Calculating Kinetic Energy for Colliding Carts

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Homework Help Overview

The discussion revolves around a problem in dynamics involving two colliding carts, one of which is in motion while the other is stationary. The focus is on calculating the velocities of the carts at minimum separation and analyzing the change in kinetic energy during the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the conservation of momentum and the relationship between the center of mass and the velocities of the carts. Some participants attempt to derive the velocities using momentum equations, while others express confusion regarding the kinetic energy calculations and the nature of the collision.

Discussion Status

There are various lines of reasoning being explored, with some participants providing insights into momentum conservation and kinetic energy calculations. However, there is no explicit consensus on the correct approach to the kinetic energy aspect, and some participants are still seeking clarity on their calculations.

Contextual Notes

Participants note the assumption of a completely elastic collision and the implications this has on the kinetic energy calculations. There is also mention of specific values and results that participants are trying to reconcile with their calculations.

crimsondarkn
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A 2.4 kg dynamics cart with a linear elastic spring attached to its front end is moving at 1.5 m/s [W] when it collides head on with a stationary 3.6 kg cart.

What is the velocity of each cart at minimum separation?





The answer is 0.6 m/s [W]

Can anyone show me the steps?
 
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crimsondarkn said:
A 2.4 kg dynamics cart with a linear elastic spring attached to its front end is moving at 1.5 m/s [W] when it collides head on with a stationary 3.6 kg cart.

What is the velocity of each cart at minimum separation?The answer is 0.6 m/s [W]

Can anyone show me the steps?
There is a very easy way to do this. Since there are no external forces, the total momentum the system does not change. What is the speed of the centre of mass? (What is the speed of the observer who sees the two cars colliding with equal and opposite momenta?)

What is the relationship between the speed of the centre of mass and the speeds relative to the centre of mass of the two cars at minimum separation?

AM
 
Oh I got it, thanks!

This is what I did..

m1v1+m2v2=m1v'1+m2v'2

m2v2 becomes zero since its stationary

m1v1=v'(m1+m2)

-3.6=v'(6)

v'=-3.6/6

v'= -0.6 m/s ---> 0.6m/s[W]

I didn't know it was completely elastic collision, that's why I was stuck.
 
I'm stuck on part c now...

Calculus the change in total kinetic energy of the system at minimum separation. Here's my work

v'=-0.6m/s

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2+1/2m2v2'^2

cross out all the 1/2s and also v2 becomes zero

m1v1^2 = m1v1'^2 + m2v2'^2

(2.4)(-1.5)^2 = (2.4)(0.6)^2 + (3.6)(0.6)^2

5.4 = 0.864 + 1.296

5.4 = 2.16

5.4-2.16 = 3.24 J

And this is where I'm stuck...


The answer at the back is -1.6 J
 
crimsondarkn said:
1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2+1/2m2v2'^2

cross out all the 1/2s and also v2 becomes zero

m1v1^2 = m1v1'^2 + m2v2'^2

Don't write this as an equation (since they are not equal!) and don't cross anything out. You have the speeds and the masses, so just calculate the initial and final KEs by plugging into the KE formula. Then compare KEi to KEf. (By "final" I mean at the point of minimum separation.)
 

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