Question

The magnitude of the momentum of a particle is increased by$100\%$. The increase in its kinetic energy is

• A. $100\%$
• B. $200\%$
• C. $300\%$
• D. $400\%$

Answer 1

The correct option is C

$300\%$

Step 1. Given data:

The magnitude of the momentum of a particle is increased by$100\%$.

Step 2. Formula used:

$K.E=\frac{p^2}{2m}$ (Where $K.E=$The kinetic energy of the particle, $p=$Linear momentum of the particle and $m=$mass of the particle)

The increased percentage can be calculated as,
$\frac{K.E_2-K.E_1}{K.E_1}\times 100\%$

Step 3. Calculating the increased percentage of kinetic energy,

Let us consider $p_1=100,p_2=200$

The increased percentage,

$\begin{array}{rcl}& & =\frac{K.E_2-K.E_1}{K.E_1}\times 100\%\\ & & =\frac{{\displaystyle \frac{{p_2}^2}{2m}-\frac{{p^2}_1}{2m}}}{\frac{{p_1}^2}{2m}}\times 100\%\\ & =& \frac{{\displaystyle \frac{1}{2m}}\left({p_2}^2-{p_1}^2\right)}{\frac{{p_1}^2}{2m}}\times 100\%\\ & =& \frac{\left(p_2+p_1\right)\left(p_2-p_1\right)}{{p_1}^2}\times 100\%\\ & =& \frac{\left(200+100\right)\left(200-100\right)}{{100}^2}\times 100\%\\ & =& \frac{300\times 100}{100}\%\\ & =& 300\%\end{array}$

Thus, the increased percentage$300\%$.

Hence, option C is the correct answer.