What Are the Final Angle and Force in an Elastic Collision with a Wall?

[bfusco]

Homework Statement

A ball of mass m moving with velocity v strikes a vertical wall. The angle between the ball's initial velocity vector and the wall is theta_i as shown on the diagram, which depicts the situation as seen from above. The duration of the collision between the ball and the wall is Δt, and this collision is completely elastic. Friction is negligible, so the ball does not start spinning. In this idealized collision, the force exerted on the ball by the wall is parallel to the x axis.

A)What is the final angle theta_f that the ball's velocity vector makes with the negative y
B)What is the magnitude F of the average force exerted on the ball by the wall? (Express your answer in terms of variables given in the problem introduction and/or Vix)

http://session.masteringphysics.com/problemAsset/1010992/31/MLM_e2.jpg

The Attempt at a Solution

A)= theta_i
B)i don't really know where to start. i broke down the final velocity vector into its x and y components. (Vi=velocity initial) Vix=Visin θi, Viy=-Vicosθi, Vfx=-Vfsinθf, Vfy=-Vfcosθf. since the collision is elastic i can't use ΔP=J because the momentum before is the same as after. i feel like somewhere i am suppose to use the equation J=FΔt, but i can't figure it out.

 

[tal444]

First, as the wall is exerting only a force parallel to x, you only need to worry about the horizontal velocity. Second, you do need to figure out the impulse, and to do that you need to think about this: what is the TOTAL change in velocity on the x-axis?

 

[emailanmol]

Hey,

During collisions an important point to know is that both body exert forces on each other.(which are equal amd opposite in magnitude).

However, what's more important is the fact that how these forces act.
If you keep the body in contact, the force between these two always acts along the common normal(i.e perpendicular to common surface).
Since no force acts along the common tangent(parallel to common surface.also valid only in absence of friction)
So the net momentum of each body is conserved along common tangent.(i.e momentum of individual bodies along the common tangent doesn't change)
Remember this result has more significance than law of conservation of momentum as not inly the sum is conserved but also the individual values.(valid only in absence of friction)

Along the common normal since a force acts, momentum maybe transferred from one body to another but sum still remains constant(valid even if friction is present) .

So in question 1) the answer is thera because x component doesn't change and y component reverses (why?)

Also net force acting on anybody is change in its momentum divided by time for change to take place.

Here along the common normal the sum of momentum is constant.Not individual momentum.(individual momentum is conserved parallel to common surface i.e parallel to wall, so no force acts in this direction).

So you can calculate the force on ball by calculating its chsnge of momentum in time delta t

also the coefficient of restitution is valid only along common normal

i.e

Along common normal
e=-(v2-v1)/(u2-u1) where v1,v2,u1,u2 are velocity vectors along common normal
(if you have understood the concept ask these questions to yourself.They will be helpful)
What is the value of e for elastic collisions?
What is velocity of wall before and after collision.?
What is velocity of ball?
How much force acts on wall?
What is net change in momentum of wall?
What are the values of initial and final momentums of wall

 

[emailanmol]

One more thing , you addef that since collision is elastic momentum is conserved.

Its true but I would like to point out to you that momentum is conserved even durimg inelastic collisions or partially elastic collisions.

It is the kinetic energy which changes in inelastic collision

 

[bfusco]

One more thing , you addef that since collision is elastic momentum is conserved.

Its true but I would like to point out to you that momentum is conserved even durimg inelastic collisions or partially elastic collisions.

It is the kinetic energy which changes in inelastic collision

im honestly more confused then when i started. i understand that when the ball hits the wall the force the wall exerts on the ball is perpendicular to the wall. i also understand that when the ball hits the wall the force increases until it reaches a max value and then decreases until the ball is not longer touching the wall. the work done by that system is the area under that curve, which is also equal to the area under the curve of the average value of the force.

also i understand that the momentum is conserved. and to find the net force on the body ΔP/Δt, which doesn't help because there is no change in momentum. IF i break down the figure, the momentum in the y direction doesn't even change direction, it is just the same everywhere, so if i have to use the change in momentum anywhere it has got to be the x direction because at least it changes direction. using what i know i have a hard time applying it to the formulas.

Unless of course there is something I am missing in which case i would appreciate pretty straight foward answers as it is hard for me to comprehend the giant post before as it contains much information in it which isn't directly linked to the part of the problem i am confused with.

P.S. i was posting it again because i was hoping to also get maybe a new perspective on the problem as I am sure this thread isn't going to pop up on the main page now that it is a day old, i can definitely post twice, as long as i don't spam, which 2 times is not, but don't take the wrong way, as since this is typing my tone cannot be percieved, therefor understand no disrespect is meant.

 

[tal444]

Hint: the wall must exert a force equal to the ball's original x force, then ANOTHER force in order to get it moving in the opposite direction.

 

[emailanmol]

Hey, no offence taken.:-)

See you are absolutely right when you are saying net momentum doesn't change.
That means momentum of ball + momentum of wall is constant.

But it doesn't mean that momentum of ball cannot change .
If ball loses momentum the wall gains it.
Its the sum which remains constant not the individual momentums.

What is momentum?
Its mass m x velocity vector.

Earlier the ball was traveling towards right(And down as well). now its traveling towards left(and still down ).
So momentum of ball has changed .(since velocity vector has changed)
This change has been brought about by the wall applying force on the ball


However sum of momentum of ball and wall hasnt changed.

Now force is change in momentum divided by time for the change to take place.

Write the equation for final momentum and initial momentum (of ball) in terms of velocity vector and mass and divide it by time.
You will get the force.

(Also understanding how forces and momentum work and which components change is v crucial.As the sums get complicated , solving them would require clarity in these concepts.
I suggest you to go through my bigger post which contains all constraints you can apply.
If you are confused{am sorry.I actually put in a lot of information in one post.} you can google about oblique collisions.They will help you a lot)