Further Mechanics: Momentum, Impulse and Collisions
In A-level Further Mechanics, momentum is the product of an object's mass and velocity (mv), a vector quantity that is conserved in any collision where no external force acts along the line of impact. Impulse is the change in momentum an object experiences, equal to force multiplied by the time for which it acts (impulse = Ft = mv - mu). For two smooth spheres colliding directly (moving along the same straight line), the total momentum before equals the total momentum after, and Newton's experimental law of restitution states that their speed of separation after impact equals e times their speed of approach before impact, where e is the coefficient of restitution and 0 <= e <= 1. Direct impact with a fixed surface is treated the same way, taking the surface as a body of infinite mass that does not move.
Before you start
No specific prerequisites - this is a good place to start.
Method
- State the positive direction being used before assigning a sign to any velocity in the problem; this single decision governs every equation that follows, so choose it once and use it consistently.
- Write the conservation of momentum equation for the collision: (mass 1)(velocity 1 before) + (mass 2)(velocity 2 before) = (mass 1)(velocity 1 after) + (mass 2)(velocity 2 after), using signed velocities.
- Write Newton's law of restitution: (velocity 2 after) - (velocity 1 after) = -e[(velocity 2 before) - (velocity 1 before)], which is equivalent to saying separation speed = e times approach speed.
- Solve the momentum and restitution equations simultaneously (usually by substitution) for the two unknown velocities after impact.
- For a collision with a fixed wall or surface, simplify directly to rebound speed = e x approach speed, since the wall does not move.
- For impulse questions on a single body, use impulse = m(v-u), keeping the same sign convention; remember impulse is a vector, so state its direction as well as its magnitude.
- For successive impacts (for example a ball bouncing more than once), apply the restitution law separately at each individual impact, using the speed the object actually has just before that impact.
- Check the answer makes physical sense: after the collision the bodies must be separating, not overlapping, and the value of e used or found should satisfy 0 <= e <= 1.
Worked example
Two smooth spheres A (mass 3 kg) and B (mass 2 kg) move towards each other in a straight line, A at 6 m/s and B at 2 m/s, and collide directly. The coefficient of restitution between them is 0.5. Find the speed and direction of each sphere after the collision, and the magnitude of the impulse exerted by A on B.
- Take the direction of A's initial velocity as positive, so u_A = 6 m/s and, since B moves towards A (the opposite way), u_B = -2 m/s.
- Conservation of momentum: 3(6) + 2(-2) = 3v_A + 2v_B, so 14 = 3v_A + 2v_B.
- Newton's law of restitution: v_B - v_A = -e(u_B - u_A) = -0.5(-2-6) = -0.5(-8) = 4, so v_B - v_A = 4.
- Substitute v_B = v_A + 4 into the momentum equation: 14 = 3v_A + 2(v_A+4) = 5v_A + 8, giving v_A = 6/5 = 1.2 m/s, and v_B = 5.2 m/s.
- Check: both velocities are positive (both spheres now move in A's original direction) with v_B > v_A, so B moves away from A and they do not overlap; momentum check 3(1.2)+2(5.2)=3.6+10.4=14 confirms the arithmetic.
- Impulse exerted by A on B = change in momentum of B = 2(v_B - u_B) = 2(5.2-(-2)) = 2(7.2) = 14.4 N s, in A's original direction of motion.
Practice questions
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Q1Define momentum, including its units.Show answer
Answer: Momentum is the product of an object's mass and its velocity (p=mv); it is a vector quantity, in the direction of the velocity, measured in kg m/s (equivalently N s).
Q2State Newton's experimental law of restitution in words.Show answer
Answer: For two bodies in direct collision, the speed of separation immediately after collision equals e (the coefficient of restitution) times the speed of approach immediately before collision, where 0 <= e <= 1.
Q3A ball of mass 0.5 kg travelling at 8 m/s hits a fixed wall at right angles and rebounds with coefficient of restitution 0.75. Find the impulse exerted by the wall on the ball.Show answer
Answer: Rebound speed = 0.75 x 8 = 6 m/s. Taking the direction toward the wall as positive, impulse = m(v-u) = 0.5(-6-8) = -7 N s, so the impulse has magnitude 7 N s, directed away from the wall.
Q4Two particles of equal mass collide directly and coalesce (move off together) after impact. State the value of e for this collision, with a reason.Show answer
Answer: e=0, because coalescing means the particles share the same velocity after impact, so their speed of separation is 0; since separation speed = e x approach speed and the approach speed is not zero, e must be 0.
Q5Particle P (mass 4 kg, velocity 5 m/s) collides directly with particle Q (mass 6 kg, at rest). Given that momentum is conserved and the velocities after collision, v_P and v_Q, satisfy v_Q - v_P = 3, find v_P and v_Q.Show answer
Answer: Momentum: 4(5)+6(0)=20=4v_P+6v_Q. Substituting v_Q=v_P+3 gives 20=4v_P+6(v_P+3)=10v_P+18, so v_P=0.2 m/s and v_Q=3.2 m/s.
Q6A particle of mass 2 kg, initially at rest, experiences a constant force of 5 N for 3 seconds. Find the impulse it receives and its velocity at the end of the 3 seconds.Show answer
Answer: Impulse = Ft = 5x3 = 15 N s; velocity = impulse/mass = 15/2 = 7.5 m/s.
Q7Explain why the coefficient of restitution e for any real collision between two solid objects must satisfy 0 <= e <= 1.Show answer
Answer: e=0 describes a perfectly inelastic collision, where the objects coalesce and there is no separation; e=1 describes a perfectly elastic collision, the fastest possible separation, since no kinetic energy is lost. A real collision always loses some energy (as heat, sound or deformation) so it cannot separate faster than a perfectly elastic collision (e<=1), and the objects cannot pass through one another, so e cannot be negative (e>=0).
Q8A ball of mass 0.2 kg falls vertically and hits the ground at 7 m/s, rebounding with e=0.6. Find the rebound speed and the impulse exerted by the ground on the ball.Show answer
Answer: Rebound speed = 0.6 x 7 = 4.2 m/s. Taking downward as positive, impulse = m(v-u) = 0.2(-4.2-7) = -2.24 N s, so the impulse has magnitude 2.24 N s, directed upward, away from the ground.
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
Two smooth spheres, P (mass 4 kg, speed 4 m/s) and Q (mass 2 kg, speed 1 m/s), move in the same direction along a straight line, with P behind Q. They collide directly, with coefficient of restitution e = 0.5 between them. (a) Find the speed of each sphere immediately after the collision. (b) Find the magnitude of the impulse exerted by P on Q during the collision.
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A ball of mass 0.4 kg is dropped and strikes horizontal ground vertically at 10 m/s. The coefficient of restitution between the ball and the ground is 0.6. The ball rebounds, rises, falls back down, and strikes the ground a second time; you may assume the ball's speed immediately before the second impact equals its speed immediately after the first impact. (a) Find the ball's speed immediately after the first impact. (b) Find the ball's speed immediately after the second impact. (c) Find the magnitude of the impulse exerted by the ground on the ball during the second impact.
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