AS Physics P1 MCQs Preparation Thread.

[h4rriet]

nov 02 question 35 plzzzz Rahma Abdelrahman h4rriet plzzz

You need to know that the potentiometer circuit shown in the question is the same as a circuit of 2 resistors in series. One resistor is the variable resistor & the other resistor is the wire of length XN.
If the resistance of the variable resistor increases, it will take more voltage and the XN resistor will take less voltage. So the voltage across XY decreases and the position of the moveable contact will have to be nearer to Y so that the resistance increases (because R is proportional to the length of the resistor).

 

[h4rriet]

?

You have to keep in mind that the points on a wave don't travel to the right, they only move up and down. A trough is travelling towards the point P, so P will go downwards. Point Q will be at the same position as point P when point P becomes a trough. Point P in the diagram has 0 displacement, so point Q too will have 0 displacement WHEN point P becomes a trough.

 

[bogus]

You need to know that the potentiometer circuit shown in the question is the same as a circuit of 2 resistors in series. One resistor is the variable resistor & the other resistor is the wire of length XN.
If the resistance of the variable resistor increases, it will take more voltage and the XN resistor will take less voltage. So the voltage across XY decreases and the position of the moveable contact will have to be nearer to Y so that the resistance increases (because R is proportional to the length of the resistor).

why does resistance have to increase? if voltage across x y decreases shouldnt it move near x as voltsge is propotional to length???

 

[h4rriet]

why does resistance have to increase? if voltage across x y decreases shouldnt it move near x as voltsge is propotional to length???

The V across XY decreases when the R of the variable resistor increases. So, for XY to take more V, it should increases it's R, and to increase its R, it should increase its length.
If the contact moves closer to X, the voltage will decrease even more. You need to get back the old V to get zero deflection once again.

 

[bogus]

The V across XY decreases when the R of the variable resistor increases. So, for XY to take more V, it should increases it's R, and to increase its R, it should increase its length.
If the contact moves closer to X, the voltage will decrease even more. You need to get back the old V to get zero deflection once again.

so u mean both the resistors should have same resistance for zero deflection... is that like a rule cuz i dont remember studying it?

 

[h4rriet]

so u mean both the resistors should have same resistance for zero deflection... is that like a rule cuz i dont remember studying it?

No, that's not what I mean. I was just giving those numbers as an example. What I meant to say is that the galvanometer showed zero deflection when the voltage across XY was, FOR INSTANCE, 6 V, and when the EMF was 12 and the variable resistor was 6 V. Then the variable resistor got 8 V across it, so the V across XY decreased. For zero deflection in this particular circuit, the V across XY must be 6 V.
It can be like this also:

The galvanometer has zero deflection when the PD across XY is 8 V. Now if the PD across variable resistor's increases the PD across XY decreases and the galvanometer has a deflection. The question says, what should be done so that the galvanometer has zero deflection once again? The PD across XY needs to increase and become 8 V again. So the R must increase. So the pointer must move closer to Y.

 

[hela]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_13.pdf Q2 Q17 Q31 Q35

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29

CAN ANYONE SOLVE
THANKS​

 

[Frankuzi]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_13.pdf Q2 Q17 Q31 Q35​

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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27​

​

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37​

​

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29​

​

CAN ANYONE SOLVE​

THANKS​

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdf
Q.2 first of all, you have to calculate the time taken to produce one complete wave which can be done by using the rate given(3000 revolution per minute)
my method:
3000 revo takes 60s
1 revo takes = 60/3000 = 0.02s
The question is asking which time base setting will separate the pulses on the screen. Notice that the screen of the c.r.o is 10cm long so we have to make full use of the screen. Which also means that your time base setting cannot be too far away from 0.02s(Calculated value) otherwise it will be very difficult to see the pulses.
(a) 1cm ---> 1s
10cm ---> 1 x 10=10s(Difference is too big! cannot show clearly on the screen)
Difference= 1 - 0.02 = 0.98s
(b) 1cm ---> 10ms
10cm ---> 10 x 10 = 100ms(0.1s) (Not too bad~ can be considered)
Difference= 0.1 - 0.02 = 0.08s
(c) 1cm ---> 100 x 10^-6s
10cm ---> 1 x 10^-3 s (THIS IS SMALLER THAN 0.02s! Cannot even show one complete pulse!)
(d) 1cm ---> 1 x 10^-6
10cm ---> 10 x 1 x 10^-6 = 0.00001s (Cannot even show one complete wave)
I believed the best answer will be B.

