Question

The momentum of a body is doubled. By what percentage does its kinetic energy increase?
1. 100 %
2. 200 %
3. 300 %
4. 400 %

Answer 1

Hint: The momentum of a body is directly proportional to its velocity, the mass being constant. The kinetic energy of a body is directly proportional to the square of the velocity of that body. So, the kinetic energy and the momentum are related to each other through velocity. So, using this, we will compute the percentage of increase in kinetic energy.
Formula used:
$$\begin{array}{rl}& KE={\displaystyle \frac{1}{2}}m{v}^2\\ & p=mv\end{array}$$

Complete answer:
From the given information, we have the data as follows.
The momentum of a body is doubled.
The formulae that we will be using to solve this problem are:
The kinetic energy of a body is,
$$KE={\displaystyle \frac{1}{2}}m{v}^2$$
Where m is the mass and v is the velocity of the body.
The momentum of a body is,
$$p=mv$$
Where m is the mass and v is the velocity of the body.
Now we will compute the value of the velocity when the momentum of a body is doubled. Let the original value of the momentum be ‘p’ and let the final value of the mometum be ‘p’’. So, we have,
$$\begin{array}{rl}& p^{\prime }=2p\\ & \Rightarrow p^{\prime }=m{v}^{\prime }\\ & \therefore p^{\prime }=m(2v)\end{array}$$
Thus, the value of the velocity of a body also doubles.
Now consider the kinetic energy of a body when the velocity is doubled. Let the original value of the kinetic energy be ‘KE’ and let the final value of the kinetic energy be ‘KE’’.
$$K{E}^{\prime }={\displaystyle \frac{1}{2}}mv{{}^{\prime }}^2$$
$$\begin{array}{rl}& \Rightarrow K{E}^{\prime }={\displaystyle \frac{1}{2}}m{(2v)}^2\\ & \Rightarrow K{E}^{\prime }={\displaystyle \frac{1}{2}}m(4v^2)\\ & \Rightarrow K{E}^{\prime }=4\left({\displaystyle \frac{1}{2}}m{v}^2\right)\\ & \therefore K{E}^{\prime }=4(KE)\end{array}$$
The increase in the value of the kinetic energy is given as follows.
$$\begin{array}{rl}& K{E}_0={\displaystyle \frac{K{E}^{\prime }-KE}{KE}}\times 100\\ & \Rightarrow K{E}_0={\displaystyle \frac{4KE-KE}{KE}}\times 100\\ & \Rightarrow K{E}_0={\displaystyle \frac{3KE}{KE}}\times 100\\ & \therefore K{E}_0=300\mathrm{\%}\end{array}$$

$$\therefore$$The increase in the value of the kinetic energy will be 300%, thus, option (3) is correct.

Note:
The kinetic energy and the momentum are not directly related to each other. Instead, both have a parameter in common, so, using that parameter we have computed the increase in the kinetic energy. The mass remains constant in both cases (kinetic energy and momentum).