Momentum (Cambridge (CIE) IGCSE Physics): Revision Note

Determine whether the tennis ball or the brick has the greater momentum.

[3]

Answer:

Step 1: Determine the momentum of the tennis ball using the momentum equation

$p=mv$

$p=0.06\times 75$

$p=4.5\mathrm{kg}\mathrm{m}/\mathrm{s}$ [1 mark]

Step 2: Determine the momentum of the brick using the momentum equation

$p=mv$

$p=3\times 1.5$

$p=4.5\mathrm{kg}\mathrm{m}/\mathrm{s}$ [1 mark]

Step 3: Compare the momentum of each object

• Both the tennis ball and the brick have the same momentum [1 mark]

• Even though the brick is much heavier than the ball, the ball is travelling much faster than the brick

A 0.20 kg ball falls downwards and strikes the ground with a speed of 15 m/s. The ball then rebounds off the ground at a speed of 12 m/s. Calculate the change in momentum of the ball.

[3]

Answer:

Step 1: State the directions that are taken as positive and negative

• Momentum is a vector so direction needs to be taken into account. Upwards will be taken as positive, and downwards will be taken as negative.

Step 2: Calculate the initial momentum of the ball

• The ball falls downwards, so the initial velocity, $u$, of the ball is $u=-15\mathrm{m}/\mathrm{s}$

• Therefore, the initial momentum is:

$p_{\mathrm{initial}}=mu$

$p_{\mathrm{initial}}=0.20\times -15$

$p_{\mathrm{initial}}=-3\mathrm{kg}\mathrm{m}/\mathrm{s}$ [1 mark]

Step 3: Calculate the final momentum of the ball

• The ball rebounds upwards, so the final velocity, $v$, of the ball is $v=12\mathrm{m}/\mathrm{s}$

• Therefore, the final momentum is:

$p_{\mathrm{final}}=mv$

$p_{\mathrm{final}}=0.20\times 12$

$p_{\mathrm{final}}=2.4\mathrm{kg}\mathrm{m}/\mathrm{s}$ [1 mark]

Step 4: Calculate the change in momentum of the ball

$∆p=p_{\mathrm{final}}-p_{\mathrm{initial}}$

$∆p=2.4-\left(-3\right)$

$∆p=5\mathrm{kg}\mathrm{m}/\mathrm{s}$ [1 mark]

You can remember momentum as mass in motion. The units of momentum are kg m/s which is the product of the units of mass (kg) and velocity (m/s). Units of N s are also acceptable for momentum.

Which direction is taken as positive is completely up to you in the exam, as long as you are consistent throughout a question. In general, the right and upwards are taken as positive, and down or to the left as negative.

The diagram shows a toy car and a toy van, just before and after the toy car collides with the toy van, which is initially at rest.

The toy car initially moves at a speed of 0.8 m/s, but this reduces to 0.1 m/s after the collision.

The mass of the toy car is 0.035 kg and the mass of the toy van is 0.085 kg.

Calculate the velocity of the toy van when it is pushed forward by the collision.

[4]

Answer:

Step 1: State the principle of the conservation of momentum

Total momentum before a collision = total momentum after a collision 

Step 2: Calculate the total momentum of the car and van before the collision

Momentum:  $p=mv$

• Initial momentum of the toy car:

$p_{\mathrm{car}}=0.035\times 0.8=0.028\mathrm{kg}\mathrm{m}/\mathrm{s}$

• Initial momentum of the toy van:

$p_{\mathrm{van}}=0$ (the van is at rest, so p = v = 0)

• Total momentum before collision:

 $p_{\mathrm{before}}=p_{\mathrm{car}}\hspace{0.17em}+p_{\mathrm{van}}$

$p_{\mathrm{before}}=0.028+0=0.028\mathrm{kg}\mathrm{m}/\mathrm{s}$ [1 mark]

Step 3: Calculate the total momentum of the car and van after the collision

• Final momentum of the toy car:

$p_{\mathrm{car}}=0.035\times 0.1=0.0035\mathrm{kg}\mathrm{m}/\mathrm{s}$

• Final momentum of the toy van:

$p_{\mathrm{van}}=0.085\times v$

• Total momentum after collision:

$p_{\mathrm{after}}=0.0035+0.085v$ [1 mark]

Step 4: Rearrange the conservation of momentum equation for the velocity of the van

$p_{\mathrm{before}}=p_{\mathrm{after}}$

$0.028=0.0035+0.085v$ [1 mark]

$0.028-0.0035=0.085v$

$v=\frac{0.028-0.0035}{0.085}$

$v=0.29\mathrm{m}/\mathrm{s}$ [1 mark]

If it is not given in the question already, drawing a diagram of before and after helps keep track of all the masses and velocities (and directions) in the conservation of momentum questions.