OCR Physics A Newtonian World 20/6/16 Unofficial mark scheme
Usual disclaimers. These are just my answers worked through quickly this morning.
They may contain errors, typos and omissions.
Didn't get chance to see my guys when they came out but the word is a bit mixed. Some liked it; others didn't.
Seems very mathematical to me  not much heat. hardly any 'explain' questions.
Heavily loaded to mechanics.
Q1 a) i) Gradient = acceleration of freefall so same for both (1)
ii) Area under graph = distance travelled.
Ball loses energy in bounce so speed after is less than speed before
so doesn't go so high (2)
I'm not sure whether you can just use the graph for i)ii) and iii) by reading off v and calculating the area. I suspect not.
b) i) suvat v^2=u^2+2as so v = () 5.8 ms1 (1)
ii) Impulse = FT = 1.2Ns = change in momentum
so new momentum = 0.13 x 5.8  1.2 = 0.754Ns (up) so v = 3.4 ms1 (2) (Doesn't match graph)
iii) suvat so V^=U^2 +2as again so s=0.59m (1)
Total: 7
Q2 a) Omnia corpore ... OK  in English Every object continues in a state of rest or uniform motion in a straight line unless acted upon by a resultant force. (1)
b) i) same magnitude; same type (2)
ii) opposite directions; act on different objects (2)
c) Nice diagram ...
i) Mass per s = vol per s x density = 3.3E4 x 25 x 1.0E3 = 8.25 kg per s (1)
ii) R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
= 902.5+169 = 1071N (3)
Total: 9
Q3 a) i) CF (1)
ii) G (1)
iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)
b) i) Inverted parabola  max KE = 50 mJ (2)
ii) 1/2 x 0.45 x v^2 = 50E3 so v = 0.471ms1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E2) = 1.50Hz
so T = 0.67s (2)
Total: 8
Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg1 (1)
b) i) Period squared is prop to radius of orbit cubed (1)
ii) (7.7/30)^2 = (9.4E3/R)^3 so R^3 = 15.18 x 9.4E3^3
so R = 2.33E4km (2)
c) mv^2/r = GMm/r^2 so v^2 = GM/r if lower orbit
if r decreases v will increase. (1)
Total: 7
Q5 Been a while since we had a binary question.
a) i) F = GM1M2/(R1+R2)^2 (1)
ii) F1 = M1v1^2/R1 = M1 (2 pi R1/T)^2 / R1 = 4 pi^2 R1 / T^2 (1)
b) Centripetal force = grav force so same on each object
M1 x (4pi^2) x R1 / T^2 = M2 x (4pi^2) x R2 / T^2
so M1/M2 = R2/R1 (2)
c) M1/M2 = R2/R1 = 3 R1+R2 = 4.8E12
so divide 4.8E12 in ratio 1:3
R1 = 1.2E12 and R2=3.6E12 (3)
d) V1 = 2 pi R1 / T = 6.08E4 ms1 (2)
e) Ok I've had an overnight rethink on this one. I'm pretty confident with this now
Its what parts A I) and ii) were intended for.
M1v1^2/r1 = GM1M2/(r1+r2)^2
so M2= v1^2 (r1+r2)^2/Gr1
= 6.08E4^2 x 4.8E12 ^2 / ( 6.67E11 x 1.2E12)
=1.06E33kg
Kepler's law of planetary motion isn't going to apply here to a binary system.
Because r = radius of orbit (r1) is not equal to r=separation of masses here. (r1+r2)
Total: 12
Q6 On to something a lot easier!
a) As falls, grav PE > KE
when hits bottom, KE > heat (2)
b) 50 x mgh = mcdT
50 x 0.025 x 9.81 x 1.2 = 0.025 x c x 4.5 so c=131 J kg1 K1 (4)
c) assume  no air resistance so all PE >KE
assume  all KE  heating le3ad shot and no heat transferred to container (2)
d) If double mass, double input energy but heat shared among 2x mas so same change in temp (2)
Total: 10
Q7 a) Ideal gas so all internal energy is translational KE
KE is prop to Kelvin temp so internal energy is prop to Kelvin temp (2)
b) i) PV = nRT
1.0E5 V = (80/0.004) x 8,81 x (21+273) so V = 489 m^3 (3)
ii) n = PV/RT = 1.3E3 x 1.4E4 / (*.81 = (27340) = 8677
so need to lose 200008677 = 11300 moles (2)
Total: 7
So there we have it. Good for maths experts; not so good for the medicos.
Good Luck Col
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OCR Physics A newtonian World 20/6/16 Unofficial Mark Scheme
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 1
 20062016 15:21
Last edited by teachercol; 21062016 at 08:59.Post rating:9 
 Follow
 2
 20062016 15:41
Thanks!

