# Question ce067

Mar 7, 2014

The momentum increases by 100 %.

Let initial kinetic energy be $K$ and momentum be $p$.

K = ½mv^2 (Equation 1)

If $K$ increases by 300 %, it increases by a factor of (300 %)/(100%) = 3.

So the new kinetic energy is

$K ' = K + 3 K = 4 K$

From Equation 1: $v = \sqrt{\frac{2 K}{m}}$

The initial momentum p = mv = m × sqrt((2K)/m) = sqrt(m^2 ×(2K)/m) = sqrt(2Km)

The new momentum $p ' = \sqrt{2 K ' m}$

So $\frac{p '}{p} = \sqrt{\frac{2 K ' m}{2 K m}}$

$\frac{p '}{p} = \sqrt{\frac{K '}{K}}$

$\frac{p '}{p} = \sqrt{\frac{4 K}{K}}$

$\frac{p '}{p} = 2$

$p ' = 2 p$

% change = (p' – p)/p × 100 % = (2p – p)/p × 100 % = 100 %#

% increase in momentum = 100%

Apr 21, 2017

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

$p = \sqrt{2 m T}$

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

$= \frac{\sqrt{\textcolor{red}{{T}_{2}}} - \sqrt{{T}_{1}}}{\sqrt{{T}_{1}}}$

$= \frac{\sqrt{\textcolor{red}{4 {T}_{1}}} - \sqrt{{T}_{1}}}{\sqrt{{T}_{1}}} = 1$ or $100 \text{%}$