Work Energy And Power Ques 24
24. If the momentum of a body is increased by $50 \%$, then the percentage increase in its kinetic energy is
[1995]
(a) $50 \%$
(b) $100 \%$
(c) $125 \%$
(d) $200 \%$
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Answer:
Correct Answer: 24.(c)
Solution:
- (c) Initial momentum $(p_1)=p$; Final momentum $(p_2)=1.5 $ $p$ and initial kinetic energy $(K_1)=K$.
Kinetic energy $(K)=\frac{p^{2}}{2 m} \propto p^{2}$
or, $\frac{K_1}{K_2}=(\frac{p_1}{p_2})^{2}=(\frac{p}{1.5 p})^{2}=\frac{1}{2.25}$
or, $K_2=2.25 $ $K$.
Therefore, increase in kinetic energy is $2.25 K-K=1.25 $ $K$ or $125 \%$.
BETA