 Follow
 3
 20062016 15:44
I don't know about 1)b)iii, because the height after the first bounce was 1.7m so it can't be 2.0m after the next one.
I got 0.6, anyone else? 
 Follow
 4
 20062016 15:45
1biii) ??????

 Follow
 5
 20062016 15:45
(Original post by ComputerMaths97)
I don't know about 1)b)iii, because the height after the first bounce was 1.7m so it can't be 2.0m after the next one.
I got 0.6, anyone else? 
 Follow
 6
 20062016 15:46
do they even teach binary system in the book.....?

 Follow
 7
 20062016 15:47
(Original post by teachercol)
OCR Physics A Newtonian World 20/6/16 Unofficial mark scheme
Usual disclaimers. These are just my answers worked through quickly this morning.
They may contain errors, typos and omissions.
Didn't get chance to see my guys when they came out but the word is a bit mixed. Some liked it; others didn't.
Seems very mathematical to me  not much heat. hardly any 'explain' questions.
Heavily loaded to mechanics.
Q1 a) i) Gradient = acceleration of freefall so same for both (1)
ii) Area under graph = distance travelled.
Ball loses energy in bounce so speed after is less than speed before
so doesn't go so high (2)
I'm not sure whether you can just use the graph for i)ii) and iii) by reading off v and calculating the area. I suspect not.
b) i) suvat v^2=u^2+2as so v = () 5.8 ms1 (1)
ii) Impulse = FT = 1.2Ns = change in momentum
so new momentum = 0.13 x 5.8  1.2 = 0.754Ns (up) so v = 3.4 ms1 (2) (Doesn't match graph)
iii) suvat so V^=U^2 +2as again so s=2.00m (1)
Total: 7
Q2 a) Omnia corpore ... OK  in English Every object continues in a state of rest or uniform motion in a straight line unless acted upon by a resultant force. (1)
b) i) same magnitude; same type (2)
ii) opposite directions; act on different objects (2)
c) Nice diagram ...
i) Mass per s = vol per s x density = 3.3E4 x 25 x 1.0E3 = 8.25 kg per s (1)
ii) R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
= 902.5+169 = 1071N (3)
Total: 9
Q3 a) i) CF (1)
ii) G (1)
iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)
b) i) Inverted parabola  max KE = 50 mJ (2)
ii) 1/2 x 0.45 x v^2 = 50E3 so v = 0.471ms1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E2) = 1.50Hz
so T = 0.67s (2)
Total: 8
Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg1 (1)
b) i) Period squared is prop to radius of orbit cubed (1)
ii) (7.7/30)^2 = (9.4E3/R)^3 so R^3 = 15.18 x 9.4E3^3
so R = 2.33E4km (2)
c) mv^2/r = GMm/r^2 so v^2 = GM/r if lower orbit
if r decreases v will increase. (1)
Total: 7
Q5 Been a while since we had a binary question.
a) i) F = GM1M2/(R1+R2)^2 (1)
ii) F1 = M1v1^2/R1 = M1 (2 pi R1/T)^2 / R1 = 4 pi^2 R1 / T^2 (1)
b) Centripetal force = grav force so same on each object
M1 x (4pi^2) x R1 / T^2 = M2 x (4pi^2) x R2 / T^2
so M1/M2 = R2/R1 (2)
c) M1/M2 = R2/R1 = 3 R1+R2 = 4.8E12
so divide 4.8E12 in ratio 1:3
R1 = 1.2E12 and R2=3.6E12 (3)
d) V1 = 2 pi R1 / T = 6.08E4 ms1 (2)
e) v1^2 = GM2/R1) so M2 = 6.65E31 kg so M1 = 3M2 = 2.0E32kg (3)
Kepler's law of planetary motion isn't going to apply here to a binary system.
Total: 12
Q6 On to something a lot easier!
a) As falls, grav PE > KE
when hits bottom, KE > heat (2)
b) 50 x mgh = mcdT
50 x 0.025 x 9.81 x 1.2 = 0.025 x c x 4.5 so c=131 J kg1 K1 (4)
c) assume  no air resistance so all PE >KE
assume  all KE  heating le3ad shot and no heat transferred to container (2)
d) If double mass, double input energy but heat shared among 2x mas so same change in temp (2)
Total: 10
Q7 a) Ideal gas so all internal energy is translational KE
KE is prop to Kelvin temp so internal energy is prop to Kelvin temp (2)
b) i) PV = nRT
1.0E5 V = (80/0.004) x 8,81 x (21+273) so V = 489 m^3 (3)
ii) n = PV/RT = 1.3E3 x 1.4E4 / (*.81 = (27340) = 8677
so need to lose 200008677 = 11300 moles (2)
Total: 7
