Question

A billiard ball collides in an elastic head-on collision with a second stationary identical ball. After the collision which of the following conditions applies to the first ball?

A) maintains the same velocity as before

B)has one half its initial velocity

C)comes to rest

D)moves in the opposite direction

E)doubles its initial velocity

Answer 1

Solution:

In the case of elastic collision in one dimension, from the conservation of linear momentum and the kinetic energy we have,

The final velocity v_1f of the first ball in terms of initial velocities of first ball v_1i and second ball v_2i and their masses m₁ and m₂ is given by,

v_1f = [(m₁ – m₂)/( m₁ + m₂)]*v_1i + [2m₂/( m₁ + m₂)]* v_2i

The final velocity v_2f of the second ball in terms of initial velocities of first ball v_1i and second ball v_2i and their masses m₁ and m₂ is given by,

v_2f = [2m₁/( m₁ + m₂)]*v_1i + [- (m₁ – m₂)/( m₁ + m₂)]*v_2i

In the case of the billiard balls, the second ball in initially stationary, that is v_2i = 0

Hence above two expressions become,

v_1f = [(m₁ – m₂)/( m₁ + m₂)]*v_1i

v_2f = [2m₁/( m₁ + m₂)]*v_1i

Now the billiard balls have same mass, m₁ = m₂ = m. Putting this above expressions we get,

v_1f = [(m – m)/( m + m)]*v_1i

v_2f = [2m/( m + m)]*v_1i

Thus,

v_1f = 0

v_2f = v_1i

That is the final velocity of the first ball is zero after the collision and the final velocity of the second ball is equal to the initial velocity of the first ball.

Hence the correct choice is,

(C) Comes to rest.