Question

If the linear momentum is increased by 50% then kinetic energy will be increased by
$$\begin{gathered} \text{A}\text{. 50\% } \\ \text{B}\text{. 100\% } \\ \text{C}\text{. 125\% } \\ \text{D}\text{. 25\% } \end{gathered}$$

Answer 1

Hint:To solve this question we have to use the relation between linear momentum and kinetic energy and then use a simple calculation of percentage to solve this question easily.

Complete step-by-step answer:
We know
Linear momentum is denoted by ‘P’ and relation between linear momentum and kinetic energy is as:
$K.E={\displaystyle \frac{p^2}{2m}}$ here we can see 2m is constant because 2 is constant and mass is always constant so kinetic energy will be proportional to square of linear momentum.
That means when we change linear momentum kinetic energy will be changed.
Now as given in the question
Linear momentum is increased by 50% that means now P will be equal to
$$\begin{gathered} P^{\prime }=P+50\mathrm{\%}\text{ of }P \\ {P}^{\prime }=P+{\displaystyle \frac{50}{100}}\times p \\ {P}^{\prime }=P+{\displaystyle \frac{P}{2}}={\displaystyle \frac{3P}{2}} \end{gathered}$$
Kinetic energy is directly proportional to square of linear momentum
And here new linear momentum is ${\displaystyle \frac{3}{2}}P$ so new kinetic energy will be
$K.E^{\prime }={\displaystyle \frac{{\left({\displaystyle \frac{3}{2}}P\right)}^2}{2m}}={\displaystyle \frac{9}{4}}\left({\displaystyle \frac{P^2}{2m}}\right)={\displaystyle \frac{9}{4}}K.E$
Hence increased kinetic energy is $\left({\displaystyle \frac{9}{4}}-1\right)K.E={\displaystyle \frac{5}{4}}K.E$
Hence increased percentage is ${\displaystyle \frac{5}{4}}\times 100=125\mathrm{\%}$
Hence option C is the correct option.

Note: -Whenever we get this type of question the key concept of solving is we have to have knowledge of relation between linear momentum and kinetic energy and using new linear momentum we have to find new kinetic energy and hence find increase in kinetic energy from which increase in percentage of kinetic energy.Kinetic energy can also be written in terms of velocity i.e $K.E={\displaystyle \frac{1}{2}}m{v}^2$.But we know linear momentum $p=mv$ simplifying this and substituting in earlier equation we get same equation only $K.E={\displaystyle \frac{p^2}{2m}}$.