Question

A mass $3m$, initially at rest at the origin explodes into three fragments of equal mass. Two of the fragments have speed $\nu$ each and move perpendicular to each other. The third fragment will move with a speed

• A. $\frac{\nu}{\sqrt{2}}$
• B. $\frac{\nu}{2}$
• C. $\nu$
• D. $\nu \sqrt{2}$

Hint:

In an explosion, the vector sum of momenta of all fragments must be zero.

Answer 1

The Correct Option is D

To solve this problem, we need to use the principle of conservation of momentum. Initially, the mass \(3m\) is at rest, meaning its total initial momentum is zero. After the explosion, the total momentum of the system must still be zero since there are no external forces acting on it.

The mass \(3m\) explodes into three equal masses of mass \(m\) each. Let's denote these masses as \(m_1\), \(m_2\), and \(m_3\).

According to the problem:

• Two fragments (say \(m_1\) and \(m_2\)) move perpendicularly to each other with a speed \(\nu\).
• The vectors of their momenta will be perpendicular, and thus their resultant momentum will be given by the vector sum.
• The third fragment (\(m_3\))'s speed and direction are what we need to find.

Using vector addition of momentum:

The momentum of \(m_1\) is \(m \nu\) in one direction (say the x-axis), and the momentum of \(m_2\) is \(m \nu\) in an orthogonal direction (y-axis).

The resultant momentum of \(m_1\) and \(m_2\) is given by the Pythagorean theorem:

\(|p_{\text{resultant}}| = \sqrt{(m \nu)^2 + (m \nu)^2} = \sqrt{2} m \nu\)

For the total momentum of the system to remain zero, the third fragment \(m_3\) must have an equal momentum in the opposite direction of the resultant momentum of \(m_1\) and \(m_2\). Therefore, the magnitude of the momentum of \(m_3\) must be:

\(|p_3| = \sqrt{2} m \nu\)

The speed \(v_3\) of the third fragment can be determined by equating the magnitudes of the momenta:

\(m_3 v_3 = \sqrt{2} m \nu\)

Given \(m_3 = m\), this simplifies to:

\(m \cdot v_3 = \sqrt{2} m \nu \Rightarrow v_3 = \nu \sqrt{2}\)

Thus, the third fragment will move with a speed \(\nu \sqrt{2}\).

Therefore, the correct option is \(\nu \sqrt{2}\).