Work Energy And Power Question 173

Question: A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it. The kinetic energy of the composite block will be [MP PET 2002]

Options:

A) $ \frac{1}{2}mv^{2}\times \frac{m}{(m+M)} $

B) $ \frac{1}{2}mv^{2}\times \frac{M}{(m+M)} $

C) $ \frac{1}{2}mv^{2}\times \frac{(M+m)}{M} $

D) $ \frac{1}{2}Mv^{2}\times \frac{m}{(m+M)} $

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Answer:

Correct Answer: A

Solution:

By conservation of momentum, $ mv+M\times 0=(m+M)V $

Velocity of composite block $ V=( \frac{m}{m+M} )v $

K.E. of composite block $ =\frac{1}{2}(M+m)V^{2} $

$ =\frac{1}{2}(M+m){{( \frac{m}{M+m} )}^{2}}v^{2}=\frac{1}{2}mv^{2}( \frac{m}{m+M} ) $



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