Conservation of Energy of cart Problem

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A 1.2 kg cart slides down a frictionless ramp, converting gravitational potential energy into kinetic energy, reaching a speed of 5.94 m/s before colliding with a stationary 2.0 kg cart cushioned by a spring. The maximum compression of the spring during the elastic collision is 2.0 cm. To find the spring constant (k), the conservation of energy principle can be applied, using the equation 1/2mv^2 = 1/2kx^2. Additionally, considering the center of mass reference frame can help analyze the collision, as momentum is conserved and the velocities of both carts will be identical at maximum compression. The calculated spring constant is 6.6 x 10^4 N/m.
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A 1.2 kg cart slides eastward down a frictionless ramp from a height of 1.8 m and then onto a horizontal surface where it has a head-on elastic collision with a stationary 2.0 kg cart cushioned by an ideal Hooke’s law spring. The maximum compression of the spring during the collision is 2.0 cm. (5.4)

  1. (a) Determine the spring constant.

    There were more parts to this question, however, I just need help with this portion of it.

    Law of Conservation of Energy:

    Et1 = Et2

    Force of Spring formula: F= kx

    Kinetic Energy, Gravitational Potential Energy, Elastic Potential Energy formulas will come in handy too.

    So in terms of solving this, I though I should find the final speed of the 1.2 kg cart which will end up being the initial velocity right before the collision takes place. I got 5.94 m/s for that (mgh = 1/2mv^2).

    Using this speed, how should i incorporate it to get the "k" value? I could use conservation of energy for the second block... (1/2mv^2 = 1/2kx^2)

    But I would need the speed for block 2, which is tricky because it was stationary initially..

    The answer for the k value is 6.6x10^4 N/m.

    Let me know what you guys suggest!
 
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I just realized this should have been posted in the homework thread, if there is a moderator here, please feel free to move it to the appropriate homework thread for Physics.

Thanks!
 
Here's an idea: Consider using the centre of mass reference frame to study the collision.
 
Momentum is conserved and their velocities will be identical when the spring is fully compressed. That should give you another equation to work with. You could consider it as a totally inelastic collision up until the spring starts to return energy.
 
sophiecentaur said:
Momentum is conserved and their velocities will be identical when the spring is fully compressed. That should give you another equation to work with. You could consider it as a totally inelastic collision up until the spring starts to return energy.
That's giving quite a lot away!
 
Yes. I realize that but I didn't write any equations out for him / her.

PS I am not mean enough to be a H/H. :wink:
 
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