Initially a girl on roller skates is at rest on a smooth horizontal pavement - NSC Physical Sciences - Question 4 - 2018 - Paper 1

The girl-roller-skate combination will move TO THE LEFT after the parcel is thrown. This can be explained using Newton's third law, which states that for every action, there is an equal and opposite reaction.

Let the mass of the girl be denoted as $m_{girl}$. The total mass of the system before throwing the parcel is equal to the mass of the girl plus the mass of the roller skates:

$m_{total} = m_{girl} + m_{roller ext{-}skates}$

After the parcel is thrown, we have: $m_{total} = 8 ext{ kg} + 2 ext{ kg} = 10 ext{ kg}$

Using conservation of momentum: After throwing the parcel: $m_{total} v_{total} = m_{parcel} v_{parcel} + m_{girl} v_{girl}$

Substituting known values: $10 imes 0.6 = 8 imes 4 + m_{girl} (velocity ext{ of girl})$ Solving for $m_{girl}$ yields approximately 5.13 kg.

Impulse can be calculated using the formula:

$ext{Impulse} = riangle p = m(v_{final} - v_{initial})$

Where:

• $m$ is the total mass of the roller-skate combination (2 kg).
• $v_{final} = 0.6 ext{ m/s}$ (after throwing).
• $v_{initial} = 0 ext{ m/s}$ (at rest).

Substituting values: $ext{Impulse} = 2 imes (0.6 - 0) = 1.2 ext{ kg m/s}.$ Therefore, the magnitude of the impulse is 1.2 kg m/s.

The change in momentum experienced by the parcel is equal to the impulse applied to it. Since the impulse has already been calculated as 1.2 kg m/s, the change in momentum for the parcel is also 1.2 kg m/s.