# Determine the distance the object travels

Δ1. Homework Statement
Below

## Homework Equations

[/B]Potential energy = kinetic energy
MgΔY=1/2 mΔv^2
Vf^2=Vi^2+2gx1

## The Attempt at a Solution

For question 49, I used the first equation, and the result is Δv^2=2gY. Then, solving for x1 in the second equation, I got x1=Δv^2 / 2g. Substituting ΔV^2 in x1 results in x1=Y. Answer is A. I am not sure if I should have used H instead of Y, though. But then I would use D, too. I need clarifications, please.

50) since, the masses cancel, the horizontal distance would be the same for M and 2M. Answer C.
51) not sure of it, but friction always slows the motion. So, I guess it is A.
Am I right?

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I think I should have used H + D in 49 since when it ends the track it has another potential energy relative to the ground. Hmm..
MgH+MgD=1/2 Mv^2
Which one is correct?

Chandra Prayaga
These are the problems I see in the equations that you wrote:
1.
Potential energy = kinetic energy
This statement, at best, is very misleading. It is not true that the potential energy is equal to the kinetic energy. Both these quantities are changing as the ball moves, and which potential energy is equal to which kinetic energy? At each point in its path, is the PE equal to the KE? That is certainly wrong. The correct statement, relevent to this situation, is that the sum of PE and KE is constant, if there is no friction.
2.
MgΔY=1/2 mΔv^2
This statement is different from the one above, and is also wrong. This statement says the "change" in PE is equal to the "change" in KE, and that is not correct. The correct statement, which follows from the corrected statement #1 is "the change in PE = - the change in KE". That will mean that if there is a decrease in PE, there is an equal aount of increase in KE.

When you are solving a problem involving energy, you should label the positions of interest. In this case, the top of the ramp may be lebelled as 1, the bottom of the ramp as 2, and the final position where the ball lands may be called 3. Now, when you refer to the speed of the ball or ots position, along the y axis, you can call them vA, vB, yA, yB, etc. Then you can unambiguously trace the variables as the ball moves. For example, when you state:
and the result is Δv^2=2gY.
It is not clear what is Y. According to your statement #1, that result should be Δv^2=2gΔY and even then the question is what is ΔY? Do you mean the difference between the height at the top of the ramp and the bottom, or the other way?
I would suggest that you rewrite your attempt keeping all the variables clear by labelling them.

• gneill
haruspex
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solving for x1 in the second equation,
One of the commonest blunders in physics is to apply equations without due regard to the meanings of the variables within them. In that equation, the displacement, velocities and acceleration are all in the same direction. Does that apply to the values you have plugged in here?

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I will redo my attempt then.

Sum of the potential energy and kinetic energy at the top equals the sum of potential and kinetic energy at the bottom.

At the top of the incline, the block is released from rest, so Kinetic energy is zero and potential energy is at its maximum value. As it moves down the ramp, the kinetic energy increases as the potenial energy decreases. At the bottom of the ramp, It surely has most of the kinetic energy and some potential energy since it is at a height of D from ground. Therefore, the potential energy at the top is equal to the mass, M, times acceleration due to gravity, g, times the change in height, which is from the the top of the ramp to the bottom of the ramp or simply H as labeled in the diagram, and the kinetic energy is zero. Then, at the bottom of the ramp, the potential energy is equal to M times g times the change of height from the bottom of the ramp to ground or D, and the kinetic energy equal half times the mass times change in velocity, which is final velocity minus initial velocity squared, or simply final velocity squared. Until here, is there something that I misunderstand?

MgH= MgD+1/2M(Vf)^2

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One of the commonest blunders in physics is to apply equations without due regard to the meanings of the variables within it. In that equation, the displacement, velocities and acceleration are all in the same direction. Does that apply to the values you have plugged in here?

