Work Energy And Power Question 304
Question: A body of mass M splits into two parts $ \alpha $ and $ (1-\alpha ) $ M by an internal explosion which generates kinetic energy T. After explosion if the two parts move in the same direction as before, their relative speed will be -
Options:
A) $ \sqrt{\frac{T}{(1-\alpha )M}} $
B) $ \sqrt{\frac{2T}{\alpha (1-\alpha )M}} $
C) $ \sqrt{\frac{T}{2(1-\alpha )M}} $
D) $ \sqrt{\frac{2T}{(1-a)M}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the speed of the body before explosion be u.
After explosion, if the two parts move with velocities $ u_1 $
and $ u_2 $ in the same direction, then according to conservation of momentum,
Or $ Mu_1+( 1-\alpha )Mu^{2}=Mu $
The kinetic energy T liberated during explosion is given
by $ T=\frac{1}{2}\alpha Mu_1^{2}+\frac{1}{2}( 1-\alpha )Mu_2^{2}-\frac{1}{2}Mu^{2} $
$ =\frac{1}{2}\alpha Mu_1^{2}+\frac{1}{2}( 1-\alpha )Mu_2^{2}-\frac{1}{2M} $
$ {{[ \alpha Mu_1+( 1-\alpha )Mu_2 ]}^{2}} $
$ =\frac{1}{2}M\alpha ( 1-\alpha )[ u_1^{2}+u_2^{2}-2u_1u_2 ] $
$ {{( u_1-u_2 )}^{2}}=\frac{2T}{\alpha ( 1-\alpha )M} $
$ \Rightarrow ( u_1-u_2 )=\sqrt{\frac{2T}{\alpha ( 1-\alpha )M}} $
