Collision of Two Carts with Springs Attached

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SUMMARY

The discussion centers on the collision of two carts, Cart A (300 Kg, 24 m/s) and Cart B (100 Kg, stationary), with springs exerting a constant repulsive force of 600 N upon contact. The key principle to solve for the final velocities post-collision is the conservation of momentum, which indicates that Cart A will transfer its momentum to Cart B after the springs compress. The user calculated that after two seconds, Cart A's velocity decreases to 16 m/s, while Cart B gains a velocity of 48 m/s, assuming complete momentum transfer. The solution is based on the assumption of elastic collision and internal forces from the springs.

PREREQUISITES
  • Understanding of conservation of momentum in elastic collisions
  • Basic knowledge of Newton's second law (F=ma)
  • Familiarity with Hooke's law and spring mechanics
  • Ability to analyze systems with internal and external forces
NEXT STEPS
  • Study the principles of elastic and inelastic collisions in physics
  • Learn about the calculation of spring constants and their role in collision dynamics
  • Explore advanced topics in momentum transfer during collisions
  • Investigate the effects of varying mass and velocity on collision outcomes
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Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for practical examples of momentum conservation in action.

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Homework Statement


Two carts collide.
Cart A has a mass of 300 Kg and a velocity of 24 m/s.
Cart B has a mass of 100 Kg and is stationary.
Both carts have 20 m springs on them. Assume that the two carts exert no force on each other until the springs touch at a separation of d= 40 m, and then exerts a repulsive force of 600 N on the other.


Homework Equations


I'm supposed to find the velocities, momentums and displacements of the carts after the collision but the problem is, there is no spring constant.
My understanding of Hooke's law leads me to believe that as springs compress, their repulsive force increases, so why are the springs repulsive forces constant, it just doesn't make any sense to me what so ever.


The Attempt at a Solution


I attempted to solve it by assuming that the cart will slow down at a rate of 4 m/s^2 because F= ma and since F is 1200 and m is 300, a must be 4.
Problem is, I can't calculate the amount of compression of the spring because there is no spring constant and I don't know the velocities of the other cart... It's just so confusing.
 
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Welcome to PF,

Advice: forget about the springs for a second. Can you think of a much more *general* physical principle that applies in this situation? One that would allow you to solve for the final velocities in terms of the initial ones?

Hint: if you consider your "system" to be the two carts, then the spring forces are all internal forces in this system. There are *no* external forces acting on the system.
 
Thank you.
Well I can think of conservation of momentum in elastic collisions which means that the first object will stop completely and the second object will gain all the momentum and move at a speed of 72 m/s.
I already attempted to solve it this way and assumed that the spring is completely compressed. What I got was that the velocity of Cart A one second after the springs touch is 20 m/s and after 2 seconds it's 16 m/s.
After two second the compression is zero so the remaining momentum (4800 kg x m/s) is transferred entirely to Cart B giving it an initial velocity of 48 m/s and an acceleration of 12 m/s^2 for 40 m.
Am I safe to assume that this solution is correct or do I need to revise it.
Thank you
 

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