Question

If the linear momentum is increase by 50%. what is the change in KE

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Let the initial momentum be "P".

And initial Kinetic energy be "K".

$\huge{\bold{\underline{Explanation:-}}}$

Momentum is increased to 50 %.

So, new momentum will become

$\large{P' = P + 0.5 P}$

$\large{ \therefore P' = 1.5 P}$

and, Let the new Kinetic energy be K'.

Now,

Initial Kinetic energy.

$\large{\bold{K = \frac{P^2}{2m} -----(1)}}$

and, Final Kinetic energy.

$\large{\bold{K' = \frac{P'^2}{2m} -----(2)}}$

Divide (1) and (2)

$\large{ \frac{K}{K'} = \frac{P^2 \times \cancel{2m}}{P'^2 \times \cancel{2m}}}$

$\large{ \frac{K}{K'} = \frac{P^2}{P'^2}}$

Substituting the values.

$\large{ \frac{K}{K'} = \frac{P^2}{{(1.5 \: P)}^2}}$

$\large{ \frac{K}{K'} = \frac{1}{2.25}}$

$\large{ K' = 2.25 \: K}$

% change in Kinetic energy.

$\large{\bold{ \% \: Change \: in \: K.E = \frac{K' - K}{K} \times 100 }}$

Substituting the values.

$\large{ \% \: Change \: in \: K.E = \frac{2.25 K - K}{K} \times 100}$

$\large{ \% \: Change \: in \: K.E = \frac{1.25 \cancel{K}}{ \cancel{K} \times 100}}$

$\large{ \% \: Change \: in \: K.E = 1.25 \times 100}$

$\huge{\boxed{\boxed{ \% \: Change \: in \: K.E = 125 \%}}}$

So, Kinetic energy is increased by 125 %.