Q.17. The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv

Q.18. All calculations. First thing you have to do is to find the resistance of the relay(voltage=16v and current=0.6A) so resistance=16/0.6 = 26.66666(80/3)Ohm. Next we will look at the second scenario where we have two wires being connected to the power supply and also the relay now. So we are given resistance of wire in 1m (0.0050Ohm) and we also know that the distance between the relay and the power supply is 800m so we have to calculate the new resistance of the wire: New resistance = 0.0050 x 800 = 4Ohm. We are not finished yet~ Since we have two of these wires so the resistance has to be multiplied by 2 (4 x 2 = 8Ohm) so overall resistance of 2 wires will be 8Ohm. In order to solve this question you have to imagine it's a series circuit so the current will still be 0.60A. Total resistance in the circuit = Resistance in the 2 wires + Resistance of the relay( (80/3) + 8 = 104/3Ohm) and finally we can use the formula R=V/I to determine the output e.m.f where your R=104/3Ohm and I=0.6A. So final calculation: V= (104/3) x 0.6 = 20.8V) and the answer is C.

Q.35. This is another tricky question. The voltmeter reading will be constant. Why? Because we are not measuring the voltage of the potentiometer P only but we are measuring the voltage for the whole circuit(see the connection of the voltmeter) so even though you slide the potentiometer along x to y it will not have any effect. However, if we are measuring the voltage of the potentiometer only, then the voltage will change when you slide the slider along x to y. So the answer will be A.

Phew~ I hope it's not too complicatedand reply if you have any doubts. Thanks~

 

[hela]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdf
Q.2 first of all, you have to calculate the time taken to produce one complete wave which can be done by using the rate given(3000 revolution per minute)
my method:
3000 revo takes 60s
1 revo takes = 60/3000 = 0.02s
The question is asking which time base setting will separate the pulses on the screen. Notice that the screen of the c.r.o is 10cm long so we have to make full use of the screen. Which also means that your time base setting cannot be too far away from 0.02s(Calculated value) otherwise it will be very difficult to see the pulses.
(a) 1cm ---> 1s
10cm ---> 1 x 10=10s(Difference is too big! cannot show clearly on the screen)
Difference= 1 - 0.02 = 0.98s
(b) 1cm ---> 10ms
10cm ---> 10 x 10 = 100ms(0.1s) (Not too bad~ can be considered)
Difference= 0.1 - 0.02 = 0.08s
(c) 1cm ---> 100 x 10^-6s
10cm ---> 1 x 10^-3 s (THIS IS SMALLER THAN 0.02s! Cannot even show one complete pulse!)
(d) 1cm ---> 1 x 10^-6
10cm ---> 10 x 1 x 10^-6 = 0.00001s (Cannot even show one complete wave)
I believed the best answer will be B.

Q.17. The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv

Q.18. All calculations. First thing you have to do is to find the resistance of the relay(voltage=16v and current=0.6A) so resistance=16/0.6 = 26.66666(80/3)Ohm. Next we will look at the second scenario where we have two wires being connected to the power supply and also the relay now. So we are given resistance of wire in 1m (0.0050Ohm) and we also know that the distance between the relay and the power supply is 800m so we have to calculate the new resistance of the wire: New resistance = 0.0050 x 800 = 4Ohm. We are not finished yet~ Since we have two of these wires so the resistance has to be multiplied by 2 (4 x 2 = 8Ohm) so overall resistance of 2 wires will be 8Ohm. In order to solve this question you have to imagine it's a series circuit so the current will still be 0.60A. Total resistance in the circuit = Resistance in the 2 wires + Resistance of the relay( (80/3) + 8 = 104/3Ohm) and finally we can use the formula R=V/I to determine the output e.m.f where your R=104/3Ohm and I=0.6A. So final calculation: V= (104/3) x 0.6 = 20.8V) and the answer is C.

