Understanding Acceleration in Curling: An Analysis of Momentum in the Laboratory

In summary, the curling stone was sent down a rink and its distance and time traveled was recorded. The problem is that we need to find the acceleration of the stone. We could use any of the equations of uniform acceleration, but for this task we will use the equation that relates initial velocity, acceleration, and time. Once we have the initial velocity, we can use the equation with final velocity to find the acceleration. Finally, we need to solve for acceleration.
  • #1
The_big_dill
33
0
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!
 
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  • #2
The_big_dill said:
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
PhanthomJay said:
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
The_big_dill said:
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?
 
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  • #5
The_big_dill said:
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
The_big_dill said:
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
gneill said:
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
The equation you want to use does, in fact, work. It's a combination of two equations already.
 
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  • #10
p21bass said:
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
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
p21bass said:
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
gneill said:
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
The_big_dill said:
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
gneill said:
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 :)
 

1. What is momentum and how is it measured in a laboratory setting?

Momentum is a measure of an object's tendency to keep moving in the same direction with the same speed. In a laboratory setting, momentum is measured by multiplying an object's mass by its velocity.

2. How is momentum conserved in a laboratory experiment?

Momentum is conserved in a laboratory experiment through the law of conservation of momentum, which states that the total momentum of a closed system remains constant in the absence of external forces.

3. What are some common methods for analyzing momentum in a laboratory?

Some common methods for analyzing momentum in a laboratory include using a ballistic pendulum, measuring the change in momentum of colliding objects, and using force sensors and motion detectors to track and calculate momentum.

4. Can momentum be negative in a laboratory analysis?

Yes, momentum can be negative in a laboratory analysis. Negative momentum indicates that the object is moving in the opposite direction of the chosen positive direction. However, the magnitude of the negative momentum is still calculated the same way as positive momentum.

5. How does the conservation of momentum apply to real-world situations?

The conservation of momentum applies to real-world situations in many ways, such as in collisions between objects, motion of objects in space, and even in sports and daily activities. Any time there is a transfer of momentum between objects, the total amount of momentum remains constant.

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