What Are the Velocities and Forces in This Elastic Collision?

AI Thread Summary
In an elastic collision involving a 0.24kg object moving at 3.5m/s and a 0.40kg object moving at 2.1m/s, the velocities after the collision were calculated using conservation of momentum and energy principles. The initial momentum was determined to be 1.68 kg*m/s, leading to final velocities of -1.75m/s for the first object and 3.15m/s for the second. There was confusion regarding the derivation of the equation "-v2i + v1i = -v1f + v2f" and the signs of the final velocities. Participants sought clarification on how to calculate the average force exerted during the 0.010s contact time. The discussion emphasized the importance of correctly applying momentum and energy conservation in solving such problems.
chrisfnet
Messages
38
Reaction score
0

Homework Statement



A 0.24kg object traveling rightward at 3.5m/s collides head-on and elastically with a 0.40kg object traveling leftward at 2.1m/s.

What are their velocities after the collision?

If the objects are in contact for 0.010s during the collision, what is the magnitude of the average force that they exert on each other?

Homework Equations



-v2i + v1i = -v1f + v2f

The Attempt at a Solution



Initial Momentum = 0.24kg(3.50m/s) + 0.40kg(2.1m/s)
Initial Momentum = 1.68 kg*m/s

1.68kg*m/s = 0.24kg * v1f + 0.40kg*v2f

* Now I find a substitution. *

-v2i + v1i = -v1f + v2f

-v2i + v1i = -2.1m/s + 3.5m/s = 1.40m/s
1.40m/s = -v1f + v2f
v2f = v1f + 1.40m/s

* Now I substitute it back into the above equation. *

1.68kg*m/s = 0.24kg*v1f + 0.40kg*(v1f + 1.40m/s)
1.68kg*m/s = 0.24kg*v1f + 0.40kg*v1f + 0.56kg*m/s
1.12kg*m/s = 0.64kg*v1f
-1.75m/s = v1f (Negative to indicate direction.)

* Now I substitute v1f into the original equation. *

1.68kg*m/s = 0.24kg * v1f + 0.40kg*v2f
1.68kg*m/s = 0.24kg * 1.75m/s + 0.40kg*v2f
v2f = 3.15m/s

I have no idea how to figure out the force exerted on each other. Could someone help here, and also verify that the above answers are correct? Thanks!
 
Physics news on Phys.org
To find the velocities of the objects after collision, you have to use the conservation of momentum and energy.
 
rl.bhat said:
To find the velocities of the objects after collision, you have to use the conservation of momentum and energy.

How would I integrate the conservation of energy?
 
I am trying to solve a similar problem and I have a question.

How did you get this one: "-v2i + v1i = -v1f + v2f" ?

how is this derived? How did ou determine "-v1f + v2f" where v1f is negative if mass is not part of the equation?
 
I am trying to solve a similar problem and I have a question.

How did you get this one: "-v2i + v1i = -v1f + v2f" ?

how is this derived? How did ou determine "-v1f + v2f" where v1f is negative if mass is not part of the equation?

Besides, if v1f is negative, then shouldn't v1f in this equation be negative as well?
"1.68kg*m/s = 0.24kg * v1f + 0.40kg*v2f"

If so, your answer will come out different.
 

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
Elastic head-on collision problem
Replies
8
Views
2K
Two dimensional Elastic Collision
Replies
2
Views
1K
Solve Elastic Collision: 2D Homework
Replies
4
Views
2K
Glancing elastic collision, and center of mass
Replies
17
Views
4K
Conservation of momentum - with understanding
Replies
2
Views
2K
What Happens to Sphere Velocities After an Elastic Collision?
Replies
1
Views
4K
Zero momentum reference frame and an inelastic collision
Replies
4
Views
3K
Elastic glancing collision problem
Replies
4
Views
3K
Elastic Collision Formula: Solving for Final Velocity and Mass Ratios
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top