Two billiard balls have velocities of 2.0 m/s and -1.0 m/s when they meet in an elastic head-on collision.What is the final velocity of the first ball after collision?
A)-2.0 m/s
B)-1.0 m/s
C)-0.5 m/s
D)+1.0 m/s

Question

Quizplus · Accepted Answer

In an elastic head-on collision, the velocity of the second ball after collision would be the same as the velocity of the first ball before collision (since they have the same mass). Therefore, the second ball will have a velocity of -2.0 m/s after collision. Using conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'
where m1 and m2 are the masses of the two balls, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities.

Plugging in the given values:

m1(2.0 m/s) + m2(-1.0 m/s) = m1v1' + m2(-2.0 m/s)

Assuming that the masses are the same, we can cancel out the m1 and m2 terms:

2.0 m/s - 1.0 m/s = v1' - 2.0 m/s

Solving for v1':

v1' = -1.0 m/s

Therefore, the final velocity of the first ball after collision is -1.0 m/s.

Two Billiard Balls Have Velocities of 2