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Step-by-Step Solution
Step 1: Identify the masses and initial velocities
• Let the first ball have mass m_1 = 10\,\text{kg} and initial velocity v_1 = 10\sqrt{3}\,\text{m/s} along the x -axis.
• The second ball has mass m_2= 20\,\text{kg} and is initially at rest.
Step 2: Note the final scenario
• After collision, the first ball of mass 10\,\text{kg} comes to rest.
• The second ball of mass 20\,\text{kg} breaks into two equal pieces, each of mass 10\,\text{kg} .
• One piece moves along the y -axis with speed 10\,\text{m/s} .
• The other piece moves at speed 20\,\text{m/s} at an angle \theta with respect to the x -axis.
Step 3: Write down the total initial momentum
Since the second ball is at rest initially, the total initial momentum of the system is only due to the first ball:
P_{\text{initial}} = m_1 \,v_1 = 10 \,\text{kg} \times 10\sqrt{3}\,\text{m/s} = 100\sqrt{3}\,\hat{i}\,\text{kg·m/s}.
Hence, in component form:
P_{\text{initial},x} = 100\sqrt{3}, \quad P_{\text{initial},y} = 0.
Step 4: Express the final momentum in components
Denote the two pieces from the second ball as piece A and piece B, each of mass 10\,\text{kg} .
Piece A moves along the y -axis with speed 10\,\text{m/s} .
Piece B moves with speed 20\,\text{m/s} at an angle \theta from the x -axis.
• Momentum of piece A:
\vec{p}_A = (10\,\text{kg})\times(10\,\text{m/s}) = 100\,\text{kg·m/s}
purely in the y direction. In component form:
p_{A,x} = 0,\quad p_{A,y} = 100.
• Momentum of piece B:
\vec{p}_B = (10\,\text{kg})\times(20\,\text{m/s}) = 200\,\text{kg·m/s}
at an angle \theta from the x -axis. In component form:
p_{B,x} = 200\cos\theta,\quad p_{B,y} = 200\sin\theta.
Step 5: Apply conservation of linear momentum
Total momentum before collision must equal total momentum after collision. Let us write the equations for each component:
(a) x -component:
P_{\text{initial},x} = P_{\text{final},x}
100\sqrt{3} = 200 \cos\theta.
\therefore \cos\theta = \frac{100\sqrt{3}}{200} = \frac{\sqrt{3}}{2}.
This gives us \theta = 30^\circ or 150^\circ or other angles co-terminal with these cos values. We will confirm the correct angle by checking the y -component.
(b) y -component:
P_{\text{initial},y} = P_{\text{final},y}
0 = p_{A,y} + p_{B,y} = 100 + 200\sin\theta.
\therefore 200\sin\theta = -100
\sin\theta = -\frac{1}{2}.
Thus, \theta must be an angle where \sin\theta < 0 . This corresponds to angles in the fourth or third quadrant. The most common principal angle with \sin\theta = -\frac{1}{2} and \cos\theta = \frac{\sqrt{3}}{2} is \theta = 330^\circ.
Step 6: Conclude the value of \theta
Because the system must have zero net momentum in the y -direction after collision (to match the initial y -component of zero), and we found \sin\theta = -\frac{1}{2} with \cos\theta = \frac{\sqrt{3}}{2} , the angle of the second piece from the positive x -axis is:
\boxed{330^\circ.}
Therefore, the required angle \theta to the nearest integer is 330^\circ.
