Edexcel A-level Physics Paper 1 Unofficial MS
12/3/18 Update: I, as per forum rules, will not be distributing the papers I acquired after my exams. However, if people sat them for a mock or otherwise attained them, and would like some worked examples I'd be more than happy to help if you send me a PM, granted I can find the papers when i return home from Uni at the end of this term (27/3/18).
These are not all correct, if you think I am wrong tell me and i'll change!
MCQ:
C - Joule per coulomb
B - 1.0 kgm/s
A - R/2
C - mgh
D - 4W
D - a little below 6V
D - Lights instantly with same brightness
C - Most a's go straight through | atom is mainly empty space
B - graph with vertical line down and one other line pointing up and to the left.
A - 0.5A
11. Graph starts at origin, increases by a non constant amount (decreasing acceleration) when a force/acceleration applied by cyclist, until drag = forward acceleration where the graph levels off.
b) Not all work is converted to Ek. The work done by the cyclist is converted into the kinetic and gravitational potential in the pedals, which is converted into kinetic energy in the wheels via the chain. Some E lost to environment as sound etc.
12) 14.2 * sin(37) = 8.55m/s
b) h=2.29m
c)s=ut+1/2at^2, in x plane a=0, so s=ut = (14.2*cos(37))*1.98 = 22.45m
13) Magnetic field acts parallel to dees and to current flowing through dees, this creates a force on the charged proton. This acts as a centripetal force. as r=mv/bq, and the momentum of the proton is increasing, the radius increases. The proton accelerates between the dees. This is because the electrostatic forces of the dee's provide a resultant force on the proton in the air. The polarity of each dee swaps every half turn, to ensure successive acceleration.
b)Proton beam stops delivering energy after the tumour (doesn't continue to penetrate, X-rays do). Peak energy is delivered to the tumour, rather than the surface of the skin.
c)Proton beam's could be more effective at treating tumours so could save lives. X-rays can cause more tumours as the energy penetrates farther.
14)Motion out of page
b)R=pl/A=1.6(0.6*10^-3)/(28*10^-3*0.5*10^-3)=68.57, I=V/R= 1.5/68.57= 0.0218A = 21.8mA
c) F = BIl = 0.4 * 21.8*10^-3 * 28*10^-3 = 2.45*10-4N
15) Pd in capacitor starts with a portion of V and decreases to as it charges. Pd in resistor starts with the rest of V and increases as the capacitor charges, to 5V.
b) f=20, so T=0.05s, labels every 10 small squares 50,100,150,200. Parabolic increase in 'boxes', exponential decrease out.
16) SDU, S = -1/3, total is 0, so 2 others must = +1/3, only combination is DU, so SDU.
b) 1116 * ((1.6*10^-19)/(3*10^8)^2) * 10^6 = 1.984*10^-27
c) First decay feasible, second decay not feasible as baryon number not conserved, third feasible, fourth feasible, fifth not feasible as charge not conserved
d)High E protons have high momentum, they slow in the atmosphere so some Ek converted to mass-energy so heavier lambda can be made.
e) /\^0 cannot be observed as it has no charge, so doesn't ionize liquid hydrogen in bubble chambers. Some of it's decay products can as they are charged and hence leave a trace. We can use information about the decay products to predict information about /\^0 as charge, momentum and energy must be conserved
17) R = pl/A, only l is changing, so R is proportional to l. as l increases, R increases. As R increases, V increases, as V=IR and current is uniform.
b) If 1.5V = 2m, then 0.64V = 0.853m, v = s/t = 0.853/1.5s = 0.569m/s
c) v=u+at, 1.5 = 0 + 2a => a = 0.75m/s/s. Vert. component of a = gsin(x), x = 4.3 degrees, so within the bounds.
18) P=W/t
W=E=mgh
P=mgh/t
h/t=v
P=mgv
v=wr
P=mgwr
P=mgr *w
mgr = moment
w= angular velocity
b) from pos > neg, not sure on the lines of equipotential, I drew peanut shaped lines.
ii) F=Q^2/r^2*k = 8.99*10^9 * ((0.1*10^-6)^2/(0.05)^2) = 0.036N
c) F(car) = 833.3N
F(Vacuum) = 17.5N
F(pond pump) = 0.03125N
Hence pond pump is only suitable appliance.
