A 500 kg cart is rolling to the right at 1.3 m/s. a 60 kg man is standing on the right end of the cart. what is the speed of the cart if tha man suddenly starts running to the left with a speed 10.0 m/s relative to the cart

Answers

Answer 1

Answer:

P1 = 1.3 (500 + 60) = 728 kg-m      total momentum to right at start

P2 = (v2 - 10) 60 + 500 v2

total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart

728 = 560 v2 - 600

v2 = 1328 / 560 = 2.37 m/s    new speed of cart

Check:

After:    p2 for cart = 500 * 2.37 = 1186

p1 for man = (2.37 - 10) * 60 = -458

P2 = p1 + p2 = 728       total momentum unchanged


Related Questions

What is First Aid.



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Answers

Answer:

First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.

Explanation:

A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.

Answers

Answer:

  1.5 m/s

Explanation:

Conservation of momentum means the momentum of the system before the collision is the same as after.

The before, after momentum of each ball is ...

  5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)

  10 kg ball: (10 kg)(0 m/s), (10 kg)(v)

The sum of the "before" products is the same as the sum of the "after" products:

  (5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v

  (10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides

  v = (15 kg·m/s)/(10 kg) = 1.5 m/s

The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.

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