AS Physics P1 MCQs Preparation Thread.

h4rriet

h4rriet

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Rahma Abdelrahman said:
Please can u help with the following:
All from June 2011 P11 , may be some of them are a bit silly but I need help in them..
Q9, 15, 16, 25, 31 and 34
Click to expand...

9. The initial momentum = mv, so the final too must = mv. That rules out C & D.
I1g5RpO.jpg

15. GPE at the top = 2 x 3 x 9.81 = 58.86 J. WD against friction = 5 x 7 = 35 J. 58.86 - 35 = 23.86 J. 23.86 = 1/2 x 2 x v^2. Solve for v.
16. P =Fv = 80 x 9.81 x (50/100).
25.
Sppwg1F.jpg

31. Q=It. Q=10. Qn=10. n=10/(1.6x10^-19).
34. Let the cube have sides of length L.
nCod2q6.jpg
 
h4rriet

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hela said:
can anyone solve
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf Q 10 AND Q33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q33
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf Q29
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf Q11
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q14

THANKS FOR YOUR HELP
Click to expand...
10. The man's motion is similar to the motion of a ball thrown upwards, except that when you throw a ball upwards, there is no horizontal component for the velocity (the ball moves only up and down, not sideways). The man's motion is a parabolic trajectory (projectile motion); the horizontal component of the velocity is always the same. Only the vertical changes. C must be ruled out because it shows velocity being 0 m/s at one instant; that isn't possible because there is a non-zero horizontal component and it's constant. A can be ruled out because it shows velocity increasing and then decreasing; velocity can't increase as the man jumps upwards because there is deceleration (the force of gravity acts opposite to the jump). B & D remain. They're almost the same; not sure why D's the answer.
33. The distance travelled will be found using the suvat equations. The acceleration will have to be found. Since the plates are parallel, E (the electric field strength) is uniform. F=Eq=ma. Let q of the proton be 1 and of the alpha particle be 2.
Proton = E1=ma, therefore a=E/m.
Alpha particle = E2=ma, therefore a=2E/m.
So you see, the acceleration of the alpha particle is twice the acceleration of the proton.
s=1/2at^2. s is proportional to a and t. If a is times two, so too will be s.

33. A is wrong; charged particles aren't provided by the supply. B is wrong; current is not the speed, but the amount of charge per time. D is wrong; a wire with a smaller diameter will have a greater resistance, and therefore the electrons will move slowly through it.
29. No idea.

11. The initial momentum = 2mu-mu=mu. You have to work out the momentum of A, B, C & D. 3 of them will be mu, the other will be something different. That'll be inconsistent with the principle of conservation of momentum (because the final wouldn't = the initial).

14. 0.3 x 2sin50 (the vertical component is the only one that will have an effect on the rule; the horizontal goes right through the pivot).
 
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h4rriet said:
10. The man's motion is similar to the motion of a ball thrown upwards, except that when you throw a ball upwards, there is no horizontal component for the velocity (the ball moves only up and down, not sideways). The man's motion is a parabolic trajectory (projectile motion); the horizontal component of the velocity is always the same. Only the vertical changes. C must be ruled out because it shows velocity being 0 m/s at one instant; that isn't possible because there is a non-zero horizontal component and it's constant. A can be ruled out because it shows velocity increasing and then decreasing; velocity can't increase as the man jumps upwards because there is deceleration (the force of gravity acts opposite to the jump). B & D remain. They're almost the same; not sure why D's the answer.
33. The distance travelled will be found using the suvat equations. The acceleration will have to be found. Since the plates are parallel, E (the electric field strength) is uniform. F=Eq=ma. Let q of the proton be 1 and of the alpha particle be 2.
Proton = E1=ma, therefore a=E/m.
Alpha particle = E2=ma, therefore a=2E/m.
So you see, the acceleration of the alpha particle is twice the acceleration of the proton.
s=1/2at^2. s is proportional to a and t. If a is times two, so too will be s.

33. A is wrong; charged particles aren't provided by the supply. B is wrong; current is not the speed, but the amount of charge per time. D is wrong; a wire with a smaller diameter will have a greater resistance, and therefore the electrons will move slowly through it.
29. No idea.

