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Collision question

  • Thread starter radagast_
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  • #1
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Homework Statement



http://img518.imageshack.us/img518/7105/76py8.gif [Broken]

Homework Equations



The Attempt at a Solution


I don't want a solution, just guidelines.. I don't really know how to approach it.
I was thinking about an equation of conservation of momentum which will give me 2 equations (x, y) and conservation of energy (because of the ellastic collision) which will give me another equation. All in all - 3 equations, but I have 4 variables: alpha, v0, and the ball and object's speeds after the collision.

What am I missing?
Thank you.
 
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Answers and Replies

  • #2
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anyone? any clue will be helpful...
 
  • #3
tiny-tim
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Hint 1: What does "elastically" mean? What does that enable you to write?

Hint 2: There's also a force - the force from the wall. How can you keep that out of your calculations? :smile:
 
  • #4
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Hint 1: What does "elastically" mean? What does that enable you to write?

Hint 2: There's also a force - the force from the wall. How can you keep that out of your calculations? :smile:
1. it means conservation of energy => it gives me an equation as I've written on the original post.

2. I know there's a force (normal?). that's my problem - combining it with the momentum equations. it's external, right? so there's no conservation of momentum. how can I put this in an equation?

thanx.
 
  • #5
tiny-tim
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1. it means conservation of energy => it gives me an equation as I've written on the original post.
Yes … but you've not actually written the equation …

2. I know there's a force (normal?). that's my problem - combining it with the momentum equations. it's external, right? so there's no conservation of momentum. how can I put this in an equation
Newton's second law: force = (rate of) change of momentum.

So you only get conservation of momentum when the force is zero.

But Newton's second law is three-dimensional: it's a different equation for every direction.

So can you see a direction in which there's zero force? :smile:
 
  • #6
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ok...

1.
ub=ball speed after
ut=table speed after

m(v0)^2=m(ub*cos(beta))^2+M(ut)^2

that's what I know.
There's a normal force from the table (because of the floor), right?
but how can I put the kinetic energy in an equation with the normal force?
 
  • #7
tiny-tim
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Take components along the slope

There's a normal force from the table (because of the floor), right?
That's right - the "instantaneous" impulse from the table is perpendicular to it :smile:(but only, of course, because the question specifies that "there are no friction forces at all").

but how can I put the kinetic energy in an equation with the normal force?
You're absolutely right! You can't! :smile:

But you can put the momentum in an equation with the normal force, provided that you do it only in the direction of the slope - because in that direction the component of the normal force is zero!

So … what is the momentum equation in the direction of the slope? :smile:
 
  • #8
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But you can put the momentum in an equation with the normal force, provided that you do it only in the direction of the slope - because in that direction the component of the normal force is zero!
Im sorry, I don't understand: how can I put it with the normal force?
practically: the momentum is m*v (meter*kg/second) and the normal is (meter*kg/second^2)... I can't place them together in an equation...
 
  • #9
tiny-tim
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the momentum is m*v (meter*kg/second) and the normal is (meter*kg/second^2)... I can't place them together in an equation...
Good point … but collisions produce an impulse (an "instantaneous" force), which is force x time.

Im sorry, I don't understand: how can I put it with the normal force?
Just write out this (putting in the numbers):
So … what is the momentum equation in the direction of the slope? :smile:
and you'll see what I'm getting at. :smile:
 
  • #10
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Just write out this (putting in the numbers):

and you'll see what I'm getting at. :smile:
well,
I get a component (cos(alpha) or cos(beta-alpha)) from every speed, v0, ub (ball after) and ut (table after)...
still three missing speeds (plus angle alpha) in one equation. I wrote it, I can't see what I can make out of it
 
  • #11
tiny-tim
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well,
I get a component (cos(alpha) or cos(beta-alpha)) from every speed, v0, ub (ball after) and ut (table after)...
still three missing speeds (plus angle alpha) in one equation. I wrote it, I can't see what I can make out of it
Hi radagast!

Hint: what forces are on the block (the object)?

So which direction will it move in?

And does that help you with the momentum equation? :smile:
 
  • #12
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Hi radagast!

Hint: what forces are on the block (the object)?

So which direction will it move in?

And does that help you with the momentum equation? :smile:
Thanks for the answers...
I'd appreciate it if you'd be more specific.. I am not following you.
Also, isn't there conservation of momentum on the x axis? how does that get along with the momentum parallel to the slope?
 
  • #13
tiny-tim
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Also, isn't there conservation of momentum on the x axis? how does that get along with the momentum parallel to the slope?
There is conservation of momentum in every direction - the x-axis, the y-axis, along the slope, perpendicular to the slope …

We just choose whatever direction is easier - in this case, it's along-the-slope! :smile:

Thanks for the answers...
I'd appreciate it if you'd be more specific.. I am not following you.
radagast, you keep not answering my questions.

That means I have no idea how far you're getting.

