A boy on a 1.9 kg skateboard initially at rest tosses a (n) 7.8 kg jug of water in the forward direction. if the jug has a speed of 3.2 m/s relative to the ground and the boy and skateboard move in the opposite direction at 0.65 m/s, find the boy's mass. answer in units of kg.

For this case we first think that the skateboard and the child are one body.

We have then:

1 = jug

2 = skateboard + boy

By conservation of the linear amount of movement:

M1V1i + M2V2i = M1V1f + M2V2f

Initial rest:

v1i = v2i = 0

0 = M1V1f + M2V2f

Substituting values

0 = (7.8) (3.2) + (M2) ( - 0.65)

0 = 24.96 + M2 (-0.65)

-24.96 = (-0.65) M2

M2 = (-24.96) / ( - 0.65) = 38.4 kg

Then, the child's mass is:

M2 = Mskateboard + Mb

Clearing:

Mb = M2-Mskateboard

Mb = 38.4 - 1.9

Mb = 36.5 Kg

answer:

the boy's mass is 36.5 Kg