Question

Hint: Use the Law of Conservation of Energy F) A 440 kg mine rolls at a speed of 0.5 m/'s on a horizontal track as shown. A 150 kg chunk of coal with a speed of 0.8 m/s leaves a chute and lands inside the mine car. 150 kg 25.0° 0.80 m/s 440 kg0.50 m/s a) Find the velocity of the coal-car system after the coal comes to rest in the car, (7 pts) b) Prove that the collision is inelastic using energy relationships. (3 pts)

Answer 1

By conversation of linear momentum in x direction

150 x 0.8 x cos(25) + 440 x 0.5 = {150+440} x v

So, v = 0.557 m/s

b) total energy before collision = 1/2 (m1v1² + m2v2²)

= 1/2 x (150 x 0.8²+ 440 x 0.5²)

= 206 J

Energy after collision = 1/2 (m1+m2) v²

= ½(150+440)x 0.557²

= 91.52 J

Since the energy is lost during collision it is an inelastic collision