Question

A $50g$ bullet moving with a velocity of $10m{s}^{-1}$ gets embeded into a $950g$ stationary body. The loss in kinetic energy of the system will be

• A. $95\%$
• B. $100\%$
• C. $5\%$
• D. $50\%$

Answer 1

Answer: (A)

Applying principal of conservation of linear momentum velocity of the system $(v)$ is
$m_1{v}_1=\left(m_1+m_2\right)v$
$\Rightarrow v=\frac{m_1{v}_1}{m_1+m_2}=\frac{50\times 10}{(50+950)}=\frac{1}{2}m{s}^{-1}$
Initial $KE$ ,
$E^2=\frac{1}{2}{m}_1{v}_1^2$
$=\frac{1}{2}\times \left(\frac{50}{1000}\right)\times 10^2=2.5J$
Final $KE$,
$E^2=\frac{1}{2}\left(m_1^2+m_2^2\right)v^2$
$=\frac{1}{2}\frac{(50+950)}{1000}\times \frac{1}{2}=0.125J$
Percentage loss in $KE$
$\frac{E_1-E_2}{E_1}\times 100=\frac{2.5-0.125}{2.5}=95\%$