Answer:
The final speed of the 8-ball if the cue ball continues to move at 0.5 m/s in the same direction after collision is approximately 1.6 m/s
Explanation:
The given values in the question are;
The mass of the given 8-ball, mβ = 0.16 kg
The initial speed of the 8-ball, vβ = 0 m/s
The final speed of the 8-ball = vβ
The mass of the cue ball, mβ = 0.17 kg
The initial speed of the cue ball, vβ = 2 m/s
The final speed of the cue ball, vβ = 0.5 m/s
By the law of conservation of linear momentum, we have that the total momentum in an isolated system is constant (always)
Therefore;
The total initial momentum = The total final momentum
The total initial momentum = mβ Γ vβ + mβ Γ vβ = The total final momentum = mβ Γ vβ + mβ Γ vβ
Substituting the known values, gives;
0.16 Γ 0 + 0.17 Γ 2 = 0.16 Γ vβ + 0.17 Γ 0.5
0.16 Γ vβ = 0.16 Γ 0 + 0.17 Γ 2 - 0.17 Γ 0.5 = 0.255
β΄ vβ = 0.255/0.16 = 1.59375 β 1.6
The final speed of the 8-ball, given that that cue ball continues to move at 0.5 m/s in the same direction = vβ β 1.6 m/s