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
