Question

A body of mass \(M\) at rest explodes into three pieces, in the ratio of masses 1:1:2. Two smaller pieces fly off perpendicular to each other with velocities of 30 m/s and 40 m/s respectively. The velocity of the third piece will be:

• A. 15 m/s
• B. 25 m/s
• C. 35 m/s
• D. 50 m/s

Hint:

- When solving explosion problems, apply the conservation of momentum in both the horizontal and vertical directions.
- Use vector addition to find the resultant velocity.
- The negative signs indicate opposite directions but do not affect magnitude.

Answer 1

The Correct Option is B

To solve this problem, we must use the principle of conservation of momentum. When the body explodes, the total momentum of the system should remain zero as it was initially at rest.

Let's designate the masses of the three pieces after the explosion as follows:

• The first piece has a mass \(m_1 = \frac{M}{4}\), flying with velocity \(v_1 = 30 \, \text{m/s}\).
• The second piece has a mass \(m_2 = \frac{M}{4}\), flying with velocity \(v_2 = 40 \, \text{m/s}\).
• The third piece has a mass \(m_3 = \frac{M}{2}\), and its velocity is denoted by \(v_3\).

Since the first two pieces move perpendicular to each other, we can use vector addition to find the resultant momentum that the third piece must balance.

The momentum of the system must be zero, so the momentum of the third piece should be equal and opposite to the vector sum of momenta of the two smaller pieces.

Let's calculate the vector resultant of the momenta of the first two pieces:

• Momentum of first piece: \( p_1 = m_1 \cdot v_1 = \frac{M}{4} \times 30 = \frac{30M}{4} \).
• Momentum of second piece: \( p_2 = m_2 \cdot v_2 = \frac{M}{4} \times 40 = \frac{40M}{4} \).

The momentum vectors are perpendicular to each other, so we use the Pythagorean Theorem:

p_{\text{net}} = \sqrt{\left(\frac{30M}{4}\right)^2 + \left(\frac{40M}{4}\right)^2}

Calculating:

p_{\text{net}} = \sqrt{\frac{900M^2}{16} + \frac{1600M^2}{16}} p_{\text{net}} = \sqrt{\frac{2500M^2}{16}} p_{\text{net}} = \frac{50M}{4} = \frac{25M}{2}

The third piece, with a mass of \(m_3 = \frac{M}{2}\), must have a momentum equal and opposite to this, thus:

• Momentum of the third piece: \(p_3 = m_3 \cdot v_3 = -\frac{25M}{2}\)
• \(\frac{M}{2} \cdot v_3 = -\frac{25M}{2}\)

Solving for \(v_3\), we get:

v_3 = 25 \, \text{m/s}

Therefore, the velocity of the third piece is 25 m/s, which matches the correct answer option.