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Physics: Post your doubts here!

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sagar65265

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Salman917 said:
when will be the force of gravitation between 2 bodies will be zero n y?????????????
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It depends on the two bodies involved.

For example, if you have a circular ring (or a symmetrical one, it can even be an oval for that matter) somewhere in space and a particle is placed at it's center, the gravitational force on it is Zero. As that particle is moved away from the ring, the force of attraction increases at first, then decreases, and decreases until it reaches Zero at Infinity.

This is because, at the center of the ring, all the matter surrounding that point is symmetrically distributed; since segment of the ring exerts the same force on any particle at that center-point and the entire construct is symmetrical, you basically get pairs of forces cancelling each other out (segments of the ring at diametrically opposite points exert a net force of Zero; since the entire construct is like this, the net force there is also Zero).

As the particle moves away, the forces decrease due to increasing distance; the gravitational force follows an inverse-square law, so the force decreases with distance, and it decreasing very fast.

Another (very non-intuitive) example is a spherical shell; suppose you have a spherical shell (a hollowed out sphere would be a good way of describing it) and inside that shell you place a particle, it doesn't matter where you place that particle; as long as that particle is inside the hollow, the gravitational force on it will be Zero. There's very little physical explanation for that I can think of, aside from "the calculus works out in that way".
The beginning of this HyperPhysics page gives a lovely explanation that's really worth reading, and the calculus after that should give you a good idea of just how powerful the methods of calculus are, although that derivation is DEFINITELY NOT needed for CIE AS & A Levels:

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell2.html

And this last one applies for ANY two or more objects - as the distance between them increases, the gravitational forces exerted by each on the other also decrease, thus reaching (but never equaling) zero at a separation distance of Infinity. Usually it is so negligible at large values of separation distance that it has no tangible effect and can be stated as practically zero.
So, it's rather interesting to imagine that a star several dozens times heavier than our sun, in some far-away galaxy, is exerting a gravitational force on you and every particle around you, but is too weak to actually show!

Hope this helped!
Good Luck for all your exams!
 
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Thought blocker

Thought blocker

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ZaqZainab said:
1 would be D because of the formula = Efficiency= Useful Output/Input
2 would be C from the formula work done= Pressure*change in volume this formula is also given in the data booklet
change in volume is (V1-V2)
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In 1) Efficiency = (output/input)*100 so, where is *100 ?? :(
 
usama321

usama321

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Thought blocker said:
View attachment 38385
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PE lost from P to Q is converted to KE. Thus the 50J at P is converted to KE, that is 50J. It also has initial KE of 5J, so total becomes 55J. 10J is lost in work against friction, so the rest that is left is 45J.
Thought blocker said:
Answer is B or D ?
Explain please

View attachment 38388
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The answer should be B

At the greatest height, the velocity of the ball would be 0. Thus this option can be ruled out.

For B, as we are neglecting air resistance, no energy is lost while doing work against friction. Thus, all the initial KE is converted to PE, and back to KE as it falls. This should be correct.

For C, the principle of conservation of momentum applies to a system, not just the ball. We would have considered this if it had stated that the momentum of the ball and the earth remained constant.

For D, we know that the ball is decelerating. Thus, it does not cover the same amount of distance every second. As PE depends on height, it therefore can not increase uniformly
 
Thought blocker

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usama321 said:
PE lost from P to Q is converted to KE. Thus the 50J at P is converted to KE, that is 50J. It also has initial KE of 5J, so total becomes 55J. 10J is lost in work against friction, so the rest that is left is 45J.


The answer should be B

At the greatest height, the velocity of the ball would be 0. Thus this option can be ruled out.

For B, as we are neglecting air resistance, no energy is lost while doing work against friction. Thus, all the initial KE is converted to PE, and back to KE as it falls. This should be correct.

For C, the principle of conservation of momentum applies to a system, not just the ball. We would have considered this if it had stated that the momentum of the ball and the earth remained constant.

