A) twice
B) thrice
C) four times
D) same
Correct Answer: B
Solution :
\[{{E}_{1}}=\frac{1}{2}m\,(v_{2}^{2}-v_{1}^{2})\] \[\therefore \] \[{{E}_{1}}=\frac{1}{2}m\,[{{(20)}^{2}}-{{(10)}^{2}}]\] \[\Rightarrow \] \[{{E}_{1}}=\frac{1}{2}m\,(400-100)\] \[\Rightarrow \] \[{{E}_{1}}=\frac{1}{2}m\,\times 300=150\,m\] ?(i) \[{{E}_{2}}=\frac{1}{2}m\,[{{(10)}^{2}}-0]\] \[\Rightarrow \] \[{{E}_{2}}=\frac{1}{2}m\,\times 100=50\,m\] ?(ii) From Eqs. (i) and (ii), we have \[{{E}_{1}}=3{{E}_{2}}\]
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