AS Physics P1 MCQs Preparation Thread.

[lee mee..]

F(net)= W = mg= (120 - 80) x 9.81 = 400
Fnet = ma = (120 + 80) x a = 400 a=2
kinematics equation
V^2= u^2 + 2as
v^2 = 0 + 2x2x9
v = 6

Thank u !
Can u help me with q15 of the same paper.http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf

And Q14 ,18 and 21 0f http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf

 

[6Astarstudent]

Thank u !
Can u help me with q15 of the same paper.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

And Q14 ,18 and 21 0f http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf

for s12 qp11

#15
L is moved further to the right, so there is a bigger clockwise moment, meaning there need to be a bigger anticlockwise moment to counter it.
note the system is in equilibrium so there must be no net force and no net torque
immediately eliminate A and B because there is a horizontal H force that is unbalanced
now the logic comes for C, D.
for D, the reaction is shifted closer to the right (you can consider centre of mass shifted closer to right), so clockwise moment decreases, anticlockwise moment increases. Thus balances the system. so D is correct

for W12 qp 11

#14
initial momentum = final momentum
momentum = mv
0.005 x 200 = (0.005 + 0.095) v
v = 10 m/s
use kinematics equation
v^2 = u^2 + 2gs
0 = 10^2 - 2 x 9.81 x s
s = 100/(2x9.81) = 5.1 answer = A

#18
Ek = (mv^2)/2 so Ek is proportional to v^2
Ek = cv^2 where c is a constant
4Ek = c(new v^2)
so new v must be 2v because 2^2 = 4

#21
efficiency = output/input
output power = VI = 200 x 6000 = 1200000
input = E(potential)/t = mgh/t = 500x300x 9.85 = 1477500
efficiency = 1200000/1477500 = 82%

 

[x-gamer-x]

for w11
#16: initial Ek = (mv^2)/2
since in a collision momentum must be conserved, initial momentum = mv . final = 2m x (v/2)
final Ek = (m (0.5v)^2)/2 = (mv^2)/4
so loses 50% of original Ek

for s11
#4
amplitude is 4x1.5= 6
period = 4 x 5ms = 20ms

#14
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2

#16
Power = Force x velocity
Force = weight = 80 x 9.81=784.8
Power = 784.8 x0.5 = 392.4W = 0.392kW

#25
ok, this one is hard to explain but
75Hz is the first harmonic, 1 node 1 antinode (1/4 wavelength)
for 2nd harmonic, it is (3/4 wavelength) so 75x3 = 225
3rd harmonic = (5/4 wavelength) so 75x5 = 375

#35
ratio of resistance is 1:2:5 and we know that ratio is current is inverse of that
so must be 1:1/2:1/5 = 10:5:2
so 5x 5/17 = 1.47

w10 qp 12

#12
torque = force x perpendicular distance to centre of pivot = 200 x 0.25 = 50Nm

#15
Ek = (mv^2)/2
since it is at steady speed, no net force so
Weight (mg) = retarding force (kv)
mg = kv
v = mg/ksubstitute this to Ek formula
Ek = (m(mg/f)^2)/2 = m^3 x g^2 / 2k^2

Thanks a lot bro
may we all get straight As

 

[lee mee..]

for s12 qp11

#15
L is moved further to the right, so there is a bigger clockwise moment, meaning there need to be a bigger anticlockwise moment to counter it.
note the system is in equilibrium so there must be no net force and no net torque
immediately eliminate A and B because there is a horizontal H force that is unbalanced
now the logic comes for C, D.
for D, the reaction is shifted closer to the right (you can consider centre of mass shifted closer to right), so clockwise moment decreases, anticlockwise moment increases. Thus balances the system. so D is correct

for W12 qp 11

#14
initial momentum = final momentum
momentum = mv
0.005 x 200 = (0.005 + 0.095) v
v = 10 m/s
use kinematics equation
v^2 = u^2 + 2gs
0 = 10^2 - 2 x 9.81 x s
s = 100/(2x9.81) = 5.1 answer = A

#18
Ek = (mv^2)/2 so Ek is proportional to v^2
Ek = cv^2 where c is a constant
4Ek = c(new v^2)
so new v must be 2v because 2^2 = 4

#21
efficiency = output/input
output power = VI = 200 x 6000 = 1200000
input = E(potential)/t = mgh/t = 500x300x 9.85 = 1477500
efficiency = 1200000/1477500 = 82%

Thank youu sooo much
but... i didn't get q18

 

[6Astarstudent]

Thank youu sooo much
but... i didn't get q18

hm
ok original v is A, new v is B
Ek = A^2
4Ek = B^2
so A^2 = (B^2)/4
square root both sides
A = B/2
B = 2A
so velocity is doubled

 

[lee mee..]

hm
ok original v is A, new v is B
Ek = A^2
4Ek = B^2
so A^2 = (B^2)/4
square root both sides
A = B/2
B = 2A
so velocity is doubled

oohhh.... Thanks alooot

 

[Sajeeb Chowdhury]

i th
i think its C .. by my own made up theories

and what r they..... ??

