What is the maximum compression of the spring xmax during the collisio

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SUMMARY

The discussion centers on a physics problem involving a 4.80 kg cart with a spring constant of 3200 N/m colliding with a stationary 1.60 kg cart. The key equations used are the potential energy of the spring (PE_spring = 1/2*kx^2) and the kinetic energy of the carts (KE = 1/2*mv^2). The main conclusion is that momentum must be conserved during the collision, which means that the speed of both carts cannot be zero at maximum spring compression. Instead, the carts must have a shared velocity at that instant, necessitating the use of conservation of momentum to solve for their speeds.

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  • Understanding of conservation of momentum principles
  • Familiarity with kinetic and potential energy equations
  • Knowledge of spring mechanics and Hooke's Law
  • Ability to solve algebraic equations involving square roots
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Elleboys
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Homework Statement


A 4.80kg cart has a spring with spring constant 3200N/m attached to its front, parallel to the ground. This cart rolls at 4.40m/s toward a stationary cart with mass 1.60kg .


Homework Equations


PEspiring = 1/2*kx^2
KE = 1/2*mv^2


The Attempt at a Solution


I was setting up the equations to be 1/2*kx^2 = 1/2*mv^2
So I got x = sqrt(mv^2/k)
But as I put it into mastering physics, it said 'If all the kinetic energy is converted to spring potential energy, then the speed of both carts is zero and the momentum is zero. This means the momentum is not conserved. Think about the motion. When the spring is compressed the most, the carts are the closest together. How are their motions related at this point?'
I don't quite understand what it is talking about so please help
 
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Hello Elleboys.

The hint they are giving you is that momentum must be conserved as well as energy.

Can you use conservation of momentum to deduce the speed of the each cart at the instant the spring is compressed the most? (The second cart does not remain stationary - it is free to move.)
 
Elleboys said:

Homework Statement


A 4.80kg cart has a spring with spring constant 3200N/m attached to its front, parallel to the ground. This cart rolls at 4.40m/s toward a stationary cart with mass 1.60kg .


Homework Equations


PEspiring = 1/2*kx^2
KE = 1/2*mv^2


The Attempt at a Solution


I was setting up the equations to be 1/2*kx^2 = 1/2*mv^2
So I got x = sqrt(mv^2/k)
But as I put it into mastering physics, it said 'If all the kinetic energy is converted to spring potential energy, then the speed of both carts is zero and the momentum is zero. This means the momentum is not conserved. Think about the motion. When the spring is compressed the most, the carts are the closest together. How are their motions related at this point?'
I don't quite understand what it is talking about so please help

In any system with no external forces acting upon it, linear momentum is always conserved. So you need a conservation of momentum equation here.

What the hint is telling you is to consider what happens when the spring is at its maximum compression. This only happens for an instant (before this the spring is compressing, and after it, it's expanding). But during that instant, the spring can be modeled as a rigid link connecting the two carts. What happens when you set two carts connected by a rigid rod in motion? What's their velocity relative to each other?
 

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