Question

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

(A) √3/2 ν

(B) 2√2/3ν

(C) 3/4ν

(D) 3/√2 v

Answer 1

To find the speed of the second block after elastic collision with the first, we use conservation of momentum and kinetic energy, resulting in a speed of (2√{2}/3)ν.

Step-by-step explanation:

When two objects of equal mass M collide elastically, conservation of momentum and energy apply. In an elastic collision, both momentum and kinetic energy are conserved. Considering the first block moves at speed v/3 after the collision at an angle θ, we want to find the second block's speed.

The total momentum before the collision is Mν, all in the horizontal direction since the second block is at rest. After the collision, the first block has a momentum of M(v/3) in a direction at an angle θ. The second block's momentum will be in a direction perpendicular to the first block's momentum due to the conservation of angular momentum in elastic collisions between equal masses, thus making the angle between their velocities 90 degrees.

Using the Pythagorean theorem on the components of the initial and final momentum vectors and the fact that the collision is elastic (kinetic energy is conserved), we can solve for the second block's speed. Since the total initial momentum is Mν and it's the same in the final state, we have:

Mν = sqrt[(M(v/3))^2 + (Mv')^2].

Solving for v', we get v' = √{8/9}ν or v' = (2√{2}/3)ν which corresponds to answer choice (B).