OCR A Mechanics 13/1/2009 :answers

Here are my answers

OCR Physics A - Mechanics 13/1/2009

Q1 a) vector = physical quantitiy with magnitude and direction (1)
b) vectors: acceleration, displacment, weight (1)
c) i) A constant acceleleration from t=0 to t=4s
constant velocity from t=4s to t=10s (2)
ii) distance = area under graph or s=1/2(u+v)t
= 1/2 x 4 x (10+24)
= 68m (2)
iii) 1) same velocity when lines cross t = 1.2s (1)
2) acc of A = (v-u)/t = (24-10)/4 = 3.5 ms-2
so Sa= 10t +1/2x3.5xt^2
sb = 14t
sa=sb when overtakes
10t + 1/2 x 3.5 x t^2 = 14t
1/2 x 3.5 t = 14
t = 2.3 s (2) (a lot of work for 2 marks)
Total (9)

Q2 a) arrow is vertically down (1)
b) Gravitational potential energy is converted to kinetic energy (2)
c) There are no forces acting horizontally if we ignore air resistance (1)
d) vert motion ; s = ut + 1/2 at^2 with u=0
t^2 = 1.3/(1/2x9,81)
t = 0.51s (1)
e) vert v = u + at = 0 + 9.81 x 0.5 = 5.0ms-1
v = sqrt( vvert^2 + vhoriz^2)
= sqrt(5.0^2 + 7.0^2)
= 8.6ms-1 (3)
Total (8)

Q3 a) 1N = 1 kg x 1ms-2 or IN is the force needed to give a mass of 1kg an acceleration of 1ms-2 (1)
b) m increases due to relativistic effects (1)
c) i) Resultant force = 200 - 80 = 120N
a = F/m = 120/900 = 0.13ms-2 (2)
ii) Air resistance increases with speed so resultant force and acc are not constant.(1)
d) T - mg = ma
T = 72( 9,81+1.4)
= 807N (3)
Total (8)

Q4 a) Torque of a couple = one of forces x perpendicular distance between them (1)
b) Both are force x distance
c) i) Moment = Fd
= 6.0x 0.40 = 2.4Nm (1)

ii) The perpendicular distance becomes zero at bottom of swing (1)
d) Let plate hang vertically. Attach plumbline to P. Draw a vertical line through P
Repeat from a different point. G is where lines cross ( or use balance method) (3)
e) F x 3.5E-2 = 18 x14E-2 + 60 x 32E-2 ( can work in N cm and leave out conversion factor E-2)
F = 621N (3)
Total (10)

Q5 a) Vh = 20 cos 38 = 15.8N (1)
Vv = 20 sin 38 = 12.3N (1)
b) i) Resolve vertically W = 2 x 20 sin 38 = 24.6N (2)
ii) Mass = weight / g = 24.6 / 9,81 = 2.51kg
Density = mass/volume = 2.51/2.9E-4 = 8650 kgm-3 ( looks like steel) (3)
Total (7)

Q6 a) Stopping distance = distance trvelled from when see obstacle to when carcomes to rest (1)
b) Velocity. The faster you go, the bigger the braking dist (prop to v^2)
Friction between tyres and road -> more friction, shorter braking dist
How hard you press the brake pedal -> more force, shorter braking dist
(any two) (4)
c) GPS: Satellites send out coded timing radio pulses
Receiver calculates time difference.
Receiver calculates dist from each satellite by d = ct
Receiver has ephemera of satellite positions at each time
Receiver uses trilateration / triangulation to work out position on surface of earth
Receiver gives result as latitude and longitude. (4)
Total (9)

Q7 a) Elastic or strain energy (1)
b) i) strain = ext / original length = 0.35E-3 / 1.2 (convert mm to m) = 2.9E-4 (1)
ii) stress = strain x E = 2.9E-4 x 1.9E11 = 5.51E7 Pa
F = stress x cross sectarea = 5.51E7x1.4E-7 = 7.71N (2)
c) i) 1) 1nm = 1E-9m (1)
2) Plastic - does not return to original size and shape when deforming forces are removed (1)
ii) ratio = 60 / 1.2 = 50x
iii) stronger / lighter / stiffer (2)
Total 9

Col

Thanks!

