Two billiard balls of equal mass undergo a perfectly elastic head-on collision. One balls initial velocity was 2.00 i m/s and the other ball’s initial velocity was – 3.00 i m/s. What are the velocities of both balls after the collision?

0.5m/s both moving in the negative x direction.

Explanation:

Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of momentum of bodies after collision. The two bodies moves with a common velocity after collision.

Momentum = mass × velocity

Given two billiard balls of equal mass

For the ball of mass M1 moving with velocity of 2.00m/s, its momentum before collision is expressed as:

Momentum = M1 × 2

= 2M1

For the ball of mass M2 moving with velocity of -3m/s, its momentum before collision is expressed as:

Momentum = M2×(-3)

= -3M2

Momentum of the bodies after collision is expressed as:

Momentum = (M1+M2)V where v is the common velocity.

Using the principle above:

2M1-3M2 = (M1+M2)V

Since the two balls have equal masses, M1 = M2 = M the equation becomes;

2M-3M = (M+M)V

-M = 2MV

-1 = 2V

V = -1/2

V = -0.5im/s

The velocities of the balls after collision is the same and is equal to 0.5m/s both moving in the negative x direction.

do you have a formula possibly