So there we have it. Good for maths experts; not so good for the medicos.
Good Luck Col
Posted from TSR Mobile 
 Follow
 8
 20062016 15:48
(Original post by Kevinwong)
You're wrong about the planets the answer is 1x10^33
Posted from TSR Mobile 
 Follow
 9
 20062016 15:49
(Original post by teachercol)
OCR Physics A Newtonian World 20/6/16 Unofficial mark scheme
Usual disclaimers. These are just my answers worked through quickly this morning.
They may contain errors, typos and omissions.
Didn't get chance to see my guys when they came out but the word is a bit mixed. Some liked it; others didn't.
Seems very mathematical to me  not much heat. hardly any 'explain' questions.
Heavily loaded to mechanics.
Q1 a) i) Gradient = acceleration of freefall so same for both (1)
ii) Area under graph = distance travelled.
Ball loses energy in bounce so speed after is less than speed before
so doesn't go so high (2)
I'm not sure whether you can just use the graph for i)ii) and iii) by reading off v and calculating the area. I suspect not.
b) i) suvat v^2=u^2+2as so v = () 5.8 ms1 (1)
ii) Impulse = FT = 1.2Ns = change in momentum
so new momentum = 0.13 x 5.8  1.2 = 0.754Ns (up) so v = 3.4 ms1 (2) (Doesn't match graph)
iii) suvat so V^=U^2 +2as again so s=2.00m (1)
Total: 7
Q2 a) Omnia corpore ... OK  in English Every object continues in a state of rest or uniform motion in a straight line unless acted upon by a resultant force. (1)
b) i) same magnitude; same type (2)
ii) opposite directions; act on different objects (2)
c) Nice diagram ...
i) Mass per s = vol per s x density = 3.3E4 x 25 x 1.0E3 = 8.25 kg per s (1)
ii) R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
= 902.5+169 = 1071N (3)
Total: 9
Q3 a) i) CF (1)
ii) G (1)
iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)
b) i) Inverted parabola  max KE = 50 mJ (2)
ii) 1/2 x 0.45 x v^2 = 50E3 so v = 0.471ms1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E2) = 1.50Hz
so T = 0.67s (2)
Total: 8
Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg1 (1)
b) i) Period squared is prop to radius of orbit cubed (1)
ii) (7.7/30)^2 = (9.4E3/R)^3 so R^3 = 15.18 x 9.4E3^3
so R = 2.33E4km (2)
c) mv^2/r = GMm/r^2 so v^2 = GM/r if lower orbit
if r decreases v will increase. (1)
Total: 7
Q5 Been a while since we had a binary question.
a) i) F = GM1M2/(R1+R2)^2 (1)
ii) F1 = M1v1^2/R1 = M1 (2 pi R1/T)^2 / R1 = 4 pi^2 R1 / T^2 (1)
b) Centripetal force = grav force so same on each object
M1 x (4pi^2) x R1 / T^2 = M2 x (4pi^2) x R2 / T^2
so M1/M2 = R2/R1 (2)
c) M1/M2 = R2/R1 = 3 R1+R2 = 4.8E12
so divide 4.8E12 in ratio 1:3
R1 = 1.2E12 and R2=3.6E12 (3)
d) V1 = 2 pi R1 / T = 6.08E4 ms1 (2)
e) v1^2 = GM2/R1) so M2 = 6.65E31 kg so M1 = 3M2 = 2.0E32kg (3)
Kepler's law of planetary motion isn't going to apply here to a binary system.
Total: 12
Q6 On to something a lot easier!
a) As falls, grav PE > KE
when hits bottom, KE > heat (2)
b) 50 x mgh = mcdT
50 x 0.025 x 9.81 x 1.2 = 0.025 x c x 4.5 so c=131 J kg1 K1 (4)
c) assume  no air resistance so all PE >KE
assume  all KE  heating le3ad shot and no heat transferred to container (2)
d) If double mass, double input energy but heat shared among 2x mas so same change in temp (2)
Total: 10
Q7 a) Ideal gas so all internal energy is translational KE
KE is prop to Kelvin temp so internal energy is prop to Kelvin temp (2)
b) i) PV = nRT
1.0E5 V = (80/0.004) x 8,81 x (21+273) so V = 489 m^3 (3)
ii) n = PV/RT = 1.3E3 x 1.4E4 / (*.81 = (27340) = 8677
so need to lose 200008677 = 11300 moles (2)
Total: 7
So there we have it. Good for maths experts; not so good for the medicos.
Good Luck Col 
 Follow
 10
 20062016 15:49
For 2)b)i anyone else put "both forces occur at the same time" or something like that as one of the ways the paired forces are the same?

