Question

An object at rest suddenly explodes into three parts of equal masses. Two of the away at right angles to each other with equal speed of 10 m/s. The speed of the just after the explosion will be: 

• A. 10 m/s
• B. 20 m/s
• C. 2\(\sqrt{10}\) m/s
• D. 0
• E. 10√2 m/s

Answer 1

Answer: (E)

Given:

• An object at rest explodes into three equal masses (\( m \)).
• Two parts move at right angles with speed \( 10 \, \text{m/s} \) each.

Step 1: Momentum Conservation

Initial momentum = 0 (object at rest).

Let the third part have velocity \( \vec{v}_3 \). Momentum conservation requires:

\[ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \]

Given \( \vec{p}_1 = m \cdot 10 \, \hat{i} \) and \( \vec{p}_2 = m \cdot 10 \, \hat{j} \), the third momentum must satisfy:

\[ \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -m (10 \, \hat{i} + 10 \, \hat{j}) \]

Thus, the speed of the third part is:

\[ v_3 = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \text{m/s} \]

The speed of the third part is \( 10\sqrt{2} \, \text{m/s} \).

Answer 2

1. Apply the principle of conservation of momentum:

Since the object is initially at rest, the total momentum before the explosion is zero. The total momentum after the explosion must also be zero. Momentum is a vector quantity.

2. Set up the momentum equations:

Let the three parts have equal mass m. Let the velocities of the first two parts be \(v_1 = 10\hat{i}\) m/s and \(v_2 = 10\hat{j}\) m/s (since they move at right angles to each other with equal speeds). Let the velocity of the third part be \(v_3\). The conservation of momentum equation is:

\[mv_1 + mv_2 + mv_3 = 0\]

Since the masses are equal, we can simplify:

\[v_1 + v_2 + v_3 = 0\]

3. Solve for the velocity of the third part:

\[v_3 = -(v_1 + v_2) = -(10\hat{i} + 10\hat{j}) = -10\hat{i} - 10\hat{j}\]

4. Calculate the speed of the third part:

The speed is the magnitude of the velocity vector:

\[|v_3| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200} = 10\sqrt{2} \, \text{m/s}\]