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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=12

p2=2p1

Now putting all the values and finding percentage change in momentum,


Δpp1=p2p1p1×1002p1p1p1×100=100%


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