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Step-by-Step Solution
1. Identify the Known Information
• Mass of sphere A = $m_1$ (initially at rest).
• Mass of sphere B = $m_2$ (initial velocity $v$ along the +x-axis).
• Velocity of B after collision = $\frac{v}{2}$ (perpendicular to the original x-direction, i.e., along the y-axis).
• Velocity of A after collision = unknown magnitude $v_A$ at some angle $\theta$ to the x-axis.
• We are given that the final direction of A is $\theta = \tan^{-1}\bigl(-\tfrac{1}{2}\bigr)$.
2. Apply Conservation of Momentum
Since there are no external forces, the total momentum before collision equals the total momentum after collision. We apply this in both the x and y directions.
2.1 In the x-direction:
Before collision, total $x$-momentum = $m_1 \times 0 + m_2 \times v = m_2\,v.$
Let the components of $\vec{v}_A$ (velocity of A after collision) be $\bigl(v_A \cos\theta,\,v_A \sin\theta\bigr).$ Then, the final $x$-momentum is:
$m_1 \, v_A \cos\theta + m_2 \times 0 \quad \bigl(\text{since B's final velocity has no }x\text{-component}\bigr).$
Hence, conservation in the $x$-direction gives:
$m_2\,v = m_1\,v_A \cos\theta. \quad (1)$
2.2 In the y-direction:
Before collision, the total $y$-momentum is zero, as both A and B have no $y$-component initially. After collision:
• A has $m_1 \, v_A \sin\theta$ in the y-direction,
• B has $m_2 \,\frac{v}{2}$ in the y-direction.
Thus, conservation in the $y$-direction gives:
$0 = m_1\,v_A \sin\theta + m_2 \left(\tfrac{v}{2}\right). \quad (2)$
3. Solve for the Angle
From equation (2):
$m_1\,v_A \sin\theta = -\,m_2 \,\tfrac{v}{2}.$
Divide this expression by equation (1) to eliminate $v_A$:
$\dfrac{m_1\,v_A \sin\theta}{m_2\,v/2} \; \Big/ \; \dfrac{m_1\,v_A \cos\theta}{m_2\,v}
\;=\; \dfrac{-\,m_2\,\tfrac{v}{2}}{m_2\,v} \;\Big/\; \dfrac{m_2\,v}{m_2\,v}\,.$
This simplifies to:
$\dfrac{\sin\theta}{\cos\theta} = \tan\theta = \dfrac{-\frac{1}{2}}{1} = -\,\dfrac{1}{2}.$
Hence,
$\theta = \tan^{-1}\Bigl(-\tfrac{1}{2}\Bigr).
$
The negative sign indicates that the velocity vector of A is below the x-axis (assuming the positive y-axis is in the perpendicular direction chosen for B after collision).
4. Final Answer
Therefore, the direction of mass A after the collision makes an angle
$\theta = \tan^{-1}\bigl(-\tfrac{1}{2}\bigr)$
with the x-axis.
