Sally and Bob are riding in bumper cars! Sally is moving with a velocity of 3.2 m/s to the left when she hits and bounces off Bob who is moving with a velocity of 4.5 m/s to the right. After the collision, Sally is moving to the right with a velocity of 3.9 m/s. The total mass of Sally and her cart is 120 kg, and the total mass of Bob and his cart is 145 kg. What is Bob’s velocity after the collision? m/s (nearest hundredth)

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.