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AS Physics P1 MCQs Preparation Thread.

N.M

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in elastic collisions kinetic energy is always conserved
so, the sum of velocity b4 collision must equal that after collision
that means
u1+v1=u2+v2
but at the same time you need to keep a reference direction
lets say towards the right is positive
so the eqn for this case will be ux+(-vx)=(-uy) + vy
now you can simply rearrange to get the correct answer that is A

hope its clear now
 
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There is just one problem, though. For question 37, the answer is A.
Oh, sorry. I mixed this up with another question of the same type so I didn't fully read the question. Disregard that explanation I wrote.

In 37, the voltmeter is connected in parallel with X and Y, and you know that in parallel to voltage is always the same, regardless of where the potentiometer is. So it would be A. Sorry for that. :|
 
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Oh, sorry. I mixed this up with another question of the same type so I didn't fully read the question. Disregard that explanation I wrote.

In 37, the voltmeter is connected in parallel with X and Y, and you know that in parallel to voltage is always the same, regardless of where the potentiometer is. So it would be A. Sorry for that. :|
That's alright. Thank you very much. :)
 

N.M

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actually i dont seem to understand this..

can u elaborate

well the thing is that even i did not understand it that well n sort of memorized it
i guess you have to memorize some answers with a vague explanation otherwise physics mcqs will drive you crazy!!
i m sorry =(
may be some body else might have a clear understanding n can explain it
 

XPFMember

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well the thing is that even i did not understand it that well n sort of memorized it
i guess you have to memorize some answers with a vague explanation otherwise physics mcqs will drive you crazy!!
i m sorry =(
may be some body else might have a clear understanding n can explain it
leosco95 replied to this..
he said..x and y will be stretched..and that is tension...z will compress..that is compression
 
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Q9)

Initial m = Final m

mv + 0 = 2mx (note: after the bodies stick, their masses combine. Let their combined speed be 'x').
mv = 2mx
x = v/2 (m cancel out)

Since the initial momentum is mv, the final momentum must be the same, so its mv also.

K.E after is given by,

0.5 * (2m) * (v/2)^2
= mv^2/4

So A. :)

Q10)

Change in momentum = final m - initial m

Since after impact the direction changes, final m is -mu

= -mu - mu
= -2mu

Q14) My advice for these type of questions would be to use some sort of fake value for something which isn't known. Here we can't use a fake value for u or v since we have to find their ratios which is fixed, but we can use a fake value for the height to make calculations easier. One of my friends asked me this earlier, so I drew a diagram - link. Let me know if you don't get anything.

Q31)

Q = I * t
Q = 10 * 1 = 10C

(Note that the area is given for no apparent reason, there's no area in the charge formula).

1 electron = 1.6 * 10^-19 C
x electron = 10C

x = 6.3 * 10^19 electrons.

Q35) Total R = 0.588
Total V = IR = 0.588 * 5 ~2.94

Where the ammeter is, the voltage will be 2.94 and R is 2, so I would be given by V/R = 2.94/2 = 1.47 which is 1.5A to 1 dp.
hey, thanks once again. I did understand question 14. That was very helpful. thank you. Jazakallah. may Allah bless you. :)
 

omg

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Qs. 10, 14 and 15 ! pleaseeeeeeeeeeeeeee
 

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I = V/r so total r is 2 paralal 100 kohm = 50 kohm and plus one series resistor hence total 150 k ohm hence I=6/150 = 40microAmperes now look at junction current is divided equally coz there is equal resistance in both paths so 40/2 = 20 A (A)
hope this helps :)
Oh Ohkayyy.... i got it ...that was so simple :oops:
 
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in elastic collisions kinetic energy is always conserved
so, the sum of velocity b4 collision must equal that after collision
that means
u1+v1=u2+v2
but at the same time you need to keep a reference direction
lets say towards the right is positive
so the eqn for this case will be ux+(-vx)=(-uy) + vy
now you can simply rearrange to get the correct answer that is A

hope its clear now
yep thats the correct way to do it
 
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1/2kx^2 is used 2 find d strain energy and fx is used 2 find d spring constant. But I dunno abt 1/2fx ( maybe the avg spring const or sumtin) sorry
i solved a question where they are asking abt da strain energy and i used 1/2 fx and da answer is c0ming ryt :S
 
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in elastic collisions kinetic energy is always conserved
so, the sum of velocity b4 collision must equal that after collision
that means
u1+v1=u2+v2
but at the same time you need to keep a reference direction
lets say towards the right is positive
so the eqn for this case will be ux+(-vx)=(-uy) + vy
now you can simply rearrange to get the correct answer that is A

hope its clear now
Jazakallah...
so u mean that the equation:
relative speed of approach=relative speed of separation
does not take directions into account???
 
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Qs. 10, 14 and 15 ! pleaseeeeeeeeeeeeeee
Q10)it decreases because a mass is added to it randomly while it was moving and the speed remains the same when the trap door opens and the sand comes because because no external force acts on it

Q14)it weighs 30kN s0 the distance is 10m on one side and 10 on the oder so weight is divided so answer is 15kN
 
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