 Follow
 11
 20062016 15:50
(Original post by Kevinwong)
You're wrong about the planets the answer is 1x10^33
Posted from TSR Mobile 
 Follow
 12
 20062016 15:51
(Original post by teachercol)
OCR Physics A Newtonian World 20/6/16 Unofficial mark scheme
Usual disclaimers. These are just my answers worked through quickly this morning.
They may contain errors, typos and omissions.
Didn't get chance to see my guys when they came out but the word is a bit mixed. Some liked it; others didn't.
Seems very mathematical to me  not much heat. hardly any 'explain' questions.
Heavily loaded to mechanics.
Q1 a) i) Gradient = acceleration of freefall so same for both (1)
ii) Area under graph = distance travelled.
Ball loses energy in bounce so speed after is less than speed before
so doesn't go so high (2)
I'm not sure whether you can just use the graph for i)ii) and iii) by reading off v and calculating the area. I suspect not.
b) i) suvat v^2=u^2+2as so v = () 5.8 ms1 (1)
ii) Impulse = FT = 1.2Ns = change in momentum
so new momentum = 0.13 x 5.8  1.2 = 0.754Ns (up) so v = 3.4 ms1 (2) (Doesn't match graph)
iii) suvat so V^=U^2 +2as again so s=2.00m (1)
Total: 7
Q2 a) Omnia corpore ... OK  in English Every object continues in a state of rest or uniform motion in a straight line unless acted upon by a resultant force. (1)
b) i) same magnitude; same type (2)
ii) opposite directions; act on different objects (2)
c) Nice diagram ...
i) Mass per s = vol per s x density = 3.3E4 x 25 x 1.0E3 = 8.25 kg per s (1)
ii) R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
= 902.5+169 = 1071N (3)
Total: 9
Q3 a) i) CF (1)
ii) G (1)
iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)
b) i) Inverted parabola  max KE = 50 mJ (2)
ii) 1/2 x 0.45 x v^2 = 50E3 so v = 0.471ms1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E2) = 1.50Hz
so T = 0.67s (2)
Total: 8
Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg1 (1)
b) i) Period squared is prop to radius of orbit cubed (1)
ii) (7.7/30)^2 = (9.4E3/R)^3 so R^3 = 15.18 x 9.4E3^3
so R = 2.33E4km (2)
c) mv^2/r = GMm/r^2 so v^2 = GM/r if lower orbit
if r decreases v will increase. (1)
Total: 7
Q5 Been a while since we had a binary question.
a) i) F = GM1M2/(R1+R2)^2 (1)
ii) F1 = M1v1^2/R1 = M1 (2 pi R1/T)^2 / R1 = 4 pi^2 R1 / T^2 (1)
