# Question #db458

Apr 29, 2017

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

$p = \sqrt{2 m T}$

$\frac{\Delta p}{p} = \frac{\sqrt{2 m {T}_{2}} - \sqrt{2 m {T}_{1}}}{\sqrt{2 m {T}_{1}}}$

$= \sqrt{{T}_{2} / {T}_{1}} - 1$

$= \sqrt{\frac{{T}_{1} \times 1.4}{T} _ 1} - 1 \approx 18 \text{%}$

But that only gives you the percentage increase, not the actual increase. For that you would need more info.