A) \[50%\]
B) \[-50%\]
C) \[75%\]
D) \[150%\]
Correct Answer: B
Solution :
Mass of body \[m+100%\]of\[m\] \[=m+\frac{100}{100}m=2\,\,m\] Velocity\[=v-\frac{50}{100}v=\frac{v}{2}\], \[K.E.=\frac{1}{2}m{{v}^{2}}\] \[K{{E}_{1}}\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}(2m){{\left( \frac{v}{2} \right)}^{2}}=\frac{m{{v}^{2}}}{4}=\frac{1}{2}KE\] Percentage change in \[=KE=\frac{K{{E}_{1}}-KE}{KE}\times 100=50%\] \[\therefore \]\[KE\] of the body decreases by\[50%\]
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