Question

A hockey puck slides on the ice with an initial velocity of vi1 = 5 m/s in the positive x direction. A second hockey puck is sliding to the left with an initial velocity of vi2 = 3 m/s. The two pucks collide and after the collision puck 1 has a final velocity of vf1 = 3 m/s in the positive x direction at an angle of 30 degrees above the positive x axis. Is the collision elastic or inelastic?

Please show all work.

Answer 1

Given,

The initial velocity of 1^st hockey puck, vi1 = 5m/s & final velocity after collision be vf1 = 3m/s in the positive x-direction at an angle of 30 degrees above the positive x-axis.

The initial velocity of a 2^nd hockey puck,vi2 = 3m/s.

Let the mass of hockey puck be M.

Suppose that this case is an elastic collision.

In an elastic collision, the relative velocity of approach (vi1 - vi2) before the collision is equal to the velocity of separation(vf2 - vf1) after the collision.

  In an inelastic collision, the relative velocity of approach (vi1 - vi2) before the collision is greater than the velocity of separation(vf2 - vf1) after the collision.

And we are considering this case to be an elastic collision. The collision to be elastic collision we know that the kinetic energy and momentum should be conserved.

Now, Kinetic Energy given as

1/2Mvi1² + 1/2 Mvi2² = 1/2 Mvf1² + 1/2Mvf2²

vi1² + vi2²  = vf1² + vf2²

vf2² = vi1² + vi2² - vf1² = (5)² + (3)² - (3)²

vf2² = (5)²

vf2 = 5 m/s

now , velocity of approach = vi1 - vi2 = 5 - 3 = 2m/s

velocity of sepration = vf2 - vf1 = 5 - 3 = 2m/s.

Thus, the velocity of approach = velocity of separation = 2m/s

Thus from the discussion, we conclude that the case is an elastic collision in 2 dimension.