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  4. If the kinetic energy of a body is increased by 300 %, the momentum of the body is increased by

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Q. If the kinetic energy of a body is increased by 300 %, the momentum of the body is increased by

AMUAMU 1996

Solution:

: We know that $ {{p}^{2}}= $ link where p denotes momentum and k denotes kinetic energy of the body. Let initial $ K.E=100A: $ $ \therefore $ Finial $ K.E=400A; $ $ \therefore $ $ p_{1}^{2}=2m(100k) $ $ p_{2}^{2}=2m(400k) $ Fractional change in momentum $ =\frac{{{p}_{2}}-{{p}_{1}}}{{{p}_{1}}} $ % change in momentum $ =\frac{({{p}_{2}}-{{p}_{1}})}{{{p}_{1}}}\times 100 $ $ =\left( \frac{{{p}_{2}}}{{{p}_{1}}}-1 \right)\times 100\left( \sqrt{\frac{4}{1}}-1 \right)\times 100 $ $ =(2-1)\times 100 $ $ \therefore $ Increase in momentum = 100%