Thank you very much - I got some wrong answers but mostly I got all right equivalent to yours :smile:

Reply 3

LearningMath

14

**** I wish my teacher told us the answers / posted them on studentroom. :p:

Reply 4

Arbitrage

(F x 3.5E-2 = 18 x14E-2 + 60 x 32E-2)

I got (F x 3.2E-1 = 18 x14E-2 + 60 x 32E-2)
and got a final answer of (21.72/.32)

How many marks do you think I will lose from all three (Obviously the Answer, but how many method)

Reply 5

username250504

crap i got 25/60 is that an E?

Reply 6

JackPearson

9

From that it looks as though I did pretty well.

I didn't use the same phrases on the trilateration question though and got the 8.6 a different way. Also got 3 ii) wrong.

And also, I put mass of the car and road conditions to affect the braking distance, will I not get marks for that?

Reply 7

37652

yeah i wrote wet roads and small tyre treads both with reduced friction resulting in greater braking distance. doesnt that count as 2 factors?

Reply 8

Spongebob*No*Pants

13

Is there any chance you could upload the paper col?

Reply 9

Robbie10538

9

teachercol

Here are my answers

OCR Physics A - Mechanics 13/1/2009

Q1 a) vector = physical quantitiy with magnitude and direction (1)
b) vectors: acceleration, displacment, weight (1)

I didnt put weight?! :mad:

c) i) A constant acceleleration from t=0 to t=4s
constant velocity from t=4s to t=10s (2)
ii) distance = area under graph or s=1/2(u+v)t
= 1/2 x 4 x (10+24)
= 68m (2)
iii) 1) same velocity when lines cross t = 1.2s (1)
2) acc of A = (v-u)/t = (24-10)/4 = 3.5 ms-2
so Sa= 10t +1/2x3.5xt^2
sb = 14t
sa=sb when overtakes
10t + 1/2 x 3.5 x t^2 = 14t
1/2 x 3.5 t = 14
t = 2.3 s (2) (a lot of work for 2 marks)

I put down 2.29 with a lot of irrelivant working. Will I get 2 marks for 2.29?

Total (9)

Q2 a) arrow is vertically down (1)
b) Gravitational potential energy is converted to kinetic energy (2)
c) There are no forces acting horizontally if we ignore air resistance (1)
d) vert motion ; s = ut + 1/2 at^2 with u=0
t^2 = 1.3/(1/2x9,81)
t = 0.51s (1)
e) vert v = u + at = 0 + 9.81 x 0.5 = 5.0ms-1
v = sqrt( vvert^2 + vhoriz^2)
= sqrt(5.0^2 + 7.0^2)
= 8.6ms-1 (3)
Total (8)

Q3 a) 1N = 1 kg x 1ms-2 or IN is the force needed to give a mass of 1kg an acceleration of 1ms-2 (1)
b) m increases due to relativistic effects (1)
c) i) Resultant force = 200 - 80 = 120N
a = F/m = 120/900 = 0.13ms-2 (2)
ii) Air resistance increases with speed so resultant force and acc are not constant.(1)
d) T - mg = ma
T = 72( 9,81+1.4)
= 807N (3)
Total (8)

Q4 a) Torque of a couple = one of forces x perpendicular distance between them (1)
b) Both are force x distance
c) i) Moment = Fd
= 6.0x 0.40 = 2.4Nm (1)

ii) The perpendicular distance becomes zero at bottom of swing (1)
d) Let plate hang vertically. Attach plumbline to P. Draw a vertical line through P
Repeat from a different point. G is where lines cross ( or use balance method) (3)
e) F x 3.5E-2 = 18 x14E-2 + 60 x 32E-2 ( can work in N cm and leave out conversion factor E-2)
F = 621N (3)
Total (10)

Q5 a) Vh = 20 cos 38 = 15.8N (1)
Vv = 20 sin 38 = 12.3N (1)
b) i) Resolve vertically W = 2 x 20 sin 38 = 24.6N (2)
ii) Mass = weight / g = 24.6 / 9,81 = 2.51kg
Density = mass/volume = 2.51/2.9E-4 = 8650 kgm-3 ( looks like steel) (3)
Total (7)

Q6 a) Stopping distance = distance trvelled from when see obstacle to when carcomes to rest (1)

Is thinking distance + braking distance ok?

b) Velocity. The faster you go, the bigger the braking dist (prop to v^2)
Friction between tyres and road -> more friction, shorter braking dist
How hard you press the brake pedal -> more force, shorter braking dist
(any two) (4)

I put: whether effects it. If the road is icy/wet there will be less friction so the braking distance will be more. The condition of the brakes effects it. The better quality the brakes are, the less the braking distance will be because there is more friction.

