Physics: Post your doubts here!

[Princess Raven]

3:
simple. the speed must have units m/s
in (a) √(gλ)=(m²/s²)^½ = m/s
no need to check rest of them
7:
O level vectors: x=v-u
v-u = at
so x=at

20:
All i can do is visualise which parts are bending in what way. X and Y are being stretched but Z is being pressed/compressed.
X and Y experience tension while Z experiences compression. just visualise.
40:
the particle which completes acceleration with the lowest speed must have had lowest acceleration as the distance is same.
F is not the same. force directly proportional to charge. acceleration = F/m so acceleration will depends on charge and mass. A ∝charge/mass
so find charge to mass ratio. the one with smallest ratio has lowest acceleration therefore lowest speed at the end.
(a)1/1=1
(b)2/4=0.5
(c)3/7=0.428...
(d)4/9=0.44...
smallest is (c)

OMG THANK U!! ur explanations rock! Seriously I get everything now.. So for the last one q40 we always do charge/mass ratio? For the lowest speed questions I mean?

 

[xXGTZXx]

The weight is 0.15N and angle with the vertical is 30. Apply .15*tan30 to get the horizontal force.

i did that turns out my calculators battery was low

 

[Thought blocker]

OMG THANK U!! ur explanations rock! Seriously I get everything now.. So for the last one q40 we always do charge/mass ratio? For the lowest speed questions I mean?

Yes

 

[Princess Raven]

Ahh I

Standing waves !

forgot to read that whoops Haha thanks could u help me with some more questions plz
O/n 2012 v12 q)17 why not A and q)34,36,37
And o/n 2009 v12 q)12,18,21,26

 

[Kamihus]

32)
The arrows point downwards which means the charge at the top is positive and the bottom is negative. So, the +Q will be attracted downwards and the -Q will be attracted upwards, spinning the rod anti-clockwise. The resultant force is zero because F = E/Q and the Q (both charges) are equal but opposite. So resultant force is zero. If the charges weren't the same in magnitude, then we would have a resultant force.

14)
sagar had explained v.well..
Suppose we take the upward direction to be positive, and the downward direction to be negative, we can again write the momentum equations and use them to find the final velocity of the system - the forces between the clay and the lead pellet are huge compared to other external forces, so even though there are external forces acting on the system during the collision, we can assume the momentum stays approximately constant.

So, the initial momentum is the momentum of the bullet alone, which is equal to mv = (5.0/1000) * 200 = 1.0 kg ms^-1 . The clay block is stationary, so it does not contribute any momentum to the system initially.
When the bullet collides with the clay, the bullet gets stuck in the clay and they both move off with the same velocity, which we'll call v(f). The mass of the lead pellet+ the clay block = (95/1000 + 5/1000) = (100/1000) = 0.1 kg.
Their final velocity = v(f)
Therefore, the final momentum of the system is 0.1 * v(f)

Since this is equal to the initial momentum, we can write (0.1 kg) * v(f) = (1.0 kg ms^-1) and so v(f) = 10 ms^-1.

This is the speed with which they rise after the collision. While the system moves upwards, gravity accelerates them downwards at a constant rate of -9.81 ms^-2. When they reach the highest point above the original position, their velocity is 0. Therefore, we can use the formula v^2 = u^2 + 2as to give us "s", which is the maximum height. So,

(0)^2 = (10)^2 +2(-9.81)s
19.62 * s = 100
s = 5.09 m = 5.1 meters = A.

You should add some index in your thread of all the difficult questions that you've answered as they are asked again and again.

 

[Thought blocker]

You should add some index in your thread of all the difficult questions that you've answered as they are asked again and again.

Bored to do that..

 

[Thought blocker]

Ahh I

forgot to read that whoops Haha thanks could u help me with some more questions plz
O/n 2012 v12 q)17 why not A and q)34,36,37
And o/n 2009 v12 q)12,18,21,26

Sure, but lemme solve some others doubt first... that they posted.

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_12.pdf
kindly help me out in
q 10 21 26 38

10)
At X, sand is stationary and truck is moving so when they stick together the speed decreases.
At Y, sand also leaves with some speed and as the total mass remains same, the speed will not change.

21)
--> extension of P is half that of Q, as you can see for more stress there is less strain for P that Q, hence statement 1 is incorrect.
--> young modulus=stress/strain for P graph is steeper than Q, means greater young modulus
--> for statement 3, before 0.1, it is a straight line

26)
As shown in the fig, at time = 18 s, phase diff btw the 2 waves = 180° or π.
1 wavelength = 360°
So, 1/8 of a wavelength = 360°/8 → 45°.

Phase diff = 180° at 18 s.
So, phase diff = 45° at x.
x = (18 * 45)/180
x = 4.5 s

Ans = B


38)
find total resistance of voltmeter and resistor connected in parallel by 2oo x 200/400=100ohms
then find the share of this resistance in total batterys voltage by :
100/500(total resistance) x 60=12V,since both voltmeter and resistor is in parallel each is having 12 V.

