Question

$A$ and $B$ are identical point masses. $A$ is released as shown in the diagram at an angle $60^\circ$ from the vertical. Find $R$ if $B$ is able to reach point $C$ after elastic impact. 

• A. $\dfrac{1}{5}$
• B. $\dfrac{1}{2}$
• C. $\dfrac{1}{3}$
• D. $\dfrac{1}{6}$

Hint:

For vertical circular motion, always remember the critical condition \[ v_{\text{bottom}}^2 = 5gR \] to just reach the top without losing contact.

Answer 1

The Correct Option is A

Concept: The solution involves:
Conservation of mechanical energy for the falling mass
Properties of a perfectly elastic collision between identical masses
Minimum speed condition to complete a vertical circular motion
Step 1: Velocity of mass $A$ just before collision Length of the pendulum is $l = 1\,\text{m}$. The vertical drop of $A$ from $60^\circ_toggle$ is: \[ h = l(1 - \cos60^\circ) = 1\left(1 - \frac{1}{2}\right) = \frac{1}{2} \] Using energy conservation: \[ mgh = \frac{1}{2}mv^2 \] \[ v^2 = 2g\left(\frac{1}{2}\right) = g \]
Step 2: Elastic collision Since the collision is perfectly elastic and masses are identical, the velocities are exchanged. Thus, mass $B$ acquires speed: \[ v_B^2 = g \]
Step 3: Condition to reach point $C$ For a particle to just reach the top of a vertical circular track of radius $R$, the minimum speed at the bottom must satisfy: \[ v^2 \ge 5gR \] Substitute $v_B^2 = g$: \[ g \ge 5gR \] Step 4: Solve for $R$ \[ R \le \frac{1}{5} \] Hence, \[ \boxed{R = \frac{1}{5}} \]