Question

A 1 kg stationary bomb is exploded in three parts having mass $1 : 1 : 3$ respectively. Part having same mass move in perpendicular direction with velocity 30 m/s, then the velocity of bigger part will be

• A. $10\sqrt2\, m/sec$
• B. $\frac{10}{\sqrt2}\,m/sec$
• C. $15\sqrt2\, m/sec$
• D. $\frac{15}{\sqrt2}\,m/sec$

Answer 1

The Correct Option is A

To solve this problem, we need to apply the principle of conservation of momentum. According to this principle, the total momentum of a system remains constant if there is no external force acting on it.

Step-by-Step Solution:

1. The bomb initially has a mass of 1 kg and is stationary. Therefore, its initial momentum is zero.
2. Upon explosion, the bomb is split into three parts in the ratio of 1:1:3. Let the masses of these parts be m_1, m_2, and m_3.
3. According to the given ratio, let:
   • m_1 = m_2 = \dfrac{1}{5} \text{ kg} = 0.2 \text{ kg}
   • m_3 = \dfrac{3}{5} \text{ kg} = 0.6 \text{ kg}
4. Parts m_1 and m_2 move in perpendicular directions with a velocity of 30 m/s. Hence, their momentum is:
   • p_1 = m_1 \times v_1 = 0.2 \times 30 = 6 \, \text{kg m/s}
   • p_2 = m_2 \times v_2 = 0.2 \times 30 = 6 \, \text{kg m/s}
5. Since m_1 and m_2 are perpendicular to each other, the resultant momentum of these two parts is given by the vector addition:
   • p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \, \text{kg m/s}
6. Since the initial momentum is zero, the momentum of the third part, p_3, must be equal and opposite to p_{12} to conserve momentum:
   • p_3 = 6\sqrt{2} \, \text{kg m/s}
7. Then, we can find the velocity of the third part (m_3 = 0.6 \text{ kg}) using:
   • v_3 = \dfrac{p_3}{m_3} = \dfrac{6\sqrt{2}}{0.6} = 10\sqrt{2} \, \text{m/s}

The velocity of the bigger part is 10\sqrt{2} \, \text{m/s}.