These are not all correct, if you think I am wrong tell me and i'll change!
MCQ:
C - Joule per coulomb
B - 1.0 kgm/s
A - R/2
C - mgh
D - 4W
D - a little below 6V
D - Lights instantly with same brightness
C - Most a's go straight through | atom is mainly empty space
B - graph with vertical line down and one other line pointing up and to the left.
A - 0.5A
11. Graph starts at origin, increases by a non constant amount (decreasing acceleration) when a force/acceleration applied by cyclist, until drag = forward acceleration where the graph levels off.
b) Not all work is converted to Ek. The work done by the cyclist is converted into the kinetic and gravitational potential in the pedals, which is converted into kinetic energy in the wheels via the chain. Some E lost to environment as sound etc.
12) 14.2 * sin(37) = 8.55m/s
b) h=2.29m
c)s=ut+1/2at^2, in x plane a=0, so s=ut = (14.2*cos(37))*1.98 = 22.45m
13) Magnetic field acts parallel to dees and to current flowing through dees, this creates a force on the charged proton. This acts as a centripetal force. as r=mv/bq, and the momentum of the proton is increasing, the radius increases. The proton accelerates between the dees. This is because the electrostatic forces of the dee's provide a resultant force on the proton in the air. The polarity of each dee swaps every half turn, to ensure successive acceleration.
b)Proton beam stops delivering energy after the tumour (doesn't continue to penetrate, X-rays do). Peak energy is delivered to the tumour, rather than the surface of the skin.
c)Proton beam's could be more effective at treating tumours so could save lives. X-rays can cause more tumours as the energy penetrates farther.
14)Motion out of page
b)R=pl/A=1.6(0.6*10^-3)/(28*10^-3*0.5*10^-3)=68.57, I=V/R= 1.5/68.57= 0.0218A = 21.8mA
c) F = BIl = 0.4 * 21.8*10^-3 * 28*10^-3 = 2.45*10-4N
15) Pd in capacitor starts with a portion of V and decreases to as it charges. Pd in resistor starts with the rest of V and increases as the capacitor charges, to 5V.
b) f=20, so T=0.05s, labels every 10 small squares 50,100,150,200. Parabolic increase in 'boxes', exponential decrease out.
16) SDU, S = -1/3, total is 0, so 2 others must = +1/3, only combination is DU, so SDU.
b) 1116 * ((1.6*10^-19)/(3*10^8)^2) * 10^6 = 1.984*10^-27
c) First decay feasible, second decay not feasible as baryon number not conserved, third feasible, fourth feasible, fifth not feasible as charge not conserved
d)High E protons have high momentum, they slow in the atmosphere so some Ek converted to mass-energy so heavier lambda can be made.
e) /\^0 cannot be observed as it has no charge, so doesn't ionize liquid hydrogen in bubble chambers. Some of it's decay products can as they are charged and hence leave a trace. We can use information about the decay products to predict information about /\^0 as charge, momentum and energy must be conserved
17) R = pl/A, only l is changing, so R is proportional to l. as l increases, R increases. As R increases, V increases, as V=IR and current is uniform.
b) If 1.5V = 2m, then 0.64V = 0.853m, v = s/t = 0.853/1.5s = 0.569m/s
c) v=u+at, 1.5 = 0 + 2a => a = 0.75m/s/s. Vert. component of a = gsin(x), x = 4.3 degrees, so within the bounds.
18) P=W/t
W=E=mgh
P=mgh/t
h/t=v
P=mgv
v=wr
P=mgwr
P=mgr *w
mgr = moment
w= angular velocity
b) from pos > neg, not sure on the lines of equipotential, I drew peanut shaped lines.
ii) F=Q^2/r^2*k = 8.99*10^9 * ((0.1*10^-6)^2/(0.05)^2) = 0.036N
c) F(car) = 833.3N
F(Vacuum) = 17.5N
F(pond pump) = 0.03125N
Hence pond pump is only suitable appliance.
