Question

When kinetic energy of a body increased by 200%. The percentage change in momentum of the body is ___%.

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Answer 1

Let initial and final momentum of the body be P and P' respectively

Then the initial kinetic energy = K = $\frac{P^2}{2m}$ ………………… (1)

If kinetic energy increases 200%, Kinetic energy = K + 2K = 3K

3k = $\frac{{P^{\prime }}^2}{2m}$ ..………………. (2)

Divide equation (2) by equation (1)

3 = $\frac{P^2}{{P^{\prime }}^2}$

${P^{\prime }}^2$ = 3$P^2$

P’ = 1.732 P

Percentage change in the momentum = $\frac{1.732P-P}{P}$ * 100 = 73.2 %.