Question

April sits at rest on a skateboard. she has a mass of 55 kg. Her friend throws her a watermelon (m=2kg) at a speed of 5 m/s. If she catches it, how fast will she, the skateboard, and the watermelon move

Answer 1

Here we can use momentum conservation

$m_1v_{1i} +m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$55*0 +2*5 = 55*v + 2*v$

$10 = 57*v$

v = 0.175 m/s

So they move together with speed 0.175 m/s.

Answer 2

Hello, for ur question we can use the conservation of momentum to solve.
Before April starts to move, her momentum(P) is 0 kgm/s because she was at rest. Only the melon was moving.
Therefore the initial momentum P1=melon’s mass*melon’s velocity=2kg*5m/s=10kgm/s
After April catches the melon and starts to move, the momentum at that point equals the sum of melon’s momentum plus April’s momentum. According to the conservation of momentum, the initial P=final P=10kgm/s
So (April’s mass+Melon’s mass)*V=10kgm/s
(55kg+2kg)*V=10kgm/s
57kg*V=10kgm/s
V= P/m(total)= 10kgms^-1 / 57kg=0.175 m/s
Hope this helps!