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

Step 1: Use momentum conservation.
No outside force acts during the collision, so the total momentum just before equals the total just after. All three masses are equal, so we can simply add velocity vectors.

Step 2: Total momentum before the hit.
Each particle heads to the centroid $G$ along its median, and these three directions are spaced $120^\circ$ apart with equal speeds. Three equal vectors at $120^\circ$ cancel, so \[ \vec v_A+\vec v_B+\vec v_C=0. \]

Step 3: List the after-collision facts.
After the hit, $A$ stops, so its velocity is $0$. $B$ retraces its path with speed $v$, so its velocity flips to $-\vec v_B$. Let $C$ have velocity $\vec v_C'$.

Step 4: Apply conservation.
\[ \vec v_A+\vec v_B+\vec v_C=0+(-\vec v_B)+\vec v_C'. \] The left side is $0$ from Step 2.

Step 5: Solve for $\vec v_C'$.
\[ 0=-\vec v_B+\vec v_C'\quad\Rightarrow\quad \vec v_C'=\vec v_B. \] So $C$ ends up moving exactly like $B$ was moving originally.

Step 6: Describe the motion.
$B$ was heading toward $G$, that is along $BG$, with speed $v$. So $C$ moves with speed $v$ along $BG$.
\[ \boxed{v\ \text{along } BG} \]