Easy -1.2 This is a straightforward application of conservation of momentum with all information given directly. Students simply need to write momentum before = momentum after, substitute the given values, and solve a single linear equation for the unknown mass. No problem-solving insight required beyond recognizing this as a standard momentum conservation scenario.
1 A child, of mass 48 kg , is initially standing at rest on a stationary skateboard. The child jumps off the skateboard and initially moves horizontally with a speed of \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The skateboard moves with a speed of \(16 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the opposite direction to the direction of motion of the child. Find the mass of the skateboard. [0pt]
[3 marks]
1 A child, of mass 48 kg , is initially standing at rest on a stationary skateboard. The child jumps off the skateboard and initially moves horizontally with a speed of $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The skateboard moves with a speed of $16 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the opposite direction to the direction of motion of the child. Find the mass of the skateboard.\\[0pt]
[3 marks]
\hfill \mbox{\textit{AQA M1 2015 Q1 [3]}}