A) 50%
B) 100%
C) 125%
D) 200%
Correct Answer: C
Solution :
[c] Initial momentum \[\left( {{p}_{1}} \right)=p,\] Final momentum \[\left( {{p}_{2}} \right)= 1.5\]p and initial kinetic energy\[\left( {{K}_{1}} \right)=K\]. Kinetic energy \[\left( K \right) = \frac{{{p}^{2}}}{2m}\propto {{p}^{2}}\] Or, \[\frac{{{K}_{1}}}{{{K}_{2}}} = {{\left( \frac{{{p}_{1}}}{{{p}_{2}}} \right)}^{2}}={{\left( \frac{p}{1.5p} \right)}^{2}}=\frac{1}{2.25}\] or, \[{{K}_{2}}=2.25K.\] Therefore, increase in kinetic energy is 2.25 \[KK=1.25\text{ }K\text{ }or\text{ }125%.\]
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