 

[IR Germany]

question 15: http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_1.pdf

 

[sagar65265]

question 15: http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf

In this question, the first thing to note is that the water levels equalize; the net result of this is that the water level on either side will be equal to h/2 .
The second thing to note is that there is actually only a relatively small change in the situation; this change can be viewed in two ways:

i) if the volume of water is divided into two equal halves, each with mass (m/2) , then we can plainly take the top half of the water column in X before the tap is opened and dump it into the bottom of container Y. This is the same as the final situation, with equal depths of water in each container.

ii) if the volume of water is again divided into two equal halves, then we can take the bottom half, let it through the tap into container Y while holding the top half stationary, and finally let the original top half drop into the bottom of the container X. Again, this is the same as the final situation.

In each case, the calculation yields the same result.

Situation (i):

Consider the center of mass of the bottom half, mass (m/2). In the final situation, the position of the center of mass of this half has not changed. Therefore, it cannot have gained any potential energy. So we can say that the change in potential energy here is zero.

Now consider the top half. The center of mass of this section drops a height of (h/2) and moves an unknown distance to the side to find itself in container Y. Since the displacement to the side is perpendicular to the force of gravity on that portion of water, the work done by gravity in that movement is zero. However, since the center of mass drops in height by a distance (h/2) - this can be confirmed by checking the change in height for the bottom of that half which ends up on the bottom on container Y as well that the top of that half which goes from height (h) to height (h/2) - gravity does work on mass (m/2) equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

Situation (ii):

If the bottom half is shifted to the right, it only moves sideways; gravity has no component in the direction of displacement so it cannot do any work on the water. Again, the work done here is zero.

However, if we consider the top half that has mass (m/2) and drops to the bottom of container X, then we can see that it drops a height (h/2) and so the work done by gravity must be equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

which is the same answer as before.

Hope this helped!
Good luck for your exams!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_1.pdf

question 16

This problem is an equilibrium problem; in the Physics sense of the word, there are two condition we can apply here:

i) The sum of all net forces on the object is equal to zero.
AND
ii) The sum of all net torques on the object is equal to zero.

The first rule is more helpful here; consider first the vertical forces on the door:

The hinge force and the force of gravity (the weight of the object) both have a downwards component; no matter that the force of gravity directly faces downwards, but the general idea is that both those forces are trying to pull/push the door downwards (in the diagram view).
However, the door is not accelerating, so there cannot be a net vertical force! Therefore, we take note of the tension force T. The vertical upwards component of this force has to balance the vertical downwards components of the other forces, and so we can say that the vertical component of the tension force is surely greater than the vertical components of the gravitational force and the pivot force put together, and since the vertical component of the gravitational force is equal to the magnitude of the gravitational force, the tension is surely greater than the gravitational force.

So, we can rule out for sure A and B! Apparently, according to them, the tension is less in magnitude than the weight!

The last thing we need to consider is the horizontal direction; we can say for sure that the horizontal push (to the right) of the hinge force H is equal in magnitude to the horizontal pull (to the left) of the tension force.

Now, if we trust the diagram and take it that the angle made by the force H with the door is smaller than the angle made by the force T with the door, then direct inspection tells us that the Cosine ratio of the angle H makes is higher than the Cosine ratio of the angle T makes; this means, that the force H doesn't need to be so strong to balance the force T, since it has the advantage of a higher Cos ratio:

H * cos(angle H makes with door) = T * cos(angle T makes with door)

Since the cos ratio of T is lower than the cos ratio of H, we can say that

T = H * (something greater than one)

or that T is greater than H.

Hope this helped!
Good Luck for your exams!

 

[IR Germany]

In this question, the first thing to note is that the water levels equalize; the net result of this is that the water level on either side will be equal to h/2 .
The second thing to note is that there is actually only a relatively small change in the situation; this change can be viewed in two ways:

i) if the volume of water is divided into two equal halves, each with mass (m/2) , then we can plainly take the top half of the water column in X before the tap is opened and dump it into the bottom of container Y. This is the same as the final situation, with equal depths of water in each container.

ii) if the volume of water is again divided into two equal halves, then we can take the bottom half, let it through the tap into container Y while holding the top half stationary, and finally let the original top half drop into the bottom of the container X. Again, this is the same as the final situation.

