Question

Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a

second identical ball (Ball B). After the collision Ball A continues to move

in the same direction at 2 m/s. What is the magnitude of the velocity for

Ball B after the collision?

Before Collision:

10 m/s

A

After Collision:

2 m/s

O

Answer 1

Answer:

6m/s

Explanation:

Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.

Using the expression

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity after collision

Substitute the given values in the formula

0.35(10)+0.35(2) = (0.35+0.35)v

3.5+0.7 = 0.7v

4.2 = 0.7v

v = 4.2/0.7

v = 6m/s

Hence the magnitude of the velocity for Ball B after the collision is 6m/s