A 44 kg child jumps off a 2.2 kg skateboard that was moving at 8.0 m/s. The skateboard comes to a stop as a result. Find the speed at which the child jumped from the board. Show all your work. Assume a frictionless, closed system.

Answer 1

original momentum = 46.2 * 8

Final momentum = original momentum
= 44 * v - 2.2 * 0
so
46.2 * 8 = 44 * v

Answer 2

Apply the law of conservation of momentum.
The child must have jumped on while traveling in the opposite direction
m*8.0 m/s + M*Vchild = (M + m) *Vfinal = 0

m is the skateboard mass and M is the child's mass.

Solve for Vchild

Vchild = -(m/M)*8.0 m/s

Answer 3

The child jumped off :)

Answer 4

whoops

Answer 5

i needz hlep