Question

Three balls $A$, $B$ and $C$ of equal mass are moving in the same order along the same straight line on a smooth horizontal surface. Initially, ball $A$ moves towards right with a velocity of $6\text{ ms}^{-1}$, ball $B$ moves towards left with a velocity of $4\text{ ms}^{-1}$ and ball $C$ moves towards right with a velocity of $3\text{ ms}^{-1}$. If the coefficient of restitution between any two balls is $0.8$, then the relative velocity of balls $B$ and $C$ after collision is

• A. $1.6\text{ ms}^{-1}$
• B. $0.8\text{ ms}^{-1}$
• C. $3.2\text{ ms}^{-1}$
• D. $2.4\text{ ms}^{-1}$

Hint:

For multi-body collisions on a line, always process the impacts step-by-step in chronological order based on which balls are moving toward each other.

Answer 1

The Correct Option is A

Step 1: Set the directions.
Take rightward as positive. Then A $=+6$, B $=-4$, C $=+3$ m s$^{-1}$. The order on the line is A, then B, then C, all equal mass.

Step 2: Which pair hits first?
A moves right at $6$ toward B, and B moves left at $4$ toward A. They rush together, so A and B collide first.

Step 3: Rule for equal masses.
For equal masses with restitution $e$, the velocities after are $v_{1}=\tfrac{(1-e)}{2}u_{1}+\tfrac{(1+e)}{2}u_{2}$ and $v_{2}=\tfrac{(1+e)}{2}u_{1}+\tfrac{(1-e)}{2}u_{2}$, with $e=0.8$.

Step 4: Collision of A and B.
Putting $u_{1}=6,\ u_{2}=-4$ gives A $=-3$ m s$^{-1}$ and B $=+5$ m s$^{-1}$. Now B chases C.

Step 5: Collision of B and C.
Now $u_{1}=5$ (B), $u_{2}=3$ (C). This gives B $=3.2$ m s$^{-1}$ and C $=4.8$ m s$^{-1}$.

Step 6: Find the relative speed of B and C.
\[ |v_{C} - v_{B}| = |4.8 - 3.2| = 1.6 \text{ m s}^{-1} \]So the answer is option 1. (This also equals $e$ times their approach speed: $0.8\times 2 = 1.6$.)
\[ \boxed{1.6 \text{ m s}^{-1}} \]