Determining 'g' Using a Free-Fall Method - PRACTICAL - A Level Physics
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- Опубликовано: 7 фев 2025
- In this video I go through an AQA Physics A Level Required Practical that uses a free-fall method experiment to calculate the acceleration due to gravity, g. (This is AQA Required Practical 3.) This is also the OCR A Level Physics PAG 1.1 Practical.
Dropping a steel ball bearing through light gates can be used to find the acceleration due to gravity. Recording the time taken for it to fall as you then change the distance between the gates allows a graph to be plotted and the value of 'g' to be determined.
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Lewis
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Me: Gotta do the experiment to get to the accurate value of g.
Systematic and Random Errors: Nope
You can avoid the systematic error by using a better electromagnet which does not retain magnetism when turned off and random errors by keeping your eye level perpendicular to point of reading, along with using a metre ruler with low uncertainty.
im so fucked
with u man🤣🤣🤣🤣🤣
REALLL 😭🙏🏻🙏🏻
Bismillah and Alhamdulillah
As Salaam alaikum brothers and sisters ❤!
I got -16.67, many thanks
never thought id need this video for university too
Mayur Which degree is that?
GCSE and A Level Physics Online can you please send me the table of your results to compare against mine also your graph which is fully plotted. This would be very much appreciated
@@abdirahman4642 lmao nice try
Can you upload all the core practicals and excess ones for the new specifications please?
Amazing video
This is a lifesaver! Thank you!
I suppose the function of each light gate is to measure the velocity, the first one is used for measuring the initial and the second one is for final velocity, then we use v^2=u^2 + 2aS to find the acceleration.
I think hes using the light gate as a stopwatch to then plot a graph with x axis as time
So basically the graph will work out the acceleration for you because youre seeing the change in velocity (S/T) -y axis by the time (T) - x axis
No he mentioned that the distance 'x' was to reach the terminal velocity, so meaning there wont be a change in speed anymore. So using v2 = u2 + 2as wouldnt be useful. The point of the light gate is to act as a timer between the point at which it reached TermV and the point it reaches the Bottom
Sorry bud, but this does not work as V is a variable whose value we cannot calculate instanteously. Rather, we must use S = ut + 1/2at^2 assuming that u is 0 and that you know the displacement S.
Can you upload videos of all the practicals required for AS Physics pls!!
Ah V I have! In the video description there is a link to all the required practicals for AS.
i'm going through a unit 1 physics paper and there's a question on practicals! for unit 1, will they only ask questions on the core practicals listed on your site?
this is for the IAL spec btw, I thought practical questions are supposed to be on unit 3 but they're appearing on unit 1?
You have to know the core practicals - these are slightly different for each exam board so make sure you read the specification thoroughly.
But they could throw in a few other practicals - but these will be about something you will have learnt about and will be about your ability to answer practical questions or interpret data.
You can hear all the flat earthers scream in pain.
Is this only for aqa or for all specifications?
What is g for no initial velocity for a object that just falls and for a object that's thrown up into the air with an initial velocity upwards in free fall?
i didnot understand the part where u wrote 2u, please clarify.
still need help?
Thanks for this video; much appreciated sir. I have a question: you mentioned using a cube is better than a sphere, why is that?
I think it's more more likely to stay on the electromagnet.
Are you working with uplearn sir?
They licensed some of my videos for their physics content!
When reading off gradients and y-intercepts from graphs, how many sig figs/decimals places can I quote my answer to ?
What one advantage and disadvantage of making the distance between the light gate as large as possible?
😂thanks for the reply even tho my exams were 2 years ago😂😂
How do you know u tho, as the cube will have accelerated before passing through the first light gate?
Is this similar to how we got the value of 9.806... for g on Earth? Or did that come about in a completely different way?
Quiet smelly poo
Could we use the equation v = u + at to plot the graph?
You dropping from distance you can do I think but you have to measure time them
And ik I'm late
there's a mistake in the video, notice that in the equation '2S/t=2u+at' t cannot equal 0, so in the graph, u should plot it out.
No 2s/t cannot reach 0, hence why the graph does not start from the origin.
thank you so much for this
is there any sources of error that might occur during this experiment
make sure the stand cannot topple over by clamping it securely
also make sure you are measuring distance between the two lightgates from eye level .
