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Q.
A $50 \,g$ bullet moving with a velocity of $10\, ms ^{-1}$ gets embeded into a $950\, g$ stationary body. The loss in kinetic energy of the system will be
Work, Energy and Power
Solution:
Applying principal of conservation of linear momentum velocity of the system $(v)$ is
$ m_{1} v_{1} =\left(m_{1}+m_{2}\right) v $
$\Rightarrow v =\frac{m_{1} v_{1}}{m_{1}+m_{2}}=\frac{50 \times 10}{(50+950)}=\frac{1}{2} \,ms ^{-1} $
Initial $KE$ ,
$ E^{2} =\frac{1}{2} m_{1} v_{1}^{2}$
$=\frac{1}{2} \times\left(\frac{50}{1000}\right) \times 10^{2}=2.5\, J$
Final $KE$,
$E^{2} =\frac{1}{2}\left(m_{1}^{2}+m_{2}^{2}\right) v^{2} $
$=\frac{1}{2} \frac{(50+950)}{1000} \times \frac{1}{2}=0.125\, J$
Percentage loss in $KE$
$\frac{E_{1}-E_{2}}{E_{1}} \times 100=\frac{2.5-0.125}{2.5}=95 \%$