(edited 6 years ago)
Scroll to see replies
I think it is R/2 and 4W, light is not instant as there is a difference in distance the electrons have to travel. Work done is not a form of energy, which was the error for that one I think. 12b was about 2.x m. You've got a good answer for 13. The 1st decay isn't feasible i think due to strangeness. For the production of the lambda particle, you need a high energy due to the high rest mass of the lambda particle, and it is pair production, so you need a high energy. Last question you need to multiply the forces by 2 (2 charged particles) - didn't get that one either :s.
(edited 6 years ago)
Original post by Toffo132
These are not all correct, if you think I am wrong tell me and i'll change!
MCQ:
C - Joule per coulomb
B - 1.0 kgm/s
C - 2R
C - mgh
C - 2W
D - a little below 6V
D - Lights instantly with same brightness
C - Most a's go straight through | atom is mainly empty space
B - graph with vertical line down and one other line pointing up and to the left.
C - 1.5V
11. Graph starts at origin, increases by a non constant amount (decreasing acceleration) when a force/acceleration applied by cyclist, until drag = forward acceleration where the graph levels off.
b) Not all work is converted to Ek. The work done by the cyclist is converted into the kinetic and gravitational potential in the pedals, which is converted into kinetic energy in the wheels via the chain. Some E lost to environment as sound etc.
12) 14.2 * sin(37) = 8.55m/s
b) mgh=1/2mv^2 => gh = 1/2v^2 so h = v^2/2g, 8.55^2/(2*9.81) = 3.73m
c)s=ut+1/2at^2, in x plane a=0, so s=ut = (14.2*cos(37))*1.98 = 22.45m
13) Magnetic field acts parallel to dees and to current flowing through dees, this creates a force on the charged proton. This acts as a centripetal force. as r=mv/bq, and the momentum of the proton is increasing, the radius increases. The proton accelerates between the dees. This is because the electrostatic forces of the dee's provide a resultant force on the proton in the air. The polarity of each dee swaps every half turn, to ensure successive acceleration.
b)Proton beam stops delivering energy after the tumour (doesn't continue to penetrate, X-rays do). Peak energy is delivered to the tumour, rather than the surface of the skin.
c)Proton beam's could be more effective at treating tumours so could save lives. X-rays can cause more tumours as the energy penetrates farther.
14)Motion out of page
b)R=pl/A=1.6(0.6*10^-3)/(28*10^-3*0.5*10^-3)=68.57, I=V/R= 1.5/68.57= 0.0218A = 21.8mA
c) F = BIl = 0.4 * 21.8*10^-3 * 28*10^-3 = 2.45*10-4N
15) Pd in capacitor starts with a portion of V and decreases to as it charges. Pd in resistor starts with the rest of V and increases as the capacitor charges, to 5V.
b) f=20, so T=0.05s, labels every 10 small squares 50,100,150,200. Parabolic increase in 'boxes', exponential decrease out.
16) SDU, S = -1/3, total is 0, so 2 others must = +1/3, only combination is DU, so SDU.
b) 1116 * ((1.6*10^-19)/(3*10^8)^2) * 10^6 = 1.984*10^-27
c) First decay feasible, second decay not feasible as baryon number not conserved, third feasible, fourth feasible, fifth not feasible as charge not conserved
d)High E protons have high momentum, they slow in the atmosphere so some Ek converted to mass-energy so heavier lambda can be made.
e) /\^0 cannot be observed as it has no charge, so doesn't ionize liquid hydrogen in bubble chambers. Some of it's decay products can as they are charged and hence leave a trace. We can use information about the decay products to predict information about /\^0 as charge, momentum and energy must be conserved
17) R = pl/A, only l is changing, so R is proportional to l. as l increases, R increases. As R increases, V increases, as V=IR and current is uniform.
b) If 1.5V = 2m, then 0.64V = 0.853m, v = s/t = 0.853/1.5s = 0.569m/s
c) v=u+at, 1.5 = 0 + 2a => a = 0.75m/s/s. Vert. component of a = gsin(x), x = 4.3 degrees, so within the bounds.
