Question

A 56.0-kgkg person holding two 0.900-kgkg bricks stands on a 2.60-kgkg skateboard. Initially, the skateboard and the person are at rest. The person now throws the two bricks at the same time so that their speed relative to the person is 15.0 m/s

What is the recoil speed of the person and the skateboard relative to the ground, assuming the skateboard moves without friction?

Answer 1

Given :-

mass of person = mp = 56 kg

mass of bricks = mb = 2 * 0.9 = 1.8 kg

mass of skateboard = ms = 2.60 kg

speed of bricks = Vb = 15 m/s

let us assume that speed of person and skateboard is Vps

now as the whole system was at rest initially so

initial momentum = 0

final momentum = mb * -Vb + ( mp + ms ) * Vps

so from momentum conservation,

initial momentum = final momentum

0 = mb * -Vb + ( mp + ms ) * Vps

mb * Vb = ( mp + ms ) * Vps

1.80 * 15 = ( 56 + 2.60 ) * Vps

Vps = 0.461 m/s

feel free to drop comment if you have any doubt.