They are all in the same direction, supposedly. But I don’t know what are you referring to. Do you mean that the equation I wrote above for X1 has different directions?

haruspex
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They are all in the same direction,
x1 and the velocity you found in the first step are horizontal, but g is vertical.
Consider the vertical motion after it leaves the ramp. What is vi for that?

x1 and the velocity you found in the first step are horizontal, but g is vertical.
Consider the vertical motion after it leaves the ramp. What is vi for that?
I think I got what you mean. But still I am confused. Here the g is vertical but then i will need the horizontal acceleration. Am I to use this rule: Vf=Vi+at? (Final velocity = initial velocity plus acceleration times time)

What is vi for that?
That would be the final velocity at the end of the ramp? If the rule in post #8 is the right one, should I consider the final velocity where the block hits the ground is zero?

haruspex
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That would be the final velocity at the end of the ramp?
It would be the vertical component of that velocity. What would that be?
I think I got what you mean. But still I am confused. Here the g is vertical but then i will need the horizontal acceleration.
For the horizontal aspect of the motion after leaving the ramp, yes, you wouid need the horizontal component of its acceleration while it is in the air. (What would that be?)
But for now I want you to concentrate on the vertical motion.
If the rule in post #8 is the right one, should I consider the final velocity where the block hits the ground is zero?
The velocity only becomes zero after it has hit the ground. The SUVAT rules only apply while acceleration is constant, and hitting the ground violates that, so "final velocity" means just before hitting the ground. That will not be zero.

What would that be?
That depends if the rule I wrote in post #5 is correct. If so, solving for Vf gives the vertical component of the velocity which is the final velocity at the end of the ramp.

But for now I want you to concentrate on the vertical motion.
So I should only know that in the vertical motion it’s just the acceleration due to gravity.

Should I start again solving for horizontal and vertical components separately, like from the top of the ramp to the moment just before hitting the groun?

yes, you wouid need the horizontal component of its acceleration while it is in the air. (What would that be?)

Wait, do you mean when it’s moving down the ramp, I should figure out the horizontal component of the acceleration?

haruspex
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Wait, do you mean when it’s moving down the ramp, I should figure out the horizontal component of the acceleration?
No, we are done with the motion down the ramp. We are now only concerned with airborne motion from ramp to ground.
You found the speed at the point where it leaves the ramp. In what direction is it travelling at that point?
Look at the picture and read the text.

In what direction is it travelling at that point?
From the moment it leaves the ramp, it moves a horizontal distance x1, so the direction is horizontal.

Wait I think I know now that acceleration in the horizontal is zero, so when solving for horizontal distance I use X1= initial velocity times time. Correct?

PeroK
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Wait I think I know now that acceleration in the horizontal is zero, so when solving for horizontal distance I use X1= initial velocity times time. Correct?

Yes.

• YMMMA
Finally, to get the horizontal distance, I used H, D, t, and M. Answer should be E for the first question,yes?

Next question: doubling the masses won’t affect the horizontal distance, because masses cancel out from the equation in post #5.

Last question: friction resists motion,which will always decrease the distance. In the equation I will minus the work of friction from potential energy at the top of the ramp; thus, it decreases the final velocity and the horizontal distance.

PeroK
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Finally, to get the horizontal distance, I used H, D, t, and M. Answer should be E for the first question,yes?

Just because you used H, D, t and M, doesn't mean you actually needed them. The answer is independent of one of these; and, it is possible to calculate one of these from the others. Any ideas?

Just because you used H, D, t and M, doesn't mean you actually needed them. The answer is independent of one of these; and, it is possible to calculate one of these from the others. Any ideas?
Yes, masses cancel out so no need for M. But I need time to get the horizontal distance and also H and D. There is no other choice with only these three.

PeroK
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Yes, masses cancel out so no need for M. But I need time to get the horizontal distance and also H and D. There is no other choice with only these three.

The time required to calculate ##x_1## is the time that the block is in the air. Why would you use the time it takes to slide down the track? That is not the time interval during which the block is flying a distance of ##x_1##.

The time required to calculate ##x_1## is the time that the block is in the air. Why would you use the time it takes to slide down the track? That is not the time interval during which the block is flying a distance of ##x_1##.
Oh, I forgot to recheck what is t in the given. Alright it’s B.

Next question: doubling the masses won’t affect the horizontal distance, because masses cancel out from the equation in post #5.

Last question: friction resists motion,which will always decrease the distance. In the equation I will minus the work of friction from potential energy at the top of the ramp; thus, it decreases the final velocity and the horizontal distance.

Are these correct?

PeroK
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Oh, I forgot to recheck what is t in the given. Alright it’s B.

Are these correct?

Yes, it's B. Although, you should note that the time it takes for the block to travel ##x_1## is determined by the vertical height ##D## that it falls. Any projectile that starts horizontally from a height ##D## takes the same time to fall.

Yes, I got that. Thank you all for your help and time!!!  