Newton's laws and momentum – A-level Physics exam questions with mark scheme

1Standard4 marks

Trolley A, of mass 0.60 kg, moves at 2.5 m s$^{-1}$ towards stationary trolley B, of mass 0.90 kg. The trolleys collide and stick together.

Calculate the velocity of the trolleys after the collision and show whether the collision is elastic or inelastic. [4]

$0.60 \times 2.5 = (0.60 + 0.90)v$(1)
$v = 1.0$ m s$^{-1}$(1)
kinetic energy before = $\frac{1}{2}\times0.60\times2.5^2 = 1.9$ J; after = $\frac{1}{2}\times1.50\times1.0^2 = 0.75$ J(1)
kinetic energy is not conserved (decreases), so the collision is inelastic(1)
Answer: 1.0 m s$^{-1}$; KE falls from 1.9 J to 0.75 J, so inelastic
Examiner tip: Momentum is conserved in every collision; only kinetic energy decides elastic vs inelastic — you must calculate both KE values.

2Standard5 marks

The graph shows how the force on a ball of mass 0.16 kg, initially at rest, varies with time while it is struck by a bat.

(a) Calculate the speed of the ball as it leaves the bat. [3]

(b) A cricket player 'follows through' with the bat. Explain how this increases the speed of the ball. [2]

(a) impulse = area under graph = $\frac{1}{2}\times10\times10^{-3}\times600$(1)
(a) impulse = 3.0 N s (= change in momentum)(1)
(a) $v = \dfrac{3.0}{0.16} = 19$ m s$^{-1}$ allow 18.75; ecf(1)
(b) the bat stays in contact with the ball for longer (increases contact time) allow the force acts on the ball for a longer time(1)
(b) so the impulse $F\Delta t$ / change in momentum of the ball is greater (so greater final speed)(1)
Answer: (a) 19 m s$^{-1}$ (b) longer contact time → greater impulse → greater change in momentum
Examiner tip: Change ms to s before finding the area — the impulse is in N s.

3Hard5 marks

A ballistic pendulum is used to measure the speed of a pellet. A pellet of mass 12 g is fired horizontally into a wooden block of mass 1.50 kg hanging from light strings. The pellet embeds in the block, which swings upwards so that its centre of mass rises by 0.18 m.

(a) Calculate the speed of the pellet before it hits the block. [4]

(b) Explain why the speed of the pellet cannot be found by equating its initial kinetic energy to the gain in gravitational potential energy of the block and pellet. [1]

(a) after the collision, $\frac{1}{2}mv^2 = mgh$ so $v = \sqrt{2 \times 9.81 \times 0.18}$(1)
(a) $v = 1.88$ m s$^{-1}$(1)
(a) conservation of momentum: $0.012\,u = (0.012 + 1.50) \times 1.88$(1)
(a) $u = 240$ m s$^{-1}$ allow 236.7–237 m s$^{-1}$ (g = 9.8 or 9.81, more s.f.); allow ecf from their speed after the collision(1)
(b) the collision is inelastic: (most of) the pellet's kinetic energy is transferred to internal (thermal) energy / deformation / sound, so kinetic energy is not conserved in the collision allow 'kinetic energy is not conserved in the collision' or 'collision is inelastic so energy is lost to heat/sound'; reject 'energy is not conserved'(1)
Answer: (a) 240 m s$^{-1}$ (b) collision is inelastic, so kinetic energy is not conserved in it
Examiner tip: Two stages, two principles: energy conservation for the swing, momentum conservation for the collision — never energy for the collision.

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