Questions on momentum and impulse

Question 1

Calculate the momentum of a 2.0 kg trolley moving at 8.0 m s⁻¹, and state its unit.

Accepted Answer

p = mv = 2.0 × 8.0 (1) p = 16 kg m s⁻¹ (or N s) (1)

Question 2

State the principle of conservation of linear momentum, including the condition under which it applies.

Accepted Answer

In a collision or explosion, the total momentum before equals the total momentum after (1), provided no external resultant force acts on the system (1).

Question 3

Define the impulse of a force and state how it is related to momentum.

Accepted Answer

Impulse = force × time for which it acts, FΔt (1). It equals the change in momentum of the object, Δ(mv) (1).

Question 4

A car travels round a bend at constant speed. Explain why its momentum changes.

Accepted Answer

Momentum is a vector, directed along the velocity (1). On the bend the direction of the velocity changes, so the momentum changes even though its magnitude is constant. (1)

Question 5

State the quantity represented by the area under a force-time graph.

Accepted Answer

The impulse of the force, equal to the change in momentum it produces. (1)

Question 6

State the difference between an elastic collision and an inelastic collision, and name the quantity that is conserved in both.

Accepted Answer

In an elastic collision kinetic energy is conserved; in an inelastic collision some kinetic energy is transferred to other forms (1). Momentum is conserved in both. (1)

Question 7

A constant force of 20 N acts for 3.0 s on a 4.0 kg object initially at rest. Calculate (a) the impulse delivered and (b) the final velocity of the object.

Accepted Answer

(a) Impulse = FΔt = 20 × 3.0 (1) Impulse = 60 N s (1) (b) Impulse = Δ(mv) = mv, so v = 60/4.0 (1) v = 15 m s⁻¹ (1)

Question 8

A 2.0 kg trolley moving at 3.0 m s⁻¹ collides with a stationary 1.0 kg trolley and they move off together. Calculate their common velocity.

Accepted Answer

Momentum is conserved: 2.0 × 3.0 = (2.0 + 1.0)v (1) v = 6.0/3.0 (1) v = 2.0 m s⁻¹ (1)

Question 9

For the collision in the previous question, show that kinetic energy is not conserved by calculating the kinetic energy before and after, and state the type of collision.

Accepted Answer

Before: ½ × 2.0 × 3.0² = 9.0 J (1) After: ½ × 3.0 × 2.0² = 6.0 J (1) Kinetic energy falls by 3.0 J (1) The collision is inelastic (1)

Question 10

A 4.0 kg gun fires a 0.040 kg bullet at 200 m s⁻¹. Assuming the gun is free to move, calculate its recoil velocity.

Accepted Answer

Total momentum starts at zero: 0 = m_gunv_gun + m_bulletv_bullet (1) v_gun = −(0.040 × 200)/4.0 (1) v_gun = −2.0 m s⁻¹, i.e. 2.0 m s⁻¹ backwards (1)

Question 11

A squash ball of mass 60 g hits a wall at 15 m s⁻¹, at right angles to it, and rebounds along the same line at 9.0 m s⁻¹. The contact time is 7.5 ms. Calculate (a) the change in momentum of the ball and (b) the average force the wall exerts on it.

Accepted Answer

(a) Taking the rebound direction as positive, the velocity changes from −15 to +9.0 m s⁻¹, a change of 24 m s⁻¹ (1). Δp = 0.060 × 24 = 1.44 kg m s⁻¹ (1). (b) F = Δp/Δt = 1.44/0.0075 (1) = 192 N, directed away from the wall. (1)

Question 12

During a collision, the force on a trolley rises linearly from zero to a peak of 3.0 kN in 20 ms and then falls linearly back to zero in a further 20 ms. Determine the impulse given to the trolley.

Accepted Answer

Impulse = area under the force-time graph (1) = ½ × 0.040 × 3000 (1) = 60 N s. (1)

Question 13

A fire hose directs a horizontal jet of water at a vertical wall. Water of total mass 4.2 kg strikes the wall each second, travelling at 15 m s⁻¹, and does not rebound. Calculate the force the water exerts on the wall.