Q.35. This is another tricky question. The voltmeter reading will be constant. Why? Because we are not measuring the voltage of the potentiometer P only but we are measuring the voltage for the whole circuit(see the connection of the voltmeter) so even though you slide the potentiometer along x to y it will not have any effect. However, if we are measuring the voltage of the potentiometer only, then the voltage will change when you slide the slider along x to y. So the answer will be A.

Phew~ I hope it's not too complicatedand reply if you have any doubts. Thanks~

VERY HELPFUL THANKS
CAN U SOLVE PLEASE
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29
CAN ANYONE SOLVE
THANKS

 

[h4rriet]

VERY HELPFUL THANKS
CAN U SOLVE PLEASE
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29
CAN ANYONE SOLVE
THANKS

13. You can make a closed triangle of sides 3 cm, 4 cm and 5 cm. Forces in equilibrium can be represented by a closed triangle.
27. WD=qV=the same for both.

13. The acceleration of the box = the acceleration of the 2 kg mass. You must consider them to be one thing. Act like the whole system is one 10 kg mass. 2x9.81 acts downwards, 6 N acts upwards. a=F/m.
22. Volume of one molecule = mass given / density given. Cube root of answer = molecular spacing.
29. v=f x lambda; v=1/t x lambda, so t = lambda/v. Number of wavefronts from XY to P = 3.
33. A is wrong; charged particles are not provided by the supply. B is wrong, current is not a measure of the speed of electrons; it is the number of coulombs passing per second; C is wrong, a larger diameter means a smaller resistance, true, but the charged particles don't move fast because there's a larger volume; D is correct, a smaller diameter means a larger resistance and so a larger voltage across the wire and so the speed increases.
37. You can redraw the circuits so that they become clear, because they deliberately drew it so that it is confusing:

18.

29. A stationary wave in a closed end tube always has an antinode at the open end, so the fundamental mode will be 1/4th of a lambda. A stationary wave in an open end tube too always has an antinode, this time at both ends, so the fundamental mode will be 1/2 a lambda. For tube P: If 1 lambda = 20cm, 1/4 lambda = x. x = 5. That's a multiple of 35, so stationary waves can be formed in P. For tube Q: If 1 lambda = 20cm, 1/2 lambda = x. x = 10. That's a multiple of 50, so stationary waves can be formed in Q.

 

[Frankuzi]

VERY HELPFUL THANKS
CAN U SOLVE PLEASE
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29
CAN ANYONE SOLVE
THANKS

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdf
Q.13. In order to solve this question, you must imagine three person trying to push an object in any directions. And the minimum resultant force will be when the object is remain at stationary. So at stationary or equilibrium, the resultant force applied by the 3 person will be zero and you can apply the same logic here. 3 forces applied to the object can have a minimum resultant force of 0N.

Q.27. My understanding of this question is that in order to do work, you will have to push a unit positive charge from P to the positive charge or must move along the electric field line produced by the positive charge at the centre. (You can try to draw the electric field lines here. It's pretty simple. just draw a radial pattern and the arrow should be pointing out because it's a positive charge). Since it's moving from P to Q, so basically there is no change in the magnitude of E. (Remember: E=V/d where d=r) Which also means that there is no change in the voltage and you can link voltage with work done by using the formula: V= W.D/Q where Q is the charge. So when E and d/r are the same, the voltage will be the same so obviously the work done at positions P and Q would be the same too. So net work = 0 so no work is done.

