OCR Physics A Mechanics 24/5/16 My answers Watch

teachercol
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OCR A G48 Mechanics 24/5/6

This felt pretty straightforward - especially after that breadth paper.
My A2 student resitting it were pretty happy - although they made a few mistakes.

Usual disclaimers - these are just my answers - done in 16 mins - they may contain errors and/or typos.

Here we go

Q1 a) acc = rate of change of velocity (with respect to time) (1)
b) it has magnitude and direction (1)
c) 1) acc is increasing (1)
2) constant (negative) acceleration (1)
ii) Area under graph is less (1)
d) suvat gives t = 0.154s (2)
Total : 7

Q2 a) Aristotle - heavier objects fall faster - applies when air resistance taken into account
Galileo _ all objects have same acceleration - applies in a vacuum. (2)
b) i) 2 from velocity / cross sectional area / viscosity (1)
ii) Initial acc = acceleration of free fall / g / 9.81 ms-2
because drag force =0 (v+0 ) so only force acting is weight so mg = ma (2)
iii) Exponential fall reaching zero at 10s (2)
iv) mg - D = ma so D = mg + ma = 545N (3)
Total : 10

Q3 a) Ep = mgh = 4.66E4J (1)
b) Ek = 1/2 mv^2 = 8.55E4J (1)
c) work is done by engine increasing KE (1)
d) Work done = Ek-Ep = 3.89E4J = Fd so F = 324N (2)
e) suvat vertically gives t = 1.39s (2)
horiz distance = 41.7m so 23 cars ( not allowing for height of car) (2)
Total: 9

Q4 a) Resultant force is zero because acc = 0 (constant velocity) (2)
b) moment = force x perpendicular distance from point to line of force (1)
c) Not couple because same direction / not same size (1)
d) Y x 1.3 = 180 x 0.60 + 720 x 0.40 so Y = 305N (3)
e) X decreases.
Distance from Y decreases so moment is less
so moment due to X must be less so X must decrease. (2)
Total : 9

Q5 a) i) Bounce back (and break neck) (1)
ii) Airbag deflates gradually so drivers head slows down to rest over longer time
so acc (= dv/dt) is less if dt in longer so force is less. (3)
b) x is prop to u squared (1)
thinking distance = 0.6 x 30 = 18m
braking distance is 72m (=8 x9)
so stopping distance = 90m (3)
c) Locus of possible positions based on distance from each satellite. (1)
receiver calculates time difference between its own clock and time signal received from satellite.
distance from satellite = time difference x speed of light (2)
Total : 11

Q6 a) k = force per unit extension (1)
same force extension halves
k = F /x so doubles. (1)
b) Measure thickness with micrometer in several places and average.
(Measure width)
Add masses one at a time until breaks
Area = width x average thickness
Breaking stress = breaking mass x g / area (3)
c) i) elastic / constant Young's modulus / obeys Hooke's law (1)
ii) permanently stretched / doesn't return to original length / past elastic limit (1)
iii) No - not straight line through origin at this point (1)
Total : 8

Q7 a) Total weight = 2.8E4 x 9.81 so for 1 cable T = 2.8E4 x 9.814 = 6.87E4N
stress = F/ A = 1.53E8 Pa
strain = stress / E = 7.27E-4
so ext = strain x L = 23.3 mm (4)
b) Increases - because T (=mg +ma) increases. (2)
Total : 6

Hopefully that adds up to 60.

Hard to say what the grade boundaries will be. The only people taking this will be A2 resits so you expect them to do better than average.
They OUGHT to put the grade boundaries in line with previous years and ignore the change of population.
We'll see.

Good Luck

Col
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Agent47_
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Thanks so much teachercol
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ChanE
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Thank you! I think I got most of these so things are looking good!
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EpicNameBro
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Nice thanks for this. Pretty sure I only got around 42 though =(, still a big improvement from last year though
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AlistairMcDonald
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Damn, 50/60. if i couldn't remember my answer i marked it as wrong though... Hopefully an A!

Thank you very much Teachercol!
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speed1✈️✈️
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Awesome thank you teachercol!!!! think I got round about 53-55, which is a dramatic improvement from my D grade last year

Only thing is for last question I put 23.4 mm instead of 23.3mm for some reason, would that cost me a mark?. Also for the air bag question I discussed how it increased time of impact and linked that to rate of change of momentum (F=p/t from G484), is that acceptable? And finally for the GPS question I said something along the lines of: the circle is the range over which the signal of microwaves travel from satellite to car and back, is that right?
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Daribig
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Q3 a) was lost in Ep, so isnt is (34.5-9.5)(9.81)(120) that gives you 29430 J. ?
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Daribig
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For 1(c)(2) i put non-uniform acceleration. This is the same as acceleration increasing right?
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RBoss
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That 5ai) question though. Am the only one who nearly started laughing at the thought of the driver bouncing to and from the airbag to the headrest? xD

Thanks though, hoping for 55 marks
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Daribig
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3) d) didnt u have to use suvats . V^2 = U^2 + 2as. Rearranged to 30^2 / (2)(120) = a = 3.75
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Daribig
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(Original post by Daribig)
3) d) didnt u have to use suvats . V^2 = U^2 + 2as. Rearranged to 30^2 / (2)(120) = a = 3.75
Assuming the driver was at rest so u=0
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DjP9
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(Original post by Daribig)
Assuming the driver was at rest so u=0
yh that's what I did
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Super199
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What was q2bi about?
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teachercol
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(Original post by Daribig)
3) d) didnt u have to use suvats . V^2 = U^2 + 2as. Rearranged to 30^2 / (2)(120) = a = 3.75
Not constant acc (curved ramp) so cant use suvat
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teachercol
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(Original post by Daribig)
Q3 a) was lost in Ep, so isnt is (34.5-9.5)(9.81)(120) that gives you 29430 J. ?
No its X to Y which is 25m
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teachercol
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(Original post by speed1✈️✈️)
Awesome thank you teachercol!!!! think I got round about 53-55, which is a dramatic improvement from my D grade last year

Only thing is for last question I put 23.4 mm instead of 23.3mm for some reason, would that cost me a mark?. Also for the air bag question I discussed how it increased time of impact and linked that to rate of change of momentum (F=p/t from G484), is that acceptable? And finally for the GPS question I said something along the lines of: the circle is the range over which the signal of microwaves travel from satellite to car and back, is that right?
Not sure if 23.4 acceptable.
Momentum argument is fine
GPS explanation sounds wrong to me.
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teachercol
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(Original post by najibahmed)
question 2bii . How did parachutist have an initial acceleration of 9.81ms-2 . Really?? I am soooo confused
No drag when v=0
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Daribig
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Wasnt X to Y 120 metres? Sorry if im wrong
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RBoss
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(Original post by Daribig)
Wasnt X to Y 120 metres? Sorry if im wrong
That was the horizontal length; 25 was vertical
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Daribig
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So if the vertical length from x to y is 25 m, then the gpe lost from x to y would be 25*9.81*mass of motorcyclist?
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