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 $



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