COM CYU 15

1. Problem Visualization

Three balls $A, B, C$ are connected by elastic threads. Ball $A$ is projected with velocity $u$ parallel to the line connecting $B$ and $C$. The problem can be visualised as treating the interactions sequentially as elastic collisions between pairs.

2. Method: Sequential Elastic Impacts

The threads are “elastic and almost inextensible”. This models the jerk as an elastic collision ($e=1$) along the direction of the thread. We analyze the propagation of impulse step-by-step.

Step 1: Impulse between A and B

Initially, ball $A$ moves with velocity $u$ parallel to $BC$. Ball $B$ is stationary. The string $AB$ connects them.
The component of $A$’s velocity along the string $AB$ drives the impulse. Let the angle between velocity $u$ (horizontal) and string $AB$ be $\theta$.

• Velocity of approach along string: $v_{app} = u \cos \theta$.
• We treat this as a 1D elastic collision between mass $m_1$ (moving) and mass $m_2$ (stationary).

Using the formula for velocity transfer in elastic collision ($v_2 = \frac{2m_1 u_1}{m_1+m_2}$): $$ v_{B, \parallel} = \frac{2 m_1}{m_1 + m_2} (u \cos \theta) $$ This is the velocity $B$ acquires along the direction of string AB.

Step 2: Impulse between B and C

Now ball $B$ acts as the projectile impacting ball $C$ via string $BC$.
Velocity of $B$ is directed along $AB$. We need the component of this velocity along the string $BC$ (horizontal).
Since the angle between string $AB$ and string $BC$ is $\theta$, the projection is: $$ v_{B \to C} = v_{B, \parallel} \cos \theta $$

Now, we have a 1D elastic collision between mass $m_2$ (moving with $v_{B \to C}$) and mass $m_3$ (stationary).
Velocity acquired by $C$: $$ v_C = \frac{2 m_2}{m_2 + m_3} v_{B \to C} $$

3. Calculation

Substituting the expression for $v_{B \to C}$: $$ v_C = \frac{2 m_2}{m_2 + m_3} \left[ v_{B, \parallel} \cos \theta \right] $$ Substitute $v_{B, \parallel}$: $$ v_C = \frac{2 m_2}{m_2 + m_3} \left[ \left( \frac{2 m_1 u \cos \theta}{m_1 + m_2} \right) \cos \theta \right] $$ $$ v_C = \frac{4 m_1 m_2 u \cos^2 \theta}{(m_1 + m_2)(m_2 + m_3)} $$

$$ v_C = \frac{4 (1)(1) (6) (0.5)^2}{(1+1)(1+2)} $$ $$ v_C = \frac{24 \times 0.25}{2 \times 3} $$ $$ v_C = \frac{6}{6} = 1.0 \, \text{m/s} $$