3.4 Momentum — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

A resultant force of 500 N acts on a mass of 200 kg. Calculate the acceleration.
A car slows from 20 m/s to 5.0 m/s in 3.0 s. Calculate its acceleration.

Answers

$a = \dfrac{F}{m} = \dfrac{500}{200} = 2.5\,\text{m/s}^2$ .
$a = \dfrac{\Delta v}{t} = \dfrac{5.0 - 20}{3.0} = -5.0\,\text{m/s}^2$ , so the deceleration is $5.0\,\text{m/s}^2$ .

If question 2 went wrong, remember that a decrease in velocity gives a negative acceleration.

Practise with help

1. A ball of mass 0.15 kg moves at 20 m/s. Calculate its momentum.

Hint

Momentum is the mass multiplied by the velocity.

Next step

Write $p = mv$ , then multiply 0.15 kg by 20 m/s.

Full working

\begin{aligned} p &= m v \\ &= 0.15 \times 20 \\ &= 3.0\,\text{kg m/s} \end{aligned}

2. A lorry of mass 8000 kg travels at 25 m/s. Calculate its momentum.

Hint

The mass is already in kilograms and the velocity is in metres per second, so you can multiply them.

Next step

Write $p = mv$ , then work out $8000 \times 25$ .

Full working

\begin{aligned} p &= m v \\ &= 8000 \times 25 \\ &= 200000\,\text{kg m/s} \\ &= 2.0 \times 10^5\,\text{kg m/s} \end{aligned}

3. A trolley of mass 2.0 kg moves at 3.0 m/s and hits a stationary trolley of mass 1.0 kg. The two trolleys stick together. Calculate their common velocity.

Hint

The total momentum before the collision equals the total momentum after it. The stationary trolley adds no momentum before the collision.

Next step

Work out the momentum before, $2.0 \times 3.0$ , then divide by the combined mass of 3.0 kg.

Full working

\begin{aligned} 2.0 \times 3.0 &= (2.0 + 1.0) \times v \\ 6.0 &= 3.0 v \\ v &= 2.0\,\text{m/s} \end{aligned}

4. A car of mass 1000 kg travels at 20 m/s and stops in 0.25 s. Calculate the force on the car.

Hint

The force equals the rate of change of momentum. The change in velocity is from 20 m/s to zero.

Next step

Write $F = \dfrac{m\Delta v}{\Delta t}$ , then use $m = 1000$ , $\Delta v = 20$ and $\Delta t = 0.25$ .

Full working

\begin{aligned} F &= \frac{m \Delta v}{\Delta t} \\ &= \frac{1000 \times 20}{0.25} \\ &= \frac{20000}{0.25} \\ &= 80000\,\text{N} \\ &= 8.0 \times 10^4\,\text{N} \end{aligned}

5. Explain how a seat belt reduces the chance of injury in a crash.

Hint

Think about the change in momentum of the passenger, and about the time it takes.

Next step

Say what happens to the time taken for the passenger to stop, and what that does to the force.

Full working

The passenger has a fixed change in momentum, set by the mass and the speed. A seat belt stretches a little during the crash, so the passenger takes a longer time to stop. A longer time means a smaller force, because $F = \dfrac{m \Delta v}{\Delta t}$ . The belt also spreads the force over a larger area of the body. Both effects reduce the chance of injury. This answer is a model answer for you to compare with your own wording.

6. Two skaters stand still and push against each other. The skater of mass 50 kg moves to the left at 1.2 m/s. Calculate the velocity of the skater of mass 75 kg.

Hint

Before the push, both skaters are stationary, so the total momentum is zero.

Next step

After the push the total momentum is still zero, so the momenta of the two skaters are equal in size and opposite in direction. Divide the momentum of the first skater by 75 kg.

Full working

The momentum of the first skater is

p = m v = 50 \times 1.2 = 60\,\text{kg m/s} \text{ to the left}

The total momentum must stay zero, so the second skater has $60\,\text{kg m/s}$ to the right.

\begin{aligned} 75 \times v &= 60 \\ v &= \frac{60}{75} \\ &= 0.80\,\text{m/s} \end{aligned}

The skater of mass 75 kg moves to the right at 0.80 m/s.