Collision of Two Carts with Springs Attached

In summary, the collision between the two carts results in both carts gaining momentum (4800 kg x m/s) and moving at a speed of 48 m/s.
  • #1
metal_maniac
4
0

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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  • #2
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.
 
  • #3
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
 

1. What is the Collision of Two Carts with Springs Attached?

The Collision of Two Carts with Springs Attached is a physics experiment that involves two carts connected by springs, which are used to simulate collisions between objects. The experiment is used to study the conservation of momentum and energy during collisions.

2. How does the experiment work?

In this experiment, two carts are connected by springs and placed on a track. One cart is given an initial velocity and collides with the stationary cart, causing the springs to compress. The carts then rebound and continue to oscillate until the energy is dissipated.

3. What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the principles of conservation of momentum and energy during collisions. It also allows for the analysis of elastic and inelastic collisions and the calculation of various physical quantities such as velocity, momentum, and kinetic energy.

4. What are the key components of this experiment?

The key components of this experiment include two carts of equal mass, springs, a track, and a motion sensor. The carts should be equipped with sensors to measure velocity, and the motion sensor is used to collect data for analysis.

5. What are some real-life applications of this experiment?

This experiment has real-life applications in the fields of engineering, physics, and vehicle safety. It can be used to study and improve the design of car airbags, crumple zones, and other safety features. It is also used in the study of collisions between particles in nuclear physics and astrophysics.

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