Projectile motion with a bounce

[lax1113]

Hey guys,
I have a problem that is a decently complex projectile motion. Well, not really, but i can't figure one thing out, which makes the whoel thing a lot harder. I am not sure how to describe this without a picture, but, here goes... an object is at Y height. The object is shot out of a cannon at 60 degrees and falls to a triangle ramp. The object hits the triangle on its hypotenuse, and bounces off ina perfectly elastic collision. The triangle's angles are 25degrees, 90, and 65. The right angle and 65 degree angle are directly below the original point of the object being fired, and the 25 is on the other side. {{{W=65 degrees, Q=25degree, p= projectile object}}


________
_____/__\
____/____\
(p)/______\
|
|
|
|
|
W\
xx\
xxx\
xxxx\ --- I know this picture
xxxxx\ --- Looks terrible
xxxxxx\ --- but the object
xxxxxxx\3 --- hits the triangle
xxxxxxxx\ --- on the side above Q
xxxxxxxxQ\ ---- designated by ''3''
---------- and has a perfectly elastic collison

is it possible to find the angle at which the object will bounce off of the triangle ramp? If anyone can actually understand this, i will be amazed. I think that its going to be 30 degrees, because it is a 60 degree angle that it is falling at, and two objects will make a 90degree angle if no spin/ other force is added, but any help would be cool.

 

[alphysicist]

Hi lax1113,

Hey guys,
I have a problem that is a decently complex projectile motion. Well, not really, but i can't figure one thing out, which makes the whoel thing a lot harder. I am not sure how to describe this without a picture, but, here goes... an object is at Y height. The object is shot out of a cannon at 60 degrees and falls to a triangle ramp. The object hits the triangle on its hypotenuse, and bounces off ina perfectly elastic collision. The triangle's angles are 25degrees, 90, and 65. The right angle and 65 degree angle are directly below the original point of the object being fired, and the 25 is on the other side. {{{W=65 degrees, Q=25degree, p= projectile object}}


________
_____/__\
____/____\
(p)/______\
|
|
|
|
|
W\
xx\
xxx\
xxxx\ --- I know this picture
xxxxx\ --- Looks terrible
xxxxxx\ --- but the object
xxxxxxx\3 --- hits the triangle
xxxxxxxx\ --- on the side above Q
xxxxxxxxQ\ ---- designated by ''3''
---------- and has a perfectly elastic collison

is it possible to find the angle at which the object will bounce off of the triangle ramp? If anyone can actually understand this, i will be amazed. I think that its going to be 30 degrees, because it is a 60 degree angle that it is falling at, and two objects will make a 90degree angle if no spin/ other force is added, but any help would be cool.

Is there more information given in the problem? I was wondering how you knew that the object impacts the ramp at a 60 degree angle. It leaves the cannon at a 60 degree angle, but as it falls its velocity angle will change direction. Did you already calculate the impact angle relative to the incline? (I hope I'm understanding the situation correctly!)

As for the angle, think about what happens to a ball that impacts the floor at some angle to the horizontal. If the collision is elastic, frictionless floor, etc., what angle to the horizontal does it leave the floor at? In particular, what direction is the force from the floor?

 

[lax1113]

Hey,
thanks for the reply, and thanks for attempting to understand my picture/explanation. I believe that you have read it right, and understood it. With that being said, all the info given was the velocity, angle of departure, height from the floor, and the angles in the triangle. I can't remember the height, it was a problem we were doing in class. Anyway, it seems to me that a ball hitting a ramp goes in a degree that when added with the original angle makes 90 degrees, similar to how two pool balls will make a 90 degree angle if they have no spin. If this is not correct, it will not surprise me, since we have not learned this yet, but i would love to have an explanation as to why this is wrong, and what the right answer would be. If i didnt supply enough info, I am sorry, but that's all i could remember from the problem.


----- forgot to say, no i thought that the angle would leave at 60degrees, end up doing a parabola type deal, so being a straight line for an instance, then fall at the same rate as it was reduced, which would make it a 60 degree angle again? make any sense? or... is this wrong?... any advice.

thanks again sir,
later

 

[alphysicist]

Hey,
thanks for the reply, and thanks for attempting to understand my picture/explanation. I believe that you have read it right, and understood it. With that being said, all the info given was the velocity, angle of departure, height from the floor, and the angles in the triangle. I can't remember the height, it was a problem we were doing in class. Anyway, it seems to me that a ball hitting a ramp goes in a degree that when added with the original angle makes 90 degrees, similar to how two pool balls will make a 90 degree angle if they have no spin. If this is not correct, it will not surprise me, since we have not learned this yet, but i would love to have an explanation as to why this is wrong, and what the right answer would be. If i didnt supply enough info, I am sorry, but that's all i could remember from the problem.

The 90 degree result is for a special case: an glancing elastic collision between two objects with equal mass; if I remember correctly, it also had to be in a reference frame in which one object is initially at rest. So that result does not apply here.

Instead, break up the initial and final velocities of the object into components that are perpendicular and parallel to the incline. (By "initial" I mean the velocity just before impact with the ramp, and by "final" I mean the velocity just after impact.) Think about what direction the force of collision is on, and apply the impulse-momentum relationship and conservation of kinetic energy.

----- forgot to say, no i thought that the angle would leave at 60degrees, end up doing a parabola type deal, so being a straight line for an instance, then fall at the same rate as it was reduced, which would make it a 60 degree angle again? make any sense? or... is this wrong?... any advice.

If you're talking about the angle the velocity makes with the horizontal, it would be 60 degrees when it reached the same vertical level as its launch height (that is, on the way back down after reaching its maximum height). But it keeps falling past that point to reach the ramp, and so the velocity vector would be greater than 60 degrees with the horizontal when it reaches the ramp.

Once you find the direction of impact velocity, you can use geometry to find the angle it makes with the ramp.

 

[lax1113]

Alphysicist,
I am in no way doubting your knowledge, or saying that you are wrong, but although this was supposed to be an advanced problem, I am pretty sure that everyone in my class was supposed to be able to answer it, whether or not they did. What i mean is that we have the forumala's/knowledge in our heads, just whether or not we could put it together. Then again, nobody at all was able to get it, so maybe not. I am just not sure if you are using things that we havne't learned yet because it is more accurate, or if i left out some information that would have helped you in solving it. With that being said, i think i am just going to wait until the explanation in class, but thank you very much for the attempt. I most likely left out a piece of the problem that screwed it up.
thanks again alpha