Velocity of cart launched by spring

In summary, the conversation discusses the problem of finding the speeds of two carts pushed towards each other by a compressed spring. The solution involves using the equations for kinetic energy and spring force, and using conservation of momentum to solve for both cart speeds. The final step is to substitute one variable into the other equation and solve for the desired speed.
  • #1

TG3

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Homework Statement


A massless spring of spring constant 20 N/m is placed between two carts. Cart 1 has a mass M1 = 5 kg and Cart 2 has a mass M2 = 3 kg. The carts are pushed toward one another until the spring is compressed a distance 1.3 m. The carts are then released and the spring pushes them apart. After the carts are free of the spring, what are their speeds?

Homework Equations


K = 1/2 MV^2
F = -kX

The Attempt at a Solution


Once I find out how to solve for one I'll know how to solve for both, so let's just deal with the 5 kg cart.
Force in the Spring = 1.3 x 20 = 26
26 = 1/2 x (5) x V^2
10.4 = V^2
3.22 = V
Wrong. I suspect I made this much simpler than it really is... where did I go wrong?
 
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  • #2
Once I find out how to solve for one I'll know how to solve for both, so let's just deal with the 5 kg cart.
That statement is incorrect, you have to deal with both of them at the same time.

So think of them in terms of Energy,

1/2kx^2 = 1/2m1v1^2 + 1/2m2v2^2
 
  • #3
What is the x in 1/2 kx^2? Is that the distance the spring is stretched? That can't be right, but I'm not sure what else to put in there...
 
  • #4
yes, x is the distance a spring is compressed (from equilibrium)
 
  • #5
Okay, so:
1/2 x 20 x 1.3^2 = 1/2 (5) v1 ^2 + (3) v2 ^2
16.9 = 1/2 5 v1^2 + 1/2 3 v2^2
33.8 = 5 v1^2 + 3 v2^2

Also, 5v1 = 3v2 because of conservation of momentum.
So substituting 5/3 V1 in for V2 will hopefully result in being able to solve for V1.

33.8 = 5V1^2 + 3 (5/3V1)^2 right?
I don't see that I'm any closer to solving for V1...
 
  • #6
TG3 said:
Okay, so:
1/2 x 20 x 1.3^2 = 1/2 (5) v1 ^2 + (3) v2 ^2
16.9 = 1/2 5 v1^2 + 1/2 3 v2^2
33.8 = 5 v1^2 + 3 v2^2

Also, 5v1 = 3v2 because of conservation of momentum.
So substituting 5/3 V1 in for V2 will hopefully result in being able to solve for V1.

33.8 = 5V1^2 + 3 (5/3V1)^2 right?
I don't see that I'm any closer to solving for V1...

Right. You can substitute one variable into the other equation which will allow you to solve for the other variable. (substituting in v1 solves for v2 and vice versa). You have one equation only with V1, so just solve it for V1. It's really just algebra at this point, you've got the physics down pat.
 

1. What is the velocity of a cart launched by a spring?

The velocity of a cart launched by a spring depends on several factors such as the mass of the cart, the spring constant, and the amount of compression of the spring. It can be calculated using the equation v = √(2kx/m), where v is the velocity, k is the spring constant, x is the amount of compression, and m is the mass of the cart.

2. How does the mass of the cart affect the velocity?

The mass of the cart directly affects the velocity of a cart launched by a spring. The heavier the cart, the slower the velocity will be, and vice versa. This is because a heavier cart requires more force from the spring to accelerate it to a certain velocity.

3. What is the spring constant and how does it impact the velocity?

The spring constant is a measure of the stiffness of the spring and is represented by the letter k. It is a constant value that determines how much force is needed to compress or stretch the spring by a certain amount. A higher spring constant means that the spring is stiffer and will require more force to compress, resulting in a higher velocity of the cart.

4. How does the amount of compression of the spring affect the velocity?

The amount of compression of the spring also directly affects the velocity of the cart. The more the spring is compressed, the more potential energy it stores. When the spring is released, this potential energy is converted into kinetic energy, resulting in a higher velocity of the cart.

5. Can the velocity of a cart launched by a spring be increased by using a stronger spring?

Yes, using a stronger spring with a higher spring constant can result in a higher velocity of the cart. This is because a stronger spring can provide more force to accelerate the cart, resulting in a higher velocity. However, it is important to consider other factors such as the mass of the cart and the amount of compression of the spring as well.

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