Question

The linear momentum of a body is increased by 50%, the kinetic energy will be increased by
A.)150%
B.)50%
C.)100%
D.)125%

Answer 1

Hint: Using the formula for momentum and kinetic energy, derive an expression such that kinetic energy is equated in terms of the momentum of the object. Then, for the change of momentum find the corresponding change in the kinetic energy of the object and find the percent change.

Formula used:
Formula of momentum:
$p=mv$

Formula of kinetic energy:
$\text{KE}=\dfrac{1}{2}m{{v}^{2}}$

Formula for percent change:
$\text{Percent change}=\dfrac{{{Q}_{new}}-{{Q}_{old}}}{{{Q}_{old}}}\times 100\%$

Complete step-by-step answer:
Let an object be in motion, the mass of the object is expressed as $m$, whereas the velocity of the object is expressed as $v$.

Momentum $p$ for the object is defined as the quantity of its motion. It is mathematically expressed as mass times velocity.

$p=mv$ … (1)

And the kinetic energy $\text{KE}$ is defined as the energy of the object due to the virtue of its motion. Mathematically, it is expressed as:

$\text{KE}=\dfrac{1}{2}m{{v}^{2}}$ … (2)

We will derive a relationship between the kinetic energy and the momentum.
Multiplying and dividing equation (2) by m:

& \text{KE}=\dfrac{1}{2}m{{v}^{2}}\dfrac{m}{m}=\dfrac{1}{2}\dfrac{{{(mv)}^{2}}}{m} \\\ & \text{KE}=\dfrac{{{p}^{2}}}{2m} \\\ \end{aligned}$$… (3) Let the initial value of momentum be ${{p}_{o}}$. Therefore, the value of initial kinetic energy is given by: $$\text{K}{{\text{E}}_{o}}=\dfrac{{{p}_{o}}^{2}}{2m}$$… (4) It is given that the momentum is increased by 50%. Which means the new momentum $\begin{aligned} & p'={{p}_{o}}+{{p}_{o}}\times \dfrac{50}{100} \\\ & p'=\dfrac{3}{2}{{p}_{o}} \\\ \end{aligned}$ Substituting the new value of momentum in equation (3), gives: $$\text{KE }\\!\\!'\\!\\!\text{ }=\dfrac{{{(p')}^{2}}}{2m}=\dfrac{{{(\dfrac{3}{2}{{p}_{o}})}^{2}}}{2m}=\dfrac{9{{p}_{o}}^{2}}{8m}$$… (5) A percent change in a quantity is defined as the ratio of the difference in the new and the old quantity to the value of the old quantity times hundred. Expression is given by: $\text{Percent change}=\dfrac{{{Q}_{new}}-{{Q}_{old}}}{{{Q}_{old}}}\times 100\%$ Therefore, for the initial and the final values of kinetic energy, the percent change is given by: $$\begin{aligned} & \text{ }\\!\\!\%\\!\\!\text{ change}=\dfrac{\text{KE }\\!\\!'\\!\\!\text{ }-\text{K}{{\text{E}}_{\text{o}}}}{\text{K}{{\text{E}}_{\text{o}}}}\times 100\% \\\ & \Rightarrow \dfrac{\dfrac{9{{p}_{o}}^{2}}{8m}-\dfrac{{{p}_{o}}^{2}}{2m}}{\dfrac{{{p}_{o}}^{2}}{2m}}\times 100=\dfrac{\dfrac{5{{p}_{o}}^{2}}{8m}}{\dfrac{{{p}_{o}}^{2}}{2m}}\times 100=100\times \dfrac{5}{4}\% \\\ & \text{ }\\!\\!\%\\!\\!\text{ change}=\text{125 }\\!\\!\%\\!\\!\text{ } \\\ \end{aligned}$$ Therefore, the answer to this question is option D. Note: It is an assumption that the mass of the object remains constant throughout the motion of the particle. Because, if the mass of the object also varied, then using equation (2) further would have caused problems until the change in the mass was also specified in the question.