Answer:
Bob's velocity after the collision is approximately 1.38 m/s to the left
Explanation:
The given parameters are;
The velocity with which Sally is moving to the left, v₁ = 3.2 m/s
The total mass of Sally an her cart, m₁ = 120 kg
The velocity with which Bob is moving to the right, v₂ = 4.5 m/s
The total mass of Bob and his cart, m₂ = 145 kg
The final velocity with which Sally is moving to the right after collision, v₃ = 3.9 m/s
Let v₄ represent Bob's velocity after the collision and taking the convention that motion to the right is positive, we have;
The total initial momentum = m₁·v₁ + m₂·v₂ = 120×(-3.2) + 145×4.5
The total final momentum = m₁·v₃ + m₂·v₄ = 120×3.9 + 145×v₄
By the principle of conservation of linear momentum, we have;
The total initial momentum = The total final momentum
∴ m₁·v₁ + m₂·v₂ = m₁·v₃ + m₂·v₄
From which we have;
120×(-3.2) + 145×4.5 = 120×3.9 + 145×v₄
268.5 = 120×3.9 + 145×v₄
145×v₄ = 268.5 - 120×3.9 = -199.5
∴ v₄ = -199.5/145 ≈ -1.38
Bob's velocity after the collision = v₄ ≈ -1.38 m/s which gives;
Bob's velocity after the collision ≈ 1.38 m/s to the left.
