Car 1 is pushed at a velocity of 5m/s

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Car "1" is pushed at a velocity of 5m/s

Car "1" is pushed at a velocity of 5m/s toward Car "2" (car 2 is not moving).
When they hit, Cart 1 bounces back a .5m/s and car 2 moves at 5.5m/s.
Is that possible?

Why or why not?

I think that it's not possible, but i'm not sure why. Maybe b/c energy was gained somehow for car 2 to bounce back faster, which is not possible.

How do i prove that?
 
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Answers and Replies

  • #2
chroot
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Conservation of linear momentum.

m1v1 + m2v2 = constant

- Warren
 
  • #3
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Originally posted by chroot
Conservation of linear momentum.

m1v1 + m2v2 = constant

- Warren
What is the constant?
 
  • #4
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The constant is total momentum. In other words, although v1 and v2 change after the collision, the above formula is equal to what it equalled using the velocities before the collision: total momentum is conserved.
 
  • #5
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so would it be (5*5)+(5*-0.5)=5*5.5+0, 22.5=27.5 ???
Is that the proof that it is not possible?
 
  • #6
chroot
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Originally posted by marshall4
What is the constant?
Whatever it was at the start of the problem is what it will always be. That's what 'constant' means.

- Warren
 
  • #7
chroot
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Originally posted by marshall4
so would it be (5*5)+(5*-0.5)=5*5.5+0, 22.5=27.5 ???
Is that the proof?
Not quite. Try this:

Initially, total momentum is (assuming the cars have the same mass):

m1v1 + m2v2 = m * 5 + m * 0 = 5m

In the end, if the cars did what the problem said, the total momentum would be

m1v1 + m2v2 = m * -0.5 + m * 5.5 = 5m

which is the same answer. The total momentum is the same before and after, so the result is plausible. The actual velocities that would result in an experiment of this sort would depend upon the "elasticity" of the collision, which is something you'll probably learn about later.

- Warren
 
  • #8
russ_watters
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That is possible if car 1 is heavier than car 2.
 
  • #9
chroot
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Actually, lol, you're right russ. Sure, this proposed collision conserves momentum, but it does not conserve energy (assuming the cars have the same mass).

*slaps forehead*

- Warren
 
  • #10
nautica
Momentum before must equal moment after

5. m/s equals 5.5 m/s + - 0.5 m/s

Nautica
 

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