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Step-by-Step Solution Using Conservation of Linear Momentum
Step 1: Understand the Initial Conditions
We have a stationary bomb of total mass 1 kg that explodes into three pieces. Their masses are in the ratio 1 : 1 : 3. Hence, the smaller pieces each have mass $m_1 = m_2 = \frac{1}{5}\,\text{kg}$ and the larger piece has mass $m_3 = \frac{3}{5}\,\text{kg}$. Since the bomb is initially at rest, the total initial momentum is zero.
Step 2: Assign Directions and Velocities
After the explosion, the two smaller pieces each move at 30 m/s in perpendicular directions. Let one piece move along the x-axis with velocity $v_1 = 30\,\text{m/s}$ and the other piece move along the y-axis with velocity $v_2 = 30\,\text{m/s}$. We need to find the velocity $\vec{v}_3$ of the bigger piece of mass $\frac{3}{5}\,\text{kg}$.
Step 3: Write the Momentum Conservation Equation
Because the total momentum before the explosion is zero, the vector sum of the momenta of the three pieces after the explosion must also be zero. Denote the momentum of each piece by mass times velocity. Therefore,
$0 \;=\; m_1 \vec{v}_1 \;+\; m_2 \vec{v}_2 \;+\; m_3 \vec{v}_3.
$
Substitute the known values:
$m_1 = \frac{1}{5}\,\text{kg}$, $\vec{v}_1 = 30\,\hat{i}$ m/s
$m_2 = \frac{1}{5}\,\text{kg}$, $\vec{v}_2 = 30\,\hat{j}$ m/s
$m_3 = \frac{3}{5}\,\text{kg}$, $\vec{v}_3$ = unknown
Step 4: Express Momentum of Smaller Pieces
The total momentum imparted by the smaller pieces is:
$
m_1 \vec{v}_1 + m_2 \vec{v}_2
= \frac{1}{5}\times 30\,\hat{i} + \frac{1}{5}\times 30\,\hat{j}
= 6\,\hat{i} + 6\,\hat{j}.
$
Step 5: Solve for the Larger Piece’s Momentum
From momentum conservation, the larger piece’s momentum must be equal and opposite to the sum of the momenta of the two smaller pieces:
$
m_3 \vec{v}_3 = -\bigl(m_1 \vec{v}_1 + m_2 \vec{v}_2\bigr)
= -(6\,\hat{i} + 6\,\hat{j})
= -6\,\hat{i} - 6\,\hat{j}.
$
Hence,
$
\vec{v}_3 = \frac{-6\,\hat{i} - 6\,\hat{j}}{m_3}
= \frac{-6\,\hat{i} - 6\,\hat{j}}{\frac{3}{5}}
= -6\,\hat{i} - 6\,\hat{j} \times \frac{5}{3}
= -10\,\hat{i} - 10\,\hat{j}.
$
Step 6: Calculate the Magnitude of the Bigger Part’s Velocity
The magnitude of $\vec{v}_3$ is
$
v_3 = \sqrt{(-10)^2 + (-10)^2}
= \sqrt{100 + 100}
= \sqrt{200}
= 10\sqrt{2}\;\text{m/s}.
$
This matches option 1, $10 \sqrt{2}$ m/s.
