Question

A mass of $400g$ and a mass of $100g$ have same KE, the ratio of their momentum will be
\[\begin{align}
  & \text{A}\text{. 2:1} \\
 & \text{B}\text{. 1:2} \\
 & \text{C}\text{. 1:3} \\
 & \text{D}\text{. 3:1} \\
\end{align}\]

Answer 1

Hint: The Kinetic energy of an object with mass $m$ and velocity $v$ is given by the half product of mass and square of the velocity. The momentum an object with mass $m$ and velocity $v$ is given by the product of mass and velocity. Calculate the kinetic energy of the given masses and equate them to get the mass and velocity product i.e. momentum of the two masses. Divide them to get the ratio of their momentum.
Formulas used:
The Kinetic energy of an object with mass $m$ and velocity $v$ is
$E=\dfrac{1}{2}m{{v}^{2}}$
The momentum of an object with mass $m$ and velocity $v$ is
$P=mv$

Complete answer:
Let the masses of the objects be
${{m}_{1}}=400g=0.4kg\text{ and }{{m}_{2}}=100g=0.1kg$
Let the mass ${{m}_{1}}$is moving with velocity ${{v}_{1}}$and mass ${{m}_{2}}$is moving with velocity ${{v}_{2}}$. Then their momentum is given by ${{P}_{1}}={{m}_{1}}{{v}_{1}}\text{ and }{{P}_{2}}={{m}_{2}}{{v}_{2}}$.
Let the kinetic energy of the mass ${{m}_{1}}$is ${{E}_{1}}$and of mass ${{m}_{2}}$is ${{E}_{2}}$then
${{E}_{1}}=\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}\text{ and }{{E}_{2}}=\dfrac{1}{2}{{m}_{2}}v_{2}^{2}$
But according to the question the kinetic energy of both the masses is equal so
$\begin{align}
  & {{E}_{1}}={{E}_{2}} \\
 & \Rightarrow \dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}=\dfrac{1}{2}{{m}_{2}}v_{2}^{2} \\
\end{align}$
Dividing and multiplying ${{m}_{1}}$in left hand side and dividing and multiplying ${{m}_{2}}$on right hand side of above equation we get
$\begin{align}
  & {{m}_{1}}{{v}_{1}}^{2}\times \dfrac{{{m}_{1}}}{{{m}_{1}}}={{m}_{2}}v_{2}^{2}\times \dfrac{{{m}_{2}}}{{{m}_{2}}} \\
 & \Rightarrow \dfrac{m_{1}^{2}v_{1}^{2}}{{{m}_{1}}}=\dfrac{m_{2}^{2}v_{2}^{2}}{{{m}_{2}}} \\
\end{align}$
But ${{P}_{1}}={{m}_{1}}{{v}_{1}}\text{ and }{{P}_{2}}={{m}_{2}}{{v}_{2}}$, so the above equation takes the form
$\begin{align}
  & \Rightarrow \dfrac{P_{1}^{2}}{{{m}_{1}}}=\dfrac{P_{2}^{2}}{{{m}_{2}}} \\
 & \Rightarrow \dfrac{P_{1}^{2}}{P_{2}^{2}}=\dfrac{{{m}_{1}}}{{{m}_{2}}} \\
 & \Rightarrow \dfrac{{{P}_{1}}}{{{P}_{2}}}=\sqrt{\dfrac{{{m}_{1}}}{{{m}_{2}}}} \\
\end{align}$
Given that ${{m}_{1}}=0.4kg\text{ and }{{m}_{2}}=0.1kg$
So
 $\begin{align}
  & \dfrac{{{P}_{1}}}{{{P}_{2}}}=\sqrt{\dfrac{{{m}_{1}}}{{{m}_{2}}}}=\sqrt{\dfrac{0.4}{0.1}}=\sqrt{4}=2 \\
 & \Rightarrow {{P}_{1}}:{{P}_{2}}=2:1 \\
\end{align}$

So the correct option is A.

Additional Information:
Kinetic energy of a body is the energy possessed by the body by virtue of its motion. It is defined as the work which needs to be done to accelerate the body of a given mass from rest to its required velocity. Momentum is the property of a moving mass that resists the changes in its velocity and is defined by the product of mass and the velocity.

Note:
Note that the momentum is a Vector quantity and the kinetic energy is a vector quantity. In S.I units the momentum is measured in $kg.m{{s}^{-1}}$. And the kinetic energy is measured in Joule. Kinetic energy only associated with the moving body. So the energy which is associated with a body with respect to its position is called potential energy.