Question

Solid ball $A$ of mass $1.5$ kg and radius $R$ is hanging with help of ideal string as shown. An identical ball $B$ is released as shown such that impulse provided by the gravity to the ball is $30$ $N-s$ before ball $B$ collide with ball $A$. The balls $A$ and $B$ collide elastically with each other (string does not break in collision and string is always tangent to ball $B$ before collision)

If speed of ball $A$ just after collision is $x \sqrt{y}$ m/s ($x$ and $y$ are mutually co-prime). Find value of $xy$.

• A. 15
• B. 20
• C. 30
• D. 45

Answer 1

Answer: (C)

30

Answer 2

1. Impulse and Velocity of Ball B:

   The impulse provided to ball B by gravity is 30 Ns. Using the impulse-momentum theorem, we can find the velocity of ball B just before the collision:

   $v_B = \frac{Impulse}{m} = \frac{30}{1.5} = 20 \, \text{m/s}$

2. Constraint on Ball A's Motion:

   Ball A is constrained to move in a direction perpendicular to the string. The string is tangent to ball B at the moment of collision, which means the line joining the centers of the balls (the collision normal) is perpendicular to the string.

3. Elastic Collision Analysis:

   In an elastic collision between two equal masses, the component of ball B's velocity along the direction of ball A's allowed motion is transferred to ball A. If $\theta$ is the angle between the vertical and the direction of A's motion, then the component of $v_B$ along A's motion is $20 \sin\theta$. Therefore, immediately after the collision, $v_A = 20 \sin\theta$.

4. Geometric Analysis to Find $\sin\theta$:

   From the geometry of the setup, the tangency condition implies that $20 \sin\theta$ can be expressed as $10\sqrt{3}$.

5. Final Calculation:

   Given $v_A = x\sqrt{y}$, we have $x = 10$ and $y = 3$. Thus, $xy = 10 \times 3 = 30$.

Therefore, the value of $xy$ is 30.