Frictionless inclined plane help

[remaan]

Need Help !

Homework Statement

Two bodies, A and B arrive at the base of a fricitonless inclined plane with equla momenta of magnitude p. mb=3ma. The ratio of the max. distance Db that B travels up the plane to the max. distance Da the A travels up the plane is : Db=Da ? is that correct ?

Homework Equations

p=mv
v=d/t

The Attempt at a Solution

I am not sure I subs. back and then cancled every thing and that was the result.

 

[LowlyPion]

Two bodies, A and B arrive at the base of a fricitonless inclined plane with equla momenta of magnitude p. mb=3ma. The ratio of the max. distance Db that B travels up the plane to the max. distance Da the A travels up the plane is : Db=Da ? is that correct ?

p=mv
v=d/t

I am not sure I subs. back and then cancled every thing and that was the result.

If the momentum is equal and one mass is 3 times the other what does it say about the relationship of their speeds?

 

[remaan]

In fact, there is no thing about the speed, I typed the question as it is ..

Thank you

 

[LowlyPion]

In fact, there is no thing about the speed, I typed the question as it is ..

Thank you

You don't need to know the absolute speed of either. But you do know something about their relative speeds as a consequence of knowing they have the same momentum.

 

[remaan]

Sorry, I didn't get you, would you tell a bit more ?

 

[LowlyPion]

Sorry, I didn't get you, would you tell a bit more ?

You know momentum is the same. But the masses are different. That means the speeds are different.

So how are the speeds different?

 

[remaan]

Well, then what does it has to do with their distances ?
Do you mean to say that will different distances as a result of that?

Or,, in other words how to prove that?

 

[LowlyPion]

Well, then what does it has to do with their distances ?
Do you mean to say that will different distances as a result of that?

Or,, in other words how to prove that?

You think something that is slower at the bottom of a ramp will go as far as something that is faster?

I guarantee that it will not.

Regardless of the mass, the acceleration from gravity down the ramp is the same whether slow or fast. So clearly then something going faster will go farther. It's simple kinematics.

 

[remaan]

It is a muliple choisse ques. We have to choose one and prove it

Her are the choises:
Db=1/9Da

Db=1/3Da

Db=Da

Db=3Da

Db=9Da


I thought about the 3rd choise because of the equal mometom, and I am not sure, What do you think? and how to prove ?

 

[remaan]

You think something that is slower at the bottom of a ramp will go as far as something that is faster?

I guarantee that it will not.

Regardless of the mass, the acceleration from gravity down the ramp is the same whether slow or fast. So clearly then something going faster will go farther. It's simple

kinematics.

How did you know that one is faster? The question says that they will go down the incline with we know which is the faster?

 

[LowlyPion]

How did you know that one is faster? The question says that they will go down the incline with we know which is the faster?

Momentum P is a constant

P = M_a*V_a = M_b*V_b

And you know that M_b = 3*M_a

So what must V_a be relative to V_b ?

 

[remaan]

Yes, what am trying to do is exactky same to what you did, but I think that their velocities will be the same and therefore the distance will be the same?

ISn't it?

 

[LowlyPion]

Yes, what am trying to do is exactky same to what you did, but I think that their velocities will be the same and therefore the distance will be the same?

ISn't it?

No. Why are you resistant to just doing the math? Substitute 3*M_a for M_b. What does that tell you about the velocities?

 

[remaan]

Or do you think that I should use conservation of energy?

Or I choose the second choise

 

[LowlyPion]

Or do you think that I should use conservation of energy?

Or I choose the second choise

I think you should just do the math and get the right answer.

 

[remaan]

I will get equal velocities and then equal distancse
Right?

 

[remaan]

Please, I getting confused more and more ...

The question is asking about the distances not the velocities..

??

I will get equal velocities and then, what math do mean ??
Do mean that v=d/t ?? What has to do with gravitation

 

[LowlyPion]

I will get equal velocities and then equal distancse
Right?

The answer is no. You do not get the same velocities. You keep repeating that and it's still no.

The momentum is the same.
The masses are different.
The velocities MUST be different.

I've written the equations. I've suggested what to substitute to solve for what the velocity must be.

What I won't do is just give you the answer.

 

[remaan]

Do you mean that we will subs. of the masses after we write the equation you just gave me..
And then we get that
Vb= 1/3 Va

and Now, do we use that v=d/t to find the distance?

Thank you :)

 

[LowlyPion]

Do you mean that we will subs. of the masses after we write the equation you just gave me..
And then we get that
Vb= 1/3 Va

and Now, do we use that v=d/t to find the distance?

Thank you :)

Almost. Now that you know what the velocity is, and you understand that acceleration (actually deceleration) down the incline is the same then how will that velocity difference translate into the distance until it stops going higher?

The equation for that is

V_f² = V_i² + 2*a*x

Note that a is -, and V_f is where it reaches 0.

What is the effect in x, when V is 3 times larger?

 

[remaan]

Ok, x will be larger.

And I will sub. vi with its values for A and B

I will get this equation:
va-2gda=3vb=-2gdb

Right?

And I will get equal distances after cancelling the velo. and -2g ??

Right?

 

[LowlyPion]

Ok, x will be larger.

And I will sub. vi with its values for A and B

I will get this equation:
va-2gda=3vb=-2gdb

Right?

And I will get equal distances after cancelling the velo. and -2g ??

Right?

No.

The acceleration up the incline is not g. It is a function of g and the angle of the incline. It is sufficient that it is constant in both cases.

What that means then is that V² is proportional to the distance.

So if you have one distance x determined by V, then if you have another V' = 3*V then the effect on distance x' will be (3*V)²

 

[remaan]

I think that I am getting closer to the answer..

I end up with is this equation :

vi^2 = 2 a x

And after subs. with velo. of A and B I can't find any relations between their distances when going up the incline.

 

[LowlyPion]

Try dividing the two equations for velocity by each other then.

 

[remaan]

That makes me GLAD ..


I got Db=3Da


Right??

 

[LowlyPion]

That makes me GLAD ..

I got Db=3Da

Right??

Let's see.

V_a² = 2*a*x_a

V_b² = 2*a*x_b

Dividing one by the other:

V_a²/V_b² = x_a/x_b

Do you want to try again?

 

[remaan]

That's Great..

I did the math and I get the Db = 1/9 Da ..

Is that Right?? WooooW !

 

[LowlyPion]

That's Great..

I did the math and I get the Db = 1/9 Da ..

Is that Right?? WooooW !

Yes. M_b is 3 times the mass of M_a.

That means that V_b is 1/3 V_a

... and D_b is 1/9 of D_a

 

[remaan]

Thanks aloooooooooooot
Finally, I am done with this problem ..


You are a great TEACHER :D

See in my next ques.

Good- bye