# Question 26e1f

May 12, 2017

We know that KE ($E$) and momentum ($p$) of a body of mass $m$ is related as follows

$E = {p}^{2} / \left(2 m\right)$

If ${E}_{i} \mathmr{and} {p}_{i}$ are respectively initial KE and momentum and ${E}_{f} \mathmr{and} {p}_{f}$ are respectively final KE and momentum then

${E}_{i} = {p}_{i}^{2} / \left(2 m\right)$

and

${E}_{f} = {p}_{f}^{2} / \left(2 m\right)$

combining these two we have

${E}_{f} / {E}_{i} = {p}_{f}^{2} / {p}_{i}^{2}$

By the given condition p_f=50%of p_i=p_i/2

so

${E}_{f} / {E}_{i} = {p}_{f}^{2} / {\left(2 {p}_{f}\right)}^{2} = \frac{1}{4}$

=>(E_f-E_i)/E_i=-3/4=-75%#

So change in KE is 75% , here negative sign denotes the decrease in KE