Question

An 8.32 kg mass moving east at 14.8 m/s on a frictionless horizontal surface collides with a 11.3 kg mass that is initially at rest. After the collision, the 8.32 kg mass moves south at 2.73 m/s. What is the magnitude of the velocity of the 11.3 kg mass after the collision

Answer 1

Use conservation of momentum.

Let p₁ and p₂ denote the initial momenta of the 8.32-kg and 11.3-kg mass, respectively. Take East and North to be the positive horizontal and vertical directions. Then

p₁ = (8.32 kg) (14.8 m/s) i ≈ (123 kg•m/s) i

p₂ = (11.3 kg) (0 m/s) = 0

Let p₁' and p₂' denote the momenta after collision. Then

p₁' = (8.32 kg) (-2.73 m/s) j ≈ (-22.7 kg•m/s) j

p₂' = (11.3 kg) (v₁ i + v₂ j)

where v₁ and v₂ are the components of the 11.3-kg mass's final velocity vector.

By conservation of momentum, we have

p₁ + p₂ = p₁' + p₂'

(123 kg•m/s) i = (-22.7 kg•m/s) j + (11.3 kg) (v₁ i + v₂ j)

(123 kg•m/s) i = (11.3 kg) v₁ i + ((11.3 kg) v₂ - 22.7 kg•m/s) j

Solve for v₁ and v₂ :

123 kg•m/s = (11.3 kg) v₁ ⇒ v₁ ≈ 10.9 m/s

0 = ((11.3 kg) v₂ - 22.7 kg•m/s) j ⇒ v₂ ≈ 2.01 m/s

Then the magnitude of the 11.3-kg mass's velocity after the collision is

√(v₁² + v₂²) ≈ 11.1 m/s