Momentum laboratory analysis

  • #1
Welcome!

I am currently doing a laboratory assignment for school and am having trouble analyzing what the number i am getting mean.

Explanation:

We sent a curling stone down a rink and recorded its distance and time traveled.

Problem:

Determine the acceleration (deceleration) of the stone...

This means that we could use any of the equations of uniform acceleration.

Attempt:

For starters, i found the velocity (v = d/t); would the velocity i found out be change in velocity, initial or final?

To find out the acceleration, which equation of uniform acceleration would be most appropriate? (a= v/t?)

Thank You!
 

Answers and Replies

  • #2
PhanthomJay
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Welcome!

I am currently doing a laboratory assignment for school and am having trouble analyzing what the number i am getting mean.

Explanation:

We sent a curling stone down a rink and recorded its distance and time traveled.

Problem:

Determine the acceleration (deceleration) of the stone...

This means that we could use any of the equations of uniform acceleration.

Attempt:

For starters, i found the velocity (v = d/t); would the velocity i found out be change in velocity, initial or final?
Neither. When you use this equation, you are calculating the average velocity.
To find out the acceleration, which equation of uniform acceleration would be most appropriate? (a= v/t?)

Thank You!
acceleration is not v/t......acceleration is the (change in v)/t....but since you know distance and time, use one of the kinematic equations that relate distance with acceleration and time.
 
  • #3
Neither. When you use this equation, you are calculating the average velocity.acceleration is not v/t......acceleration is the (change in v)/t....but since you know distance and time, use one of the kinematic equations that relate distance with acceleration and time.
What would i have the velocity set as then, initial or final (that is, the velocity i calculated)
 
  • #4
gneill
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What would i have the velocity set as then, initial or final (that is, the velocity i calculated)
d/t gives an average velocity as was previously pointed out. It is not the initial velocity, nor is it the final velocity (zero in this case).

What kinematics expression do you know that relates initial velocity, acceleration, and time?
 
Last edited:
  • #5
PhanthomJay
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What would i have the velocity set as then, initial or final (that is, the velocity i calculated)
the velocity you calculated was the average velocity. I assume that you recorded the distance traveled when the stone came to a stop. Is this correct? If so, its final velocity was 0, its average velocity was what you calculated, and then its initial velocity you can determine from this info. But you still need to calculate the acceleration.
 
  • #6
So what you are telling me, is that i should find initial velocity as another step to find acceleration?

Wouldn't it be easier to use the equation with final velocity, since i already know it?
([tex]\Delta[/tex]d=(v2)([tex]\Delta[/tex]t)-1/2(a)([tex]\Delta[/tex]t)2)

If i were to find initial velocity, i could use this equation...

([tex]\Delta[/tex]d=(v1)([tex]\Delta[/tex]t)+1/2(a)([tex]\Delta[/tex]t)2)

But that would be an unnecessary step as i see it.
 
  • #7
gneill
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Wouldn't it be easier to use the equation with final velocity, since i already know it?
([tex]\Delta[/tex]d=(v2)([tex]\Delta[/tex]t)-1/2(a)([tex]\Delta[/tex]t)2)
What makes you think that you can plug any velocity you wish into the equation? The formula is:

[tex]\Delta d = v_i t + \frac{1}{2}a t^2[/tex]

[tex]v_f[/tex] is not [tex]v_i[/tex]

If i were to find initial velocity, i could use this equation...

([tex]\Delta[/tex]d=(v1)([tex]\Delta[/tex]t)+1/2(a)([tex]\Delta[/tex]t)2)

But that would be an unnecessary step as i see it.
You have two unknowns: a and vi. In mathematics, what do you usually need to solve for two unknowns?
 
  • #8
What makes you think that you can plug any velocity you wish into the equation? The formula is:

[tex]\Delta d = v_i t + \frac{1}{2}a t^2[/tex]

[tex]v_f[/tex] is not [tex]v_i[/tex]

You have two unknowns: a and vi. In mathematics, what do you usually need to solve for two unknowns?

... when did i say vf = vi?

I simply said, since the final velocity is 0, we can use that equation.

Two unknowns = 2 equations?
 
  • #9
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The equation you want to use does, in fact, work. It's a combination of two equations already.
 
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  • #10
You didn't explicitly say that vf = vi, but you put vf into the equation that specifically uses vi.

And no - you can't put it into that equation.
According to the kinematics equations i know, that is a real equation... note that there is a minus (-) instead of a (+) in that equation.
 
  • #11
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I realized what you did, sorry...

In essence, you've solved the equation vf= vi + at for vi and substituted it into the equation df = di + vit + at2/2.

Or, someone did, at least.
 
  • #12
gneill
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I realized what you did, sorry...

In essence, you've solved the equation vf= vi + at for vi and substituted it into the equation df = di + vit + at2/2.

Or, someone did, at least.
Ah. Okay, it wasn't clear that this had been done. I thought perhaps TBD (The Big Dill) was just assuming that there was a deceleration, hence the "-" sign.

So, he's got his two equations already massaged into one for the task of determining the acceleration. Job done!
 
  • #13
Ah. Okay, it wasn't clear that this had been done. I thought perhaps TBD (The Big Dill) was just assuming that there was a deceleration, hence the "-" sign.

So, he's got his two equations already massaged into one for the task of determining the acceleration. Job done!

Just to clarify, it is okay to use that equation with final velocity = 0?
 
  • #14
gneill
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20,816
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Just to clarify, it is okay to use that equation with final velocity = 0?
Yes, as long as you understand where it comes from!
 
  • #15
Yes, as long as you understand where it comes from!
Thanks a LOT!

i am so glad there is always someone, somewhere to help me out :)
 

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