Accepted Answer

Momentum of the water arriving each second = 4.2 × 15 = 63 kg m s⁻¹, all of it removed by the wall (1). The wall therefore exerts a force equal to this rate of change of momentum (1), and by Newton's third law the water exerts an equal force of 63 N on the wall. (1)

Question 14

Derive the equation E_k = p²/2m from the definitions of momentum and of kinetic energy. A hockey ball of mass 0.16 kg has a momentum of 2.4 kg m s⁻¹; use your result to calculate its kinetic energy.

Accepted Answer

p = mv and E_k = ½mv², so rearrange the first to v = p/m (1). Substituting: E_k = ½m(p/m)² = ½mp²/m² (1), which collapses to E_k = p²/2m (1). E_k = 2.4²/(2 × 0.16) = 18 J (1).

Question 15

A 3.0 kg trolley moving right at 4.0 m s⁻¹ collides head-on with a 1.0 kg trolley moving left at 2.0 m s⁻¹, and they stick together. Calculate the velocity of the combined trolleys, stating its direction.

Accepted Answer

Taking right as positive (1) (3.0 × 4.0) + (1.0 × −2.0) = (3.0 + 1.0)v (1) v = 10/4.0 (1) v = 2.5 m s⁻¹ to the right (1)

Question 16

A 1000 kg car travelling at 20 m s⁻¹ is brought to rest in a collision. Calculate the average force on the car if it stops in (a) 0.10 s and (b) 0.50 s, and comment on the role of a crumple zone.

Accepted Answer

F = Δ(mv)/Δt (1)
(a) F = (1000 × 20)/0.10 = 2.0 × 10⁵ N (1)
(b) F = (1000 × 20)/0.50 = 4.0 × 10⁴ N (1)
A crumple zone lengthens the stopping time, reducing the force for the same change in momentum (1)

Question 17

Explain, in terms of impulse and momentum, how a seatbelt and an airbag reduce the risk of injury in a crash.

Accepted Answer

The change in momentum of the occupant is fixed (1). Seatbelts and airbags increase the time over which they are brought to rest (1), and since force = change in momentum / time, a longer time means a smaller average force on the body (1).

Question 18

A 1400 kg car ran into the back of a 2100 kg van. The two vehicles locked together, and skid marks show that they moved off at 8.0 m s⁻¹ immediately after the impact. The van driver states that the van was stationary when it was hit. The speed limit on the road is 13.5 m s⁻¹. Assuming the van driver's statement is correct, deduce whether the car was exceeding the speed limit.

Accepted Answer

Momentum is conserved: 1400u = (1400 + 2100) × 8.0 (1). u = 3500 × 8.0/1400 (1) = 20 m s⁻¹ (1). This is well above 13.5 m s⁻¹, so the car was exceeding the limit. (1)

Question 19

Two trolleys, of mass 1.2 kg and 0.80 kg, are held at rest on a smooth track with a compressed spring between them. When the spring is released, the 0.80 kg trolley moves off at 0.90 m s⁻¹. Calculate (a) the velocity of the 1.2 kg trolley and (b) the total kinetic energy after the release, and state where this energy came from.

Accepted Answer

(a) The total momentum stays zero: 1.2v = 0.80 × 0.90, so v = 0.60 m s⁻¹ in the opposite direction (1). (b) KE = ½ × 0.80 × 0.90² + ½ × 1.2 × 0.60² (1) = 0.32 + 0.22 = 0.54 J (1). It was transferred from the elastic potential energy stored in the compressed spring. (1)

Question 20

A rocket in deep space, far from any planet, fires its engine and accelerates. A student claims this is impossible because there is no air for the exhaust gases to push against. Explain, in terms of momentum, why the rocket does accelerate.

Accepted Answer

The engine pushes the exhaust gas backwards, giving it momentum backwards (1). No external force acts on the system of rocket and exhaust, so total momentum is conserved (1). The rocket therefore gains an equal amount of momentum forwards (1). Equivalently, by Newton's third law the expelled gas pushes forwards on the rocket: nothing needs to push on the surrounding air. (1)