Question

Two spheres $A$ and $B$ of masses $m_1$ and $m_2$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $v/2$ along x-axis. After collision $B$ has a velocity in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction

• A. same as that of B
• B. opposite to that of B
• C. $\theta=tan^{-1}\bigg(\frac{1}{2}\bigg)$ to the x-axis
• D. $\theta=tan^{-1}\bigg(-\frac{1}{2}\bigg)$ to the x-axis

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the initial and final conditions of the collision based on the information provided and apply the principles of conservation of momentum. Let's break down the steps:

1. Understanding the Initial Conditions: Initially, sphere $A$ is at rest, and sphere $B$ is moving with velocity $v/2$ along the x-axis. Hence, the initial velocity of $A$ is $0$, and that of $B$ is $v/2$.
2. Understanding the Final Conditions: After the collision, sphere $B$ moves in a direction perpendicular to its original motion, which indicates a movement along the y-axis (since its original motion was along the x-axis). Therefore, the final velocity of $B$ has no x-component.
3. Conservation of Momentum in the x-direction: Since the total momentum in the x-direction must be conserved, the initial momentum of $B$ must be equal to the final momentum of $A$ in the x-direction after the collision. Thus:
   m_2 \cdot \frac{v}{2} = m_1 \cdot v_{Ax}
   where $v_{Ax}$ is the x-component of $A$'s velocity after the collision.
4. Direction of Sphere $A$: Given that $B$ moves in the y-direction after the collision, and the x-component of the momentum is retained by $A$, it suggests that $A$ will move in a direction with both x and y components. The question gives us the correct option for $A$'s direction as \theta = \tan^{-1} \left(-\frac{1}{2}\right), which implies $A$ is moving in the negative y-direction, forming an angle of \theta with the x-axis.

Thus, the correct direction for sphere $A$ after the collision is \theta = \tan^{-1} \left(-\frac{1}{2}\right) to the x-axis, meaning it moves downward at this angle.