Question

Two masses of 1g and 4g are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is

• A. 4 : 1
• B. $\sqrt{2}:1$
• C. 1 : 2
• D. 1 : 16

Answer 1

The correct option is C

1 : 2

Step-1: Given data:

Let the first mass be $m_1=m$

Let the second mass be $m_2=4m$

Kinetic energy of first mass = Kinetic energy of second mass = $K.E\mathit{.}$

Velocity of first mass = $V_1$

Velocity of second mass = $V_2$

Step 2: Formula Used

Kinetic energy of a particle with mass $m$ and velocity $v$ is given by: $K.E.=\frac{1}{2}m{v}^2$

Linear momentum of a particle of mass $m$ and velocity $V$ is given by: $p=mV$

Step-3: Calculation:

We have two cases as the Kinetic energy is equal for both cases.

$K.E.=\frac{1}{2}m{V_1}^2\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}i\mathit{\right)}$

$K.E.=\frac{1}{2}\left(4m\right){V_2}^2............\left(\mathrm{ii}\right)$

Dividing equation (i) with (ii) we get,

$V_1=2V_2$

$\therefore$Now we have,

$P_1=m_1{V}_1$ and $P_2=m_2{V}_2$

Dividing ${\mathrm{P}}_2\mathrm{by}{\mathrm{P}}_1$ we get

$\frac{P_2}{P_1}=\frac{m_2{V}_2}{m_1{V}_1}$

$\frac{P_2}{P_1}=\frac{m{V}_1}{m\left(2V_2\right)}$

$\frac{P_2}{P_1}=\frac{1}{2}$

The ratio of the magnitudes of their linear momenta is $\frac{1}{2}$

Therefore, option (C) is the correct answer.