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AS Physics P1 MCQs Preparation Thread.

Abu mota

Abu mota

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6Astarstudent said:
S11 qp11
#14
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2

#34
R= pl/A
and unit of l/A = m^-1 = V^(-1/3)
so R = p/(V^1/3)

#35
ratio of resistance is 1:2:5 and we know that ratio is current is inverse of that
so must be 1:1/2:1/5 = 10:5:2
so 5x 5/17 = 1.47 answer is C
Click to expand...

tyvm, but i still dont understand Q.35, is there an easier way ( i dont know how to find the inverse of a ratio)
 
Abu mota

Abu mota

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Obsidian Fl1ght said:
s_12_qp17.
Sorry for bothering u guys but again - how do we get 6mgh??
Click to expand...

first block doesnt require any energy to life so 0
second block requires u to lift it a height of h and so mgh
third block requires u to lift it a height of 2h and so 2mgh
fourth block requires u to lift it a height of 3h and so 3mgh
add them all together to get 6mgh
 
Suno Yaar

Suno Yaar

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w
6Astarstudent said:
S11 qp11
#14
this one is tricky, utilize the kinematics equation
V^2 = U^2 + 2as
so the first drop is
u^2 = 0 + 2ah
the rebound is
0 = v^2 - 2a(h/2) = v^2 - ah
so v^2/u^2 = 1/2
v/u = 1/squareroot 2

#34
R= pl/A
and unit of l/A = m^-1 = V^(-1/3)
so R = p/(V^1/3)

#35
ratio of resistance is 1:2:5 and we know that ratio is current is inverse of that
so must be 1:1/2:1/5 = 10:5:2
so 5x 5/17 = 1.47 answer is C
Click to expand...
what information about quarks is necessary to know in nuclear physixs??
 
6

6Astarstudent

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Abu mota said:
tyvm, but i still dont understand Q.35, is there an easier way ( i dont know how to find the inverse of a ratio)
Click to expand...
you can find the seperate

continuation:
W10 qp 12
#8
probably one of the toughest kinematic Q ever in AS level
s = ut + (at^2)/2
so 40 = 12u + 72a
or 40 = 6v + 18a
then use s = t(u + v)/2
so 40 = 6(u+v) u + v = 40/6 v = 40/6 - u
substitute into 40 = 6v + 18a
so 40 = 40 - 6u + 18a so you get that a:u = 3:1
then substitute back to 40 = 12u + 72a
so you get 12(3a) + 72a = 40 so 108a = 40 a =0.37 answer is A
#15
Ek = (mv^2)/2 and its at steady speed so kv = mg so v = mg/k
Ek = m(mg/k)^2)/2 = m^3g^2/2k^2

#27
λ= ax/D so a=λD/x where a is slit seperation, D is distance , x is fringe seperation
substitution gives you
a= (600 x 10^-9 x D)/(1x10^-3) and
a= (600 x 10^-9 x (D+2)/(1x10^-3)
solve D and a, and a is 6x10^-3 which is B
#34
R= pl/A so great p means greater R as current goes into the wire, the "l" value increases too so R increases dropping voltage. so answer is B, with constant decrease in value of V and sudden change in gradient everytime p changes.
#40
for this one it is definitely random not spontaneous. so its A or B
mark scheme says its B but I would have picked A. cant help you on this one
 
Abu mota

Abu mota

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6Astarstudent said:
you can find the seperate

continuation:
W10 qp 12
#8
probably one of the toughest kinematic Q ever in AS level
s = ut + (at^2)/2
so 40 = 12u + 72a
or 40 = 6v + 18a
then use s = t(u + v)/2
so 40 = 6(u+v) u + v = 40/6 v = 40/6 - u
substitute into 40 = 6v + 18a
so 40 = 40 - 6u + 18a so you get that a:u = 3:1
then substitute back to 40 = 12u + 72a
so you get 12(3a) + 72a = 40 so 108a = 40 a =0.37 answer is A
#15
Ek = (mv^2)/2 and its at steady speed so kv = mg so v = mg/k
Ek = m(mg/k)^2)/2 = m^3g^2/2k^2

#27
λ= ax/D so a=λD/x where a is slit seperation, D is distance , x is fringe seperation
substitution gives you
a= (600 x 10^-9 x D)/(1x10^-3) and
a= (600 x 10^-9 x (D+2)/(1x10^-3)
solve D and a, and a is 6x10^-3 which is B
#34
R= pl/A so great p means greater R as current goes into the wire, the "l" value increases too so R increases dropping voltage. so answer is B, with constant decrease in value of V and sudden change in gradient everytime p changes.
#40
for this one it is definitely random not spontaneous. so its A or B
mark scheme says its B but I would have picked A. cant help you on this one
Click to expand...

tyvm, i still dont understand Q.34
 
6

6Astarstudent

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Abu mota said:
tyvm, i still dont understand Q.34
Click to expand...
hm okay so basically there are 2 factors that affect the R, which drops the V.
the factors are : length of wire current has travelled through and the resistivity of the wire
since the rate of current passing through the wire is always the same under same resistivty so we know gradient will be a constant.
Since the change in resistivity is descrete not continuous so there will be sudden gradient change between P1 P2 and P2 P3
 
Abu mota

Abu mota

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6Astarstudent said:
hm okay so basically there are 2 factors that affect the R, which drops the V.
the factors are : length of wire current has travelled through and the resistivity of the wire
since the rate of current passing through the wire is always the same under same resistivty so we know gradient will be a constant.
Since the change in resistivity is descrete not continuous so there will be sudden gradient change between P1 P2 and P2 P3
Click to expand...

tyvm, i understand it now.
 
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