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A bag of mass M hangs by a long massless...

A bag of mass `M` hangs by a long massless rope. A bullet of mass in, moving horizontally with velocity `u`, is caught in the bag. Then for the combined (bag `+` bullet) system, just after collision

A

Momentum is `mMv//(M + m)`

B

kinetic energy is `(1//2)Mv^(2)`

C

Momentum is `mv`

D

kinetic energy is `m^(2)v^(2)//2(M + m)`

Text Solution

Verified by Experts

The correct Answer is:
C, D

`COLM : m u + 0 = (m + M)v rArr v = ((m)/(M + m))u`
`KE` after collision
`= (1)/(2) (m + M) xx ((m)/(m + M))^(2) u^(2) = (m^(3)u^(2))/(2(m + M))`
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