A man standing on a trolley pushes another identical a trolley (both trolleys are at rest on a rough surface), are set in motion and stop alter some time so that they If the ratio of mass of first trolley with man to mass of second trolley is `3`, then find the ratio of the stopping distances of the second trolley to that of the first trolley. (Assume coefficient of friction to be the same for both the trolleys)

Answer

9

Explanation

Trolleys gain momentum due to the force applied by man which will be internal force for the system of trolleys and man and there is no other external force. Here we assume that man applies force for a very short time, during which effect o friction can be neglected.
Momentum jst before pushing=momentum just after pushing.

`0=3mv_(1)-mv_(2)impliesv_(1)=(v_(2))/3`
From work energy theorem for indicidual trolleys
`f_(1)S_(1)=1/23mv_(1)^(2), f_(2)S_(2)=1/2mv_(2)^(2)`
Here `f_(1)=mu3mg, f_(2)=mumg`
Solve to get `(S_(2))/(S_(1))=((v_(2))/(v_(1)))^(2)=9`