In each case, the calculation yields the same result.

Situation (i):

Consider the center of mass of the bottom half, mass (m/2). In the final situation, the position of the center of mass of this half has not changed. Therefore, it cannot have gained any potential energy. So we can say that the change in potential energy here is zero.

Now consider the top half. The center of mass of this section drops a height of (h/2) and moves an unknown distance to the side to find itself in container Y. Since the displacement to the side is perpendicular to the force of gravity on that portion of water, the work done by gravity in that movement is zero. However, since the center of mass drops in height by a distance (h/2) - this can be confirmed by checking the change in height for the bottom of that half which ends up on the bottom on container Y as well that the top of that half which goes from height (h) to height (h/2) - gravity does work on mass (m/2) equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

Situation (ii):

If the bottom half is shifted to the right, it only moves sideways; gravity has no component in the direction of displacement so it cannot do any work on the water. Again, the work done here is zero.

However, if we consider the top half that has mass (m/2) and drops to the bottom of container X, then we can see that it drops a height (h/2) and so the work done by gravity must be equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

which is the same answer as before.

Hope this helped!
Good luck for your exams!
Thank you so much, it helps me

 

[IR Germany]

In this question, the first thing to note is that the water levels equalize; the net result of this is that the water level on either side will be equal to h/2 .
The second thing to note is that there is actually only a relatively small change in the situation; this change can be viewed in two ways:

i) if the volume of water is divided into two equal halves, each with mass (m/2) , then we can plainly take the top half of the water column in X before the tap is opened and dump it into the bottom of container Y. This is the same as the final situation, with equal depths of water in each container.

ii) if the volume of water is again divided into two equal halves, then we can take the bottom half, let it through the tap into container Y while holding the top half stationary, and finally let the original top half drop into the bottom of the container X. Again, this is the same as the final situation.

In each case, the calculation yields the same result.

Situation (i):

Consider the center of mass of the bottom half, mass (m/2). In the final situation, the position of the center of mass of this half has not changed. Therefore, it cannot have gained any potential energy. So we can say that the change in potential energy here is zero.

Now consider the top half. The center of mass of this section drops a height of (h/2) and moves an unknown distance to the side to find itself in container Y. Since the displacement to the side is perpendicular to the force of gravity on that portion of water, the work done by gravity in that movement is zero. However, since the center of mass drops in height by a distance (h/2) - this can be confirmed by checking the change in height for the bottom of that half which ends up on the bottom on container Y as well that the top of that half which goes from height (h) to height (h/2) - gravity does work on mass (m/2) equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

Situation (ii):

If the bottom half is shifted to the right, it only moves sideways; gravity has no component in the direction of displacement so it cannot do any work on the water. Again, the work done here is zero.

However, if we consider the top half that has mass (m/2) and drops to the bottom of container X, then we can see that it drops a height (h/2) and so the work done by gravity must be equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

which is the same answer as before.

Hope this helped!
Good luck for your exams!
Thank you so much, it helps me

 

[IR Germany]

In this question, the first thing to note is that the water levels equalize; the net result of this is that the water level on either side will be equal to h/2 .
The second thing to note is that there is actually only a relatively small change in the situation; this change can be viewed in two ways:

i) if the volume of water is divided into two equal halves, each with mass (m/2) , then we can plainly take the top half of the water column in X before the tap is opened and dump it into the bottom of container Y. This is the same as the final situation, with equal depths of water in each container.

ii) if the volume of water is again divided into two equal halves, then we can take the bottom half, let it through the tap into container Y while holding the top half stationary, and finally let the original top half drop into the bottom of the container X. Again, this is the same as the final situation.

In each case, the calculation yields the same result.

Situation (i):

Consider the center of mass of the bottom half, mass (m/2). In the final situation, the position of the center of mass of this half has not changed. Therefore, it cannot have gained any potential energy. So we can say that the change in potential energy here is zero.

Now consider the top half. The center of mass of this section drops a height of (h/2) and moves an unknown distance to the side to find itself in container Y. Since the displacement to the side is perpendicular to the force of gravity on that portion of water, the work done by gravity in that movement is zero. However, since the center of mass drops in height by a distance (h/2) - this can be confirmed by checking the change in height for the bottom of that half which ends up on the bottom on container Y as well that the top of that half which goes from height (h) to height (h/2) - gravity does work on mass (m/2) equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

Situation (ii):

If the bottom half is shifted to the right, it only moves sideways; gravity has no component in the direction of displacement so it cannot do any work on the water. Again, the work done here is zero.

