Question

A mine car (mass = 440 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 150-kg chunk of coal has a speed of 0.80 m/s when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car.

Answer 1

Mass of the mine car = m₁ = 440 kg

Mass of the chunk of coal = m₂ = 150 kg

Initial velocity of the mine car = V₁ = 0.5 m/s

Initial velocity of the chunk of coal = V₂ = 0.8 m/s

Angle of velocity of the chunk of coal = $\theta$ = 25^o

Initial velocity of the chunk of coal in the horizontal direction = V_2x = V₂Cos$\theta$ = (0.8)Cos(25) = 0.725 m/s

Final velocity of the car-coal system = V₃

The final velocity will be in the horizontal direction.

Conserving the momentum in the horizontal direction.

m₁V₁ + m₂V_2x = (m₁ + m₂)V₃

(440)(0.5) + (150)(0.725) = (440 + 150)V₃

V₃ = 0.557 m/s

Speed of the car-coal system after the coal has come to rest in the car = 0.557 m/s