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Momentum and impulse questions
Force applied for a time changes momentum by exactly that much. When two bodies push on each other the pushes are equal and opposite, so with no external impulse the total momentum is unchanged.
6 original questions · 25 marks · the momentum and impulse notes · Further Mechanics 1
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Define impulse, state its units, and explain why a rebound needs a larger impulse than being stopped dead.
Worked answer
Impulse is force times the time it acts, measured in newton seconds, and equals the change in momentum. Stopping a body removes its momentum; reversing it removes that momentum and supplies the same again the other way, so the change is the sum of the two speeds rather than just the first. B1 for the definition, B1 for newton seconds, B1 for the rebound explanation.A ball of mass 0.2 kg travelling at 12 m/s is struck straight back at 18 m/s. The contact lasts 0.03 s. Find the impulse and the average force.
Worked answer
Taking the outgoing direction as positive, u = −12 and v = 18. Impulse = 0.2(18) − 0.2(−12) = 3.6 + 2.4 = 6 N s. Average force = 6/0.03 = 200 N. M1 for the change in momentum with signs, A1 for the impulse, A1 for the average force.A sphere of mass 5 kg moving at 3 m/s catches up with one of mass 4 kg moving at 1 m/s in the same direction. After the collision the 5 kg sphere moves at 1.8 m/s. Find the velocity of the other.
Worked answer
Take the original direction of motion as positive and draw a before and after diagram. Momentum before = 5(3) + 4(1) = 19 N s. After the collision 5(1.8) + 4v = 19, so 4v = 19 − 9 = 10 and v = 2.5 m/s in the same direction.
M1 for conservation of momentum, A1 for 19 N s before, M1 for the equation after the collision, A1 for 2.5 m/s.
The struck sphere has sped up and the striking one has slowed, and 2.5 > 1.8 means the spheres have separated. Check that, since an answer with the rear sphere still faster than the front one is impossible.A stationary object of mass 8 kg explodes into two pieces of mass 3 kg and 5 kg. The 3 kg piece moves off at 6 m/s. Find the velocity of the other piece.
Worked answer
Total momentum before is zero, since nothing is moving. After: 3(6) + 5v = 0, so 5v = −18 and v = −3.6 m/s: the 5 kg piece moves at 3.6 m/s in the opposite direction. B1 for zero momentum before, M1 for the momentum equation after, A1 for the magnitude, A1 for the direction. An explosion is conservation of momentum with the masses grouped the other way round.Explain, in terms of impulse, why a longer contact time reduces the force in a collision, and give one everyday application.
Worked answer
The change in momentum is fixed by the speeds before and after, so the impulse is fixed. Since impulse is force times time, spreading the same impulse over a longer contact reduces the average force in proportion. B1 for the impulse being fixed, B1 for the inverse relation between force and time, B1 for an example. Crumple zones, airbags, thick landing mats and bending the knees on landing all work this way.A ball of mass 2 kg is dropped from rest at a height of 5 m and rebounds to a height of 1.8 m. It is in contact with the ground for 0.05 s. Find the impulse the ground exerts on it and the average force, first ignoring the weight during the impact and then allowing for it. Take g = 9.8 m/s².
Worked answer
Speed on landing = √(2 × 9.8 × 5) = 9.90 m/s downwards. Rebound speed = √(2 × 9.8 × 1.8) = 5.94 m/s upwards.
Taking upwards as positive, the change in momentum is 2(5.94) − 2(−9.90) = 11.88 + 19.80 = 31.7 N s upwards.
Ignoring the weight, the average force is 31.68/0.05 = 634 N.
Allowing for it, the weight acts downwards throughout the contact and contributes an impulse of 2(9.8)(0.05) = 0.98 N s downwards. The ground must supply 31.68 + 0.98 = 32.66 N s, an average force of 32.66/0.05 = 653 N.
M1 for a landing speed from the drop, A1 for 9.90 m/s, A1 for 5.94 m/s, M1 for the change in momentum with signs, A1 for 31.7 N s, A1 for 634 N, M1 for including the impulse of the weight, A1 for 653 N.
The two answers differ by about 3%. Over so short a contact the weight is normally neglected, and either answer stands provided the assumption is stated. Momentum is not conserved for the ball alone here, because the ground is an external body.
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