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alukav5142 [94]

2 years ago

8

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

1 answer:

Karolina [17]2 years ago

6 0

Answer:

Explanation:

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

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