Question

If momentum is increased by 300% what will be the change in the % of kinetic energy

Answer 1

Answer:

1500%

As

$k = {p}^{2} / 2m$

Let original momentum be P

New Momentum = p + 300% of p

= p+ 3p=4p

As

$k = {p}^{2} / 2m$ is original kinetic energy

New kinetic energy=

$k = {4p}^{2} / 2m$

=$k = 16p/ 2m$

% Change =

$final momemtum - initilal momentum / initial momentum * 100$

=

$((16 {p}^{2} \div 2 )\: \: - ({p}^{2} \div 2m)) \div ({p}^{2} \div 2m) \: \: \: \: \: \: \times \: \: \: \: \: 100$

=1500%

Thanks for asking.