A) Equal
B) 4 times
C) 2 times
D) 3 times
Correct Answer: D
Solution :
Kinetic energy for first condition \[=\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})=\frac{1}{2}m({{20}^{2}}-{{10}^{2}})\] \[=150\,\,mJ\] \[KE\]for second condition \[=\frac{1}{2}m({{10}^{2}}-{{0}^{2}})=50\,\,mJ\] \[\therefore \] \[\frac{{{(KE)}_{I}}}{{{(KE)}_{II}}}=\frac{150\,\,m}{50\,\,m}=3\]
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