b) Centripetal force = grav force so same on each object
M1 x (4pi^2) x R1 / T^2 = M2 x (4pi^2) x R2 / T^2
so M1/M2 = R2/R1 (2)
c) M1/M2 = R2/R1 = 3 R1+R2 = 4.8E12
so divide 4.8E12 in ratio 1:3
R1 = 1.2E12 and R2=3.6E12 (3)
d) V1 = 2 pi R1 / T = 6.08E4 ms1 (2)
e) v1^2 = GM2/R1) so M2 = 6.65E31 kg so M1 = 3M2 = 2.0E32kg (3)
Kepler's law of planetary motion isn't going to apply here to a binary system.
Total: 12
Q6 On to something a lot easier!
a) As falls, grav PE > KE
when hits bottom, KE > heat (2)
b) 50 x mgh = mcdT
50 x 0.025 x 9.81 x 1.2 = 0.025 x c x 4.5 so c=131 J kg1 K1 (4)
c) assume  no air resistance so all PE >KE
assume  all KE  heating le3ad shot and no heat transferred to container (2)
d) If double mass, double input energy but heat shared among 2x mas so same change in temp (2)
Total: 10
Q7 a) Ideal gas so all internal energy is translational KE
KE is prop to Kelvin temp so internal energy is prop to Kelvin temp (2)
b) i) PV = nRT
1.0E5 V = (80/0.004) x 8,81 x (21+273) so V = 489 m^3 (3)
ii) n = PV/RT = 1.3E3 x 1.4E4 / (*.81 = (27340) = 8677
so need to lose 200008677 = 11300 moles (2)
Total: 7
So there we have it. Good for maths experts; not so good for the medicos.
Good Luck Col 
 Follow
 13
 20062016 15:51
Teachercol please revise your answer 1b iii). An object that has lost kinetic energy will NOT rebound at a height higher than its initial height above the ground before the bounce. It is 0.6m. Peace.

 Follow
 14
 20062016 15:51
I'm pretty sure the mass of S2 is 1.03*10^33 right?c

 Follow
 15
 20062016 15:52
(Original post by ComputerMaths97)
I don't know about 1)b)iii, because the height after the first bounce was 1.7m so it can't be 2.0m after the next one.
I got 0.6, anyone else? 
 Follow
 16
 20062016 15:52
(Original post by tombrunt45)
Well I got 1.74x10^33 what makes you so sure?
Posted from TSR Mobile 
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 17
 20062016 15:53

 Follow
 18
 20062016 15:57

 Follow
 19
 20062016 16:00

 Follow
 20
 20062016 16:00
(Original post by ComputerMaths97)
I don't know about 1)b)iii, because the height after the first bounce was 1.7m so it can't be 2.0m after the next one.
I got 0.6, anyone else?
I'll edit it.Post rating:1
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Updated: June 22, 2016
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