Question

A bomb at rest explodes into three pieces in the ratio of masses 2 : 2 : 3. The identical masses fly off perpendicular to each other with 18 m/s. Find speed of the third piece.

• A. \(12\sqrt{2}\) m/s
• B. \(12/\sqrt{2}\) m/s
• C. 12 m/s
• D. 18 m/s

Hint:

For explosion problems where the initial object is at rest, the momentum of one piece must be equal in magnitude and opposite in direction to the vector sum of the momenta of all other pieces.
Setting up a Cartesian coordinate system aligned with the perpendicular velocities is the easiest way to solve such problems.

Answer 1

The Correct Option is A

In this problem, we need to determine the speed of the third piece of a bomb that explodes into three pieces. Initially, the bomb is at rest, meaning the total momentum before the explosion is zero.

Let's denote the masses of the three pieces as \( m_1 \), \( m_2 \), and \( m_3 \) with the given mass ratio \( 2:2:3 \). We assign a common factor \( k \) such that: 

• \(m_1 = 2k\)
• \(m_2 = 2k\)
• \(m_3 = 3k\)

The identical masses \( m_1 \) and \( m_2 \) move perpendicularly with a speed of 18 m/s. The directions are perpendicular, so we can use the concept of conservation of momentum in two dimensions, considering each direction separately.

Step-by-step solution:

1. The initial total momentum before the explosion is zero as the bomb is at rest.
2. The momentum of the system must still be zero after the explosion due to the conservation of momentum.
3. Calculate the individual momentum of each piece after the explosion:
   • Momentum of first piece: \(p_1 = m_1 \times 18 = 2k \times 18\)
   • Momentum of second piece: \(p_2 = m_2 \times 18 = 2k \times 18\)
4. The two components (perpendicularly) are \(p_1 = 36k\) and \(p_2 = 36k\). Since these are perpendicular, the resultant momentum due to these two pieces is:
   • \(p_{12} = \sqrt{(36k)^2 + (36k)^2} = 36k\sqrt{2}\)
5. Equating with the third piece's momentum for balance:
   • \(p_3 = m_3 \times v_3 = 3k \times v_3\)
   • Using conservation of momentum: \(3k \times v_3 = 36k\sqrt{2}\)
   • Solving for \(v_3\):
     • \(v_3 = \frac{36k\sqrt{2}}{3k} = 12\sqrt{2} \text{ m/s}\)

Therefore, the speed of the third piece is \(12\sqrt{2}\) m/s.