Question

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio \(2:2:1\). If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is

• A. \(v\)
• B. \(\sqrt2v\)
• C. \(2\sqrt2v\)
• D. \(3\sqrt2v\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Explosions are internal events where internal forces drive the fragments apart.
Since no external net force acts on the shell, the total linear momentum of the system must be conserved.
Initial momentum is zero, so the vector sum of the final momenta of all fragments must also be zero.
Key Formula or Approach:
Conservation of momentum:
\[ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) \]
Magnitude: \(p_3 = |\vec{p}_1 + \vec{p}_2|\).
Step 2: Detailed Explanation:
1. Assign masses:
The mass ratio is \(2:2:1\). Let the mass of the lighter fragment be \(m_3 = m_0\).
Then the other two fragments have masses \(m_1 = 2m_0\) and \(m_2 = 2m_0\).
(Note: \(m = 2m_0 + 2m_0 + m_0 = 5m_0\)).
2. Set up fragment momenta:
The two equal masses (\(2m_0\)) fly in perpendicular directions (say along x and y axes) with speed \(v\).
\(\vec{p}_1 = (2m_0)v \hat{i}\)
\(\vec{p}_2 = (2m_0)v \hat{j}\)
3. Calculate total momentum of first two fragments:
\(\vec{p}_{12} = \vec{p}_1 + \vec{p}_2 = 2m_0 v \hat{i} + 2m_0 v \hat{j}\)
Magnitude: \(p_{12} = \sqrt{(2m_0 v)^2 + (2m_0 v)^2} = \sqrt{4m_0^2 v^2 + 4m_0^2 v^2} = \sqrt{8m_0^2 v^2} = 2\sqrt{2} m_0 v\).
4. Find speed of third fragment:
The third fragment's momentum must be equal and opposite to \(\vec{p}_{12}\) for the net sum to be zero.
\(p_3 = p_{12} \implies m_0 \times v_3 = 2\sqrt{2} m_0 v\).
Canceling \(m_0\): \(v_3 = 2\sqrt{2} v\).
Step 3: Final Answer:
The speed of the lighter fragment is \(2\sqrt{2}v\).