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Question

# If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

A
100%
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B
150%
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C
265%
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D
73.2%
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Solution

## The correct option is B: 100%We know,Kinetic energy is given as K.E=12mv2 From momentum, p=mv K.E=12m2v2m=1p22m So,p=√2×K.E.×m Initial kinetic energy, =K1 Initial momentum, =p1 Final momentum, =p2 After increment kinetic energy will be, =K1+300%ofK1=4K1 So, ratio of momentum in both condition will be,p1p2=√K14K1=√14=12p2=2p1 Now putting all the values and finding percentage change in momentum,Δpp1=p2−p1p1×100⟹2p1−p1p1×100=100%

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