IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: joel on August 22, 2010, 07:04:35 am

Title: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: joel on August 22, 2010, 07:04:35 am: Can some one explain the answers please...
Especially the last question because E = V/D...

I have attached a file containing the questions.

1. May/June 2002 Question 9 - Correct Answer B
2. May/June 2002 Question 28 - Correct Answer C
3. October/November 2002 Question 11 - Correct Answer A
4. October/November 2002 Question 18 - Correct Answer D
5. October/November 2002 Question 35 - Correct Answer D
6. May/June 2003 Question 5 - Correct Answer C
7. May/June 2003 Question 35 - Correct Answer C
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: nid404 on August 22, 2010, 07:16:09 am: in an hour's time
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: nid404 on August 22, 2010, 07:24:26 am: 1) Momentum and kinetic energy both are conserved

Momentum before collision= mv+ m(-v) = 0

So momentum after collision will also be zerio

Kinetic energy before collision= 1/2 mv2 + 1/2 m (-v)²= mv²
This is conserved since it's an elastic collision.

Hence B
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: nid404 on August 22, 2010, 07:31:16 am: 2)

there's a formula for such type of question

$phase difference=$ $(2\pi/\lambda)X path difference$

for destructive interference the phase difference should be $\pi$

so if you rearrange the eqn (i substituted phase difference as $\pi$

$\lambda$ = $(2\pi/ \pi)x path difference$

path difference= S2X-S1X

so you get

$\lambda$ = $(2) x S2X-S1X$
$\lambda)/2=S2X-S1X$
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: nid404 on August 22, 2010, 07:42:43 am: 3) elastic collision kinetic energy is conserved.

Quote from: ~Ahana~ on May 07, 2010, 09:24:38 pm
Q11) this is gonna be long...

In an elastic collision...momentum as well as kinetic energy is conserved,,
conservation of momentum
total momentum before collision=total momentum after collision.
mu1+mu2=mv1+mv2

conservation of k.e
1/2mu1²+ 1/2mu2²= 1/2mv1²+1/2mv2²

cancel all the halfs first, then cancel all the 'm' terms.

in steps...here's how u do it

m(u1-v1)=m(v2-u2) 1)
m(u1²-v1²)= m( v2²-u2²) 2)

since (u1²-v1²)= (u1-v1)(u1+v1)
& ( v2²-u2²)=(v2-u2)(v2+u2)

divide the 2nd eqn by the first
m(u1²-v1²) m( v2²-u2²)
---------------------------------= ------------------------------------
m(u1-v1) m(v2-u2)

m [(u1-v1)(u1+v1)]=[m(v2-u2)(v2+u2)]
------------------- ------------------
m(u1-v1) m(v2-u2)

and then cancelling gives
(u1+v1)=(v2+u2)
or
u1-u2=v2-v1

now coming back to the question.
u1-u2 in this case will be positive because u for the other body is in the opp direction
u1-(-u2)=u1+u2
v2-v1 will be as it...both move in the same direction after collision
so u1+u2=v2-v1
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: nid404 on August 22, 2010, 07:43:20 am: Rest when I get back home.
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: S.M.A.T on August 22, 2010, 07:52:19 am: 4)G.P.E=mgh=(1.3*10^9)*9.81*2=2.55*10^10 J
Therefore output power for one day is

Power=Energy/time=(2.55*10^10)/(1*24*60*60)=2.95*10^5 which is approximately 300kW.
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: S.M.A.T on August 22, 2010, 08:34:04 am: 5)Initially the emf of the cell(in the bottom circuit) is equal to the potential difference in wire XN so galvanometer shows no deflection.But when the resistance of variable resistor is increased the p.d across variable resistor increases so the p.d across XY decreases.(because total p.d in the(top) circuit remain the same).Now the potential difference across XN is less than the emf of the cell so the movable contact must be move towards the Y to increase the p.d across XN.so when the p.d become equal to emf of the cell the galvanometer again shows zero deflection.

Summarizing,

Initial
Emf of the cell in the bottom circuit=pd across XN.
After increasing resistance of variable resistor,the p.d across XY decreases.
Now p.d across XN is less than before,so the movable contact must be move towards the Y to increase the p.d across XN.
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: S.M.A.T on August 22, 2010, 09:21:43 am: 7)The answer would be c because in the region between two parallel charge plate,the equipotential lines are equally spaced.So as the distance 'd' increase voltage increases proportionally so 'E' remain constant as E=V/d.In the region between two parallel charge plate the electric field strength is always constant
Title: Re: CIE AS LEVEL PHYSICS MULTIPLE CHOICE DOUTS
Post by: nid404 on August 22, 2010, 10:22:41 am: Thanks +rep