On a frictionless surface, a block of mass M moving at speed υ collides elastically with another block of the same mass M which is initially at rest. After the collision, the first block moves at an angle θ to its initial direction and has a speed υ/3. The second block's speed after the collision is:

In an elastic collision: Ki = Kf where i and f refer to initial and final, K refers to kinetic energy of the particle.

By conservation of momentum:
Initial momentum = Mυ
Final momentum = M(υ/3)cosθ + Mυ2
where υ2 is the speed of the second block after collision.
Therefore, Mυ = M(υ/3)cosθ + Mυ2
υ = (υ/3)cosθ + υ2

By conservation of energy:
(1/2)Mυ^2 = (1/2)M(υ/3)^2 + (1/2)Mυ2^2
υ^2 = (υ^2/9) + υ2^2

Solving for υ2 in terms of υ and θ:
υ2^2 = υ^2 - (υ^2/9) = (8/9)υ^2
υ2 = (√8/3)υ = (2√2/3)υ
Therefore, the speed of the second block after collision is (2√2/3)υ