A) 100%
B) 95%
C) 5%
D) 50%
Correct Answer: B
Solution :
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\[KE=\frac{\Delta K}{K}\times 100%\] \[=\frac{23.75\times {{10}^{6}}}{25\times {{10}^{6}}}\times 100\] \[=95.00%\]
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