# Relationship between distance traveled and momentum

• abpandanguyen
In summary: Try again …In summary, the question asks for the ratio of the maximum distance traveled by two bodies, A and B, on a frictionless inclined plane with equal momenta and different masses. Using the equation v^2 = 2ad, we can find that the ratio of the maximum distances traveled by B and A is 1/3. This is because the momenta are equal, so the ratio of the velocities of B and A is 1/3, and the ratio of the distances traveled is equal to the ratio of their velocities squared. The correct answer is D) db = (1/3)da.
abpandanguyen

## Homework Statement

Two bodies, A and B, arrive at the base of a frictionless inclined plane with equal momenta of magnitude p. If massb = 3massa, the ratio of the maximum distance that B travels up the plane to the maximum distance A travels up the plane is...
with choices
A) db = d a
B) db = 9da
C) db = 3da
D) db = (1/3)da
E) db = (1/9)da

## Homework Equations

mv = p
not sure what else

## The Attempt at a Solution

Knowing they arrive at the base of this inclined plane with the same momentum, I'm thinking that massa is traveling at a faster velocity than massb. With this, I am assuming that massa is going to travel up the inclined plane farther than massb, but I do not really know where to start in terms of finding the actual ratio of the distance traveled between the two masses. Please and thank you for helping >_<

Welcome to PF!

Hi abpandanguyen! Welcome to PF!

What principle would you use to find the maximum height, given the initial speed?

Alright, so I'm not exactly sure what the name of the principle is, but I used the equation

(vf)2 = (vi)2 + 2as(sf-si)

and plugged in a random but reasonable incline to get an acceleration. With this and other initial velocities I plugged in just for the sake of using numbers to get a better idea of what I was doing, I found that E is the best answer.

The numbers I used were:
massa = 2 kg
massb = 6 kg
so respectively I gave them initial velocities of 6 m/s and 2 m/s and the incline I used was 5o and I got db = 2.34m vs da = 21.07m

Yes? No? D:<

You're missing the point of this exercise.

Basically, you're using the equation v2 = 2as (or v2 = 2gh).

a is the same for both masses, so you don't need to know what a is, nor m.

you can find s1/s2 just from v1/v2

Okaaay, that makes sense to me now. Thank you very much :D

I guess I still don't see the answer, yet, I think it is Db=1/3Da, can anyone confirm this?

thanks

samsorz said:
I guess I still don't see the answer, yet, I think it is Db=1/3Da, can anyone confirm this?

thanks

Hi samsorz!

(are you the same person as abpandanguyen?)

Write the equation for Db, and divide it by the equation for Da

what do you get?

no, i am not the same person, but probably same class :)

assuming that s1/s2 is the same thing as D1b/D2b, then... since you said the velocity is just the v1/v2, and velocity isn't affected by a change in mass, are they equal then?

no, this is abpandanguyen
I got E as my answer. I am pretty sure that is right, unless I just screwed up my equation somewhere. Thanks again btw Tiny Tim.

Welcome to PF!

Hi everyone! Welcome to PF, samsorz!
samsorz said:
no, i am not the same person, but probably same class :)

assuming that s1/s2 is the same thing as D1b/D2b, then... since you said the velocity is just the v1/v2, and velocity isn't affected by a change in mass, are they equal then?

Sorry, I'm not following you … they start with different velocities, but with equal momenta …

write the equation for Db, and divide it by the equation for Da

what do you get?

ok, see if this is better

Pb= mv m=2, v=6, p = 12
Pa=3mv m=1, v=4, p = 12

6/4, 3/2?, I am getting more and more lost :(

EDIT: V^2=2d?, so 36 and 16?

samsorz said:
ok, see if this is better

Pb= mv m=2, v=6, p = 12
Pa=3mv m=1, v=4, p = 12

6/4, 3/2?, I am getting more and more lost :(

EDIT: V^2=2d?, so 36 and 16?

Hi samsorz!

(try using the X2 and X2 tags just above the Reply box )

If you wrote proper equations, you wouldn't get into this mess!

Write it all out in full (and show us) …

what is the equation relating pa and va?

what is the equation relating pb and vb?

so what is the equation relating va and vb?

what is the equation relating da and va?

what is the equation relating db and vb?

so what is the equation relating da and db?

what is the equation relating pa and va? pa=mava

what is the equation relating pb and vb? pb=3mavb

so what is the equation relating va and vb? pa/va=pb/3vb

EDIT: the rest I am having trouble finding

samsorz said:
so what is the equation relating va and vb? pa/va=pb/3vb

No, you haven't finished that part yet …

what is va/vb ?

The next step is the relation between v and d.

You wrote v2 = 2d, which is the right idea, but you've left something out.​

so what is the equation relating va and vb? Va/Vb=[Pa/ma]/[Pb/3ma]

I'm not sure what I am leaving out

samsorz said:
so what is the equation relating va and vb? Va/Vb=[Pa/ma]/[Pb/3ma]

I'm not sure what I am leaving out

Well, the question tells you the momenta (Pa and Pb) are equal,

so Va/Vb=[Pa/ma]/[Pb/3ma = 1/3

Carry on.

By carry on do you mean there's more parts? or is that it lol

TY

samsorz said:
By carry on do you mean there's more parts? or is that it lol

TY

I meant …
tiny-tim said:
what is the equation relating da and va?

what is the equation relating db and vb?

so what is the equation relating da and db?

hmm … i think misunderstood when you wrote …
samsorz said:
I'm not sure what I am leaving out

you meant in v2 = 2d, didn't you?

That was supposed to be one of the standard constant acceleration equations …

so where's the acceleration?

a is the same for both masses, so you don't need to know what a is, nor m.

You said that earlier so I thought we neglected it

samsorz said:
tiny-tim said:
a is the same for both masses, so you don't need to know what a is, nor m.
You said that earlier so I thought we neglected it

Yes, we don't need to know what a is …

but that's no excuse for writing an equation that's wrong!

## 1. What is the relationship between distance traveled and momentum?

The relationship between distance traveled and momentum is that the momentum of an object is directly proportional to the distance it travels. This means that as an object travels a greater distance, its momentum also increases.

## 2. How does distance affect momentum?

Distance affects momentum by directly impacting the speed of the object. The greater the distance an object travels, the faster it will be moving, and therefore have a greater momentum.

## 3. Is momentum only influenced by distance traveled?

No, momentum is influenced by both distance traveled and the mass of the object. The larger the mass of an object, the greater its momentum will be, regardless of the distance traveled.

## 4. How can the relationship between distance traveled and momentum be represented mathematically?

The mathematical representation of the relationship between distance traveled and momentum is given by the equation p = mv, where p is momentum, m is mass, and v is velocity (which is directly related to distance traveled).

## 5. How does distance traveled affect the transfer of momentum between two objects?

Distance traveled plays a crucial role in the transfer of momentum between two objects. The longer the distance over which two objects interact, the more time they have to exchange momentum and therefore the greater the transfer of momentum will be.

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