A-Level Physics Paper 1 doubts (MCQ)

[krtcobain82]

Guys I need explanations for the following questions.

1) may/june 07 Paper 1

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s07_qp_1.pdf

Question no 37 and 40.

2) may/june 09 Paper 1

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s09_qp_1.pdf

Question no 5, 7 .

[Ghost Of Highbury]

Guys I need explanations for the following questions.

1) may/june 07 Paper 1

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s07_qp_1.pdf

Question no 37 and 40.

2) may/june 09 Paper 1

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s09_qp_1.pdf

Question no 5, 7 .

Q37) This is a tricky one. Electrical resistance of P - R - pl/A
Let's assume the cross-sectional area of the resistor P is A
Thus, R = px/A

Resistance of Q
Now, the resistance is the combined resistance of all the wires
Say for instance, the resistance one of those wires is px/a (a < A, p, because they're of the same material)
So as the wires are in parallel, the combined resistance => 1/r = 1/r1 + 1/r2 + 1/r3 + ...
= a/px + a/px + a/px + ....
1/r = na/px

r = px/na

Now, As the volumes of the materials used is the same for both resistors, and their length is the same, the cross-sectional area of resistor P
should be the sum of all the cross-sectional areas of the wires in resistor Q
Meaning, na = A
thus r = px/A = Resistance of resistor P

Answer : C

40) The final velocity depends on the acceleration. The one with the least acceleration has the lowest final velocity (v^2 = 2as)
Acceleration, a = F/m
F= EQ
E is constant because V and d are constant.
a = QE/m
E is constant, thus, a is lowest, when the Q/m of the particle is the lowest. That is with C, Li, 3/7
Answer : C
--
5) Best to use elimination in these type of sums. The final displacement can't be zero, thus, you are left with C or D
It can't be C because it doesn't show "constant terminal velocity"
Answer : D

7) Well, to be honest, i thought the answer was C. Well, i'm not sure of A, but that's the answer. The second law deals with force being
the rate of change of velocity times mass, and the first law is about inertia (unless acted upon external force one).

[ashwinkandel]

For no. 7 I think its like this. According to first law, the body remains in its state or uniform motion or rest until acted by an external force. If we take this mathematically then F=ma. In absence of external force the equation becomes 0=ma>> or a=0. a=0 tells us that either the body is moving in constant speed or is not moving at all.

[Ghost Of Highbury]

For no. 7 I think its like this. According to first law, the body remains in its state or uniform motion or rest until acted by an external force. If we take this mathematically then F=ma. In absence of external force the equation becomes 0=ma>> or a=0. a=0 tells us that either the body is moving in constant speed or is not moving at all.

F=ma is the mathematical expression for the second law (Force is the rate of change of momentum). I don't think it's for the first.

[ashwinkandel]

Yes, F=ma is the second law. For that question, no. 7, the solution as you said is option A (The first law follows from the second law.) and i tried to show that thing.

[NidZ- Hero]

Some one xeplain me the steps

[elemis]

Give me a few minutes.

[elemis]

You can write two formulas from the given data :

S₁ = ut +0.5at²

S₁+S₂ = u(2t) +0.5a(2t)²

S₁+S₂ = 2ut + 2at²

Now, notice all the answers have S1, S2 and time.... Initial velocity has been eliminated. So we know we have to play around with the second equation.

Using the first equation :

ut = S₁ -0.5at²

Substitute into the second one :

S₁+S₂ = 2*(S₁ -0.5at²) + 2at²

Simplifying :

S₁+S₂ = 2S₁ - at² +2at²

S₂-S₁ =  at²

(S₂-S₁)/t² = a

Answer = A

[Ghost Of Highbury]

Some one xeplain me the steps

s1 = ut + 0.5at^2

u = (2s1 - at^2)/2t

2. s1+s2 = 2ut + 2at^2    (Because it takes time 2t to cover a distance of s1+s2)

u = (s1+s2-2at^2)/2t

Equate both the Us

(2s1-at^2)/2t = (s1+s2-2at^2)/2t

2t gets cancelled

2s1-at^2-s1-s2+2at^2 = 0
a = (s2-s1)/t^2

[Ghost Of Highbury]

Ah, sry, my bad.

[NidZ- Hero]

Thanks a million


I have soo many other doubts can yu help

[Ghost Of Highbury]

How is it C?

[elemis]

Thanks a million


I have soo many other doubts can yu help

Lets say the 2m ball is Ball A and the m ball is Ball B.

Hence,

U_A - -U_b = 5u/3 --u/3

I'm using the idea that when there is an elastic collision the initial velocity of body A + the initial velociity of body B is equal to the final velcoity of body B + that of Body A.

Make sure to use VELCOITIES.

[elemis]

How is it C?

Sketch the wave on an axis as it would appear a few seconds later in time. Compare to the original... VOILA !!!

[Ghost Of Highbury]

Thanks a million


I have soo many other doubts can yu help

The total KE before collision, i.e 1.5mu^2, should be equal to total KE after collision.
And the momentum before collision should be the momentum after collision.

A satisfies both the conditions. Answer : A