Momentum Help -- simulate the movement of a puck

[Tasdel]

Hi there, I am doing a physics assignment. I had to use a program called tracker to simulate the movement of a puck.
So I just need some help to see if anyone can figure out what I've done wrong as something isn't making much sense. Let's jump right into it.

When the puck moves down and hits the "fence" it bounces back with the same constant velocity in the opposite direction. I calculated (when its going down towards the fence) it is going -0.7m/s. The puck had a mass off 0.025kg.

So I calculated my initial velocity or pi = mi * -vi = 0.025 * 0.7 = -0.0175
Final velocity, its going back up away from the fence. so: pi = mi * vi = 0.0175
NOW I had to calculate the change in momentum. Δp=pf - pi = 0.0175 - (-0.0175) = 0.035 kg *m/s

Is this ok so far?
The force of the puck bouncing (according to my graph and it was almost the exact same as the examples graph for my project which was around 0.37N)

The puck literally touched the fence for 0.046 seconds. (I found the initial time and final time of contact)
I had to find the Favg of the puck. so they give me the equation Fave*Δt =Δp
so I did F = Δp/Δt.
F=0.037/0.046
=0.80.
So my main questions are
1)did I calculate the Δp right?
2)why is the force I am calculating a little more than double the force? did I do the force wrong?
3)when does it matter using p=mv and p=mΔv?

Thanks so much for your time. Sorry for the silly questions.

 

[Merlin3189]

...simulate the movement of a puck.
...

When the puck moves down and hits the "fence" it bounces back with the same constant velocity in the opposite direction. I calculated (when its going down towards the fence) it is going -0.7m/s. The puck had a mass off 0.025kg.

So I calculated my initial velocity or pi = mi * -vi = 0.025 * 0.7 = -0.0175
Actually you calculated initial momentum and that's correct
Final velocity, its going back up away from the fence. so: pi = mi * vi = 0.0175
Final momentum and that's ok
NOW I had to calculate the change in momentum. Δp=pf - pi = 0.0175 - (-0.0175) = 0.035 kg *m/s

Is this ok so far? Yes.
The force of the puck bouncing (according to my graph and it was almost the exact same as the examples graph for my project which was around 0.37N) Don't really know what you're saying here? Do you mean this is the answer you were given?

The puck literally touched the fence for 0.046 seconds. (I found the initial time and final time of contact)
I had to find the Favg of the puck. so they give me the equation Fave*Δt =Δp Yes, Impulse = change in momentum
so I did F = Δp/Δt. Ok
F=0.037/0.046 Well, I think Δp was 0.035 a few seconds ago, but near enough!
=0.80. Ok
So my main questions are
1)did I calculate the Δp right? Yes
2)why is the force I am calculating a little more than double the force? did I do the force wrong? Don't understand this question? But then I couldn't understand what you were saying about the 0.37 N before.
3)when does it matter using p=mv and p=mΔv? Δx means "change in x"
p=mv says "momentum=mass x velocity" and p=mΔv says "momentum=mass x change in velocity", so here it should probably be Δp=mΔv "change in momentum = mass x change in velocity"

Thanks so much for your time. Sorry for the silly questions.

 

[Tasdel]

Haha. Thanks for the response. Yes I meant to put 0.035 sorry. I am just thinking because of this dumb graph that I was given, was saying at the time of force of impact when the puck is bouncing off the fence, it had a force of 0.37 Newtons. It could be a flawed assignment. Most of the links I was given for the assignment don't even work. Thanks for your help.

 

[Merlin3189]

As you indicated, the force varies over the period of impact. You have calculated an average force over the duration of the impact. If the 0.37N is the peak force, I can't see how that could square with an average of 0.8N.
But if the average force is 0.8N, then at some time during the impact (in fact twice), the force would have been momentarily 0.37N
Anyhow, I think your calculation is sound, so see what your tutor says.

 

[Tasdel]

As you indicated, the force varies over the period of impact. You have calculated an average force over the duration of the impact. If the 0.37N is the peak force, I can't see how that could square with an average of 0.8N.
But if the average force is 0.8N, then at some time during the impact (in fact twice), the force would have been momentarily 0.37N
Anyhow, I think your calculation is sound, so see what your tutor says.

Thanks for your time Merlin, just wanted to make sure I wasn't just missing something!

 

[haruspex]

The force of the puck bouncing (according to my graph and it was almost the exact same as the examples graph for my project

Would you mind explaining that some more? Are you saying there are two graphs showing ... what ... force as a function of time? One provided with the assignment and one you obtained in your experiment? If so, what shape are the graphs, and what does the 0.37N relate to in them?

 

[Tasdel]

Would you mind explaining that some more? Are you saying there are two graphs showing ... what ... force as a function of time? One provided with the assignment and one you obtained in your experiment? If so, what shape are the graphs, and what does the 0.37N relate to in them?

