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)} $
Show Answer
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} ) $
BETA