However, if we consider the top half that has mass (m/2) and drops to the bottom of container X, then we can see that it drops a height (h/2) and so the work done by gravity must be equal to:

W = mgh = (m/2) * g * (h/2) = mgh/4

which is the same answer as before.

Hope this helped!
Good luck for your exams!

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_12.pdf can you help with this too? question 17 please

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_12.pdf can you help with this too? question 17 please

Okay, so the concept here is that for a string or rubber band or any other elastic material that exerts a tension force, the assumption can be made that the tension is the same at any and all points on the string/rubber band. Another thing is that the force of tension on the object due to the rubber band always acts along the rubber band, i.e. acts parallel to the rubber band at the point where the rubber band is connected to the object. This tension always tries to bring the string back to it's original length.

Therefore, the forces on the object try to move the object in such a way that the rubber band is restored to a position where it has less tension, until it eventually reaches a position where it has no tension within itself. In this case, the force of tension cannot be acting from the center of each 50 mm section towards the object, because then that would accelerate the object in the leftwards direction; this will increase the tension in the rubber band, but this cannot happen, otherwise the elastic potential energy stored in the rubber band will be increasing without any external force doing work.

After all this, we can conclude that the tension acts parallel to the rubber band in the forwards direction, i.e. the tension forces on the object will in acting from X towards the center of the line XY, as shown in the attached diagram.

Now, since we know the force magnitudes and directions, we can solve the equations for the motion of the object. Since the force diagram is symmetric about the line passing along the trolley (from X in the direction of P) we can conclude that the net force in the direction from Y to Z is zero. Now, we need to find the force components in the direction along P.

Since tension is 4N throughout the rubber band, we can write:

4cos(theta) + 4cos(theta) = Net forward force

Where theta is the angle between the line of symmetry in the direction of P and the rubber band segments.
Therefore, since we have a classic 3-4-5 triangle here, where
i) the hypotenuse, XY, is 50mm
ii)half the length of YZ, is 40 mm
iii)the other side has length 30 mm by pythagoras's theorem.

Therefore, cos(theta) = (30 mm)/(50 mm) = 3/5
So Net forward force = 8cos(theta) = 8 * 3/5 = 24/5 = 4.8 Newtons = C

Hope this helped!
Good luck for all your exams!

 

[mnotanerd]

..........

 

[Moi]

Please Help.

Summer 2012- V12

Q7- (why A)
Q9- What is the difference between a curved and a straight lined graph? why A not B
Q14- why B not A
Q15- why clockwise?
Q17- ?

Winter 2012- V12

Q9-i why C not D. why curved not straight lines?
Q10- Same as above reason?
Q11- ?
Q12-?
Q16- What is a viscous Force? why zero?
Q18-?
Q23- why are molecules in ice further apart than the ones in water>
Q30-?
Q36-?
Q37-?


(i can't link the paper xtremepapers papers section not working)

Thank you

 

[moinul]

can anyone plz tell me why i cant access the p1 qp n ms section.... its coming error 522

 

[naoomi]

Please Help.

Summer 2012- V12

Q7- (why A)
Q9- What is the difference between a curved and a straight lined graph? why A not B
Q14- why B not A
Q15- why clockwise?
Q17- ?

Winter 2012- V12

Q9-i why C not D. why curved not straight lines?
Q10- Same as above reason?
Q11- ?
Q12-?
Q16- What is a viscous Force? why zero?
Q18-?
Q23- why are molecules in ice further apart than the ones in water>
Q30-?
Q36-?
Q37-?


(i can't link the paper xtremepapers papers section not working)

Thank you

summer 12

ques 7- whether the velocity is increasing or decreasing doesnt matter, if it's positive the displacement will increase and reaches its max when velocity becomes zero. When the velocity is negative, that means the object is now travelling in an opposite direction and hence the displacement will decrease.

ques 9- its not a straight line since the decrease in height is not uniform. there is an acceleration due to gravity because of which the velocity will keep on increasing and hence the height will decrease at an increasing rate.

14) the reaction force P should be perpendicular to the surface.

 

[naoomi]

Can some one please help me out with these questions from june 2013 variant 12.

3 , 6 , 9 , 13 , 17 , 18 , 20 , 22 , 27 , 35 , 36.

 

[Apuurv A Mehra]

Can some one please help me out with these questions from june 2013 variant 12.

3 , 6 , 9 , 13 , 17 , 18 , 20 , 22 , 27 , 35 , 36.

The ans will be A for the 18th question because i.e because workdone=f*s so here the force applied is the pressure and the displacement is V2-V1

the ans to the 22nd question will be B because its an hysterious curve and in hysterious curve the energy bounded between the two curves is the energy dissipated which shows an increase in the temperature of the material used