A billiard ball (Ball #1)moving at 5.00 m/s strikes a stationary ball (Ball #2)of the same mass.After the collision,Ball #1 moves at a speed of 4.35 m/s.Find the speed of Ball #2 after the collision.
A)1.25 m/s
B)1.44 m/s
C)2.16 m/s
D)2.47 m/s

Question

Quizplus · Accepted Answer

The law of conservation of momentum states that the total momentum of a system remains constant if there are no external forces acting on it. In other words, the momentum before the collision is equal to the momentum after the collision.

Let's denote the velocity of ball #2 after the collision as v. Using conservation of momentum, we can write:

(mass of ball #1)(velocity of ball #1) + (mass of ball #2)(velocity of ball #2) = (mass of ball #1)(velocity of ball #1) + (mass of ball #2)(velocity of ball #2)

where the mass of ball #1 is the same as the mass of ball #2.

Plugging in the numbers we know:

(1 kg)(5.00 m/s) + (1 kg)(0 m/s) = (1 kg)(4.35 m/s) + (1 kg)(v)

Solving for v:

v = 2.47 m/s

Therefore, the answer is D: 2.47 m/s.

A Billiard Ball (Ball #1)moving at 5