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A bullet (m(1)=25g) is fired with a velo...

A bullet `(m_(1)=25g)` is fired with a velocity 400 m/s gets embedded into a bag of sand `(m_(2)=4.975g)` suspended by a rope. The velocity gained by the bag is

A

0.2 m/s

B

3 m/s

C

4 m/s

D

2 m/s

Text Solution

Verified by Experts

The correct Answer is:
D
Initial momentum of the stone `= m_1 v`
When it explodes into two fragments, their masses will be `m_(2)` and `(m_(1) -m_(2))` .
The fragment of mass `m_2` remains at rest and the other moves with a velocity v'.
`therefore ` Momentum after collision `=m_2 xx 0 +(m_(1) -m_(2))v'`
`therefore ` By the principle of conservation of momentum
`m_1 v=(m_1 -m_2 ) v'
`therefore `v'=(m_(1)v)/(m_1 -m_2)`
`therefore ` By the principle of conservation of momentum,
`m_(1)v = (m_(1) - m_(2))v`
`therefore v , = (m_(1)v)/(m_(1) - m_(2))`
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