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

To solve this problem, let's use the law of conservation of momentum. Initially, the shell is at rest, so its total initial momentum is zero. After the explosion, the shell breaks into three fragments with their mass ratios \(2:2:1\). Let's denote the mass of the third fragment as \(m_3\) and the common mass of the other two heavier fragments as \(m_1 = m_2 = 2m_3\). 

The fragments \(m_1\) and \(m_2\) move along perpendicular directions with speed \(v\). We need to calculate the speed \(v_3\) of the third fragment \(m_3\).

To use the conservation of momentum, the total momentum in each direction should remain zero.

Vector sum of momenta in the x and y-directions:

\[ m_1v + m_2v + m_3v_3 = 0 \]

Let's assume \(m_1\) moves along the x-axis and \(m_2\) moves along the y-axis:

Along the x-axis: \[ m_1v + m_3v_{3x} = 0 \]

Along the y-axis: \[ m_2v + m_3v_{3y} = 0 \]

Since \(m_1 = m_2 = 2m_3\), substitute these into the momentum equations:

\[ 2m_3v + m_3v_{3x} = 0 \rightarrow v_{3x} = -2v \]

\[ 2m_3v + m_3v_{3y} = 0 \rightarrow v_{3y} = -2v \]

The speed of the third fragment \(v_3\) is given by combining the components:

\[ v_3 = \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{(-2v)^2 + (-2v)^2} = \sqrt{4v^2 + 4v^2} \]

\[ v_3 = \sqrt{8v^2} = 2\sqrt{2}v \]

Therefore, the speed of the third fragment is \(2\sqrt{2}v\).