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Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is :

A

\frac{1}{9}

B

\frac{8}{9}

C

\frac{4}{9}

D

\frac{5}{9}

Step-by-Step Solution

The fractional loss of kinetic energy is given by ΔKEKE=4(m1m2)(m1+m2)2\frac{\Delta KE}{KE} = \frac{4(m_1 m_2)}{(m_1 + m_2)^2}. Substituting m1=4mm_1 = 4m and m2=2mm_2 = 2m, we get 4(4m)(2m)(4m+2m)2=32m236m2=89\frac{4(4m)(2m)}{(4m + 2m)^2} = \frac{32m^2}{36m^2} = \frac{8}{9}.

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