Solving Collision Questions: Guidelines and Tips

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In summary, the conversation discusses how to approach a problem involving an elastic collision between a ball and an object on a slope. The participants discuss using equations of conservation of momentum and energy to solve for the four variables involved, but are unsure how to incorporate the normal force from the slope into the equations. The expert suggests taking components along the slope to eliminate the normal force from the calculations. They also discuss the forces on the object, and the expert clarifies that gravity can be ignored in this scenario.
  • #1
radagast_
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Homework Statement



http://img518.imageshack.us/img518/7105/76py8.gif

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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  • #2
anyone? any clue will be helpful...
 
  • #3
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
tiny-tim said:
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
radagast_ said:
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
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
Take components along the slope

radagast_ said:
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
tiny-tim said:
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
radagast_ said:
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):
tiny-tim said:
So … what is the momentum equation in the direction of the slope? :smile:
and you'll see what I'm getting at. :smile:
 
  • #10
tiny-tim said:
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
radagast_ said:
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
tiny-tim said:
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
radagast_ said:
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
tiny-tim said:
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
Ignore gravity

radagast_ said:
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
Ignore gravity

radagast_ said:
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
Ignore gravity

radagast_ said:
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
Ignore gravity

radagast_ said:
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
radagast_ said:
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
Ignore gravity

radagast_ said:
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
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)
 
Last edited:
  • #22
radagast_ said:
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:
 

1. How can I determine the direction of motion after a collision?

To determine the direction of motion after a collision, you can use the law of conservation of momentum. This states that the total momentum before a collision is equal to the total momentum after the collision. By calculating the momentum of each object before and after the collision, you can determine the direction of motion.

2. Do the masses of the colliding objects affect the outcome of a collision?

Yes, the masses of the colliding objects do affect the outcome of a collision. The larger the mass of an object, the more momentum it has. This means that a collision involving objects with different masses will result in a different outcome compared to a collision with objects of the same mass.

3. Can elastic collisions occur in real-life scenarios?

Yes, elastic collisions can occur in real-life scenarios. An example of an elastic collision is a game of billiards, where the balls bounce off each other without losing any kinetic energy. However, it is important to note that most collisions in the real world are not completely elastic and some energy is usually lost in the form of heat or sound.

4. How can I calculate the speed of an object after a collision?

To calculate the speed of an object after a collision, you can use the formula v = p/m, where v is the velocity, p is the momentum, and m is the mass. You will need to calculate the momentum of the object before and after the collision and use the law of conservation of momentum to determine the final velocity.

5. What are some tips for solving collision questions?

Some tips for solving collision questions include drawing diagrams to visualize the situation, applying the laws of conservation of momentum and energy, and breaking down the problem into smaller, more manageable steps. It is also important to carefully consider the direction and speed of the objects before and after the collision, as well as any external forces that may be present.

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