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

To calculate the mass of Skater A, we can use the formula for momentum:

Given:

• Momentum, $p = 200 \, \text{kg m/s}$
• Velocity, $v = 3.2 \, \text{m/s}$

Using the formula:

$p = mv$

We can rearrange the formula to find mass:

$m = \frac{p}{v}$

Substituting the values:

$m = \frac{200}{3.2} \approx 62.5 \, \text{kg}$

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

When Skater A collides with Skater B and they move off together, the principle of conservation of momentum applies.

Before the collision, the total momentum is:

$\text{Total momentum before} = \text{momentum of Skater A} + \text{momentum of Skater B} \qquad (Skater B \text{ is stationary})$

Since Skater B is stationary, its momentum is 0. Thus:

$\text{Total momentum before} = 200 \, \text{kg m/s} + 0 = 200 \, \text{kg m/s}$

After the collision, the two skaters move together with a combined mass of $(m_A + m_B)$.

Let $v_f$ be their final velocity:

$\text{Total momentum after} = (m_A + m_B) v_f$

Setting the total momentum before equal to the total momentum after gives:

$200 = (62.5 + m_B)v_f$

This means that as Skater A collides with Skater B, Skater A's velocity decreases and Skater B's velocity increases in order to conserve the total momentum.