A) 50%
B) 100%
C) 125%
D) 150%
Correct Answer: C
Solution :
Relation between kinetic energy (KE) and momentum (p) is, \[KE=\frac{{{p}^{2}}}{2m}\] Initial kinetic energy \[K{{E}_{1}}=\frac{\vec{p}_{2}^{2}}{2m},\] Similarly, final kinetic energy \[K{{E}_{2}}=\frac{\vec{p}_{1}^{2}}{2m}\] \[\therefore \] \[\frac{{{E}_{2}}-{{E}_{1}}}{{{E}_{1}}}=\frac{p_{2}^{2}/2m-p_{1}^{2}/2m}{p_{1}^{2}/2m}\] ie. Percentage increase in KE \[=\frac{p_{2}^{2}-p_{1}^{2}}{p_{1}^{2}}\times 100\] Now, let, \[{{p}_{1}}=100\] then \[{{p}_{2}}=150\] So, % increase in KE \[=\frac{{{(50)}^{2}}-{{(100)}^{2}}}{{{(100)}^{2}}}\times 100\] or % increase in \[E=125%\]
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