Question

A boy on a 2kg skateboard initially at rest tosses an 8kg jug of water towards the right. If the jug has a speed of 3 m/s to the right and the boy and the skateboard move in the opposite direction at 0.60 m/s. What is the boys mass?

i understand that this could be an explosive problem.I just don't why there are 3 masses and if there's one that i should ignore.

Answer 1

`38 Kg`

Explanation:

See,here the boy,the jug in his hand and the skate board all were initially at rest,then he threw the jug towards his right,as a result he attributed some momentum to the jug,due to this,he along with his skateboard started moving to his left,due to conservation of linear momentum.

Suppose,the boy had a mass of `m`,

So,we can say initial momentum of the system was `(2+m+8)*0 =0` (as the system was at rest)

Final momentum of the system became, `(8*3) +(m+2)*(-0.60) Kg.ms^-1` (considering right sided direction to be positive)

So,from conservation of momentum,we can say,

`0 =(8*3) + (m+2)*(-0.60)`

Solving we get, `m=38 Kg`