v_A+3.6\) and \(v_A=2.0\) m s⁻¹, in the original direction
(b) Impulse on B = change in B's momentum \(=0.9(4-2)=\) 1.8 N s in the direction of motion. On A: \(0.6(2-5)=-1.8\), so 1.8 N s opposite to the direction of motion
Sense check: after the impact A (2 m s⁻¹) is slower than B (4 m s⁻¹), so they separate. If your answer had A still faster than B, A would have to pass through B — impossible.
What is Newton's law of restitution?
Momentum gives one equation, but a collision has two unknown velocities. The second equation describes how much the particles bounce apart. The speed of approach is how fast the gap between them was closing before the impact; the speed of separation is how fast it opens afterwards. Experiments show that, for two given materials, separation ÷ approach is roughly constant. This constant is the coefficient of restitution \(e\).
Particle 1 is behind particle 2 (in the positive direction), \(u\) are velocities before and \(v\) velocities after, all signed. This is not in the formulae booklet: learn it. Both brackets must be positive: \(u_1-u_2>0\) means they really do approach, and \(v_2-v_1\ge 0\) means they separate.
Two smooth spheres P and Q, of equal radius and masses 2 kg and 3 kg, move towards each other along the same straight line on a smooth horizontal table. Just before they collide, P has speed 5 m s⁻¹ and Q has speed 2 m s⁻¹. The coefficient of restitution between them is ½. (a) Find the velocities of P and Q immediately after the collision. (b) Find the kinetic energy lost in the collision.
(a) Take P's direction as positive, so \(u_P=5\), \(u_Q=-2\). Momentum: \(2(5)+3(-2)=2v_P+3v_Q\), so \(2v_P+3v_Q=4\) (1). Restitution, with approach speed \(5-(-2)=7\): \(v_Q-v_P=\tfrac12\times 7=3.5\) (2). Substitute \(v_Q=v_P+3.5\) into (1): \(5v_P+10.5=4\), so \(v_P=-1.3\) m s⁻¹, \(v_Q=2.2\) m s⁻¹. P rebounds at 1.3 m s⁻¹ and Q rebounds at 2.2 m s⁻¹: both directions of motion are reversed.
(b) Before: \(\tfrac12(2)(5^2)+\tfrac12(3)(2^2)=25+6=31\) J. After: \(\tfrac12(2)(1.3^2)+\tfrac12(3)(2.2^2)=1.69+7.26=8.95\) J. Loss \(=31-8.95=\) 22.05 J
Write the restitution equation the same way every time: "(velocity of the front particle after) − (velocity of the back particle after) = \(e\) × (velocity of the back particle before − velocity of the front particle before)". With signed velocities this never goes wrong, even when a particle reverses.
How much kinetic energy is lost in an impact?
Momentum is always conserved in a collision; kinetic energy is not, unless \(e=1\). The difference becomes heat, sound and permanent deformation. To find the loss, work out total kinetic energy before − total kinetic energy after. Kinetic energy is a scalar, so add \(\tfrac12 mv^2\) for each particle; signs of velocities do not matter once squared.
Algebraically, for masses \(m_1\) and \(m_2\) with approach speed \(w\), the loss is
\[\tfrac12\cdot\frac{m_1m_2}{m_1+m_2}\cdot(1-e^2)\,w^2 .\]In worked example 4: \(\tfrac12\times\tfrac65\times\tfrac34\times 49=22.05\) J, as found. This shows why \(0\le e\le 1\): if \(e\) were greater than 1 the "loss" would be negative and the collision would create kinetic energy from nowhere.
A particle A of mass \(2m\) moves with speed \(u\) on a smooth horizontal plane and collides directly with a particle B of mass \(m\), which is at rest. The coefficient of restitution between A and B is \(e\). (a) Show that the speed of B immediately after the collision is \(\dfrac{2u(1+e)}{3}\), and find the velocity of A. (b) Show that the kinetic energy lost is \(\tfrac13 mu^2(1-e^2)\). (c) Given that a quarter of the kinetic energy is lost, find \(e\).
(a) A's direction positive. Momentum: \(2mu=2mv_A+mv_B\), so \(2u=2v_A+v_B\). Restitution: \(v_B-v_A=eu\). Eliminate \(v_B\): \(2u=3v_A+eu\), so \(v_A=\dfrac{u(2-e)}{3}\) and \(v_B=\dfrac{u(2-e)}{3}+eu=\dfrac{2u(1+e)}{3}\), as required. Since \(e\le 1\), \(2-e>0\): A always carries on in its original direction.
