A child of mass `4 kg` jumps from cart `B` to cart A and then immediately back to cart `B`. The mass of each cart is `20 kg` and they are initially at rest. In both the cases the child jumps at `6 m//s` relative to the cart. If the cart moves along the same line with negligible friction with the final velocities of `V_(B)` and `V_(A)`, respectively, find the ratio of `6V_(B)` and `5v_(A)`.

Answer

1

Explanation

All the velocities shown in diagrams are w.r.t ground.
After first jump:

`20v_(1)=4v_(2)` and `v_(1)+v_(2)=6`(given)
Solve to get `v_(1)=1m//s, v_(2)=5m//s`
When child arrives on `A`:

`(20+4)v_(3)=4v_(2)impliesv_(3)=5/6m//s`
After the second jump:

`v_(4)+v_(A)=6, 24v_(3)=20v_(A)=25/6m//s`
When child arrives on `B`:

`24v_(B)=4v_(4)+20v_(1)`
`implies24v_(B)=4(25/6)+20xx1impliesv_(B)=55/36m//s`
Now `(6v_(B))/(5v_(A))=(6x55xx6)/(36x5xx11)=1`