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# OCR Physics A newtonian World 20/6/16 Unofficial Mark Scheme

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1. 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.

Q1 a) i) Gradient = acceleration of free-fall 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 ms-1 (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 ms-1 (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 = 3.3E-4 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 = 50E-3 so v = 0.471ms-1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E-2) = 1.50Hz
so T = 0.67s (2)
Total: 8

Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E-11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg-1 (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 ms-1 (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 / ( 6.67E-11 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 kg-1 K-1 (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 = (273-40) = 8677
so need to lose 20000-8677 = 11300 moles (2)
Total: 7

So there we have it. Good for maths experts; not so good for the medicos.

Good Luck Col
2. Thanks!
3. 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?
4. 1biii) ??????
5. (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?
same
6. do they even teach binary system in the book.....?
7. (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.

Q1 a) i) Gradient = acceleration of free-fall 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 ms-1 (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 ms-1 (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 = 3.3E-4 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 = 50E-3 so v = 0.471ms-1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E-2) = 1.50Hz
so T = 0.67s (2)
Total: 8

Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E-11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg-1 (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 ms-1 (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 kg-1 K-1 (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 = (273-40) = 8677
so need to lose 20000-8677 = 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
8. (Original post by Kevinwong)

Posted from TSR Mobile
????????
9. (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.

Q1 a) i) Gradient = acceleration of free-fall 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 ms-1 (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 ms-1 (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 = 3.3E-4 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 = 50E-3 so v = 0.471ms-1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E-2) = 1.50Hz
so T = 0.67s (2)
Total: 8

Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E-11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg-1 (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 ms-1 (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 kg-1 K-1 (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 = (273-40) = 8677
so need to lose 20000-8677 = 11300 moles (2)
Total: 7

So there we have it. Good for maths experts; not so good for the medicos.

Good Luck Col
In what way does Kepler's third not apply? I have not seen that answer with anyone I've asked (Q.5) I can't say I agree with your answer here, but thanks for the quick first efforts!
10. 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?
11. (Original post by Kevinwong)

Posted from TSR Mobile
Well I got 1.74x10^33 what makes you so sure?
12. (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.

Q1 a) i) Gradient = acceleration of free-fall 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 ms-1 (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 ms-1 (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 = 3.3E-4 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 = 50E-3 so v = 0.471ms-1
iii) max v = 2 pi f A so F = 0.471 /( 2 x pi x 5.0E-2) = 1.50Hz
so T = 0.67s (2)
Total: 8

Q4 a) i) g = GM/R^2 so M = 3.7 x (3.4E6)^2 / 6.67E-11 = 6.41E23 kg (2)
ii) If double r, then g is 4x less so g= 0.925 Nkg-1 (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 ms-1 (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 kg-1 K-1 (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 = (273-40) = 8677
so need to lose 20000-8677 = 11300 moles (2)
Total: 7

So there we have it. Good for maths experts; not so good for the medicos.

Good Luck Col
Thanks a lot! Is M1 suppose to be 10^31 or 10^32?
13. 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.
14. I'm pretty sure the mass of S2 is 1.03*10^33 right?c
Attached Images

15. (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?
so did i !
16. (Original post by tombrunt45)
Well I got 1.74x10^33 what makes you so sure?
I'm certain I am

Posted from TSR Mobile
17. (Original post by Kevinwong)
I'm certain I am

Posted from TSR Mobile
I said 'What' makes you so sure not 'How certain are you'
18. (Original post by Kevinwong)
I'm certain I am

Posted from TSR Mobile
how do you calculate that value????????
19. (Original post by Kevinwong)
I'm certain I am

Posted from TSR Mobile
Saying you're certain means nothing because I was equally certain when I left the exam, now however I recognize I might be wrong. What is it because we have similar answers, so there must be a small thing I missed. (Or you missed)
20. (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?
Quite right - big typo - I got 0.59m.
I'll edit it.

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