Question

A 53.4 kg girl is at rest on 3.55 kg skateboard. She jumps forward at 1.32 m/s. What is the velocity of the skateboard afterword?

Answer 1

M = mass of the girl = 53.4 kg

m = mass of skateboard = 3.55 kg

V' = velocity of the combination of girl and skateboard before she jumps forward = 0 $\frac{m }{s}$

V = velocity of the girl forward = 1.32 $\frac{m }{s}$

v = velocity of the skateboard afterward = ?

Using conservation of momentum

(M + m) V' = MV + m v

inserting the values

(53.4 + 3.55) (0) = (53.4) (1.32) + (3.55) v

v = - 19.86 $\frac{m }{s}$