I have a question about the significance of u, if the ball is being dropped, won't it have 0 initial velocity, or do we take the initial speed of when it passes the first light gate?
Since the initial time doesn't start until it passes through the first gate, that is our initial reference point sonthatbis what we consider to be u. V=0 at some negative time before u
@@markm4952 so how would we find u?
I got g = 14.6 when I did this xD
Seems good enough
I did better with a rock and 3 rulers. Bitch
Bro I got -0.5
I got 8.8🤧🤧🤧
NO WAY I GOT 14.62 LOL. I’m 7 years late though
Why should x be constant in this experiment?
how do measure the initial velocity?
It's 0 because it starts at rest
@@amirpenkar947 no because then 2u=0 which isnt the case for this mans graph
Does it matter at what velocity the object emters the first light gate?
it s easiest if it is zero because then in the equation you can have u (initial velocity) as 0
No my friend. As long as you have got a value for the initial velocity and the final velocity, you are able to work out the acceleration using the time and the distance between the light gates.
Thank you for the helpful video. Just one thing: why do we need to rearrange the formula? (stupid question I know, but I just need to clear it up)
Umamah Rehman This is just one way of doing it, so that the gradient is exactly equal to the acceleration due to gravity.
OK that makes sense. Do we need to know why the formula is arranged in this particular way? Also, we don't need to find u for this version of the experiment, do we? As long as distance x is constant, u shouldn't affect the results?
Umamah Rehman That’s correct. You don’t have to remember why the formula is like that, but it is something you should be able to follow.
Thank you very much for helping out!
Why did you add the 2s in the equation ?
he multiplied everything by 2
I don’t get y u put a 2 infront of ut +at^2 ?
he multiplied everything by 2 to get 2s because the graph plots 2s/t
why does u double? 2:28
he got rid of the 1/2 so he needed to times everything by 2
The way i understood nothing and my test is tmrw. Pray for me
why would it be better to use a cube??
um why is it better to use a cube than a sphere?
So it dosent roll off or bounce when it hits the floor
why is a cube better than a ball?
what happened to the squared of T?????? or am I the only one confused??
he divided the whole equation by t
Actually, why didn’t you rearrange the equation to make acceleration the subject of the formula ?
Because you're plotting 2s/t (on the y axis) against t (on the x axis) so you make 2s/t the subject
I think it is because we want the acceleration to be the GRADIENT of the graph. That would not be the case if acceleration was on one of the axes. It just so happens that when you plot a graph of 2s/t (which is in itself a totally abstract concept) against the time, then the gradient equals the acceleration, which is precisely what we are trying to find in this experiment to determine the value of g. I hope I'm right?
How accurate is this experiment
Is calculating "u" important? Or I'll plot the graph and then find the y-intercept with a best fit line?
"U" will be zero as the object starts from rest, so no
You can't plot a graph from this unless you change 's' so how is 'x' constant?
You lower the bottom light gate. The top light gate, closest to the electromagnet, must be at a fixed distance from the drop height the entire time. So 'x' is kept constant.
in this wouldn't the value of initial velocity(u) be 0? leading to the equation to be s = 0.5 x a x t^2 ?
ARYAN DHAKA No, because the initial velocity 'u' is taken to be the velocity of the object at the instant it passes through the top light gate, not the instant the object is dropped (in which case, 'u' would indeed be zero as it would be dropped from rest).
Can anyone tell me how do we get the value of 'u'?
it is the initial velocity, so it is easiest if you drop the ball just above the light gate so it is zero
When we did the required practical we used s (dist.), as the independent variable, and t^2 as the dependant, giving a gradient of 2/g. Which method of taking data is better?
Cosmin Covrig depends exactly what version of the experiment you do. Many of them don't require the initial speed part of the equation since you don't have initial speed to take into account, in which case t^2 against h is fine, the graph should pass through the origin and the gradient is equal to 2/g.
As I say, it really does depend which version you're doing as to what data you need to take into account.
Multiplication of 2 in step 2nd is wrong..😐
He corrected it afterwards anyway.