Question

If momentum of a particle is decreased by $x\%$, then K.E is decreased by 36%. Find $x$.

• A. 20%
• B. 6%
• C. 12%
• D. 72%

Answer 1

The correct option is A 20%

$KE=\left(\frac{p^2}{2m}\right)$
$K=100$ ; $K^{\prime }=64$
$\frac{K^{\prime }}{K}=\frac{64}{100}=\frac{8^2}{{10}^2}={\left(\frac{p^{\prime }}{p}\right)}^2$
$p^{\prime }=0.8p$