Question

Two balls of masses $2g$ and $6g$ are moving with kinetic energy in the ratio of $3:1$. What is the ratio of their linear momenta?

• A. $1:1$
• B. $2:1$
• C. $1:2$
• D. None of these

Answer 1

The correct option is A

$1:1$

Step 1: Given Data

Mass of the first ball $m_1=2g$

Mass of the second ball $m_2=6g$

The ratio of two kinetic energies, $K_1:K_2=3:1$

Let the velocities be $v_1$ and $v_2$.

Let the linear momenta be $p_1$ and $p_2$.

Step 2: Formula Used

Kinetic energy, $K=\frac{1}{2}m{v}^2$

Linear momentum, $p=mv$

Step 3: Relation between Kinetic Energy and Momentum

We know that, kinetic energy,

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ =\frac{1}{2m}{\left(mv\right)}^2 \\ =\frac{p^2}{2m} \end{gathered}$$

Step 4: Calculate the Ratio

Given that,

$$\begin{gathered} K_1:K_2=3:1 \\ \Rightarrow \frac{1}{2}{m}_1{v_1}^2:\frac{1}{2}{m}_2{v_2}^2=3:1 \\ \Rightarrow \frac{{p_1}^2}{2m_1}:\frac{{p_2}^2}{2m_2}=3:1 \\ \Rightarrow \frac{{p_1}^2}{2g}:\frac{{p_2}^2}{6g}=3:1 \\ \Rightarrow {p_1}^2:\frac{{p_2}^2}{3}=3:1 \\ \Rightarrow {p_1}^2:{p_2}^2=3:3 \\ \Rightarrow p_1:p_2=1:1 \end{gathered}$$

Hence, the correct answer is option (A).