Question

If the momentum of a particle is increased by $20\%$, then the kinetic energy increases by

• A. $44\%$
• B. $55\%$
• C. $66\%$
• D. $77\%$

Answer 1

The correct option is A $44\%$

The relation between momentum $\left(p\right)$ and kinetic energy $\left(k\right)$ is given by,$$k=\frac{p^2}{2m}...\left(1\right)$$Let the initial momentum is $p_i$.

So, the final momentum,
$p_f=p_i+\frac{20p_i}{100}=1.2p_i$

From equation $\left(1\right)$,
$k\propto p^2$

Using this we have,
Initial kinetic energy, $k_i\propto p_i^2$
and
Final kinetic energy $k_f\propto (1.2p_i{)}^2$

Now, $\%$ change in kinetic energy,

$\Delta KE=\frac{k_f-k_i}{k_i}\times 100$

$\Rightarrow \Delta KE=\frac{(1.2p_i{)}^2-p_i^2}{p_i^2}\times 100$

$\Rightarrow \Delta KE=44\%$

Hence, option (a) is the correct answer.