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NISA [10]
3 years ago
8

A 2.3 kg cart is rolling across a frictionless, horizontal track toward a 1.5 kg cart that is held initially at rest. The carts

are loaded with strong magnets that cause them to attract to one another; thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is +4.5 m/s and the second cart's ve;ocity is -1.9 m/s. What is the total momentum of the system of the two carts at this instant? What was the velocity of the forst cart when the second cart was still at rest?
Physics
1 answer:
4vir4ik [10]3 years ago
5 0

Answer:

Part a)

P = 7.5 kg m/s

Part b)

v = 3.26 m/s

Explanation:

As we know that two carts are moving in opposite directions

so here total momentum of two carts must be given by vector subtraction

It is given as

P = m_1\vec v_1 + m_2\vec v_2

P = (2.3)(4.5) + (1.5)(-1.9)

P = 7.5 kg m/s

Now we know that total momentum of the system is conserved here as there is no external force on this system

so the speed of the first cart when second cart was at rest is given as

P = m_1 v_1 + 0

7.5 = 2.3 v

v = 3.26 m/s

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Answer:

a) α = 0.338 rad / s²  b)   θ = 21.9 rev

Explanation:

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b) let's use the relationship of rotational kinematics

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Let's reduce the angular velocity

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2 years ago
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the mirror equation is given by:

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Answer:

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