If someone on this forum helps you by asking a question, then you must answer it!

So … what forces are on the block?
 
  • #14
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So … what forces are on the block?
ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
 
  • #15
tiny-tim
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Ignore gravity

ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
hmm … I wish I'd called it the "hypotenuse" rather than the "slope". :redface:

I'm pretty sure the question intends you to ignore gravity - either x and y are coordinates on a smooth horizontal plane, or it's all in a "free-fall" frame in deep space.

Is that what was bothering you?

(This is a bad question - the examiner should specify things like that :frown:. Maybe the examiner thinks that the down direction is always called "z"?)

So, ignoring the mg, the only force is perpendicular to the hypotenuse, and so the block moves in that direction, and so its momentum along the hypotenuse is zero!

ok … so the momentum equation along the hypotenuse is … ? :smile:
 
  • #16
tiny-tim
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Ignore gravity

ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
hmm … I wish I'd called it the "hypotenuse" rather than the "slope". :redface:

I'm pretty sure the question intends you to ignore gravity - either x and y are coordinates on a smooth horizontal plane, or it's all in a "free-fall" frame in deep space.

Is that what was bothering you?

(This is a bad question - the examiner should specify things like that :frown:. Maybe the examiner thinks that the down direction is always called "z"?)

So, ignoring the mg, the only force is perpendicular to the hypotenuse, and so the block moves in that direction, and so its momentum along the hypotenuse is zero!

ok … so the momentum equation along the hypotenuse is … ? :smile:
 
  • #17
tiny-tim
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Ignore gravity

ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
hmm … I wish I'd called it the "hypotenuse" rather than the "slope". :redface:

I'm pretty sure the question intends you to ignore gravity - either x and y are coordinates on a smooth horizontal plane, or it's all in a "free-fall" frame in deep space.

Is that what was bothering you?

(This is a bad question - the examiner should specify things like that :frown:. Maybe the examiner thinks that the down direction is always called "z"?)

So, ignoring the mg, the only force is perpendicular to the hypotenuse, and so the block moves in that direction, and so its momentum along the hypotenuse is zero!

ok … so the momentum equation along the hypotenuse is … ? :smile:
 
  • #18
tiny-tim
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Ignore gravity

ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
hmm … I wish I'd called it the "hypotenuse" rather than the "slope". :redface:

I'm pretty sure the question intends you to ignore gravity - either x and y are coordinates on a smooth horizontal plane, or it's all in a "free-fall" frame in deep space.

Is that what was bothering you?

(This is a bad question - the examiner should specify things like that :frown:. Maybe the examiner thinks that the down direction is always called "z"?)

So, ignoring the mg, the only force is perpendicular to the hypotenuse, and so the block moves in that direction, and so its momentum along the hypotenuse is zero!

ok … so the momentum equation along the hypotenuse is … ? :smile:
 
  • #19
tiny-tim
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ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
hmm … I wish I'd called it the "hypotenuse" rather than the "slope". :redface:

I'm pretty sure the question intends you to ignore gravity - either x and y are coordinates on a smooth horizontal plane, or it's all in a "free-fall" frame in deep space.

Is that what was bothering you?

(This is a bad question - the examiner should specify things like that :frown:. Maybe the examiner thinks that the down direction is always called "z"?)

So, ignoring the mg, the only force is perpendicular to the hypotenuse, and so the block moves in that direction, and so its momentum along the hypotenuse is zero!

ok … so the momentum equation along the hypotenuse is … ? :smile:
 
  • #20
tiny-tim
Science Advisor
Homework Helper
25,832
249
Ignore gravity

ok.

The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?
hmm … I wish I'd called it the "hypotenuse" rather than the "slope". :redface:

I'm pretty sure the question intends you to ignore gravity - either x and y are coordinates on a smooth horizontal plane, or it's all in a "free-fall" frame in deep space.

Is that what was bothering you?

(This is a bad question - the examiner should specify things like that :frown:. Maybe the examiner thinks that the down direction is always called "z"?)

So, ignoring the mg, the only force is perpendicular to the hypotenuse, and so the block moves in that direction, and so its momentum along the hypotenuse is zero!

ok … so the momentum equation along the hypotenuse is … ? :smile:
 
  • #21
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Well, the object isn't moving perpendicular to the force: the normal is at an angle (not 90) relative to the x axis, and the object's moving on the x axis.

the momentum equation is:
(a=alpha, b=beta, ub=ballspeed after, ut=tablespeed after)

m*v0*cos(a) = m*ub*cos(b-a) + M*ut*cos(a)
 
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  • #22
tiny-tim
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Well, the object isn't moving perpendicular to the force: the normal is at an angle (not 90) relative to the x axis, and the object's moving on the x axis.
I'm confused … because you're confused … :confused:

The object moves perpendicular to the hypotenuse - in other words, it moves along the normal - so how do you get it moving on the x-axis?

Try again! :smile:
 

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