Calculating Spring Constants and Predicting Speeds in Colliding Carts

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The discussion focuses on calculating the speed of Cart B and determining its spring constant after two equal mass carts, each with a spring, are released from a compressed state. It emphasizes that linear momentum conservation is key to finding Cart B's speed without needing to know the force involved. After calculating the kinetic energy of both carts, this energy must equal the total potential energy stored in the springs before release. The potential energy of Cart A's spring can be used to find the initial potential energy of Cart B's spring, which leads to calculating its spring constant. The conversation highlights the importance of understanding energy conservation principles in solving the problem.
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Cart A and B are equal mass ( 500 kg each) and have frictionless wheels. Cart A has a spring attached to its end with a constant 2000 N/m. Cart B has a spring constant attached to its end with a different spring constant. They both sit on a track with their spring facing each other, and are pushed together and held so that each of the springs are compressed 10 mm. When released, Cart A moves away with a speed of 0.9375m/s
a. Find the speed of Cart B after the two are released
b. Predict the spring constant of Cart B's spring ( friction should be ignored and use conservation of energy for this)

This is my assignment needs to be handed in last Fri. Even though I tried so hard, I couldn't get the right answer. I handed that in, but I think my answer for the spring constant of B was wrong. I didnt use 2 dimention aspect to solve this though. I'm really confused...Can someone help me please...Thank you :smile:
 
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luvlypenguin said:
Cart A and B are equal mass ( 500 kg each) and have frictionless wheels. Cart A has a spring attached to its end with a constant 2000 N/m. Cart B has a spring constant attached to its end with a different spring constant. They both sit on a track with their spring facing each other, and are pushed together and held so that each of the springs are compressed 10 mm. When released, Cart A moves away with a speed of 0.9375m/s
a. Find the speed of Cart B after the two are released
b. Predict the spring constant of Cart B's spring ( friction should be ignored and use conservation of energy for this)

This is my assignment needs to be handed in last Fri. Even though I tried so hard, I couldn't get the right answer. I handed that in, but I think my answer for the spring constant of B was wrong. I didnt use 2 dimention aspect to solve this though. I'm really confused...Can someone help me please...Thank you :smile:

For part a) you do not need to know anything about the force that pushes the carts apart. No matter how much that force is, or how it is created, it is a force between carts. Linear momentum will be conserved, and that is all you need to know to find out haw fast cart B moves. Once you have the answer for a), you can calculate the total kinetic energy of both carts. This has to equal the total potential energy of the two springs before the carts are released. You are given enough information to calculate the energy of one spring. Use that and the total knetic energy to find the initial potential energy of the second spring. From that you can calculate the spring constant.

See if you can do part a) and at least look up the equation for the potential energy of a spring. Then check back with what you have been able to do.
 
When pushed togather tension in the springs will be same. Commpression in each spring is same (10 mm). What do you think of the spring constant if the springs are in contect. If the springs are not pushing each other and the both are between each blocks then the energy method is applied.
 
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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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