Figure 5 shows an ice skater, Skater A - AQA - GCSE Physics Combined Science - Question 3 - 2018 - Paper 2

Given that the momentum of Skater A is 200 kg m/s and the velocity is 3.2 m/s, we can use the equation from the first step:

1. Rearranging the formula gives us: $m = \frac{p}{v}$

2. Substituting the known values: $m = \frac{200 \text{ kg m/s}}{3.2 \text{ m/s}}$

3. Calculating the mass: $m \approx 62.5 \text{ kg}$

Thus, the mass of Skater A is approximately 62.5 kg.

When Skater A bumps into Skater B, the principle of conservation of momentum applies. The total momentum before the collision equals the total momentum after the collision:

1. Total momentum before collision: $p_{total, before} = p_{A} + p_{B}$ where Skater B is stationary, so: $p_{B} = 0$.

2. Thus, the momentum of the system is entirely due to Skater A: $p_{total, before} = 200 \text{ kg m/s}$

3. After the collision, both skaters A and B move together. Therefore, their combined velocity is a shared outcome of their masses and the initial momentum.

4. The velocity of Skater A will decrease because some of its momentum is transferred to Skater B, resulting in Skater B starting to move. Therefore:

   • Velocity of Skater A decreases.
   • Velocity of Skater B increases.