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Question: A bag (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and g...

A bag (mass M) hangs by a long thread and a bullet (mass m) comes horizontally with velocity v and gets caught in the bag. Then for the combined (bag + bullet) system

A

Momentum is mvMM+m\frac{mvM}{M + m}

B

Kinetic energy is mv22\frac{mv^{2}}{2}

C

Momentum is mv(M+m)M\frac{mv(M + m)}{M}

D

Kinetic energy is m2v22(M+m)\frac{m^{2}v^{2}}{2(M + m)}

Answer

Kinetic energy is m2v22(M+m)\frac{m^{2}v^{2}}{2(M + m)}

Explanation

Solution

Initial momentum = mvm v Final momentum = (m+M)V( m + M ) V

By conservation of momentum mv=(m+M)Vm v = ( m + M ) V

∴ Velocity of (bag + bullet) systemV=mvM+mV = \frac { m v } { M + m }

∴ Kinetic energy = 12(m+M)V2\frac { 1 } { 2 } ( m + M ) V ^ { 2 }

=12(m+M)(mvM+m)2\frac { 1 } { 2 } ( m + M ) \left( \frac { m v } { M + m } \right) ^ { 2 } =12m2v2M+m= \frac { 1 } { 2 } \frac { m ^ { 2 } v ^ { 2 } } { M + m }