Question

In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward at 2 m/s and the other cart rolls leftward at 1 m/s. After the carts collide, they couple (attach together) and roll together with a speed of _____. [Ignore resistive forces]

Answer 1

v = 0.5 m/s

Step-by-step explanation:

Theory :

. Conservation of momentum:

m1v1 + m2v2 = m1u1 + m2u2

. Where:

m1, m2 = masses of object colliding

v1, v2 = final velocity of collision

u1, u2 = initial velocity before collision

Facts :

. We have two equal-mass carts rolling towards each other => m1 = m2 = M

. Initial velocity of m1 = u1 = 2m/s [ rolling righward ]

. Initial velocity of m2 = u2 = -1m/s [ rolling leftward aka oposite direction ]

. Final velocity of m1 = v1 = v [same direction]

. Final velocity of m2 = v2 = v [same direction]

Calcul :

=> M * 2m/s + M(-1m/s) = Mv + Mv

= 1m/s = 2v (Dividing each term by M and adding the right side and substracting the left side terms)

= v = 0.5 m/s

Conclusion :

The speed of the carts after their collision is 0.5 m/s.