Collision/Projectile Motion Problem -NEED HELP-

[Plutonium88]

A Ball with mass Mo froms H1(Total Height),before it hits the ground it bounces off a stationary inclined plane at height (H2) with angle (Ω), it falls a distance of (H2) and travels a distance (L) over a time (∆t)
FIND L and ∆t

Image of Problem: http://s9.postimage.org/4xn9jpklb/Velocity.png

(this is my verbal expression of the question, there will be a picture attached) unfortunately there was no written part to this question.

am only having problem with one part specifically... Which is the part i assume everyone has problems on..

The collision between the inclined plane, or how exactly to find the velocity after it has been reflected by the inclined plane. (I know the angle of reflection is = to angle of incidence) so therefore angle Ω of the plane is equal to angle Ω of the velocity after collision...

I know that Px=Px' momentum is conserved in the X direction...

My Attempt:

Et1= Eg1=mgh1

Et2= Eg2 + Ek2 = mg(h1-H2) + 1/2MoV2^2

So Et1=Et2

mogh1=mog(h1-h2) +1/2MoV2^2 (masses cancel)

V2 = Sqrt 2gh1-2g(h1-h2)

(V2 is the velocity just before it collides with the plane)

Now I've Drawn a FBD (my frame of reference/axis is along the surface horizontally and vertically the positive direction is towards the normal force) of when the ball is on the inclined plane (and it hits the plane so the normal force is perpindicular to the surface)

so when i do my net forces i get

fnetx= fgsinΩ
fnety= fn -fgcosΩ=0 (therefore fn = fgcosΩ)

Now I'm curious am i confusing this part, am i supposed to take the Normal force as the one which has components?

Also when i do... this is where i get really stuck..

MOMENTUM!

p=p'

Since the inclined plane remains stationary during collison that means its vertical and horizontal compenents are both zero...

let v2x =v2cosΩ
v2y = v2sinΩ
so I am left with

0 + mov2x = 0 + mov2x'
and
0 + mov2y = 0 + mov2y'so now components are equal but when i continued with these steps and showed my teacher the answer she said it was incorrect... could someone please help me with Collision between an inclined plane and the ball and finding the velocity after the collision.. i have been working on this question for weeks, my semester for physics gr12 finished 3 weeks ago.. but i really want to answer this question.

 

[Plutonium88]

Ah also, The angle Ω is 30° and this question is to be solved in terms of variables.

 

[jedishrfu]

okay I think your omega angle s/b (pi/2 - 2*omega) right?

omega defines the plane incline relative to horizontal

if the vertical drop continues thru to the ground state then the complementary angle of the
small triangle formed is pi/2-omega

and the angle between the vertical drop and incline is (pi/2-omega)

since the ball reflects off the incline then the reflection angle relative to the incline is also (pi/2 - omega)

consequently the angle relative to the horizontal is ((pi/2 - omega) - omega)

hence (pi/2 - 2* omega)

now you should have the velocity of the ball the moment it hits the incline and the angle it takes as it bounces off the incline

 

[jedishrfu]

so I get the following eqns:

vx = V2 * cos( pi/2 - 2 * omega)
vy = V2 * sin( pi/2 - 2 * omega)

y = H2 + (vx * t) - 1/2 * g * t^2

x = V2 * cos( pi/2 - 2 * omega) * t

then set y = 0 and solve for t

plug t into x and find the distance traveled

 

[Plutonium88]

so I get the following eqns:

vx = V2 * cos( pi/2 - 2 * omega)
vy = V2 * sin( pi/2 - 2 * omega)

y = H2 + (vx * t) - 1/2 * g * t^2

x = V2 * cos( pi/2 - 2 * omega) * t

then set y = 0 and solve for t

plug t into x and find the distance traveled

Anyway you know how to do this without radiants, i haven't quite learned them yet.. or could you possibly post a drawing so i could learn from your drawing how to use radiants?

 

[Doc Al]

The collision between the inclined plane, or how exactly to find the velocity after it has been reflected by the inclined plane.

Assuming it's a perfectly elastic collision, the speed will not change during the collision.

 

[Plutonium88]

Assuming it's a perfectly elastic collision, the speed will not change during the collision.

So assuming it's not an elastic collision, what can i do to calculate the velocity after the collision, and also how to calculate the angle assuming my strategy of angle of incidence is = to angle of reflection is incorrect.

 

[Doc Al]

So assuming it's not an elastic collision, what can i do to calculate the velocity after the collision, and also how to calculate the angle assuming my strategy of angle of incidence is = to angle of reflection is incorrect.

Why would you assume it's not an elastic collision? (If it isn't, then I think you do not have enough information to solve it.)

 

[Doc Al]

Ah also, The angle Ω is 30° and this question is to be solved in terms of variables.

So what angle does the ball make with respect to the horizontal when it bounces off the incline?

 

[Plutonium88]

So what angle does the ball make with respect to the horizontal when it bounces off the incline?

The angle the ball makes after collision is 30 degrees from the horizontal

it says when it hits the surface of the plane it is perpendicular to the surfacewait I'm wrong.. the geometry says it must be the opposing angle of 60 degrees... derrr

 

[jedishrfu]

just replace the pi/2 with 90 degrees and compute the sin / cos values from that.

wrt radians, they are just another method of describing angle measure that is more natural to use in math problems especially in the mathematical series that define the values of sin and cos functions.

2π = 360
π = 180
π/2 = 90

wikipedia has a good writeup on radian measure for angles at:

http://en.wikipedia.org/wiki/Radians

 

[jedishrfu]

I was trying to say that your geometry is wrong wrt to the angle.