11. The initial momentum = 2mu-mu=mu. You have to work out the momentum of A, B, C & D. 3 of them will be mu, the other will be something different. That'll be inconsistent with the principle of conservation of momentum (because the final wouldn't = the initial).
Click to expand...
thanks a lot
 
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h4rriet said:
10. The man's motion is similar to the motion of a ball thrown upwards, except that when you throw a ball upwards, there is no horizontal component for the velocity (the ball moves only up and down, not sideways). The man's motion is a parabolic trajectory (projectile motion); the horizontal component of the velocity is always the same. Only the vertical changes. C must be ruled out because it shows velocity being 0 m/s at one instant; that isn't possible because there is a non-zero horizontal component and it's constant. A can be ruled out because it shows velocity increasing and then decreasing; velocity can't increase as the man jumps upwards because there is deceleration (the force of gravity acts opposite to the jump). B & D remain. They're almost the same; not sure why D's the answer.
33. The distance travelled will be found using the suvat equations. The acceleration will have to be found. Since the plates are parallel, E (the electric field strength) is uniform. F=Eq=ma. Let q of the proton be 1 and of the alpha particle be 2.
Proton = E1=ma, therefore a=E/m.
Alpha particle = E2=ma, therefore a=2E/m.
So you see, the acceleration of the alpha particle is twice the acceleration of the proton.
s=1/2at^2. s is proportional to a and t. If a is times two, so too will be s.

33. A is wrong; charged particles aren't provided by the supply. B is wrong; current is not the speed, but the amount of charge per time. D is wrong; a wire with a smaller diameter will have a greater resistance, and therefore the electrons will move slowly through it.
29. No idea.

11. The initial momentum = 2mu-mu=mu. You have to work out the momentum of A, B, C & D. 3 of them will be mu, the other will be something different. That'll be inconsistent with the principle of conservation of momentum (because the final wouldn't = the initial).

14. 0.3 x 2sin50 (the vertical component is the only one that will have an effect on the rule; the horizontal goes right through the pivot).
Click to expand...
can u please solve thank you
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_12.pdf Q 10
 
ahmed abdulla

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h4rriet said:
10. The man's motion is similar to the motion of a ball thrown upwards, except that when you throw a ball upwards, there is no horizontal component for the velocity (the ball moves only up and down, not sideways). The man's motion is a parabolic trajectory (projectile motion); the horizontal component of the velocity is always the same. Only the vertical changes. C must be ruled out because it shows velocity being 0 m/s at one instant; that isn't possible because there is a non-zero horizontal component and it's constant. A can be ruled out because it shows velocity increasing and then decreasing; velocity can't increase as the man jumps upwards because there is deceleration (the force of gravity acts opposite to the jump). B & D remain. They're almost the same; not sure why D's the answer.
33. The distance travelled will be found using the suvat equations. The acceleration will have to be found. Since the plates are parallel, E (the electric field strength) is uniform. F=Eq=ma. Let q of the proton be 1 and of the alpha particle be 2.
Proton = E1=ma, therefore a=E/m.
Alpha particle = E2=ma, therefore a=2E/m.
So you see, the acceleration of the alpha particle is twice the acceleration of the proton.
s=1/2at^2. s is proportional to a and t. If a is times two, so too will be s.

33. A is wrong; charged particles aren't provided by the supply. B is wrong; current is not the speed, but the amount of charge per time. D is wrong; a wire with a smaller diameter will have a greater resistance, and therefore the electrons will move slowly through it.
29. No idea.

11. The initial momentum = 2mu-mu=mu. You have to work out the momentum of A, B, C & D. 3 of them will be mu, the other will be something different. That'll be inconsistent with the principle of conservation of momentum (because the final wouldn't = the initial).

14. 0.3 x 2sin50 (the vertical component is the only one that will have an effect on the rule; the horizontal goes right through the pivot).
Click to expand...


any one help ?
 

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h4rriet

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ahmed abdulla said:
any one help ?
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35. The voltage across R1 = 5-3=2 V, across R2 = 3-2=1 V, and R3 = 2 V. You can find the resistance of the resistor using the formula for potential dividers in series. For instance, the resistance in R1 can be found using the formula R1/(R1+R2+R3) x 5 = 2.
22. k=F/e. F for one spring is W/3, so k=W/3e. So e when F = W for each spring (2W is divided by two because there are 2 springs), e=W/W/3e=3e.
 
Rahma Abdelrahman

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hela said:
can u solve
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29
THANKS
Click to expand...

18-- Here u have to use : (Change in PE=Change in KE) So, -50000=Final KE -5000 -->> So u get final as -45000.. Answer is B.
29--> Both will form Stationary waves as the wavelength is 20 cm... (Length of 2 segments) and since the two pipes have length greater than 20, they will form stationary waves when the wave is reflected.. Answer is A.. (There is not much to think of here.. )
 
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