A 2.4-kg cart is rolling along a frictionless, horizontal track towards 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 +3.5 m/s, and the second cart's velocity is -1.3 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?
To solve this problem, we need to use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after the interaction between its components (in this case, the carts).
(a) To find the total momentum of the system at the given instant, we can use the formula for momentum:
momentum = mass × velocity
The momentum of the first cart is given by:
momentum1 = mass1 × velocity1
momentum1 = 2.4 kg × 3.5 m/s
The momentum of the second cart is given by:
momentum2 = mass2 × velocity2
momentum2 = 1.3 kg × (-1.3 m/s)
To find the total momentum of the system, we can simply add the individual momenta:
total momentum = momentum1 + momentum2
total momentum = (2.4 kg × 3.5 m/s) + (1.3 kg × (-1.3 m/s))
total momentum ≈ 8.4 kg*m/s
Therefore, the total momentum of the system at this instant is approximately 8.4 kg*m/s.
(b) To find the velocity of the first cart when the second cart was still at rest, we can apply the conservation of momentum. Since the total momentum of an isolated system is conserved, we can set up the equation:
(mass1 × velocity1) + (mass2 × velocity2) = (mass1 × velocity1') + (mass2 × velocity2')
Where:
mass1 = 2.4 kg (mass of the first cart)
velocity1 = 3.5 m/s (velocity of the first cart at the given instant)
mass2 = 1.3 kg (mass of the second cart)
velocity2 = -1.3 m/s (velocity of the second cart at the given instant)
velocity1' = ? (velocity of the first cart when the second cart was still at rest)
velocity2' = 0 m/s (velocity of the second cart when it was still at rest)
Substituting the given values and variables into the equation, we get:
(2.4 kg × 3.5 m/s) + (1.3 kg × (-1.3 m/s)) = (2.4 kg × velocity1') + (1.3 kg × 0 m/s)
Simplifying and solving for velocity1', we find:
(2.4 kg × 3.5 m/s) + (1.3 kg × (-1.3 m/s)) = (2.4 kg × velocity1') + 0 kg*m/s
(8.4 kg*m/s) - (1.69 kg*m/s) = (2.4 kg × velocity1')
6.71 kg*m/s = (2.4 kg × velocity1')
velocity1' = 6.71 kg*m/s / 2.4 kg
velocity1' ≈ 2.796 m/s
Therefore, the velocity of the first cart when the second cart was at rest is approximately 2.796 m/s.