Just because the incline is omega doesn't mean the angle of the ball bouncing off is also omega

I think its actually 90 degrees - 2*omega

As an example say the incline was at 80 degrees then by your analysis it bounces off at 80 degrees to the horizontal right. But if you try the experiment it will be 70 degrees below the horizontal ie

angle = 90 degrees - 2 * 80 degrees = - 70 degrees to the horizontal

As another example say the incline is 45 degrees then by your analysis the ball bounces off at 45 degrees instead of zero degrees to the horizontal.

angle = 90 degrees - 2 * 45 degrees = 0 degrees to the horizontal

 

[Plutonium88]

I was trying to say that your geometry is wrong wrt to the angle.

Just because the incline is omega doesn't mean the angle of the ball bouncing off is also omega

I think its actually 90 degrees - 2*omega

As an example say the incline was at 80 degrees then by your analysis it bounces off at 80 degrees to the horizontal right. But if you try the experiment it will be 70 degrees below the horizontal ie

angle = 90 degrees - 2 * 80 degrees = - 70 degrees to the horizontal

As another example say the incline is 45 degrees then by your analysis the ball bounces off at 45 degrees instead of zero degrees to the horizontal.

angle = 90 degrees - 2 * 45 degrees = 0 degrees to the horizontal

thanks a lot bro this is very clear. I'm going to start practicing in radiant and get familiar with them. i appreciate this.

 

[Doc Al]

The angle the ball makes after collision is 30 degrees from the horizontal

That's correct.

it says when it hits the surface of the plane it is perpendicular to the surface


wait I'm wrong.. the geometry says it must be the opposing angle of 60 degrees... derrr

 

[Plutonium88]

c.c. I'm confused so bad.

 

[Doc Al]

c.c. I'm confused so bad.

Why is that?

The collision with the incline is elastic. The speed doesn't change and the angle of incidence equals the angle of reflection. From that point on, it's a projectile motion problem.

 

[Plutonium88]

Okay can you just go through the angle part with me... Cause i confused myself with this geometry...So if it bounces so that the object is perpindicular to the surface...

How can i prove that the angle on the plane = the angle of the velocity.

 

[Plutonium88]

okay so...

Energy:

et1=et2
mgh1=1/2mv2^2 + mg(h1-h2)

v2 = √2gh2

Momentum:

p=p'

0 + mv2 = 0 + mv2'

v2=v2'

Kinematics:
v2'x = v2'cos30
v2'y= v2'sin30

Uniform accelerated motion
H2 = v3^2 - V2'y^2
--------------
2g

V3 = √2GH2 + v2^2


H2 = v3 + v2
--------- (ΔT)
2

Δt = 2h2
------------------
√2gh2 + √(2gh2)sin30


Am i correct up to this ponit?? :O

 

[jedishrfu]

see diagram and compare to your angles

 

[Plutonium88]

see diagram and compare to your angles

okay so in order to get the angle, you compared your x axis, with the two forces of normal..

and you subtracted the two 2Alpha from 90 degrees, because that is that quadrant...


parallel lines are referring to the bottom of the plane being parallel with the axis?

And thanks a lot for spending your time man. i really appreciate this. I'm just asking a lot of questions, so that in the future if i ever encounter a similar problem i under stand things better, and geometry has never been my strong suit.

 

[Plutonium88]

see diagram and compare to your angles

also i solved the for time using the quadratic formula..

i ended up with t=[√2H2] by simplification... but this doesn't seem right in terms of units... However when i solve for my final answer...

I get L = [√3]H2

as my answer.. could you please tell me if this is correct?

 

[jedishrfu]

the (90 - 2 α) is the angle of elevation. Sometimes diagrams be misinterpreted especially if angles like 30 or 45 degrees are used. I also try 80 and 10 degrees to see if there anything I missed in my answer.

yes and your units of measure are wrong for the time. Thats a good way to check whether your solution is reasonable. I used the same method to check my eqns on the scanned paper.

I don't think I can help you any more as you really need to think about the solution. There are two problems here:
- in the first you determine the speed of the ball once it hits the incline and then you use your knowledge of
reflection to determine the angle of 90-2α.
- The second problem is given the speed of the ball, the angle of elevation and the height above the ground
you can compute how far it will travel before it hits the ground.

My initial post was because I thought I saw the source of your error and so corrected it to be (90 - 2*α)beyond that I feel I will confuse you more. Try to work it out as two problems and forget what everyone including me have written has written. On my scanned paper are the equations for the vx, vy, x and y. Use the y eqn to determine the time the ball hits the ground and then use that in the x eqn to get the dist and you will be done.

Good luck.

 

[Plutonium88]

the (90 - 2 α) is the angle of elevation. Sometimes diagrams be misinterpreted especially if angles like 30 or 45 degrees are used. I also try 80 and 10 degrees to see if there anything I missed in my answer.

yes and your units of measure are wrong for the time. Thats a good way to check whether your solution is reasonable. I used the same method to check my eqns on the scanned paper.

I don't think I can help you any more as you really need to think about the solution. There are two problems here:
- in the first you determine the speed of the ball once it hits the incline and then you use your knowledge of
reflection to determine the angle of 90-2α.
- The second problem is given the speed of the ball, the angle of elevation and the height above the ground
you can compute how far it will travel before it hits the ground.

My initial post was because I thought I saw the source of your error and so corrected it to be (90 - 2*α)beyond that I feel I will confuse you more. Try to work it out as two problems and forget what everyone including me have written has written. On my scanned paper are the equations for the vx, vy, x and y. Use the y eqn to determine the time the ball hits the ground and then use that in the x eqn to get the dist and you will be done.

Good luck.

Alright, much obliged. all of your time is greatly appreciated and i will put it too good use. Thank you.