Work Energy And Power Question 77
Question: A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K.E. of colliding body before and after collision will be [Roorkee 1982]
Options:
A) 1 : 1
B) 2 : 1
C) 4 : 1
D) 9 : 1
Show Answer
Answer:
Correct Answer: D
Solution:
K.E. of colliding body before collision $ =\frac{1}{2}mv^{2} $
After collision its velocity becomes $ {v}’=\frac{(m_1-m_2)}{(m_1+m_2)}v=\frac{m}{3m}v=\frac{v}{3} $ \ K.E. after collision $ \frac{1}{2}mv{{’}^{2}} $
$ =\frac{1}{2}\frac{mv^{2}}{9} $ Ratio of kinetic energy = $ \frac{K\text{.E}{{.} _{before}}}{K\text{.E}{{.} _{after}}}=\frac{\frac{1}{2}mv^{2}}{\frac{1}{2}\frac{mv^{2}}{9}}=9:1 $
