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

Step 1: Understanding the Question:
This is a classic explosion problem that is governed by the principle of conservation of linear momentum. Since the bomb is initially at rest, the total initial momentum is zero. Therefore, the vector sum of the final momenta of all fragments must also be zero.
Step 2: Key Formula or Approach:
Conservation of Linear Momentum: \(\vec{p}_{initial} = \vec{p}_{final}\).
For this problem, \(0 = \vec{p}_1 + \vec{p}_2 + \vec{p}_3\), where \(\vec{p} = m\vec{v}\).
Step 3: Detailed Explanation:
Let the common factor for the mass ratio be \(m\).
- Mass of the first piece, \(m_1 = 2m\).
- Mass of the second piece, \(m_2 = 2m\).
- Mass of the third piece, \(m_3 = 3m\).
The two identical masses (\(m_1\) and \(m_2\)) fly off perpendicular to each other. We can set up a coordinate system where their velocities are along the x and y axes.
- Velocity of the first piece, \(\vec{v}_1 = 18 \hat{i}\) m/s.
- Velocity of the second piece, \(\vec{v}_2 = 18 \hat{j}\) m/s.
Now, calculate their momenta:
- \(\vec{p}_1 = m_1 \vec{v}_1 = (2m)(18 \hat{i}) = 36m \hat{i}\).
- \(\vec{p}_2 = m_2 \vec{v}_2 = (2m)(18 \hat{j}) = 36m \hat{j}\).
Let the velocity of the third piece be \(\vec{v}_3\). Its momentum is \(\vec{p}_3 = m_3 \vec{v}_3 = (3m)\vec{v}_3\).
Applying the conservation of momentum:
\[ 0 = \vec{p}_1 + \vec{p}_2 + \vec{p}_3 \] \[ \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) \] \[ (3m)\vec{v}_3 = -(36m \hat{i} + 36m \hat{j}) \] We can cancel \(m\) from both sides:
\[ 3\vec{v}_3 = -36 \hat{i} - 36 \hat{j} \] \[ \vec{v}_3 = -12 \hat{i} - 12 \hat{j} \] The question asks for the speed, which is the magnitude of the velocity \(\vec{v}_3\).
\[ |\vec{v}_3| = \sqrt{(-12)^2 + (-12)^2} = \sqrt{144 + 144} = \sqrt{2 \times 144} \] \[ |\vec{v}_3| = 12\sqrt{2} \text{ m/s} \] Step 4: Final Answer:
The speed of the third piece is \(12\sqrt{2}\) m/s.