(b) KE before \(=\tfrac12(2m)u^2=mu^2\). After: \(\tfrac12(2m)\dfrac{u^2(2-e)^2}{9}+\tfrac12 m\dfrac{4u^2(1+e)^2}{9}=\dfrac{mu^2}{9}\left[(4-4e+e^2)+(2+4e+2e^2)\right]=\dfrac{mu^2(6+3e^2)}{9}\). Loss \(=mu^2-\dfrac{mu^2(2+e^2)}{3}=\) \(\tfrac13 mu^2(1-e^2)\)
(c) The fraction lost is \((1-e^2)/3\): \(\dfrac{1-e^2}{3}=\dfrac14\), so \(e^2=\dfrac14\) and \(e=\tfrac12\)
What happens when a particle hits a fixed smooth surface?
A wall or the ground does not move, so its velocity is zero before and after. Restitution reduces to rebound speed = \(e\) × impact speed, with the direction reversed. Momentum of the particle is not conserved: the surface gives it an impulse. Find that impulse with \(I=mv-mu\), remembering the change of sign.
In M2 the particle strikes the surface at right angles. The plane is smooth, so there is no sideways impulse. For a ball bouncing on the ground, use the constant-acceleration formulae between bounces and restitution at each bounce. Each rebound speed is \(e\) times the last, so each rebound height is \(e^2\) times the last.
A small ball of mass 0.15 kg is released from rest at a height of 2.5 m above a horizontal floor. The coefficient of restitution between the ball and the floor is 0.6. The ball is modelled as a particle and air resistance is ignored. (a) Find the speed of the ball just before it hits the floor, and just after. (b) Find the magnitude of the impulse of the floor on the ball. (c) Find the height reached after the first bounce. (d) Find the time from release until the ball hits the floor for the second time.
(a) \(v^2=u^2+2as\) with \(u=0\), \(a=9.8\), \(s=2.5\): \(v=\sqrt{2\times 9.8\times 2.5}=\sqrt{49}=\) 7.0 m s⁻¹ before; \(0.6\times 7.0=4.2\) m s⁻¹ after
(b) Upwards positive: \(u=-7.0\), \(v=+4.2\). \(I=0.15(4.2-(-7.0))=\) 1.68 N s (upwards)
(c) Rising from 4.2 m s⁻¹ to rest: \(0=4.2^2-2(9.8)h\), so \(h=0.90\) m
(d) Fall: \(7.0=9.8t_1\), so \(t_1=0.714\) s. Rebound flight: \(0=4.2t-4.9t^2\), so \(t=0.857\) s. Total \(=\) 1.57 s (3 s.f.)
Energy check: the kinetic energy lost at the bounce is \(\tfrac12(0.15)(7.0^2-4.2^2)=2.35\) J, which equals the drop in maximum potential energy, \(0.15\times 9.8\times(2.5-0.9)\).
Forgetting that the velocity changes sign at the bounce. The impulse is \(0.15\times(4.2+7.0)\), not \(0.15\times(7.0-4.2)=0.42\) N s, which would be the answer if the ball were merely slowed down without changing direction.
How do you decide whether there are further collisions?
Treat the impacts one at a time, in the order they happen. Each impact involves only the two bodies touching; any other particle carries on unchanged. After each impact, write down every particle's velocity with its sign, in the order the particles sit on the line.
The test: two neighbouring particles collide again only if the one behind has a greater velocity (in the positive direction) than the one in front. If the back one is slower, or moving away, they never meet again. With a wall at the front, the wall reverses the front particle; then compare its new velocity with that of the particle following it. "Show that there will be no further collisions" and "find the range of values of \(e\) for which there is a further collision" are the same test, written as an inequality.
Three particles A, B and C, of masses 2 kg, 1 kg and 3 kg, lie at rest in that order on a smooth horizontal surface, in a straight line. A is projected towards B with speed 6 m s⁻¹. The coefficient of restitution between any two of the particles is ½. (a) Find the velocities of A and B after A hits B. (b) Find the velocities of B and C after B hits C. (c) Explain why A and B collide again. (d) Find the velocities of A and B after their second collision. (e) Determine whether B and C collide again.
