A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is + 4.6 m/s, and the second cart's velocity is - 1.9 m/s.

(Indicate the direction with the sign of your answer.)

(a) What is the total momentum of the system of the two carts at this instant?

kg · m/s

(b) What was the velocity of the first cart when the second cart was still at rest?

m/s

Question

Alisa Bauer

Physics

26 May, 15:57

A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is + 4.6 m/s, and the second cart's velocity is - 1.9 m/s.

(Indicate the direction with the sign of your answer.)

(a) What is the total momentum of the system of the two carts at this instant?

kg · m/s

(b) What was the velocity of the first cart when the second cart was still at rest?

m/s

+2

Answers (1)

Know the Answer?

Doofus · Answer 1

a) Momentum of first cart = mass x velocity

= 3 x 4.6 = + 13.8 kg m / s

Momentum of second cart = 1.3 x - 1.9 = - 2.47 kg m / s

Total momentum = 13.8 - 2.47

= + 11.33 kg m / s

b)

Let the velocity of first cart be v at the moment when second cart was at rest

total momentum = 3 x v + 0 = 3 v

Applying conservation of momentum law

3 v = + 11.33

v = + 3.77 m / s