If the linear momentum of a body is increased by 50 %, the Kinetic energy will be increased by
150 %
125 %
50 %
100 %
A
50 %
B
100 %
C
150 %
D
125 %
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Solution
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If the initial momentum is p0, the initial kinetic energy is K0=p202m
When momentum is increased by 50 %, the momentum becomes p=p0+(p0×50100)=(3/2)p0
Now kinetic energy, K=p22m=(9/4)×(p20/2m)=(9/4)K0
Thus the percentage increased in kinetic energy =K−K0K0×100=[(9/4)−1]100=125 %
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