A) 48%
B) 44%
C) 40%
D) 36%
Correct Answer: B
Solution :
The kinetic energy is given by\[KE=\frac{{{p}^{2}}}{2m}\] So, \[\Delta KE=\frac{2p\Delta p}{2m}=\frac{p\Delta p}{m}\] \[\Rightarrow \] \[\frac{\Delta KE}{KE}=\frac{p\Delta p}{m}\times \frac{2m}{{{p}^{2}}}=\frac{2\Delta p}{p}\] Since momentum p increases by 20%, so the final momentum becomes 1.2 p. Hence, \[K{{E}_{final}}=\frac{{{(1.2p)}^{2}}}{2m}=1.44\,\frac{{{p}^{2}}}{2m}=1.44\,KE\] So, % change in \[KE=44%\]
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