A) \[200\,m{{s}^{-1}}\]
B) \[150\,m{{s}^{-1}}\]
C) \[400\,m{{s}^{-1}}\]
D) \[300\,m{{s}^{-1}}\]
Correct Answer: A
Solution :
Given that, \[{{m}_{1}}=20\,kg,\,\,{{m}_{2}}=4kg,\,v=600\,m{{s}^{-1}}\] According to conservation of linear momentum \[{{m}_{1}}{{v}_{1}}\,={{m}_{1}}v\,+{{m}_{2}}{{v}_{2}}\] where v is the velocity of bullet after the collision and \[{{v}_{2}}\] is the velocity of block \[0.02\,\times 600\,=0.02\,\times 4{{v}_{2}}\] Here \[{{v}_{2}}=\sqrt{2gh}=\sqrt{2\times 10\times 0.2}\,=2\,m{{s}^{-1}}\] \[0.02\,\times 600\,=0.02\,v+4\times 2\] \[v=\frac{4}{0.02}\,=200\,m{{s}^{-1}}\]
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