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}}=50\,\,g,\,\,{{\mu }_{1}}=10\,\,m/s $ For block, $ {{m}_{2}}=950\,\,g,\,\,{{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}=0.5\,\,m/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% $