(a) A–B: \(2(6)=2v_A+v_B\) and \(v_B-v_A=\tfrac12(6)=3\), so \(3v_A=9\): \(v_A=3\) m s⁻¹, \(v_B=6\) m s⁻¹
(b) B–C, with C at rest: \(1(6)=v_B+3v_C\) and \(v_C-v_B=\tfrac12(6)=3\), so \(4v_B=-3\): \(v_B=-0.75\) m s⁻¹, \(v_C=2.25\) m s⁻¹
(c) A moves at +3 m s⁻¹ towards B, and B now moves at −0.75 m s⁻¹ (back towards A): they are approaching, so A and B collide again
(d) Approach speed \(3-(-0.75)=3.75\). \(2(3)+1(-0.75)=2v_A+v_B=5.25\) and \(v_B-v_A=1.875\), so \(v_A=1.125\) m s⁻¹, \(v_B=3.0\) m s⁻¹
(e) B (3.0 m s⁻¹) is behind C (2.25 m s⁻¹) and faster: B and C collide again
Carrying on: after that fourth impact B has \(\tfrac{69}{32}\approx 2.16\) m s⁻¹ and C \(\tfrac{81}{32}\approx 2.53\) m s⁻¹. Now A (1.125) < B < C, so all three separate and there are no more collisions.
A particle P of mass \(m\) and a particle Q of mass \(3m\) lie on a smooth horizontal floor on a straight line perpendicular to a fixed smooth vertical wall. Q is at rest at distance \(d\) from the wall, and P moves towards Q, and the wall, with speed \(u\). The coefficient of restitution between P and Q is ½, and between Q and the wall it is \(f\). (a) Find the velocities of P and Q immediately after they collide. (b) Show that P and Q collide again if \(f>\tfrac13\). (c) Given \(f=\tfrac23\), find the distance from the wall of the second collision between P and Q, and the time between the two collisions.
(a) Towards the wall positive: \(mu=mv_P+3mv_Q\) and \(v_Q-v_P=\tfrac12 u\). Then \(u=4v_P+\tfrac32 u\), so \(v_P=-\dfrac{u}{8}\) (P moves away from the wall at \(u/8\)), \(v_Q=\dfrac{3u}{8}\) towards the wall
(b) Q rebounds from the wall with speed \(f\times\dfrac{3u}{8}\), moving away from the wall, the same way as P. Q catches P only if \(\dfrac{3fu}{8}>\dfrac{u}{8}\), that is \(f>\tfrac13\)
(c) With \(f=\tfrac23\), Q comes back at \(\tfrac23\times\tfrac{3u}{8}=\tfrac{u}{4}\). Q's time to the wall: \(d\div\tfrac{3u}{8}=\tfrac{8d}{3u}\); P is then \(d+\tfrac{u}{8}\times\tfrac{8d}{3u}=\tfrac{4d}{3}\) from the wall. Measuring \(t\) from Q leaving the wall: \(\tfrac{u}{4}t=\tfrac{4d}{3}+\tfrac{u}{8}t\), so \(t=\tfrac{32d}{3u}\) and the distance is \(\tfrac{u}{4}\times\tfrac{32d}{3u}=\) \(\tfrac{8d}{3}\) from the wall. Time between collisions \(=\tfrac{8d}{3u}+\tfrac{32d}{3u}=\) \(\tfrac{40d}{3u}\)
Before you start chasing distances and times, answer the yes/no question with velocities alone. Distances only matter when you are asked where or when the next impact happens; whether it happens depends only on which particle is faster.
Exam tips for M2.5
Draw "before" and "after" for each collision, with masses, a positive direction and every velocity arrow pointing that way. Most sign errors disappear once the diagram is on the page.
A direct impact always gives one momentum equation and one restitution equation. Label them (1) and (2), then solve simultaneously. If a question gives you one velocity after the impact, use momentum for the other and restitution to find \(e\).
When the answer contains \(e\), keep it as a letter throughout and use \(0\le e\le 1\) to decide directions: for example, \(u(1-3e)/4\) is negative only when \(e>\tfrac13\).
Work out the total kinetic energy before and after on separate lines and subtract. Squaring removes signs, so use speeds. A negative "loss" means an arithmetic slip, not an explosion.
In a successive-impacts question, end with a sentence that compares velocities: "A moves at 1.125 m s⁻¹ and B at 2.16 m s⁻¹ in the same direction, so A cannot catch B: no further collision." The comparison is the answer.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Kinetic energy is conserved in a collision. | Only when \(e=1\). Otherwise some becomes heat, sound and deformation. | Momentum is conserved; kinetic energy decreases unless the impact is perfectly elastic. |
| Momentum is conserved when a ball bounces off a wall. | The wall exerts an external impulse on the ball. | Use restitution for the rebound speed and \(I=mv-mu\) for the impulse. |
| Restitution uses speeds, so signs do not matter. | A particle that reverses has a negative velocity; ignoring this changes the approach or separation speed. | Write \(v_2-v_1=e(u_1-u_2)\) with signed velocities. |
| A coefficient of 0 means the particles stop. | \(e=0\) means the speed of separation is zero, not the speeds. | With \(e=0\) the particles move off together with a common velocity. |
| An impulse can be added to a velocity. | Impulse changes momentum, which includes the mass. | \(\mathbf{v}=\mathbf{u}+\mathbf{I}/m\). |
| If two particles are moving in the same direction they cannot collide again. | The one behind may be faster. | Compare velocities: a further collision happens if the back particle is faster than the front one. |
How this topic connects
M1 Momentum and impulse
The one-dimensional conservation law, coalescing particles and jerks in strings come from M1; M2 adds restitution and vectors in \(\mathbf{i}\), \(\mathbf{j}\) form.