18) *not sure how to do this one!* P=W/t, W = Fs so P = Fs/t, s=displacement, so horizontal displacement is x*theta, so P=Fx(theta)/t, theta/t = w, so P=Fxw
b) from pos > neg, not sure on the lines of equipotential, I drew peanut shaped lines.
ii) F=Q^2/r^2*k = 8.99*10^9 * ((0.1*10^-6)^2/(0.05)^2) = 0.036N
c) F(car) = 833.3N
F(Vacuum) = 17.5N
F(pond pump) = 0.03125N
Hence pond pump is only suitable appliance.
MCQ:
C - Joule per coulomb
B - 1.0 kgm/s
C - 2R
C - mgh
C - 2W
D - a little below 6V
D - Lights instantly with same brightness
C - Most a's go straight through | atom is mainly empty space
B - graph with vertical line down and one other line pointing up and to the left.
C - 1.5V
11. Graph starts at origin, increases by a non constant amount (decreasing acceleration) when a force/acceleration applied by cyclist, until drag = forward acceleration where the graph levels off.
b) Not all work is converted to Ek. The work done by the cyclist is converted into the kinetic and gravitational potential in the pedals, which is converted into kinetic energy in the wheels via the chain. Some E lost to environment as sound etc.
12) 14.2 * sin(37) = 8.55m/s
b) mgh=1/2mv^2 => gh = 1/2v^2 so h = v^2/2g, 8.55^2/(2*9.81) = 3.73m
c)s=ut+1/2at^2, in x plane a=0, so s=ut = (14.2*cos(37))*1.98 = 22.45m
13) Magnetic field acts parallel to dees and to current flowing through dees, this creates a force on the charged proton. This acts as a centripetal force. as r=mv/bq, and the momentum of the proton is increasing, the radius increases. The proton accelerates between the dees. This is because the electrostatic forces of the dee's provide a resultant force on the proton in the air. The polarity of each dee swaps every half turn, to ensure successive acceleration.
b)Proton beam stops delivering energy after the tumour (doesn't continue to penetrate, X-rays do). Peak energy is delivered to the tumour, rather than the surface of the skin.
c)Proton beam's could be more effective at treating tumours so could save lives. X-rays can cause more tumours as the energy penetrates farther.
14)Motion out of page
b)R=pl/A=1.6(0.6*10^-3)/(28*10^-3*0.5*10^-3)=68.57, I=V/R= 1.5/68.57= 0.0218A = 21.8mA
c) F = BIl = 0.4 * 21.8*10^-3 * 28*10^-3 = 2.45*10-4N
15) Pd in capacitor starts with a portion of V and decreases to as it charges. Pd in resistor starts with the rest of V and increases as the capacitor charges, to 5V.
b) f=20, so T=0.05s, labels every 10 small squares 50,100,150,200. Parabolic increase in 'boxes', exponential decrease out.
16) SDU, S = -1/3, total is 0, so 2 others must = +1/3, only combination is DU, so SDU.
b) 1116 * ((1.6*10^-19)/(3*10^8)^2) * 10^6 = 1.984*10^-27
c) First decay feasible, second decay not feasible as baryon number not conserved, third feasible, fourth feasible, fifth not feasible as charge not conserved
d)High E protons have high momentum, they slow in the atmosphere so some Ek converted to mass-energy so heavier lambda can be made.
e) /\^0 cannot be observed as it has no charge, so doesn't ionize liquid hydrogen in bubble chambers. Some of it's decay products can as they are charged and hence leave a trace. We can use information about the decay products to predict information about /\^0 as charge, momentum and energy must be conserved
17) R = pl/A, only l is changing, so R is proportional to l. as l increases, R increases. As R increases, V increases, as V=IR and current is uniform.
b) If 1.5V = 2m, then 0.64V = 0.853m, v = s/t = 0.853/1.5s = 0.569m/s
c) v=u+at, 1.5 = 0 + 2a => a = 0.75m/s/s. Vert. component of a = gsin(x), x = 4.3 degrees, so within the bounds.
