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Question

If kinetic energy of a body is increased by 300%, then percentage change in momentum will be
  1. 100%
  2. 150%
  3. 265%
  4. 73.2%


Solution

The correct option is B 150%
KE=12mv2=12m2v2m              (momentum,p=mv)=12p2m
so, p=2(KE)m
 Δpp=12ΔKEKE

given, ΔKEKE×100=300 %= 3
ΔKEKE=3
Therefore, ΔPP=12×3=1.5
ΔPP×100=150 %

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