Hel please urgent If a 0.16 kg 8-ball is hit by the 0.17 kg cue ball that is moving at a speed of 2 m/s, what is the speed of the 8-ball if the cue ball continues to move at 0.5 m/s in the same direction after the collision?

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, m₁ = 0.16 kg

The initial speed of the 8-ball, v₁ = 0 m/s

The final speed of the 8-ball = v₃

The mass of the cue ball, m₂ = 0.17 kg

The initial speed of the cue ball, v₂ = 2 m/s

The final speed of the cue ball, v₄ = 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 = m₁ × v₁ + m₂ × v₂ = The total final momentum = m₁ × v₃ + m₂ × v₄

Substituting the known values, gives;

0.16 × 0 + 0.17 × 2 = 0.16 × v₃ + 0.17 × 0.5

0.16 × v₃ = 0.16 × 0 + 0.17 × 2 - 0.17 × 0.5 = 0.255

∴ v₃ = 0.255/0.16 = 1.59375 ≈ 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 = v₃ ≈ 1.6 m/s

unbalanced force

explanation:

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