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How do momentum and impulse describe collisions, and what does the coefficient of restitution tell you?

Linear momentum and impulse, conservation of momentum, Newton's experimental law and the coefficient of restitution, and direct and oblique collisions including the impulse during impact.

A focused answer to the OCR A-Level Further Mathematics A Mechanics option content on momentum, impulse and collisions, covering linear momentum and the impulse-momentum principle, conservation of momentum in one and two dimensions, Newton's experimental law and the coefficient of restitution, and direct and oblique collisions.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. Momentum and impulse
  3. Conservation of momentum
  4. The coefficient of restitution
  5. Direct and oblique impacts
  6. Try this

What this dot point is asking

OCR's Mechanics option wants you to use linear momentum and the impulse-momentum principle, apply conservation of momentum in collisions (in one and two dimensions), use Newton's experimental law with the coefficient of restitution to relate the speeds of approach and separation, and analyse direct and oblique impacts, including impacts with a fixed surface.

Momentum and impulse

Momentum is mass times velocity, a vector quantity. A force acting over a time changes the momentum, and that change is the impulse.

Conservation of momentum

When two bodies collide and no external impulse acts during the brief impact, the total momentum is unchanged. This single equation links the velocities before and after.

The coefficient of restitution

Newton's experimental law captures how "bouncy" a collision is through the coefficient of restitution, the ratio of the speed of separation to the speed of approach. With conservation of momentum it provides the second equation needed to find both final velocities.

Direct and oblique impacts

In a direct (head-on) collision the velocities lie along one line, so conservation of momentum and restitution give two equations in the two unknown final velocities, which you solve simultaneously. In an oblique impact with a smooth fixed surface, resolve into components: the component perpendicular to the surface is reversed and scaled by ee, while the component parallel to the surface is unchanged (no impulse acts along a smooth surface). The resultant rebound speed is then recombined from the two components by Pythagoras, and the rebound angle from their ratio. The same component approach handles an oblique collision between two spheres: momentum is conserved along the line of centres and the coefficient of restitution applies to the components along that line, while the components perpendicular to the line of centres are unchanged for smooth spheres. Drawing the line of centres and resolving along and perpendicular to it is the key first step in any oblique problem.

Momentum and impulse provide the collision methods of the mechanics option, complementing the energy methods and underpinning problems that combine the two.

Try this

Q1. Find the impulse needed to change a 22 kg mass from 33 m s−1^{-1} to 77 m s−1^{-1}. [2 marks]

  • Cue. J=mv−mu=2(7)−2(3)=8J = mv - mu = 2(7) - 2(3) = 8 N s.

Q2. A ball approaches a fixed wall at 55 m s−1^{-1} and rebounds at 33 m s−1^{-1}. Find ee. [2 marks]

  • Cue. e=35=0.6e = \dfrac{3}{5} = 0.6.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20186 marksA ball AA of mass 33 kg moving at 44 m s−1^{-1} collides directly with a stationary ball BB of mass 22 kg. After the collision AA moves at 11 m s−1^{-1} in the same direction. Find the speed of BB and the coefficient of restitution.
Show worked answer →

Conservation of momentum (M1): 3(4)+2(0)=3(1)+2vB3(4) + 2(0) = 3(1) + 2v_B, so 12=3+2vB12 = 3 + 2v_B, giving vB=4.5v_B = 4.5 m s−1^{-1} (A1, A1).

The coefficient of restitution is e=speed of separationspeed of approache = \dfrac{\text{speed of separation}}{\text{speed of approach}} (M1): speed of approach =4−0=4= 4 - 0 = 4; speed of separation =vB−vA=4.5−1=3.5= v_B - v_A = 4.5 - 1 = 3.5 (A1).

So e=3.54=0.875e = \dfrac{3.5}{4} = 0.875 (A1).

Markers reward conservation of momentum, the value of vBv_B, the restitution definition, the separation and approach speeds, and the value of ee.

OCR 20226 marksA particle of mass 0.50.5 kg moving at 66 m s−1^{-1} strikes a fixed vertical wall at right angles and rebounds with coefficient of restitution 0.40.4. Find the rebound speed and the magnitude of the impulse on the particle.
Show worked answer →

The wall is fixed, so the rebound speed is ee times the approach speed (M1): v=0.4×6=2.4v = 0.4 \times 6 = 2.4 m s−1^{-1} (A1).

Impulse == change in momentum (M1). Taking the initial direction as positive, the momentum changes from 0.5×6=30.5 \times 6 = 3 to 0.5×(−2.4)=−1.20.5 \times (-2.4) = -1.2 (rebound is in the opposite direction) (A1).

Impulse =final−initial=−1.2−3=−4.2= \text{final} - \text{initial} = -1.2 - 3 = -4.2 kg m s−1^{-1} (M1), so the magnitude is 4.24.2 N s (A1).

Markers reward the rebound speed, the impulse-momentum principle, careful signs for the reversed direction, and the magnitude 4.24.2 N s.

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