 

[amal sharkawi]

A projectile is launched at 45° to the horizontal with initial kinetic energy E.

Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it
reaches its highest point?
A 0.50 E B 0.71 E C 0.87 E D E
can any one answer this plz??

 

[Thought blocker]

A projectile is launched at 45° to the horizontal with initial kinetic energy E.

Assuming air resistance to be negligible, what will be the kinetic energy of the projectile when it
reaches its highest point?
A 0.50 E B 0.71 E C 0.87 E D E
can any one answer this plz??

vertical velocity= u cos 45
horizontal velocity = u sin 45
so, initial K.E. : the vertical and horizontal components of kinetic energy = 1/2 m ( u cos 45 ) + 1/2 m ( u sin 45 ) ......... ( 1/2 m v^2 formula )

so, sin 45 = 0.71
....cos 45 = 0.71 ( check them in the calculator )

so they will give the same kinetic energy for both velocities ( horizontally and vertically )

so at the highest point, (theta) = 90 ( for the vertical velocity ) , horizontal velocity doesn't change
so final kinetic energy = 1/2 m ( u cos 90 ) + 1/2 m ( u sin 45 )..................... ( horizontal velocity is constant ) ( cos 90 is 0 )
final kinetic energy = 1/2 m ( u sin 45 ) + 0 only which is half of the initial kinetic energy.

 

[amal sharkawi]

can any one answer this plz??

 

[sagar65265]

View attachment 45021

can any one answer this plz??

This should help, if you still have any doubts, make sure to post on the forums:
https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-535#post-823687

Good Luck for all your exams!

 

[Thought blocker]

This should help, if you still have any doubts, make sure to post on the forums:
https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-535#post-823687

Good Luck for all your exams!

Link not working

 

[Thought blocker]

Question 30 http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

30)

 

[sagar65265]

Link not working

Try it now

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q12,13,14 and 22

12)
The angle that T1 makes with R is greater than the angle that T2 makes with R.
R has to be balanced by the horizontal componts of T1 and T2
Let the angle be θ
R= T1cosθ + T1cosθ and R= T2cosθ + T2cosθ
We can see that for cos,, the greater the angle.. less the value of cosθ
so in order to cancel out R the value of T1 must be greater as cos θ is less in diagram 1. Therefore T1>T2

13)
distance of upper string from centre opf Q = 100/2 = 50mm=0.05m
distance of upper string from centre of p = 150/2=75mm=0.075m
tension in upper string = torque/distance = 3 Nm / 0.05 m = 60 N
torque on P = force*distance = 60 * 0.075 = 4.5

14)
The horizontal velocity will remain the same.
The vertical velocity will be zero at that time.
So the kinetic energy will be due to horizontal velocity alone which is v*cos*45 = v √2/2
Kinetic energy at max point = 1/2 * m * (v √2/2)² = [1/2 m v² ]* 1/2 (the thing in square brackets is equal to kinetic energy at the beginning) so k(max)=0.5K(initial)
Got it ?

22)
use the formula for E
E = F L / Ax
rearrange to get the ratio x / L on one side (change in length / original length)
you'll get is x / L = F / E A : Where - (A = pi r ^2 )
= 20 / 2 * 10^(11) * pi x (2.5 * 10^(-4))^2
= 5.1 * 10^(-4)
multiply this by a 100 to get the percentage
5.1 x 10^-4 x 100 = 5.1 x 10^-2 %
So answer is B.

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w09_qp_11.pdf

Q12,13,14 and 22

Q12)

Suppose we take the threads and the picture as the system. Then, the only forces on our "system" are the weight of the system (the force of gravity on it) and the tension force from the uppermost string. In other words, the forces in both cases are

i) Weight of the system, acting downwards; in both case, the magnitude of this is equal to the weight of the picture (threads don't weigh a significant amount)
ii) Tension in the uppermost string, acting upwards; in the first case, the magnitude is R₁ and in the second case it is R₂.

Therefore, in the first case, since the picture is balanced, the forces are also balanced and (Weight of the system) = (Tension in uppermost thread). So, W = R₁.
In the second case, the picture is again balanced, so the forces are also balanced and (Weight of the system) = (Tension in the uppermost thread). So, W = R₂.

Since the weight is the same in both cases, R₁ = R₂, narrowing down our options to A or B.

Let's continue this analysis. Suppose we take the picture alone as our system, the sum of the vertical components of the tension forces on it should be equal to it's weight. In other words, since the two threads exert the same tension force, 2 * the vertical component of one tension force = weight of the picture (otherwise the picture is not balanced).