18) *not sure how to do this one!* P=W/t, W = Fs so P = Fs/t, s=displacement, so horizontal displacement is x*theta, so P=Fx(theta)/t, theta/t = w, so P=Fxw
b) from pos > neg, not sure on the lines of equipotential, I drew peanut shaped lines.
ii) F=Q^2/r^2*k = 8.99*10^9 * ((0.1*10^-6)^2/(0.05)^2) = 0.036N
c) F(car) = 833.3N
F(Vacuum) = 17.5N
F(pond pump) = 0.03125N
Hence pond pump is only suitable appliance.
I think that It should be 4W as it is 1/2CV^2 and if you double the Voltage you square that for 4x.
I also think it should be mgh/sin(theta)
Original post by Arnaez4
MCQ 3 is wrong I think? Parallel resistance can't be larger than individual resistor
Yep, my bad
Original post by TechnoEngineer
I think that It should be 4W as it is 1/2CV^2 and if you double the Voltage you square that for 4x.
I also think it should be mgh/sin(theta)
I also think it should be mgh/sin(theta)
4W is right, the change in Ep is work, so mgh is right (my teacher agree'd)
Pretty sure question 2 was in parallel no?
b is wrong in projectiles it should be 2.29
i done it by resolving vertically then horizontally and minusing first height from second
for derivation p=w/t
w=fd
p=fd/t
d=2pir ( circum of circle)
t=2pi/ omega
p=2piFr( omega)/2pi so P=Frw
b is wrong in projectiles it should be 2.29
i done it by resolving vertically then horizontally and minusing first height from second
for derivation p=w/t
w=fd
p=fd/t
d=2pir ( circum of circle)
t=2pi/ omega
p=2piFr( omega)/2pi so P=Frw
I agree it should be R/2 and 4W
thanks for this btw
thanks for this btw
Original post by glad-he-ate-her
I agree it should be R/2 and 4W
thanks for this btw
thanks for this btw
Yeah they're changed! Np
if you want any questions to look over I can email/Pm you themLast MC should be 0.5 v
when i did question 12b using SUVAT i got an answer which was 2.(something) so i think that needs correcting
Dont think the 4th Lambda equation is feasible as it violates conservation of momentum as it only produces one decay product.
Original post by glad-he-ate-her
Pretty sure question 2 was in parallel no?
b is wrong in projectiles it should be 2.29
i done it by resolving vertically then horizontally and minusing first height from second
for derivation p=w/t
w=fd
p=fd/t
d=2pir ( circum of circle)
t=2pi/ omega
p=2piFr( omega)/2pi so P=Frw
b is wrong in projectiles it should be 2.29
i done it by resolving vertically then horizontally and minusing first height from second
for derivation p=w/t
w=fd
p=fd/t
d=2pir ( circum of circle)
t=2pi/ omega
p=2piFr( omega)/2pi so P=Frw
Could you show me the working for h=2.29? i've seen a lot of people say they got that and i'm clearly dumb and can't see how!
Original post by BerkanMarasli
when i did question 12b using SUVAT i got an answer which was 2.(something) so i think that needs correcting
2.29
i did the same
Original post by Toffo132
Yeah they're changed! Np if you want any questions to look over I can email/Pm you them
if you want any questions to look over I can email/Pm you themI think Q10 is wrong. It seems you made the same mistake I initially did, which was assuming that the resistor has a p.d. of 6V, which means you're forgetting about the p.d. across the thermistor. I think you were supposed to look for values of V for each such that they add to 6, and the only option which satisfied this was D - 2V.
Original post by pjlemz
for the question 7, wouldnt there be a delay and then the brightness is same. because if its a coil, right as the current flows, theres is a change in flux linkage so through lenz law theres a back emf to oppose it?
Yeah i said the same
Original post by pjlemz
for the question 7, wouldnt there be a delay and then the brightness is same. because if its a coil, right as the current flows, theres is a change in flux linkage so through lenz law theres a back emf to oppose it?
There's no magnetic field
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