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

OCR Physics A Newtonian World 20/6/16

iii) 5pi/4 = 3.93 rad (might be SF penalties here if left in pi) (1)

Good Luck Col

you need to start your derivation from F1 equation, that's what the questions asks.

Looks like i got 45... Pushing for an A*? lol

Looking like sky high grade boundaries.

Quite right - big typo - I got 0.59m.
I'll edit it.

right could someone literally anyone respond please, i wrote as an assumption that there is no time between the inversions of the tube in the lead experiment? reckon that's okay?

How were you doing in Past Papers?

I was getting A's/ Low A*s and I got 45, I expect it to be very close to the A* boundary regardless.

The A* boundary will not be under 42 but it won't be like 49.

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 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=0.69m (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

Way to scare everyone

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 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=0.69m (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

Yeah probs going to delete the post as it's not well timed... Personally I found it to be more difficult than previous papers but I'm just judging by the reaction on twitter tbh.

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!

If this is right then I actually did better than I thought I did. Hopefully I can still get an A.

Did anyone else think that was harder than the June 2014 paper?

it cant apply to a binary system as one star doesnt orbit another they both orbit a third object, not each other

Thank you. As with most students I have a few questions to ask so if you have the time I'd greatly appreciate your opinion.

2) Would I lose 1 or 2 marks by saying R = mg - Fsin55 having got the right value for Fsin55.

3) If I left it as 5/4pi will this lose the mark?

6) For the uncertainties in question 6, will by stating something on the lines of "not all the pebbles behave the same way ie not all pebbles hit both sides of the tube at the same time, therefore are not all subjected to the same energy transfers" gain any credit for the second mark besides the obvious reasons?

Lastly, grade boundary predictions please, sir