On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle to its initial direction and has a speed v3. The second block's speed after the collision is
- 3 2v
- 32v
- 2 23v
- 34v
Answer
(3) 2 23v
Solution
In an elastic collision kinetic energy is conserved. Let the second block's speed be v2.
12Mv2 = 12M ( v3 )2 + 12Mv22
v22 = v2 - v29 = 8v29
∴ v2 = 2 23v
Answer: (3)