When a body of mass `m_(1)` moving with uniform velocity `40 ms^(-1)` collides with another body of mass `m_(2)` at rest, then the two together begin to move with uniform velocity of `30 ms^(-1)`. The ratio of the masses (i.e., `m_(1)//m_(2)`) of the two bodies will beA. `1 : 3`B. `3 : 1`C. `1 : 1.33`D. `1 : 0.75`

When a body of mass `m_(1)` moving with uniform velocity `40 ms^(-1)`

1.	When a body of mass `m_(1)` moving with uniform velocity `40 ms^(-1)` collides with another body of mass `m_(2)` at rest, then the two together begin to move with uniform velocity of `30 ms^(-1)`. The ratio of the masses (i.e., `m_(1)//m_(2)`) of the two bodies will beA. `1 : 3`B. `3 : 1`C. `1 : 1.33`D. `1 : 0.75`
Answer» Correct Answer - B Initial momentum of the system `=m_(1)xx40+m_(2) xx0=40 m_(1)` Final momentum of the system `=(m_(1)+m_(2))xx30` By the law of conservation of momentum, Initial momentum = Final momentum `40 m_(1) =(m_(1)+m_(2))30` `40 m_(1)-30 m_(1) =30 m_(2)` `10 m_(1) = 30 m_(2) implies m_(1)/m_(2) =3/1`