M1 Kinematics and M2.4 Work, energy and power
Bouncing balls need the constant-acceleration formulae between impacts, and kinetic energy losses link straight to the work–energy principle.
Further Mechanics and IAL Physics
Oblique impacts with a plane and between spheres, where only the velocity component along the line of centres is affected, build directly on this page in IAL Further Mathematics. In IAL Physics Units 1 and 4, momentum conservation, impulse and elastic versus inelastic collisions are the same ideas in a physics setting.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| momentum \(=m\mathbf{v}\) | momentum of a particle (vector) | No — learn it |
| \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\) | impulse on one particle, in \(\mathbf{i}\), \(\mathbf{j}\) form or along a line | No — learn it |
| \(m_1u_1+m_2u_2=m_1v_1+m_2v_2\) | conservation of momentum in a collision | No — learn it |
| \(v_2-v_1=e(u_1-u_2),\ 0\le e\le 1\) | Newton's law of restitution, direct impact | No — learn it |
| rebound speed = \(e\) × impact speed | particle hitting a fixed smooth surface at right angles | No — learn it |
| \(\text{KE}=\tfrac12 mv^2\) | kinetic energy lost = total before − total after | No — learn it |
| loss \(=\tfrac12\cdot\dfrac{m_1m_2}{m_1+m_2}(1-e^2)w^2\) | quick check on a kinetic energy loss (\(w\) = approach speed) | No — derive it |
| \(v^2=u^2+2as,\ v=u+at\) | motion between bounces | No — learn it |
Check your understanding
1. Why can the coefficient of restitution not be greater than 1?
The kinetic energy lost is proportional to \(1-e^2\). If \(e>1\) the loss would be negative, so the collision would create kinetic energy with no source of energy. Without an explosion this cannot happen.
2. A particle of mass 0.2 kg hits a fixed smooth wall at right angles with speed 6 m s⁻¹, and \(e=0.5\). Find its rebound speed and the impulse on it.
Rebound speed \(=0.5\times 6=3\) m s⁻¹. Impulse \(=0.2(3-(-6))=1.8\) N s, directed away from the wall.
3. Two particles collide and move off together. What is the value of \(e\)?
\(e=0\): the speed of separation is zero because they share a common velocity.
4. A particle of mass 0.5 kg moving with velocity \((2\mathbf{i}+3\mathbf{j})\) m s⁻¹ receives an impulse of \((\mathbf{i}-4\mathbf{j})\) N s. Find its new speed.
\(\mathbf{v}=\mathbf{u}+\mathbf{I}/m=(2+2)\mathbf{i}+(3-8)\mathbf{j}=4\mathbf{i}-5\mathbf{j}\), so the speed is \(\sqrt{41}=6.40\) m s⁻¹.
5. Two particles of equal mass collide directly with \(e=1\); one was at rest. What happens?
Momentum: \(u=v_1+v_2\); restitution: \(v_2-v_1=u\). So \(v_1=0\) and \(v_2=u\): the moving particle stops and the other takes its velocity.
6. A ball dropped onto a floor rebounds to 64% of the height it fell from. Find \(e\).
Rebound height is \(e^2\) times the drop height, so \(e^2=0.64\) and \(e=0.8\).
Specification checklist
Tick every box before the exam. Wording is ours; references are to the specification sections.
- (M2 4.1) I can treat momentum as a vector and use the impulse–momentum principle \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}\) with \(\mathbf{i}\) and \(\mathbf{j}\) components.
- (M2 4.1) I can apply conservation of linear momentum to a collision, and use the fact that the two particles receive equal and opposite impulses.
- (M2 4.2) I can use Newton's law of restitution for a direct impact between two particles, knowing that \(0\le e\le 1\) and what \(e=0\) and \(e=1\) mean.
- (M2 4.2) I can calculate the kinetic energy lost in an impact, numerically or in terms of \(m\), \(u\) and \(e\).
- (M2 4.2, 4.3) I can use restitution for a particle striking a fixed smooth plane surface at right angles, including bouncing on the ground.
- (M2 4.3) I can work through successive impacts of up to three particles in a line, or of two particles and a wall, and decide whether a further collision takes place.