Question

Three small particles \(A\), \(B\) and \(C\) of equal mass move with equal speed \(v\) along the medians of an equilateral triangle as shown. They collide at the centroid \(G\) of the triangle. After the collision, \(A\) comes to rest, while \(B\) retraces its path with the same speed \(v\). What is the speed and direction of motion of \(C\) after the collision?

• A. \(v\) in the direction along \(GB\)
• B. \(v\) in the direction along \(BG\)
• C. \(4v\) in the direction along \(GB\)
• D. \(2v\) in the direction along \(BG\)

Hint:

For three equal vectors directed along the medians of an equilateral triangle, \[ \vec v_A+\vec v_B+\vec v_C=0. \] This symmetry greatly simplifies momentum conservation problems.

Answer 1

The Correct Option is B

Concept: Linear momentum is conserved during collision. Since all masses are equal, we can work directly with velocity vectors.

Step 1: Find the initial resultant momentum. The three particles move toward the centroid along the three medians of an equilateral triangle. The angle between any two velocity vectors is \[ 120^\circ. \] Hence, \[ \vec v_A+\vec v_B+\vec v_C=0. \] Therefore, \[ \vec P_{\text{initial}}=0. \]

Step 2: Apply conservation of momentum after collision. After collision, \[ \vec v_A'=0. \] Particle \(B\) retraces its path with speed \(v\). Hence, \[ \vec v_B'=-\vec v_B. \] Since total momentum must remain zero, \[ \vec v_A'+\vec v_B'+\vec v_C'=0 \] \[ 0-\vec v_B+\vec v_C'=0. \] Thus, \[ \vec v_C'=\vec v_B. \]

Step 3: Interpret the result. Initially \(B\) was moving from \(B\) toward \(G\). Therefore, \[ \vec v_C' \] has magnitude \(v\) and direction along \(BG\). \[\begin{aligned} \boxed{\text{Speed }=v,\ \text{direction along }BG} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.