What percentage of kinetic energy of a moving | vedclass

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Question

Velocity after collision

$V_{2}=\frac{\left(m_{2}-m_{1}\right) u_{2}+2 m_{1} u_{1}}{m_{1}+m_{2}}$

$V_{2}=\frac{(5 m-m) 0+2 m \cdot u_{0}}{m+5 m}

Vedclass · Accepted Answer

$55.5$

Vedclass · Answer

Velocity after collision

$V_{2}=\frac{\left(m_{2}-m_{1}ight) u_{2}+2 m_{1} u_{1}}{m_{1}+m_{2}}$

$V_{2}=\frac{(5 m-m) 0+2 m \cdot u_{0}}{m+5 m}=\frac{u_{0}}{3}$

$\% \Delta K E=\frac{\frac{1}{2} 5 m\left(\frac{u_{0}}{3}ight)^{2}-0}{\frac{1}{2} m u_{0}^{2}} 	imes 100$

$=\frac{5 u_{0}^{2}}{9 u_{0}^{2}} 	imes 100=\frac{500}{9}=55.6 \%$

What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of $5$ times its mass? (Assume the collision to be head-on elastic collision)

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