Question

A 2.8-kg cart is rolling along a frictionless, horizontal track towards a 1.2-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 -2.7 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

The total momentum of the system of the two carts is 9.64 kg·m/s. The initial velocity of the first cart, when the second cart was still at rest, was also 4.6 m/s, as the total momentum of the system is conserved.

Step-by-step explanation:

To solve this problem, we will use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces are acting on it. Since the track is frictionless, we can assume no external forces act on the carts.

Part (a): Total Momentum of the System

The total momentum can be calculated as follows:

ptotal = pcart1 + pcart2

ptotal = (m1 × v1) + (m2 × v2)

ptotal =
`(2.8 kg × 4.6 m/s) + (1.2 kg × (-2.7 m/s))`

ptotal = 12.88 kg·m/s - 3.24 kg·m/s

ptotal = 9.64 kg·m/s

Part (b): Initial Velocity of the First Cart

Using the conservation of momentum before the second cart starts moving:

m1v1i + m2v2i = m1v1f + m2v2f

Since the second cart is initially at rest, v2i = 0, and:

v1i = (m1v1f + m2v2f)/m1

v1i =
`(12.88 kg·m/s) / 2.8 kg`

v1i = 4.6 m/s

Therefore, the velocity of the first cart when the second cart was still at rest was 4.6 m/s, the same as its velocity at the later time described in the problem because the total momentum of the system has not changed.

Answer 2

Answer with Explanation:

We are given that

Mass of one cart,
`m_1=2.8 kg`

Mass of second cart,
`m_2=1.2 kg`

Initial velocity of one cart,
`u_1=4.6m/s`

Initial velocity of second cart,
`u_2=-2.7 m/s`

a.Total momentum,
`P=m_1u_1+m_2u_2=2.8(4.6)+1.2(-2.7)`

`P=9.64 kgm/s`

b.Velocity of second cart,
`v_2`=0

According to law of conservation of momentum

Initial momentum=Final momentum

`9.64=2.8v_1+1.2* 0`

`v_1=(9.64)/(2.8)`

`v_1=3.44m/s`