4, m દળ ધરાવતો A પદાર્થ u વેગથી 2m દળ ધર | vedclass

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Question

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}+

Vedclass · Accepted Answer

$\frac{8}{9}$

Vedclass · Answer

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,

$\frac{\frac{1}{2} M_{2} V_{B}^{2}}{\frac{1}{2} M_{1} u_{1}^{2}}=\frac{\frac{1}{2}(2 M)\left(\frac{4}{3}\right)^{2} u^{2}}{\frac{1}{2}(4 M) u^{2}}$

$=\frac{8}{9}$

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