Question

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

• A. 100%
• B. 95%
• C. 5%
• D. 50%

Answer 1

Answer: (B)

For bullet, $m_1=50g,{\mu }_1=10m/s$ For block, $m_2=950g,u_2=0$ From law of conservation of momentum, $m_1{u}_1+m_2{u}_2=(m_1+m_2)v$ $50\times 10+950\times 0=(50+950)\times v$ $v=\frac{500}{1000}=0.5m/s$ Initial $KE=\frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}\times 50\times {(1000)}^2$ $=25\times {10}^{6}erg$ $\therefore$ Loss in $KE=\frac{1}{2}{m}_1{u}_1^2-\frac{1}{2}(m_1+m_2)v^2$ $\Delta K=\frac{1}{2}\times 50\times {(1000)}^2-\frac{1}{2}(50+950)\times {(50)}^2$ $=25\times {10}^6-1250\times 1000$ $=25\times {10}^6-1.25\times 1000$ $=23.75\times {10}^{6}erg$ $\therefore$ loss in $KE=\frac{\Delta K}{K}\times 100$ $=\frac{23.75\times {10}^6}{25\times {10}^6}\times 100$ $=95.00$