Collision of Two Carts with Springs Attached

AI Thread Summary
The discussion revolves around a collision between two carts, one moving and one stationary, with springs that exert a constant repulsive force upon contact. The participant struggles with the lack of a spring constant and the implications for calculating velocities and displacements post-collision. They initially apply Newton's second law to estimate deceleration but find it challenging to determine spring compression and the second cart's velocity. They eventually consider the conservation of momentum, concluding that Cart A slows down while Cart B accelerates after the springs compress. The participant seeks validation for their calculations and assumptions regarding the system's behavior.
metal_maniac
Messages
4
Reaction score
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.
 
Physics news on Phys.org
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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top