SATHEE: Work Energy And Power Question 77

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 $= \frac{1}{2} m v^{2}$

After collision its velocity becomes $v^{'} = \frac{(m_{1} - m_{2})}{(m_{1} + m_{2})} v = \frac{m}{3 m} v = \frac{v}{3}$ \ K.E. after collision $\frac{1}{2} m v {^{'}}^{2}$

$= \frac{1}{2} \frac{m v^{2}}{9}$ Ratio of kinetic energy = $\frac{K .E ._{b e f o r e}}{K .E ._{a f t e r}} = \frac{\frac{1}{2} m v^{2}}{\frac{1}{2} \frac{m v^{2}}{9}} = 9 : 1$

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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:

Answer:

Solution:

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