What Are the Final Velocities in a Perfectly Elastic Collision Problem?

AI Thread Summary
In a perfectly elastic collision problem involving two carts, cart1 (1 kg, initial velocity 2 m/s) collides with cart2 (2 kg, initial velocity 0 m/s). The conservation of momentum and kinetic energy equations are used to find the final velocities of both carts after the collision. The user has derived two equations: 2 = V1f + 2V2f and 4 = V1f^2 + 2V2f^2, but is struggling with the substitution method to solve for the final velocities. A suggestion is made to isolate one variable in the first equation and substitute it into the second to solve the resulting quadratic equation. This approach is expected to yield the correct final velocities for both carts.
southernbelle
Messages
35
Reaction score
0

Homework Statement


In an isolated system, cart1 (with mass = 1 kg and vi1 = 2) has a perfectly elastic collision with cart2 (with mass = 2 kg and vi2 = 0). Find the velocity of cart1 and the velocity of cart2 after the collision.
I have to solve this using kinetic energy and momentum equations.


Homework Equations


m1vi1 + m2vi2 = m1v1f + m2v2f
1/2m1vi12 + 1/2m2vi22 = 1/2m1v1f2 + 1/2m2v2f2


The Attempt at a Solution


I have gotten to this point:
2= V1F + 2V2F
4= V1F2 + 2V2F

but I cannot get the numbers to work out correctly.
Using another equation I know that V1f = 2/3 and V2f = 4/3

I am doing substitution wrong or something.
Please help!
 
Physics news on Phys.org
southernbelle said:

The Attempt at a Solution


I have gotten to this point:
2= V1F + 2V2F
This makes sense.
4= V1F2 + 2V2F
How did you get this?
 
I got that by solving the kinetic energy equation.
I mutiplied both sides by 2 to get rid of the halves.

It is actually supposed to read:

4 = Vif^2 + 2V2f^2
 
southernbelle said:
It is actually supposed to read:

4 = Vif^2 + 2V2f^2
That's good.

Take the first equation and solve for one of your variables. Then substitute that into the second equation. Solve the quadratic.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How to Calculate Speed of Head of Driver After Collision?
Replies
10
Views
2K
Question about two elastic collision formulas
Replies
6
Views
2K
What was the mistake in solving for the speeds of two marbles after a collision?
Replies
5
Views
1K
What are the velocities of the carts after a perfectly elastic collision?
Replies
2
Views
2K
Solve Elastic Collision: 2D Homework
Replies
4
Views
2K
How Do You Solve for Final Velocities in Elastic Collisions?
Replies
20
Views
2K
Question about the solution to this elastic collision
Replies
4
Views
1K
Finding two objects velocity after an elastic collision
Replies
10
Views
4K
Elastic Collision and Momentum of Ice Skaters
Replies
16
Views
3K
Elastic head-on collision problem
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top