Solving Cart-Cart Collision: Velocities & Momentum

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The discussion focuses on solving a cart collision problem involving two carts with different masses and velocities. The first cart (140 kg) moves at 24 m/s and rebounds at 7 m/s after colliding with a stationary 300 kg cart. Participants initially attempted to use kinetic energy formulas but realized that conservation of momentum is the correct approach to find the final velocity of the second cart. The final velocity of the 300 kg cart after the collision was calculated to be approximately 14.46 m/s. The conversation highlights the importance of using momentum conservation principles in collision problems.
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A cart (m1 = 140 kg) is moving to the right along a track at v1i = 24 m/s when it hits a stationary cart (m2 = 300 kg) and rebounds with a speed of v1f = 7 m/s in the opposite direction.

a) With what speed does the 300 kg cart move after the collision?

An observer moves in the same direction as the incoming cart with a speed of 10 m/s.

Using the convention that the positive direction is to the right, what are the following velocities with respect to this observer:
b) v1i,ob =
c) v2i,ob =
d) v1f,ob =
e) v2f,ob =

f) What is the total momentum of the system before the collision as seen by this moving observer?

g) What is the total momentum of the system after the collision as seen by this same observer?

So I thought that I could use 1/2mv^2 to get the momentum of cart 1 before impact, and I got 1/2(140)(24^2)= 40320. And then use 40320=300(v) to solve for be, but that didn't work. Can anyone tell me what I should be doing?




 
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r34racer01 said:
So I thought that I could use 1/2mv^2 to get the momentum of cart 1 before impact, and I got 1/2(140)(24^2)= 40320. And then use 40320=300(v) to solve for be, but that didn't work. Can anyone tell me what I should be doing?

The 140kg cart is moving also after the collision.
 
cristo said:
The 140kg cart is moving also after the collision.

Yes its moving 7m/s after the collision, but I don't get how I can use that to get the 300kg's speed after collision.
 
Never mind I realized we just nee to use conservation of momentum, so you just do (24*140)-(-7*140)/300 = 14.46 = V2f. And I was able to solve for everything else so thanks.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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