Question

A particle of mass \(4m\) explodes into three pieces of masses \(m\), \(m\), and \(2m\). The equal masses move along X-axis and Y-axis with velocities \(4\,m\,s^{-1}\) and \(6\,m\,s^{-1}\) respectively. The magnitude of velocity of the heavier mass is

• A. \(\sqrt{17}\,m\,s^{-1}\)
• B. \(2\sqrt{13}\,m\,s^{-1}\)
• C. \(\sqrt{13}\,m\,s^{-1}\)
• D. \(\dfrac{\sqrt{13}}{2}\,m\,s^{-1}\)

Hint:

If an explosion happens in a closed system, total momentum remains conserved, even if kinetic energy changes.

Answer 1

The Correct Option is C

Step 1: Apply conservation of momentum.
Initially particle is at rest, so total momentum is zero.
\[ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \]
Step 2: Momentum of first two pieces.
First mass \(m\) moves along X-axis with \(4\):
\[ \vec{p}_x = m(4)\hat{i} = 4m\hat{i} \]
Second mass \(m\) moves along Y-axis with \(6\):
\[ \vec{p}_y = m(6)\hat{j} = 6m\hat{j} \]
Step 3: Momentum of third piece must cancel these.
So for mass \(2m\):
\[ \vec{p}_3 = -(4m\hat{i} + 6m\hat{j}) \]
Magnitude:
\[ |\vec{p}_3| = m\sqrt{4^2 + 6^2} = m\sqrt{52} = 2m\sqrt{13} \]
Step 4: Find velocity of heavier piece.
\[ |\vec{p}_3| = (2m)v \]
So:
\[ 2m v = 2m\sqrt{13} \Rightarrow v = \sqrt{13} \]
Final Answer: \[ \boxed{\sqrt{13}\,m\,s^{-1}} \]