Question

A 50 g bullet moving with velocity 10 m/s strikes a block of mass 950 g at rest and gets embedded in it. The loss in kinetic energy will be

Answer 1

Initial K.E. of system = K.E. of the bullet = $\frac{1}{2}{m}_s{v}_s^2$

By the law of conservation of linear momentum

$m_s{v}_s+0=m_{sys}\times v_{sys}$

$\Rightarrow v_{sys}=\frac{m_s{v}_s}{m_{sys}}=\frac{50\times 10}{50+950}=0.5m/s$

Fractional loss in K.E. = $\frac{\frac{1}{2}{m}_s{v}_s^2-\frac{1}{2}{m}_{sys}{v}_{sys}^2}{\frac{1}{2}{m}_s{v}_s^2}$

By substituting $m_s=50\times {10}^{-3}kg,v_s=10m/s$

$m_{sys}=1kg,v_{sys}=0.5m/s$ we get

Fractional loss = $\frac{95}{100}$ $\therefore$ Percentage loss = 95 %