So, the vertical component of the tension T₁ should be equal to the vertical component of the tension T₂.
The vertical component of the tension should be equal to T₁ * cos(θ₁) OR T₂ * cos(θ₂) where θ is the angle the tension force makes with the vertical.
You can see that the angle θ₂ for T₂ is smaller than the angle θ₁ for T₁, and since cos(θ) decreases as θ increases, cos(θ₂) > cos(θ₁) (check that line again, and confirm to yourself that the inequality holds).

But if the vertical components of the tension have to be the same, T₁ has to exert a larger force to compensate for the smaller value of cos(θ₁) than T₂, therefore we can say that T₂ < T₁, which means that B is the correct answer.

I'll try posting the others later, have some stuff to deal with.

Good Luck for all your exams!

 

[Thought blocker]

Suppose the resistance of one of the resistors is R. Then the combined resistance of Q and R is given as follows:

1/R(Q + R) = 1/R(Q) + 1/R(R) = 1/R + 1/R = 2/R
Therefore, R(Q + R) = R/2.

This resistance is in series with R(P), and since they are in series we can add them straight away:

R(P + Q + R) = R + R/2 = 3R/2.

So the total resistance of the circuit is equal to 3R/2. Suppose the voltage is V, then we can see that the current = I = V/R:

I(circuit) = V/(3R/2) = 2V/3R Amperes

From this, since we know that the total power delivered is 12 Watts and the power given is equal to P = IV, we can write

P = 12 Watts = 2V/3R * V = 2V²/(3R)
So, we can write that

V²/R = 3/2 * 12 = 18
So V²/R = 18. Let's move on.

Since we know that the resistance of network Q + R = R(Q + R) = R/2, we can say that the potential difference across Q + R section is equal to V(Q + R) =
I * R(Q + R) =

V(Q + R) = 2V/3R * R/2 = V/3.

So the potential difference across the resistor called R is equal to V/3. Therefore, since P(resistor) = V(across resistor)²/R, we can write

P(R) = (V/3)²/R = V²/3²R = V²/(9R)

Since we know that V²/R = 18, we can write

P(R) = (18)/9 = 2 Watts = A.

Hope this helped!
Good Luck for all your exams!

V²/R = 3/2 * 12 = 18
So V²/R = 18. Let's move on.

Since we know that the resistance of network Q + R = R(Q + R) = R/2, we can say that the potential difference across Q + R section is equal to V(Q + R) =
I * R(Q + R) =

V(Q + R) = 2V/3R * R/2 = V/3.

So the potential difference across the resistor called R is equal to V/3. Therefore, since P(resistor) = V(across resistor)²/R, we can write

P(R) = (V/3)²/R = V²/3²R = V²/(9R)

Since we know that V²/R = 18, we can write

P(R) = (18)/9 = 2 Watts = A.
How it is 18 ? Is ke baad khuch ni aya ._.

 

[sagar65265]

V²/R = 3/2 * 12 = 18
So V²/R = 18. Let's move on.

Since we know that the resistance of network Q + R = R(Q + R) = R/2, we can say that the potential difference across Q + R section is equal to V(Q + R) =
I * R(Q + R) =

V(Q + R) = 2V/3R * R/2 = V/3.

So the potential difference across the resistor called R is equal to V/3. Therefore, since P(resistor) = V(across resistor)²/R, we can write

P(R) = (V/3)²/R = V²/3²R = V²/(9R)

Since we know that V²/R = 18, we can write

P(R) = (18)/9 = 2 Watts = A.
How it is 18 ? Is ke baad khuch ni aya ._.

The Potential Difference through the circuit is some value V. The current through the circuit, as we can see after combining the resistances, is equal to 2V/3R.

So, the current in the circuit is 2V/3R and the voltage across the total resistance is V. Therefore, by P = IV, we can write

(Total Power expended in circuit) = (Total Potential difference across resistances) * (Current in the circuit) = (2V/3R) * V = 2V²/3R.

Since the (Total Power expended in circuit) is given in the question = 12 Watts, we can say that 12 = 2V²/3R.

Multiplying both sides by 3, and dividing both sides by 2, we get

3 * 12/2 = 3 * 6 = 18 = V²/R.
So V²/R = 18.

Good Luck for all your exams!

 

[Thought blocker]

The Potential Difference through the circuit is some value V. The current through the circuit, as we can see after combining the resistances, is equal to 2V/3R.

So, the current in the circuit is 2V/3R and the voltage across the total resistance is V. Therefore, by P = IV, we can write

(Total Power expended in circuit) = (Total Potential difference across resistances) * (Current in the circuit) = (2V/3R) * V = 2V²/3R.

Since the (Total Power expended in circuit) is given in the question = 12 Watts, we can say that 12 = 2V²/3R.

Multiplying both sides by 3, and dividing both sides by 2, we get

3 * 12/2 = 3 * 6 = 18 = V²/R.
So V²/R = 18.

Good Luck for all your exams!

Lol silly me ._.