A) 200 m/s
B) 150 m/s
C) 400 m/s
D) 300 m/s
Correct Answer: A
Solution :
Key Idea: Conservation of linear momentum holds here. According to conservation of linear momentum, \[{{m}_{1}}{{v}_{1}}={{m}_{1}}v+{{m}_{2}}{{v}_{2}}\] where \[{{v}_{1}}\] is velocity of bullet before collision, v is velocity of bullet after the collision and \[{{v}_{2}}\]is the velocity of block. \[\therefore \] \[0.02\times 600=0.02v+4{{v}_{2}}\] Here, \[{{v}_{2}}=\sqrt{2gh}=\sqrt{2\times 10\times 0.2}=2\,m/s\] \[\therefore \] \[0.02\times 600=0.02v+4\times 2\] \[\Rightarrow \] \[0.02v=12-8\] \[\Rightarrow \] \[v=\frac{4}{0.02}=200\,m/s\]
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