A mass `m_(1)` moves with a great velocity. It strikes another mass `m_(2)` at rest in head-on collision. It comes back along its path with low speed after collision. Then

Options

A. `m_(1)gtm_(2)`
B. `m_(1)ltm_(2)`
C. `m_(1)=m_(2)`
D. there is no relation between `m_(1)` and `m_(2)`

Answer

B

Explanation

Conservation of linear rmomentum just before and after yields. ltbr `m_(1)vi=m_(2)v_(2)i+m_(1)v_(1)(-i)`…..i
`implies m_(1)v=m_(2)v_(2)-m_(1)v_(1)`
`e=-(v_(1)+v_(2))/(0-v)=1`
therefore `v=v_(1)+v_(2)` ………ii
[for elastic collision `e=1`]
Elaminating `v_(2)` form i and ii we obtain
`m_(1)v=m_(2)(v-v_(1))-m_(1)v_(1)`
`implies v_(1)=(m_(2)-m_(1))/(m_(1)=m_(2))v`
Since `v_(1)gt0,(m_(2)-m_(1))/(m_(1)+m_(2))gt1`
`impliesm_(2)-m_(1)gt0impliesm_(1)ltm_(2)`.