Question

A block of mass \(m\) moving on a frictionless surface at speed \(V\) collides elastically with a block of same mass, initially at rest. Now, the first block moves at an angle \(\theta\) with its initial direction and has speed \(V_{1}\). The speed of the second block after collision is

• A. \(\sqrt{V_{1}^{2}-V^{2}}\)
• B. \(\sqrt{V^{2}-V_{1}^{2}}\)
• C. \(\sqrt{V^{2}+V_{1}^{2}}\)
• D. \(\sqrt{V-V_{1}}\)

Hint:

The collision is elastic, so total kinetic energy is conserved. Equate the kinetic energy before and after and solve for the second block's speed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In an elastic collision the total kinetic energy is conserved. Both blocks have the same mass \(m\).

Step 2: Energy Before and After
Before: \(\frac{1}{2}mV^{2}\).
After: \(\frac{1}{2}mV_{1}^{2}+\frac{1}{2}mV_{2}^{2}\), where \(V_{2}\) is the speed of the second block.

Step 3: Calculation
\(V^{2}=V_{1}^{2}+V_{2}^{2}\), so \(V_{2}=\sqrt{V^{2}-V_{1}^{2}}\).

Final Answer:
The second block moves with speed equal to the square root of V squared minus V1 squared. \[ \boxed{\text{(B) }\sqrt{V^{2}-V_{1}^{2}}} \]