# Question ac65e

Mar 26, 2017

26.5%, rounded to one decimal place

#### Explanation:

Let initial velocity be ${v}_{i}$, Kinetic energy would be $= \frac{1}{2} m {v}_{i}^{2}$
Given final Kinetic energy =1/2mv_i^2xx(1+60%)#
$\implies$ final Kinetic energy $= \frac{1}{2} m {v}_{i}^{2} \times 1.60$
If final velocity is ${v}_{f}$, comparing with final kinetic energy$= \frac{1}{2} m {v}_{f}^{2}$
$\frac{1}{2} m {v}_{f}^{2} = \frac{1}{2} m {v}_{i}^{2} \times 1.60$
$\implies {v}_{f} = \sqrt{1.6} {v}_{i}$ .........(1)

Increase in momentum $= \text{Final Momentum"-"Inital Momentum}$
Increase in momentum $= m {v}_{f} - m {v}_{i}$
% Increase in momentum $= \frac{m {v}_{f} - m {v}_{i}}{m {v}_{i}} \times 100$

Using equation (1) we get
% Increase in momentum $= \frac{m \sqrt{1.6} {v}_{i} - m {v}_{i}}{m {v}_{i}} \times 100$
$\implies$ % Increase in momentum $= \left(\sqrt{1.6} - 1\right) \times 100$
$\implies$ % Increase in momentum $= 26.5$, rounded to one decimal place