Question

Three identical blocks A, B and C are placed on a horizontal frictionless surface. The blocks B and C are at rest. Block A is approaching towards B with a speed \(10\) m/s. The coefficient of restitution for all collisions is \(0.5\). The speed of block C just after collision is

• A. \(5.6\) m/s
• B. \(6\) m/s
• C. \(8\) m/s
• D. \(10\) m/s

Hint:

Apply momentum conservation and the restitution relation to the A-B collision first, then to the B-C collision.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For two equal masses with coefficient of restitution \(e\), if one is at rest, the struck block gets speed \(\dfrac{(1+e)}{2}u\) and the striking block keeps \(\dfrac{(1-e)}{2}u\).

Step 2: First Collision (A with B)
\(v_{B}=\dfrac{1+0.5}{2}\times10=7.5\) m/s.

Step 3: Second Collision (B with C)
B moves at 7.5 m/s and hits C at rest, so \(v_{C}=\dfrac{1+0.5}{2}\times7.5=5.625\) m/s.

Step 4: Result
The speed of C just after its collision is about \(5.6\) m/s.

Final Answer:
The speed of block C just after collision is about 5.6 m/s. \[ \boxed{\text{(A) }5.6\text{ m/s}} \]