Question

The momentum of a body increases by 20%. Calculate the percentage increase in its K.E.

Answer 1

Step 1: Given data

Increase in momentum = $20\%$

Step 2: Calculating initial and final momentum

Momentum is given as,

$p=mv$

Where, $m$ is mass and $v$ is velocity.

Now, let initial momentum = $p_i$

Final momentum ($p_f$) can be calculated as,

$\begin{array}{l}{p}_f=p_i+\frac{20}{100}{p}_i\\ p_f=p_i+0.2p_i\\ p_f=1.2p_i\end{array}$

Step 3: Calculating initial and final kinetic energy

The relation between kinetic energy and momentum can be represented as follows:

$KE=\frac{p^2}{2m}$

Let, initial kinetic energy = $K{E}_i=\frac{{p_i}^2}{2m}$

Since final momentum, $p_f=1.2p_i$, final kinetic energy $\left(K{E}_f\right)$, can be calculated as,

$$\begin{gathered} K{E}_f=\frac{{p_f}^2}{2m} \\ =\frac{{(1.2p_i)}^2}{2m} \\ =\frac{1.44{p_i}^2}{2m} \\ =1.44K{E}_i \end{gathered}$$

Step 4: Calculating percentage increase in kinetic energy

The percentage increase is given by,

$\%\text{increase =}\frac{\text{Final value}-\text{Initial value}}{\text{Initial value}}\times 100\%$

Substituting the initial and final value of kinetic energy is the above equation, we get,

$$\begin{gathered} \%\text{increase}=\frac{1.44K{E}_i-K{E}_i}{K{E}_i}\times 100\% \\ =\frac{0.44K{E}_i}{K{E}_i}\times 100\% \\ =0.44\times 100\% \\ =44\% \end{gathered}$$

Therefore, the percent increase in kinetic energy is $44\%$.