Question

A 2.9-kg cart is rolling along a frictionless, horizontal track towards a 1.4-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 +3.9 m/s, and the second cart's velocity is -1.2 m/s. (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?

Answer 1

9.63 kg m/s

3.32 m/s

Step-by-step explanation:

`m_1` = Mass of first cart = 2.9 kg

`m_2` = Mass of second cart = 1.4 kg

`v_1` = Velocity of first cart = 3.9 m/s

`v_2` = Velocity of second cart = -1.2 m/s

Total momentum of the system

`p=m_1v_1+m_2v_2\\\Rightarrow p=2.9* 3.9+1.4* -1.2\\\Rightarrow p=9.63\ kg m/s`

The total momentum of the system is 9.63 kg m/s

Equating the same equation with
`v_2=0`

`p=m_1v_1+m_2v_2\\\Rightarrow v_1=(P-m_2v_2)/(m_1)\\\Rightarrow v_1=(9.63-1.4* 0)/(2.9)\\\Rightarrow v_1=3.32\ m/s`

The velocity of the first cart when the second cart was still at rest is 3.32 m/s