I am not really good in explaining so please forgive me if I've made any mistakes esp the silly ones . You are welcome to ask anytime~

 

[lunafawkes]

can anyone please help me solve and understand Q 10,12,21 and 40 of s_07 qp_01?

 

[h4rriet]

can anyone please help me solve and understand Q 10,12,21 and 40 of s_07 qp_01?

10. Force = rate of change of momentum, which is p/t. The gradient of a p/t graph is the force. To find the gradient, we use y2-y1/x2-x1.
12. The collision is inelastic. The initial momentum = 12. The final too must = 12. The final mass = 6. The final v=p/m=12/6=2. The final k.e.=1/2x6x(2^2).
21. A and B is ruled out, because only transverse waves can be polarised. D is wrong because sound waves are not transverse. That leaves C.
40. a=F/m. F=Eq. E is the same for all; q & m are different. For A q is 1, B 2, C 3 and D 4. So acceleration for A=1E/1, for B=2E/4=1/2E, for C=3E/7 and D=4E/9.

 

[Frankuzi]

can anyone please help me solve and understand Q 10,12,21 and 40 of s_07 qp_01?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q.10. The formula that links momentum with force will be F x (Change in t) = (Change in momentum). So first we will determine the change in momentum. Change in momentum= P1 - P2 (Not P2 - P1 because if you look at the answers, all are P1-P2 and P1+P2 so we have no choice but to use P1 - P2). Change in time is simple which can be done by simply minus t2 by t1(t2 - t1) so the average force will be: Ave. force = (p1 - p2)/(t2 - t1). So answer will be B.

Q.12. During this collision, the momentum of the system must be conserved(As stated by the principle of conservation of linear momentum) so we will use the formula m1u1 + m2u2 = m1v1 + m2v2. By performing some calculations:
(2)(4) + (4)(1) = (4+2)(Velocity after the collision) *Note that I add the mass of both trolley together because they move off together after collision.
So velocity after the collision will be 2ms^-1.
To find K.E, use the formula: 1/2mv^2 and do calculation: 1/2(4+2)(2)^2 = 12J so final answer is B.

Q.21. This is simple. The only thing that you have to know is only transverse wave can be polarised.(That's what required in our syllabus) (a) wrong since it's a longitudinal wave. (b) Still longitudinal so wrong. (c) Correct because it's a transverse wave. (d) Sound wave cannot be a transverse wave so answer will be C.

Q.40. Question is asking which of the particular nuclei will travel at the lowest speed so it must have something to do with K.E. What I did was determine the number of protons and also the mass number in the nuclei. Mass number is labelled on the top part of the nuclei where as the proton number is labelled at the bottom part. My results (u = atomic mass constant(1.66x10^-27) given in exam)
(a) mass number = 1u
proton number = 1

(b) mass number = 4u
proton number = 2

(c) mass number = 7u
proton number = 3

(d) mass number = 9u
proton number = 4
Now I will introduce one infamous formula which is: 1/2mv^2=eV (eV=electronvolt). Why do we have to use this weird looking formula? Because this question involved the potential difference. The calculation part was quite lengthy so..
(* I will not show the V here because potential difference is the same anyway and e here is the elementary charge(1.6 x 10^-19))
(a) 1/2 (1u)v^2 = e
1/2(1.66x10^-27)v^2 = 1.6 x 10^-19
v^2 = 192771084..
v= 13884.20ms^-1

(b)1/2(4u)v^2 = 2e (*need to multiply by 2 because of 2 protons)
1/2(4(1.66x10^-27)v^2 = 2(1.6x10^-19)
v^2 = 96385542.1
v= 9817.61ms^-1

(c)1/2(7u)v^2 = 3e
1/2(7(1.66x10^-27)v^2 = 3(1.6x10^-19)
v^2 = 82616179
v=9089.344..ms^-1

(d)1/2(9u)v^2 = 4e
1/2(9(1.66 x 10^-27))v^2 = 4(1.6 x 10^-19)
v^2 = 85676037..
v=9256.135ms^-1
Phew~ Overall the answer should be C.

