3.4 Momentum — Practise

Work through these on your own. Write the positive direction at the top of each answer, and check whether the collision is elastic before you claim that kinetic energy is conserved.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A ball of mass 0.25 kg moves at $20\,\text{m}\,\text{s}^{-1}$ . Find its momentum.
A force of 600 N acts for 0.20 s. Find the impulse.
A ball of mass 0.10 kg hits a wall at $20\,\text{m}\,\text{s}^{-1}$ and rebounds at $15\,\text{m}\,\text{s}^{-1}$ . Find the change in momentum. Take the direction towards the wall as positive.
The momentum of a ball increases from $1.2\,\text{kg}\,\text{m}\,\text{s}^{-1}$ to $3.6\,\text{kg}\,\text{m}\,\text{s}^{-1}$ in 0.30 s. Find the average force on the ball.
A trolley of mass 4.0 kg moves at $2.5\,\text{m}\,\text{s}^{-1}$ and collides with a stationary trolley of mass 1.0 kg. The trolleys stick together. Find their common velocity.
A trolley of mass 2.0 kg moves at $5.0\,\text{m}\,\text{s}^{-1}$ and collides with a stationary trolley of mass 3.0 kg. They stick together. Find the kinetic energy lost in the collision.
A ball of mass 0.20 kg moves due east at $6.0\,\text{m}\,\text{s}^{-1}$ . A second ball of mass 0.30 kg moves due north at $4.0\,\text{m}\,\text{s}^{-1}$ . Find the magnitude of the total momentum of the two balls.
A force–time graph for a collision has the points $(0, 0)$ , $(0.10, 200)$ and $(0.20, 0)$ . Find the impulse.
A rifle of mass 3.5 kg fires a bullet of mass 0.015 kg at $600\,\text{m}\,\text{s}^{-1}$ . Find the recoil velocity of the rifle.
An object of mass 6.0 kg is at rest. It explodes into two pieces of mass 2.0 kg and 4.0 kg. The 2.0 kg piece moves at $9.0\,\text{m}\,\text{s}^{-1}$ . Find the velocity of the 4.0 kg piece.
Explain why an airbag reduces the injury to a driver in a crash, using the impulse–momentum relation.
Two railway wagons collide and couple. Say whether you would use the equations of motion, conservation of energy, or conservation of momentum to find the speed after the collision, and why.

Answers and concise solutions

1. The momentum

Answer: $p = mv = 0.25 \times 20 = 5.0\,\text{kg}\,\text{m}\,\text{s}^{-1}$ , in the direction of motion.

2. The impulse

Answer: Impulse $= F\Delta t = 600 \times 0.20 = 120\,\text{N}\,\text{s}$ , which is also $120\,\text{kg}\,\text{m}\,\text{s}^{-1}$ .

3. The change in momentum

Answer: $\Delta p = m(v - u) = 0.10(-15 - 20) = -3.5\,\text{kg}\,\text{m}\,\text{s}^{-1}$ , so the change is $3.5\,\text{kg}\,\text{m}\,\text{s}^{-1}$ away from the wall.

4. The average force

Answer: $F = \dfrac{\Delta p}{\Delta t} = \dfrac{3.6 - 1.2}{0.30} = 8.0\,\text{N}$ , in the direction of the change.

5. The common velocity

Answer: Total momentum before $= 4.0 \times 2.5 = 10\,\text{kg}\,\text{m}\,\text{s}^{-1}$ . Total mass after $= 5.0\,\text{kg}$ , so $v = \dfrac{10}{5.0} = 2.0\,\text{m}\,\text{s}^{-1}$ in the original direction.

6. The kinetic energy lost

Answer: Before, $E_k = \tfrac{1}{2} \times 2.0 \times 5.0^2 = 25\,\text{J}$ . The common velocity is $v = \dfrac{10}{5.0} = 2.0\,\text{m}\,\text{s}^{-1}$ , so after, $E_k = \tfrac{1}{2} \times 5.0 \times 2.0^2 = 10\,\text{J}$ . The kinetic energy lost is $25 - 10 = 15\,\text{J}$ , transferred to thermal stores and sound. The collision is inelastic.

7. The magnitude of the total momentum

Answer: The momenta are $0.20 \times 6.0 = 1.2\,\text{kg}\,\text{m}\,\text{s}^{-1}$ east and $0.30 \times 4.0 = 1.2\,\text{kg}\,\text{m}\,\text{s}^{-1}$ north. They are perpendicular, so the magnitude is $\sqrt{1.2^2 + 1.2^2} = 1.7\,\text{kg}\,\text{m}\,\text{s}^{-1}$ .

8. The impulse

Answer: The impulse is the area under the force–time graph, a triangle of base 0.20 s and height 200 N: $\tfrac{1}{2} \times 0.20 \times 200 = 20\,\text{N}\,\text{s}$ .

9. The recoil velocity

Answer: Total momentum is zero before firing, so $0 = 0.015 \times 600 + 3.5v$ . Then $v = -\dfrac{9.0}{3.5} = -2.6\,\text{m}\,\text{s}^{-1}$ . The rifle recoils at $2.6\,\text{m}\,\text{s}^{-1}$ opposite to the bullet.

10. The velocity of the larger piece

Answer: Total momentum stays at zero: $0 = 2.0 \times 9.0 + 4.0v$ , so $v = -\dfrac{18}{4.0} = -4.5\,\text{m}\,\text{s}^{-1}$ . The larger piece moves at $4.5\,\text{m}\,\text{s}^{-1}$ in the opposite direction.

11. Why the airbag reduces injury

The change in the momentum of the driver is fixed by the initial and final velocities, so the impulse F times t is fixed.
The airbag increases the contact time, so the average force on the driver is smaller, because F = delta p / delta t.

Answer: The change in momentum is fixed. The airbag increases the contact time, so the average force is smaller.

12. The method and the reason

Use conservation of momentum, because the collision involves two masses and momentum is conserved in every collision.
Kinetic energy is not conserved, because the wagons couple and some energy becomes thermal energy and sound.
The equations of motion need a constant acceleration, and a collision does not have one that is known.

Answer: Use conservation of momentum, because it holds in every collision. Kinetic energy is not conserved when the wagons couple, so an energy method would be wrong.