Question

A 13.0 kg toy train car moving at 3.40 m/s on a straight level train track, collides head-on with a second train car whose mass is 38.0 kg and was at rest on the track. If the collision is perfectly elastic and all motion is frictionless, calculate the velocities of the two cars after the​

Answer 1

Before the collision, the total momentum of the two train cars is

(13.0 kg) (3.40 m/s) + 0 = 44.2 kg•m/s

After the collision, the two cars have velocity v₁ and v₂ so that their total momentum is

(13.0 kg) v₁ + (38.0 kg) v₂

Momentum is conserved, so

(13.0 kg) v₁ + (38.0 kg) v₂ = 44.2 kg•m/s

Before collision, the total kinetic energy of the two cars is

1/2 (13.0 kg) (3.40 m/s)² + 0 = 75.14 J

and after collision,

1/2 (13.0 kg) v₁² + 1/2 (38.0 kg) v₂²

Energy is also conserved, so

1/2 (13.0 kg) v₁² + 1/2 (38.0 kg) v₂² = 75.14 J

or

(13.0 kg) v₁² + (38.0 kg) v₂² = 150.28 J

Solve the momentum equation for either v₁ or v₂ :

(13.0 kg) v₁ + (38.0 kg) v₂ = 44.2 kg•m/s

⇒ v₂ = (44.2 kg•m/s - (13.0 kg) v₁) / (38.0 kg)

Substitute this into the energy equation and solve for the other variable :

(13.0 kg) v₁² + (38.0 kg) [(44.2 kg•m/s - (13.0 kg) v₁) / (38.0 kg)]² = 150.28 J

⇒ (17.4 kg) v₁² - (30.2 kg•m/s) v₁ + 51.4 J ≈ 150.28 J

⇒ (17.4 kg) v₁² - (30.2 kg•m/s) v₁ - 98.9 J ≈ 0

Using the quadratic formula,

⇒ v₁ = 3.40 m/s or v₁ ≈ -1.67 m/s

(We ignore the second solution because that's the initial condition.)

Solve for the remaining velocity:

v₂ = (44.2 kg•m/s - (13.0 kg) (-1.67 m/s)) / (38.0 kg)

⇒ v₂ ≈ 1.73 m/s