I know it's hard~ But take a deep breath and chill yourself. You can do it~

 

[Frankuzi]

13. You can make a closed triangle of sides 3 cm, 4 cm and 5 cm. Forces in equilibrium can be represented by a closed triangle.
27. WD=qV=the same for both.

13. The acceleration of the box = the acceleration of the 2 kg mass. You must consider them to be one thing. Act like the whole system is one 10 kg mass. 2x9.81 acts downwards, 6 N acts upwards. a=F/m.
22. Volume of one molecule = mass given / density given. Cube root of answer = molecular spacing.
29. v=f x lambda; v=1/t x lambda, so t = lambda/v. Number of wavefronts from XY to P = 3.
33. A is wrong; charged particles are not provided by the supply. B is wrong, current is not a measure of the speed of electrons; it is the number of coulombs passing per second; C is correct, a larger diameter means a smaller resistance; D is wrong, a smaller diameter means a larger resistance.
37. You can redraw the circuits so that it becomes clear, because the way they deliberately drew it so that it is confusing:

18.

29. A stationary wave in a closed end tube always has an antinode at the open end, so the fundamental mode will be 1/4th of a lambda. A stationary wave in an open end tube too always has an antinode, this time at both ends, so the fundamental mode will be 1/2 a lambda. For tube P: If 1 lambda = 20cm, 1/4 lambda = x. x = 5. That's a multiple of 35, so stationary waves can be formed in P. For tube Q: If 1 lambda = 20cm, 1/2 lambda = x. x = 10. That's a multiple of 50, so stationary waves can be formed in Q.

ummmm... regarding the question 33, I think we shared a different opinion.
(a)Will be wrong because the speed of the charged particle not only depends on the e.m.f of the power supply but also the potential difference across the wire as well.

(b) Wrong because the speed of the charged particle does not depends on the current flow.

(c) Larger diameter means lower resistance so the potential difference across the wire will also be lower and since the speed of the charged particle depends on the potential difference, the charged particle will move slower because of lower P.D.

(d) Smaller diameter will result in greater resistance which will later gives bigger P.D across the wire. The bigger the P.D the faster the charged particle will move.
My answer is D.

I know it's rude but I just want to know more. So kindly explain to me your thoughts please???~~~

 

[h4rriet]

ummmm... regarding the question 33, I think we shared a different opinion.
(a)Will be wrong because the speed of the charged particle not only depends on the e.m.f of the power supply but also the potential difference across the wire as well.

(b) Wrong because the speed of the charged particle does not depends on the current flow.

(c) Larger diameter means lower resistance so the potential difference across the wire will also be lower and since the speed of the charged particle depends on the potential difference, the charged particle will move slower because of lower P.D.

(d) Smaller diameter will result in greater resistance which will later gives bigger P.D across the wire. The bigger the P.D the faster the charged particle will move.
My answer is D.

I know it's rude but I just want to know more. So kindly explain to me your thoughts please???~~~

It's OK! And you're right about D. C is wrong.
A: 'Charged particles are provided by the power supply.' That is wrong. The charged particles = the electrons. The electrons aren't provided by the power supply; they're already there in the wire. The charged particles are provided energy by the power supply. And the EMF causes the PD across the wire.
B: 'The charged particles move with the same speed because the current is the same.' Current=Q/t. Current is not speed, so the same current does not mean the same speed. The current flow is the flow of charged particles.

 

[lunafawkes]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q.10. The formula that links momentum with force will be F x (Change in t) = (Change in momentum). So first we will determine the change in momentum. Change in momentum= P1 - P2 (Not P2 - P1 because if you look at the answers, all are P1-P2 and P1+P2 so we have no choice but to use P1 - P2). Change in time is simple which can be done by simply minus t2 by t1(t2 - t1) so the average force will be: Ave. force = (p1 - p2)/(t2 - t1). So answer will be B.

