Question

A shell of mass \(M\) initially at rest suddenly explodes into three fragments. Two of these fragments are of mass \(M/4\) each, which move with velocities \(3 \, \text{m s}^{-1}\) and \(4 \, \text{m s}^{-1}\) respectively in mutually perpendicular directions. The magnitude of velocity of the third fragment is

• A. \(3.0 \, \text{m s}^{-1}\)
• B. \(2.5 \, \text{m s}^{-1}\)
• C. \(1.5 \, \text{m s}^{-1}\)
• D. \(2.0 \, \text{m s}^{-1}\)

Hint:

In explosion problems, always apply conservation of momentum and resolve momentum vectors carefully when directions are perpendicular.

Answer 1

The Correct Option is B

Step 1: Principle of conservation of momentum.
Since the shell is initially at rest, the total momentum before explosion is zero. Therefore, the vector sum of momenta of all three fragments after explosion must also be zero.
Step 2: Momentum of the first two fragments.
Momentum of first fragment: \[ p_1 = \frac{M}{4} \times 3 = \frac{3M}{4}. \] Momentum of second fragment: \[ p_2 = \frac{M}{4} \times 4 = M. \] These momenta are mutually perpendicular.
Step 3: Resultant momentum and velocity of the third fragment.
The resultant momentum of the first two fragments is \[ p = \sqrt{\left(\frac{3M}{4}\right)^2 + M^2} = \frac{5M}{4}. \] The mass of the third fragment is \[ M - \frac{M}{4} - \frac{M}{4} = \frac{M}{2}. \] Using conservation of momentum, the magnitude of velocity of the third fragment is \[ v = \frac{\frac{5M}{4}}{\frac{M}{2}} = 2.5 \, \text{m s}^{-1}. \]
Step 4: Conclusion.
The magnitude of velocity of the third fragment is \(2.5 \, \text{m s}^{-1}\).