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Body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2 m$ at rest the collision is head on and elastic in nature. After the collision the fraction of energy lost by colliding body $A$ is
$\frac{5}{9}$
$\frac{1}{9}$
$\frac{8}{9}$
$\frac{4}{9}$
Solution
The expression of final velocity is given by,
$V _{ B }=\left(\frac{ M _{2}- M _{1}}{ M _{2}+ M _{1}}\right) u _{2}+\frac{2 M _{1} u _{1}}{ M _{2}+ M _{1}}$
Here,
$M _{1}=4 M , u _{1}= u$
$M _{2}=2 M , u _{2}=0$
Therefore,
$V _{ B }=\left(\frac{2 M -4 M }{2 M +4 M }\right)(0)+\frac{2(4 M ) u }{2 M +4 M }$
$=\frac{4}{3} u$
The fraction of energy lost by colliding body $A$ is as,
$\frac{\frac{1}{2} M_{2} V_{B}^{2}}{\frac{1}{2} M_{1} u_{1}^{2}}=\frac{\frac{1}{2}(2 M)\left(\frac{4}{3}\right)^{2} u^{2}}{\frac{1}{2}(4 M) u^{2}}$
$=\frac{8}{9}$
