Question

Bert strikes a cue ball of mass 0.17 kg, giving it a velocity of 3 m/s in the x-direction. When the cue ball strikes the eight ball (mass=0.16 kg), previously at rest, the eight ball is deflected 45 degrees from the cue ball’s previous path, and the cue ball is deflected 40 degrees in the opposite direction. Find the velocity of the cue ball and the eight ball after the collision.

Answer 1

Answer:

The velocity of the cue ball after collision is -2.13 m/s and the velocity of the eight ball after collision is 2.06 m/s

Explanation:

Let m₁ = mass of cue ball = 0.17 kg, m₂ = mass of eight ball = 0.16 kg, v₁ = initial speed of cue ball = 3 m/s, v₂ = initial speed of eight ball = 0 (since it is initially at rest), v₃ = final speed of cue ball, v₄ = final speed of eight ball.

Since the cue ball and eight ball are deflected at an angle to the horizontal, we resolve their final velocity into its components in the x and y direction and then apply it to the law of conservation of momentum.

From the law of conservation of momentum,

In the x- direction

m₁v₁ + m₂v₂ = -m₁v₃cos40 + m₂v₄cos45 (the third quantity is negative since the cue ball moves 40° in the opposite direction)

0.17 × 3 = -0.17v₃cos40 + 0.16v₄cos45

0.51 = -0.130v₃ + 0.113v₄

-0.130v₃ + 0.113v₄ = 0.51 (1)

In the y - direction

0 = m₁v₃sin40 + m₂v₄sin45  (since our initial momentum in the y-direction is zero)

m₁v₃sin40 = -m₂v₄sin45

0.17v₃sin40 = -0.16v₄sin45

0.1093v₃ = -0.113v₄  

v₄ = -0.1093v₃/0.113  (2)

Substituting (2) into (1), we have

-0.130v₃ + 0.113 × -0.1093v₃/0.113  = 0.51

-0.130v₃ -0.1093v₃ = 0.51

-0.239v₃ = 0.51

v₃ = 0.51/-0.239 = -2.134 m/s ≅ -2.13 m/s

Substituting v₃ into (2), we have

v₄ = -0.1093(-2.134)/0.113 = 2.064 m/s ≅ 2.06 m/s

So, the velocity of the cue ball after collision is -2.13 m/s and the velocity of the eight ball after collision is 2.06 m/s