For D, we know that the ball is decelerating. Thus, it does not cover the same amount of distance every second. As PE depends on height, it therefore can not increase uniformly
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ty :)
 
zonaali

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sagar65265 said:
The two most important things to note for this question are as follows:

i) Since the stri
sagar65265 said:
The two most important things to note for this question are as follows:

i) Since the string is "mass-less", the tension in every part of the string is the same. In other words, the string pulls each block with the same force; it pulls the 2.0 kg block to the right with the same force as it tries to pull the 1.0 kg block upwards.

ii)Since the string is assumed to be in-extensible (in other words the length of the string cannot change, it cannot stretch at all) the acceleration magnitude of the two blocks are the same; if the 1.0 kg block moves a small distance downwards, the 2.0 kg block has to move the same distance in the horizontal direction. So, since both of them have to fall the same distance in the same time interval, they have the same velocity. From here, since any change in the speed of the 1.0 kg block during a particular time interval has to result in the SAME change in speed of the 2.0 kg block during the EXACT SAME time interval, the two accelerations at any point are the same.

Let's define the directions, first: for the 1.0 kg block the positive direction is downwards and the negative direction is upwards (it doesn't travel or accelerate sideways, so we don't need to worry about that direction). Also, for the 2.0 kg block, the positive direction is to the right and the negative direction is to the left.
We've chosen the axis so that the accelerations are all positive in sign, so that we don't have to juggle around any confusing negative values.

So, taking Newton's second law for the 1.0 kg block (Net Force = Mass * Acceleration):

(1.0 kg)(9.81 ms^-2) - T = (1.0 kg)a
9.81 - T = a

Alright, so that's the first equation. Considering the 2.0 kg block:

T = (2.0 kg)a
T = 2.0a

And there's the second one. Since there is no friction acting on the 2.0 kg block, that's it!

Now, since the two accelerations and the two tension forces are the same, we can substitute the second equation into the first as:

9.81 - 2.0a = 1.0a
9.81 = 3.0a

Therefore, a = 9.81/3 = 3.27 ms^-2

Since the acceleration here is constant and does not change with time, we can use the constant acceleration kinematics equations with:
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.50 m (since both blocks travel this distance before the 1.0 kg block hits the ground and stops)
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 + 2 * 3.27 * 0.5
v^2 = 3.27

Therefore, the final speed is √(3.27) = 1.8(08) ms^-1 = 1.8 ms^-1 = A.

Hope this helped!
Good Luck for all your exams!
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Thanks alot sagar. Its been a great help. Thank you very much. God bless.
ng is "mass-less", the tension in every part of the string is the same. In other words, the string pulls each block with the same force; it pulls the 2.0 kg block to the right with the same force as it tries to pull the 1.0 kg block upwards.

ii)Since the string is assumed to be in-extensible (in other words the length of the string cannot change, it cannot stretch at all) the acceleration magnitude of the two blocks are the same; if the 1.0 kg block moves a small distance downwards, the 2.0 kg block has to move the same distance in the horizontal direction. So, since both of them have to fall the same distance in the same time interval, they have the same velocity. From here, since any change in the speed of the 1.0 kg block during a particular time interval has to result in the SAME change in speed of the 2.0 kg block during the EXACT SAME time interval, the two accelerations at any point are the same.

Let's define the directions, first: for the 1.0 kg block the positive direction is downwards and the negative direction is upwards (it doesn't travel or accelerate sideways, so we don't need to worry about that direction). Also, for the 2.0 kg block, the positive direction is to the right and the negative direction is to the left.
We've chosen the axis so that the accelerations are all positive in sign, so that we don't have to juggle around any confusing negative values.

So, taking Newton's second law for the 1.0 kg block (Net Force = Mass * Acceleration):

(1.0 kg)(9.81 ms^-2) - T = (1.0 kg)a
9.81 - T = a

Alright, so that's the first equation. Considering the 2.0 kg block:

T = (2.0 kg)a
T = 2.0a

And there's the second one. Since there is no friction acting on the 2.0 kg block, that's it!

Now, since the two accelerations and the two tension forces are the same, we can substitute the second equation into the first as:

9.81 - 2.0a = 1.0a
9.81 = 3.0a

Therefore, a = 9.81/3 = 3.27 ms^-2

Since the acceleration here is constant and does not change with time, we can use the constant acceleration kinematics equations with:
u = 0 ms^-1
a = 3.27 ms^-2
s = 0.50 m (since both blocks travel this distance before the 1.0 kg block hits the ground and stops)
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 + 2 * 3.27 * 0.5
v^2 = 3.27

Therefore, the final speed is √(3.27) = 1.8(08) ms^-1 = 1.8 ms^-1 = A.

Hope this helped!
Good Luck for all your exams!
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