Question

Some kids are playing by rolling a skateboard across level ground beside a 0.75 m high wall. The goal is to drop from the wall vertically downward onto the skateboard as it passes by. A 45 kg kid manages to land on the 4.5 kg skateboard as it is passing by at 8 m/s, and they move off together horizontally. If the kid is able to stay on, what will be the speed of the kid and skateboard together

Answer 1

Answer:

The speed of the kid and the skateboard together is 0.723 m/s

Explanation:

Let the mass of the kid be $m_{k} = 45 kg$

Let the mass of the skateboard be $m_{s} = 4,5 kg$

The initial speed of the skateboard, $v_{s} = 8 m/s$

The initial horizontal speed of the boy as he jumps on the skateboard, $v_{k} = 0 m/s$

Let the speed of the kid and the skateboard together be v

According to the principle of conservation of momentum:

$m_{k} v_{k} + m_{s} v_{s} = (m_{k} + m_{s}) v$

Substituting the appropriate values into  the given equation

$(45*0) + (4.5*8) = (45 + 4.8) v$

36 = 49.8 v

v = 36/49.8

v = 0.723 m/s