It was a really crappy one second video of a puck bouncing off a fence.
Yes it was a force vs time graph. Right when the puck bounces off the fence, the force exerted on the puck (according to the graph) was 0.37N of force.
Here is an image of the graph. I am still confused about why my force in my calculation is off...

Please note at the top of the graph (around 0.25s) that's when the puck is bouncing off the fence. Also on my graph I used 0.37N because that was what I had, this is the demo graph from my class which used 0.42N at 0.25s.

 

[haruspex]

It was a really crappy one second video of a puck bouncing off a fence.
Yes it was a force vs time graph. Right when the puck bounces off the fence, the force exerted on the puck (according to the graph) was 0.37N of force.
Here is an image of the graph. I am still confused about why my force in my calculation is off...

Please note at the top of the graph (around 0.25s) that's when the puck is bouncing off the fence. Also on my graph I used 0.37N because that was what I had, this is the demo graph from my class which used 0.42N at 0.25s.

I'm as confused as ever.
Are there two graphs of force against time, one that goes with the video and one from your own data? If so, please post both.
What makes you think the video experiment should involve the same forces that arose in your experiment?
What do you mean by "I used 0.37N because that was what I had". Where did you get 0.37N from in connection with your experiment?

 

[Tasdel]

I'm as confused as ever.
Are there two graphs of force against time, one that goes with the video and one from your own data? If so, please post both.
What makes you think the video experiment should involve the same forces that arose in your experiment?
What do you mean by "I used 0.37N because that was what I had". Where did you get 0.37N from in connection with your experiment?

It was from the graph I made given the same information as how the demo graph was made. I followed some instructions. I did it again and thankfully this time I had 0.42N as my max force when the puck bounced off the fence. The puck video was used to make the demo graph and my graph that I am making myself.

 

[haruspex]

It was from the graph I made given the same information as how the demo graph was made. I followed some instructions. I did it again and thankfully this time I had 0.42N as my max force when the puck bounced off the fence. The puck video was used to make the demo graph and my graph that I am making myself.

So both graphs were created from the video?
Are you trying to simulate the video?
I note that in the graph the contact time was about 0.15s, three times as long as you used in your calculation. Or are the units in the graph not seconds?

 

[Tasdel]

So both graphs were created from the video?
Are you trying to simulate the video?
I note that in the graph the contact time was about 0.15s, three times as long as you used in your calculation. Or are the units in the graph not seconds?

Yes both from the same video.
Yes, I had to use a program that tracks the movement of the puck and gives a graph of the movement. Interesting.. I did pick a long time, thanks for the note! they are used in seconds yes. Thanks! I really do make dumb mistakes a lot lol.

 

[Tasdel]

Yes both from the same video.
Yes, I had to use a program that tracks the movement of the puck and gives a graph of the movement. Interesting.. I did pick a long time, thanks for the note! they are used in seconds yes. Thanks! I really do make dumb mistakes a lot lol.

After looking at it again I thought it was about 0.08 seconds of contact. 0.15 would make more sense for my average force though. The average force should be around the middle right?

 

[haruspex]

thought it was about 0.08 seconds of contact.

I read it as lasting from 0.17s to 0.33s, a duration of 0.16s.

 

[Tasdel]

I read it as lasting from 0.17s to 0.33s, a duration of 0.16s.

Thanks for your help. My question all makes sense now, I am just a buffoon!

 

[Merlin3189]

Yes, now you have the time right, it all works out ok.

Just as an aside, I notice the force time graph is approximately triangular. So whatever the time was, even if there had been no markings on the time axis, the average force is about half the peak force.
(Average height = area / base = 0.5 x base x peak height / base = 0.5 peak height)

And since I wan't sure how the graph peaked between the 0.378N and the 0.414N points - ie whether it rounded off nicely just about there (which is how it appears in the video), or had shot up to a high peak between the two sample points (as if perhaps it had hit a solid backstop) - the result of your calculation, which now gives mean force 0.22N, suggests the peak is indeed about 0.44N,

And as a general hint for future queries, once you included all the info, you see how it became much easier for people to see what's actually going on and spot any errors for you.

 

[Tasdel]

Yes, now you have the time right, it all works out ok.

Just as an aside, I notice the force time graph is approximately triangular. So whatever the time was, even if there had been no markings on the time axis, the average force is about half the peak force.
(Average height = area / base = 0.5 x base x peak height / base = 0.5 peak height)

And since I wan't sure how the graph peaked between the 0.378N and the 0.414N points - ie whether it rounded off nicely just about there (which is how it appears in the video), or had shot up to a high peak between the two sample points (as if perhaps it had hit a solid backstop) - the result of your calculation, which now gives mean force 0.22N, suggests the peak is indeed about 0.44N,

And as a general hint for future queries, once you included all the info, you see how it became much easier for people to see what's actually going on and spot any errors for you.

Totally, Thanks a lot Merlin and haruspex. I really appreciate your help and time. I'm not used to using physics forums so its been a great first experience thank you both. Perfect, also thanks for the tip about triangular calculation of graphs.