3.4 Momentum — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

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

H A car of mass 900 kg travels at 25 m/s. Calculate its momentum.
H A ball of mass 0.20 kg moves at 30 m/s. Calculate its momentum.
H A trolley of mass 5.0 kg moves at 4.0 m/s and hits a stationary trolley of mass 3.0 kg. The two trolleys stick together. Calculate their common velocity.
H In the chapter problem the bus of mass 12000 kg travels at 12 m/s and hits the stationary van of mass 8000 kg. The two move off together. Calculate their common velocity.
H A car of mass 1500 kg travels at 20 m/s and stops in 0.50 s. Calculate the force on the car.
H In a crash test, a car of mass 1500 kg travelling at 20 m/s stops in 1.5 s because the front of the car crumples. Calculate the force on the car.
H Two trolleys are pushed apart. Before the push both are stationary. The trolley of mass 2.0 kg moves to the left at 3.0 m/s. Calculate the speed of the trolley of mass 1.0 kg.
H Explain how a crumple zone on the front of a car reduces the force on the passengers in a crash.
H Explain why the total momentum of two colliding objects is conserved.
H State the equation for the momentum of an object, and the equation for the force from a rate of change of momentum. Name each quantity and its unit.
H A trolley of mass 4.0 kg moves to the right at 3.0 m/s. It collides head-on with a trolley of mass 2.0 kg moving to the left at 3.0 m/s. The two trolleys stick together. Calculate their common velocity, taking the right as positive.
H A skater of mass 60 kg moves at 2.0 m/s. Calculate the momentum of the skater.
H A van of mass 800 kg travels at 15 m/s and stops in 0.30 s. Calculate the force on the van.

Answers and concise solutions

1. Momentum of the car

Answer: $p = mv = 900 \times 25 = 22500\,\text{kg m/s} = 2.3 \times 10^4\,\text{kg m/s}$ to 2 significant figures.

2. Momentum of the ball

Answer: $p = mv = 0.20 \times 30 = 6.0\,\text{kg m/s}$ .

3. Common velocity of the trolleys

Answer: Total momentum before $= 5.0 \times 4.0 = 20\,\text{kg m/s}$ . Total momentum after $= (5.0 + 3.0) \times v$ , so $v = \dfrac{20}{8.0} = 2.5\,\text{m/s}$ .

4. Common velocity of the bus and van

Answer: Total momentum before $= 12000 \times 12 = 144000\,\text{kg m/s}$ . Total momentum after $= 20000 \times v$ , so $v = \dfrac{144000}{20000} = 7.2\,\text{m/s}$ .

5. Force on the car

Answer: $F = \dfrac{m\Delta v}{\Delta t} = \dfrac{1500 \times 20}{0.50} = 60000\,\text{N} = 6.0 \times 10^4\,\text{N}$ .

6. Force on the car with the crumple zone

Answer: $F = \dfrac{m\Delta v}{\Delta t} = \dfrac{1500 \times 20}{1.5} = 20000\,\text{N} = 2.0 \times 10^4\,\text{N}$ . The longer time gives a smaller force.

7. Speed of the second trolley

Answer: The total momentum stays zero, so the momenta are equal in size and opposite. The first trolley has $2.0 \times 3.0 = 6.0\,\text{kg m/s}$ , so the second has $6.0\,\text{kg m/s}$ and moves at $\dfrac{6.0}{1.0} = 6.0\,\text{m/s}$ to the right.

8. How a crumple zone helps

The front of the car crumples, so the car takes a longer time to stop. The change in momentum is fixed by the mass and the speed.
A longer time for the same change in momentum gives a smaller force, because $F = \dfrac{m\Delta v}{\Delta t}$ . A smaller force means less injury.

9. Why momentum is conserved

During the collision the two objects push on each other with equal and opposite forces, by Newton's third law.
The two forces act for the same time, so the momentum gained by one object equals the momentum lost by the other. The total momentum does not change.

10. The two equations

Momentum $p = mv$ , where $m$ is the mass in kilograms and $v$ is the velocity in metres per second. Momentum is measured in kg m/s.
Force $F = \dfrac{m\Delta v}{\Delta t}$ , where $m\Delta v$ is the change in momentum and $\Delta t$ is the time taken. Force is measured in newtons.

11. Common velocity after the head-on collision

Answer: Taking the right as positive, the momentum before is $4.0 \times 3.0 - 2.0 \times 3.0 = 12 - 6.0 = 6.0\,\text{kg m/s}$ . The combined mass is 6.0 kg, so $v = \dfrac{6.0}{6.0} = 1.0\,\text{m/s}$ to the right.

12. Momentum of the skater

Answer: $p = mv = 60 \times 2.0 = 120\,\text{kg m/s} = 1.2 \times 10^2\,\text{kg m/s}$ .

13. Force on the van

Answer: $F = \dfrac{m\Delta v}{\Delta t} = \dfrac{800 \times 15}{0.30} = 40000\,\text{N} = 4.0 \times 10^4\,\text{N}$ .