Question

If the kinetic energy of a body becomes 4 times then what will be the percentage increase in its momentum. Please describe the calculation.

Answer 1

Step 1. Given data:

The kinetic energy of a body becomes 4 times

${\left(KE\right)}_1$ = $4{\left(KE\right)}_{}$

Where KE is the initial kinetic energy, ${\left(KE\right)}_1$ is the final kinetic energy

Step 2. Calculation:

Relation between KE and momentum:

Let $m$= mass of the body,

$v$ = velocity of the body,

$P$= momentum exerted by the body,

Then kinetic energy ${\left(KE\right)}_1$ = $\frac{1}{2}m{v}^2$………….$\left(1\right)$

And momentum $P$=$m.v$

or $v=\frac{P}{m}$…………..$\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$, we get

${\left(KE\right)}_1=\frac{1}{2}m{\left(\frac{P}{m}\right)}^2=\frac{1}{2}\frac{P^2}{m}$

So, $P^2=2m{\left(KE\right)}_1$

or $P=\sqrt{2m{\left(KE\right)}_1}$

So $P\propto \sqrt{{\left(KE\right)}_1}$

It becomes $P\propto \sqrt{4{\left(KE\right)}_{}}=2\sqrt{{\left(KE\right)}_{}}$

Hence, if the kinetic energy of a body will increase four times, then its momentum will increase two times.

Percentage change in momentum:

If $P_I$= initial momentum, $P_f$= final momentum

Then, $P_f=2P_i$

So percentage increase = $$\begin{gathered} \frac{P_f-P_i}{P_i}\times 100=\frac{2P_i-P_i}{P_i}\times 100 \\ =100\% \end{gathered}$$

Thus, the momentum will increase $100\%$.