Physics: Post your doubts here!

[Aishayasin]

PART Cii in detail

 

[Aishayasin]

View attachment 61485
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View attachment 61486

PART Cii in detail

Please can aNyone help me with part cii in detail as i am unable to solve these type of question

 

[Thought blocker]

Please can aNyone help me with part cii in detail as i am unable to solve these type of question

These are the easiest ones.
A = Ao e^(-lambda x t) right?
We have to find A/Ao
Thus,
A/Ao = e^(-lambda x t)
we have lambda and t both given.
lambda = 0.025 years
and t = 5 years

Substitute this in the formula, and then answer will be 0.88 (2dp)

 

[nehaoscar]

Here are complete AS and A2 notes for physics 9702!
The notes include each and every point of the syllabus

https://www.xtremepapers.com/community/threads/9702-physics-notes.44847/

Please like and share if you find them helpful
Check out my signature for more A level notes and revision resources!

 

[Frankuzi]

View attachment 61396
Why is the answer B?

I think for this type of question playing around with conservation of linear momentum is the key
Let's consider a scenario where we first pour the sand into the cart, With the equation:
M(cart)*U(cart) = [M(cart) + M(sand)]*V
Rearranging the equation therefore we have:
V = [M(cart)*U(cart)]/[M(cart) + M(sand)] * U(cart)
This in return makes V less than U(cart) so the speed at X will decrease when being compared to its original state before adding the sand in.

Let us continue with the another scenario at Y,
similarly we will be using the conservation of linear momentum to solve for this,
When comparing scenario at X and Y,
[M(cart) + M(Sand)]*V =M(cart)*V ' (This is the case where the sand is dropped from the cart),
therefore making V ' the subject we have,
V' = [M(cart) + M(Sand)]*V/ [M(cart)]

Using the two underlined equation, simply play around with the equation by substituting the V,
Simplified and you will get V' = U which in the end the velocity remained unchanged as it was from the
beginning .

My way of approaching this type of question is usually in the sense of mathematical interpretation so there might be other way of explaining or visualizing this but feel free to correct or ask me if there is any problem with the explaination

 

[Salman Dhillon]

Anyone having complete As and A level formulae list of physics?

 

[Thought blocker]

Anyone having complete As and A level formulae list of physics?

Mr. Salman,

Here is my awesome thread on physics, chemistry and math notes compilation. Look at the image below, that two yellow files in physics section has your formulas.

 

[sadafas]

can anyone please upload detailed notes for Aslevel physics?

 

[Thought blocker]

can anyone please upload detailed notes for Aslevel physics?

Click the first link in my signature in blue colour and go to physics section also see at last uploaded files related to physics of first post in that thread.

 

[drowning-in-studies]

where can i possibly get paper 5 solved papers?

 

[Thought blocker]

where can i possibly get paper 5 solved papers?

http://cieoandalevelnotes.blogspot.in/

 

[drowning-in-studies]

http://cieoandalevelnotes.blogspot.in/

thanks

 

[i_try9621]

Need help with b) ii). It's Q1 from 9702/41/M/J/10

 

[DragonCub]

View attachment 61541 Need help with b) ii). It's Q1 from 9702/41/M/J/10

 

[i_try9621]

View attachment 61545

Why should the stone weight matter here? Shouldn't the centripetal force and the Tension be equal?

 

[DragonCub]

Why should the stone weight matter here? Shouldn't the centripetal force and the Tension be equal?

When talking about circular motion, the centripetal force should always be the net force on the object. This is why we need to take into account every single force on the stone, including its weight.
Why necessarily net force? Because Newton's Second Law demands that the F in F = ma be the net force, in every case; and in circular motion, centripetal acceleration and centripetal force also obeys Newtons Second Law.

 

[i_try9621]

When talking about circular motion, the centripetal force should always be the net force on the object. This is why we need to take into account every single force on the stone, including its weight.
Why necessarily net force? Because Newton's Second Law demands that the F in F = ma be the net force, in every case; and in circular motion, centripetal acceleration and centripetal force also obeys Newtons Second Law.

So the weight also acts towards the centre?

 

[DragonCub]

So the weight also acts towards the centre?

Not always.

The weight always acts downward.
And since:
1. the centripetal force (F_c) is constantly changing direction as the object changes position
2. F_c = F + F_g ( F is the force exerted by the glue)
So...
The required magnitude of F is also changing.
And the largest F required corresponds to the "bottom" instant, where F_c is upwards and F_g is downwards (opposite to each other).

 

[i_try9621]

Not always.

The weight always acts downward.
And since:
1. the centripetal force (F_c) is constantly changing direction as the object changes position
2. F_c = F + F_g ( F is the force exerted by the glue)
So...
The required magnitude of F is also changing.
And the largest F required corresponds to the "bottom" instant, where F_c is upwards and F_g is downwards (opposite to each other).

As you mentioned above in the first reply, that the gravity of the stone is exerted on the glue. Could that be the reason why the weight should be included here ? Or am I looking at it the wrong way?

 

[DragonCub]

As you mentioned above in the first reply, that the gravity of the stone is exerted on the glue. Could that be the reason why the weight should be included here ? Or am I looking at it the wrong way?

I think I explained that a bit vaguely at first.

Let's start it again from the beginning. First, two things hold true at any moment:
1. F_c = F + F_g
2. F_g always acts downward

Now, since we know that a force has directions, that may affect its sign (positive / negative), let's suppose that, on the vertical axis, upward is positive, downward is negative.
F_g always acts downward, so we can say that F_g = - 3.0 N.
I know the question says that the stone's gravity is 3.0 N, without the negative sign. It actually means that the gravity has a magnitude of 3.0 N.

Here, the magnitude of something means the modulus, or absolute value, or that thing. For gravity, its magnitude means | F_g |.
We can also say that the magnitude of a force does not consider the force's direction, so it doesn't have a +/- discretion.

But once we start to consider directions, we need to use +/- signs to designate them. And surely you can designate downward as positive - either way is okay. But let's take upward as "+" for this time.

We know that F_c always acts towards the centre. And let's suppose that F_c has a magnitude of 10 N.
Then, when the stone is at the top, F_c is acting downward (centre is below the stone). At this moment, F_c = -10 N.
Based on F_c = F + F_g, we can get: -10 = F + (-3) --> F = -7 N then F has a magnitude of 7 N.

When the stone is at the bottom, F_c is acting upward (centre is above the stone). Then F_c = +10 N.
Again, using F_c = F + F_g, --> +10 = F + (-3) --> F = +13 N --> magnitude of F is 13 N.

You can see, at the top and the bottom, the gravity has opposite effects on the required magnitude of F.
This is why we need to take into account the stone's gravity.



P.S. We can also discuss on the moments when the stone is horizontally level with the centre (centripetal force acting horizontally), but that will involve the horizontal axis as well, and make things more complicated.
But if you are curious enough, we can discuss this in a separate chat.