Why Isn't Momentum Conserved on All Axes in a Bouncing Ball Scenario?

  • Thread starter Thread starter ZxcvbnM2000
  • Start date Start date
  • Tags Tags
    Law Momentum
AI Thread Summary
Momentum is conserved in a closed system only when no external forces act on it. In the bouncing ball scenario, the ground exerts a reaction force on the ball, which is considered an external force, causing the ball's momentum to change upon impact. When analyzing collisions, both momentum and energy conservation must be considered, particularly along the line of impact and perpendicular to it. The confusion often arises from misunderstanding how the system is defined; including the ground in the system allows for momentum conservation. Ultimately, momentum is conserved in all directions when the entire system, including all interacting bodies, is accounted for.
ZxcvbnM2000
Messages
62
Reaction score
1
Hello everyone , here we go :

I understand that momentum is always conserved as long as no external forces act on a given system . I also understand that the law of restitution applies to the axis perpendicular to the line of impact between two objects .However i have a problem .

For example . A ball hits the ground at an angle of 45 degrees and bounces up at an angle of 30 degrees.

So from the law of restitution : e= - ( Vball*sin30 - Vground)/(Uball*sin45 - U ground) , the ground does not move therefore Uground = Vground=0 so

e = - Vball*sin30/(Uball*sin45).

My question is , why can't we apply momentum conservation on the y-axis ? Is it because the reaction force when hitting the ground is considered an external force ?

I am very confused please explain :S

Thank you !
 
Physics news on Phys.org
ZxcvbnM2000 said:
My question is , why can't we apply momentum conservation on the y-axis ? Is it because the reaction force when hitting the ground is considered an external force ?
Exactly. If you look at the ball itself, its direction reverses when it bounces. So obviously, momentum of the ball is not conserved--it's being smacked by the ground!

If you expand your 'system' to be 'ball + ground/earth', then momentum will be conserved again. (The force between ground and ball would then be an internal force.)
 
So my restitution formula is wrong and the only thing i have in my "system " is the ball , no ground etc . So it should be e= - V*sin30/(u*sin45) but e can't be negative ...argh ! :S
 
Let's assume that two balls(m1=m2=m ) collide with one another .

The first ball is traveling at a speed u while the second ball is stationary.The first ball strikes the second one at an angle θ to the line of impact.If the coefficient of restitution is e , find the angle at which the second ball travels after the impact.

What i don't understand in this case is why isn't momentum conserved both along the line of impact and on the axis perpendicular to it as well ? I mean , both balls are part of our system so there are no external forces ?!

Could you please solve this exercise and explain each step thoroughly so i can finally understand ? Thank you very much !
 
ZxcvbnM2000 said:
What i don't understand in this case is why isn't momentum conserved both along the line of impact and on the axis perpendicular to it as well ? I mean , both balls are part of our system so there are no external forces ?!
Why would you think that momentum isn't conserved in all directions?
 

Thread '1:1 versus 2:1 Mechanical Advantage debate'

The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
View full post »

Thread 'Newton's first law?'

Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
View full post »
Thread 'Beam on an inclined plane'

Thread 'Beam on an inclined plane'

Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
View full post »

Similar threads

B Conservation of momentum and conservation of energy details
Replies
30
Views
2K
B How to characterize a rubber ball moving horizontally and bouncing off of a vertical wall?
Replies
8
Views
2K
B Why is KE not conserved when momentum is?
2
Replies
53
Views
4K
I Conservation of angular momentum during collisions
Replies
3
Views
3K
I Forward momentum of mass due to inertia
Replies
4
Views
1K
Momentum is not conserved with elastic bouncing ball?
Replies
2
Views
2K
B Why do objects bounce?
Replies
6
Views
1K
Momentum of a System and External Forces
Replies
5
Views
1K
Why is friction minimal in a non-ideal ball bounce situation?
Replies
10
Views
2K
Linear momentum in oblique collisions and generally
Replies
8
Views
3K

Hot Threads

Recent Insights

Back
Top