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A body of mass m moving with velocity v ...

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

A

`1:1`

B

`2:1`

C

`4:1`

D

`9:1`

Text Solution

Verified by Experts

The correct Answer is:
D
KE of colliding bodies before collision ` = 1/2 mv^3`
After collision the mass = m + 2m = 3m
velocity becomes ` V. = ( (m_1 - m_2)/(m_1 + m_2) )v = (mv)/(3m) = v/3`
KE after collision ` = 1/2 m ( V/3)^2 = 1/9 (1/2 mv^2) `
` (KE_("before"))/(KE_("after")) = (1/2 mv^2)/(1/9 (1/2 mv^2) ) = 9 : 1`
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  • A particle of mass 'm' moving with velocity 'v' collides with a mass m_(2) at rest, then they get embedded. At the instant of collision, velocity of the system

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