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Physics: Post your doubts here!

Suchal Riaz

Suchal Riaz

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usama321

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Suchal Riaz said:
Very simple.
Find the minimum area for which the rod doesn't break then subtract that from the total area of wire to find the maximum area of bubble.
This might help you: View attachment 38254
Click to expand...
Ohk, thanks :) For the bubble to have the maximum area, the wire needs to have the minimum area, and the rest is occupied by the bubble. Thanks :)
 
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unique111

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What is the concept behind potentiometer? What are the formulae?
 

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Muskan Achhpilia

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Thank you!!!

Can you please help me in the following question too...
upload_2014-4-3_16-56-30.png

Then the question is-
A second ball is thrown from point P with the same velocity as the ball in the diagram. For this ball, air resistance is not negligible. This ball hits the wall and rebounds.On Fig. 2.1, sketch the path of this ball between point P and the point where it first hits
the ground.

The answer given is
smooth curve with ball hitting wall below original
smooth curve showing rebound to ground with correct reflection at wall

I am confused whether it is rebounding then hitting wall or hitting the wall then rebounding...

Thanks a tonne!
 

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unique111

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*the bottom of the ladder rests on rough ground where there is friction. The top of the ladder is at a height h above the ground and the foot .....
 

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unique111

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Muskan Achhpilia said:
Thank you!!!

Can you please help me in the following question too...
View attachment 38258

Then the question is-
A second ball is thrown from point P with the same velocity as the ball in the diagram. For this ball, air resistance is not negligible. This ball hits the wall and rebounds.On Fig. 2.1, sketch the path of this ball between point P and the point where it first hits
the ground.

The answer given is
smooth curve with ball hitting wall below original
smooth curve showing rebound to ground with correct reflection at wall

I am confused whether it is rebounding then hitting wall or hitting the wall then rebounding...

Thanks a tonne!
Click to expand...

It hits the wall below the original point first.
 
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sagar65265

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TheStallion-Reborn said:
Saad Mughal usama321 AbbbbY asma tareen been stuck for over hours now!! Any help would be really appreciatedView attachment 38272 !View attachment 38271
Click to expand...

For Question 13, there are a few important points to be noticed in the question:

i) the string is light, so we can assume that the magnitude of the tension for in the string is the same at all points along the string;

Suppose the string were indeed heavy, each part of the string would have a particular mass, and since the string itself is accelerating, there would have to be a net force on each one of those parts. However, the only forces on the string would be gravity and the 2 tension forces, one from each side. So, for a horizontal part of the string, the tension would have to be different at each point (otherwise no individual part could accelerate, and so the rope on a whole won't accelerate either.

ii)all surfaces is frictionless, so we need not make any assumptions concerning the magnitude of any friction forces and so this simplifies the question a lot.

Now onto Newton's Laws:

Taking the 2.0 kg mass into account, we need only look at the horizontal forces , since there is no vertical acceleration of that mass.
The only horizontal force acting on the mass is the Tension in the string (which, as noted before, is constant throughout the length of the string).

Therefore, T = (2.0) * a

In other words, our positive direction for the 2.0 kg mass is towards the right, and the tension force is positive since it acts towards the right.

Taking the 1.0 kg mass into account, we need only look at the vertical forces, since there is no horizontal/ sideways acceleration of the 1.0 kg mass.
The only 2 horizontal forces acting on the mass are the Tension in the string (same value as in the previous equation) and it's weight (= 1.0 * 9.81 = 9.81 Newtons).

Therefore, since Tension acts in the direction opposite to the acceleration and the weight acts in the direction of the acceleration,

-T + 9.81 = (1.0) * a

In other words, for the 1.0 kg mass, our positive direction is downwards. Therefore, the tension is negative and the weight in positive (in terms of sign value).

The acceleration of each object is the same, since the string can be assumed to be rigid and in-extensible (cannot be stretched), so any movement of one mass will the mirrored in the movement of the other mass, the only difference being the direction.

Solving these equations, we get a = 9.81 / 3 = 3.27 ms^-2

Since this acceleration is constant, we can apply the constant acceleration equations for either object. Suppose we take the 2.0 kg mass, it will have traveled 0.5 meters when the other object hits the ground, it started from rest, and it moves with the acceleration calculated above, so

u = 0 ms^-1
a = 3.27 ms^-2
s = 0.5 m
v = ?

v^2 = u^2 + 2 * a * s
v^2 = 0 +3.27
So, the velocity v = 1.81 ms^-1 = A

For the next question, we can use several theories and laws, but lack of available information constricts our choices like so:

i) We can use Newton's Laws, but we do not know what the magnitudes of the forces that act between the two objects during the collision are; we also do not know how long the collision lasts, so we cannot answer this question using Newton's Laws.

ii) We can use the Work-Energy theorem, but there too we don't know the magnitudes of the forces acting during the collision and we do not know the distance the object move during the collision. So, we cannot use the Work-Energy theorem either.

iii) We can use the Principle of Conservation of Momentum, which does allow us to answer this question; the reason for this is that we do not need to know the forces acting between the objects involved if we take them both as our system.

So, taking both the pellet and the block as the system, we have:

Initial Momentum = (200 ms^-1) * (0.005 kg) = 1 kg ms^-1

Since no external forces act on the system during the collision (gravity is a rather negligible force compared to the forces acting between the two objects; it does cause some change, but gravity has rather little significance during the collision), momentum is assumed to be conserved. Not only that, since the pellet is embedded in the block, the two objects move as one mass, with a common velocity.

Therefore,

Final Momentum = Initial Momentum = [(0.095 + 0.005) kg] * [(final velocity) ms^-1)

0.1 * final velocity = 1

Therefore, final velocity = 10 ms^-1

During the collision we can assume that the objects don't move much upwards, and only start moving at 10 ms^-1 after the collision is complete. Therefore, since the only force acting on the system of the two objects is gravity = -0.981 Newtons, we can use the equations of constant acceleration motion:

u = 10 ms^-1
v = 0 ms^-1 (at the top of it's ascent)
a = -9.81 ms^-2

so v^2 = u^2 + 2as
0 = 100 - 2 * 9.81 * s
Therefore, s = 5.1 meters = A

Hope this helps!
Best of luck for your exams!
 
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