Q.12. During this collision, the momentum of the system must be conserved(As stated by the principle of conservation of linear momentum) so we will use the formula m1u1 + m2u2 = m1v1 + m2v2. By performing some calculations:
(2)(4) + (4)(1) = (4+2)(Velocity after the collision) *Note that I add the mass of both trolley together because they move off together after collision.
So velocity after the collision will be 2ms^-1.
To find K.E, use the formula: 1/2mv^2 and do calculation: 1/2(4+2)(2)^2 = 12J so final answer is B.

Q.21. This is simple. The only thing that you have to know is only transverse wave can be polarised.(That's what required in our syllabus) (a) wrong since it's a longitudinal wave. (b) Still longitudinal so wrong. (c) Correct because it's a transverse wave. (d) Sound wave cannot be a transverse wave so answer will be C.

Q.40. Question is asking which of the particular nuclei will travel at the lowest speed so it must have something to do with K.E. What I did was determine the number of protons and also the mass number in the nuclei. Mass number is labelled on the top part of the nuclei where as the proton number is labelled at the bottom part. My results (u = atomic mass constant(1.66x10^-27) given in exam)
(a) mass number = 1u
proton number = 1

(b) mass number = 4u
proton number = 2

(c) mass number = 7u
proton number = 3

(d) mass number = 9u
proton number = 4
Now I will introduce one infamous formula which is: 1/2mv^2=eV (eV=electronvolt). Why do we have to use this weird looking formula? Because this question involved the potential difference. The calculation part was quite lengthy so..
(* I will not show the V here because potential difference is the same anyway and e here is the elementary charge(1.6 x 10^-19))
(a) 1/2 (1u)v^2 = e
1/2(1.66x10^-27)v^2 = 1.6 x 10^-19
v^2 = 192771084..
v= 13884.20ms^-1

(b)1/2(4u)v^2 = 2e (*need to multiply by 2 because of 2 protons)
1/2(4(1.66x10^-27)v^2 = 2(1.6x10^-19)
v^2 = 96385542.1
v= 9817.61ms^-1

(c)1/2(7u)v^2 = 3e
1/2(7(1.66x10^-27)v^2 = 3(1.6x10^-19)
v^2 = 82616179
v=9089.344..ms^-1

(d)1/2(9u)v^2 = 4e
1/2(9(1.66 x 10^-27))v^2 = 4(1.6 x 10^-19)
v^2 = 85676037..
v=9256.135ms^-1
Phew~ Overall the answer should be C.

I know it's hard~ But take a deep breath and chill yourself. You can do it~

Thankyou very much, yes 21 was simple , in my hurry i overlooked the fact that sound waves are not transverse! , and thanks for the last question especially! , aand in momentum i get really confused as to when do we add the momentum and when do we subtract them, cuz sometimes its the opposite ( you know same direction, u subtract, different directions u add) so momentum's not my best topic, but thank you anyways!

 

[lunafawkes]

10. Force = rate of change of momentum, which is p/t. The gradient of a p/t graph is the force. To find the gradient, we use y2-y1/x2-x1.
12. The collision is inelastic. The initial momentum = 12. The final too must = 12. The final mass = 6. The final v=p/m=12/6=2. The final k.e.=1/2x6x(2^2).
21. A and B is ruled out, because only transverse waves can be polarised. D is wrong because sound waves are not transverse. That leaves C.
40. a=F/m. F=Eq. E is the same for all; q & m are different. For A q is 1, B 2, C 3 and D 4. So acceleration for A=1E/1, for B=2E/4=1/2E, for C=3E/7 and D=4E/9.

Thnak you!

 

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nov 02 question 35 plzzzz Rahma Abdelrahman h4rriet plzzz

ANYONE good in the subject, please helppp me out! - explain the attached questions

 

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This is the second question,