Is that ok?

c) GPS: Satellites send out coded timing radio pulses
Receiver calculates time difference.
Receiver calculates dist from each satellite by d = ct
Receiver has ephemera of satellite positions at each time
Receiver uses trilateration / triangulation to work out position on surface of earth
Receiver gives result as latitude and longitude. (4)
Total (9)

Q7 a) Elastic or strain energy (1)

Elastic Potential energy; is that the same thing??

b) i) strain = ext / original length = 0.35E-3 / 1.2 (convert mm to m) = 2.9E-4 (1)
ii) stress = strain x E = 2.9E-4 x 1.9E11 = 5.51E7 Pa
F = stress x cross sectarea = 5.51E7x1.4E-7 = 7.71N (2)
c) i) 1) 1nm = 1E-9m (1)
2) Plastic - does not return to original size and shape when deforming forces are removed (1)
ii) ratio = 60 / 1.2 = 50x
iii) stronger / lighter / stiffer (2)
Total 9

I got the second part of the first question wrong (1 mark). APart from the there are 13 marks worth where I dont know whether my answers will count or I just cant remember my answers.

Reply 10

DrVas

teachercol

Here are my answers

OCR Physics A - Mechanics 13/1/2009
Col

Thanks for F&M

Reply 11

Arbitrage

teachercol

Here are my answers

OCR Physics A - Mechanics 13/1/2009

Q1 a) vector = physical quantitiy with magnitude and direction (1)
b) vectors: acceleration, displacment, weight (1)
c) i) A constant acceleleration from t=0 to t=4s
constant velocity from t=4s to t=10s (2)
ii) distance = area under graph or s=1/2(u+v)t
= 1/2 x 4 x (10+24)
= 68m (2)
iii) 1) same velocity when lines cross t = 1.2s (1)
2) acc of A = (v-u)/t = (24-10)/4 = 3.5 ms-2
so Sa= 10t +1/2x3.5xt^2
sb = 14t
sa=sb when overtakes
10t + 1/2 x 3.5 x t^2 = 14t
1/2 x 3.5 t = 14
t = 2.3 s (2) (a lot of work for 2 marks)
Total (9)

Q2 a) arrow is vertically down (1)
b) Gravitational potential energy is converted to kinetic energy (2)
c) There are no forces acting horizontally if we ignore air resistance (1)
d) vert motion ; s = ut + 1/2 at^2 with u=0
t^2 = 1.3/(1/2x9,81)
t = 0.51s (1)
e) vert v = u + at = 0 + 9.81 x 0.5 = 5.0ms-1
v = sqrt( vvert^2 + vhoriz^2)
= sqrt(5.0^2 + 7.0^2)
= 8.6ms-1 (3)
Total (8)

Q3 a) 1N = 1 kg x 1ms-2 or IN is the force needed to give a mass of 1kg an acceleration of 1ms-2 (1)
b) m increases due to relativistic effects (1)
c) i) Resultant force = 200 - 80 = 120N
a = F/m = 120/900 = 0.13ms-2 (2)
ii) Air resistance increases with speed so resultant force and acc are not constant.(1)
d) T - mg = ma
T = 72( 9,81+1.4)
= 807N (3)
Total (8)

Q4 a) Torque of a couple = one of forces x perpendicular distance between them (1)
b) Both are force x distance
c) i) Moment = Fd
= 6.0x 0.40 = 2.4Nm (1)

ii) The perpendicular distance becomes zero at bottom of swing (1)
d) Let plate hang vertically. Attach plumbline to P. Draw a vertical line through P
Repeat from a different point. G is where lines cross ( or use balance method) (3)
e) F x 3.5E-2 = 18 x14E-2 + 60 x 32E-2 ( can work in N cm and leave out conversion factor E-2)
F = 621N (3)
Total (10)

Q5 a) Vh = 20 cos 38 = 15.8N (1)
Vv = 20 sin 38 = 12.3N (1)
b) i) Resolve vertically W = 2 x 20 sin 38 = 24.6N (2)
ii) Mass = weight / g = 24.6 / 9,81 = 2.51kg
Density = mass/volume = 2.51/2.9E-4 = 8650 kgm-3 ( looks like steel) (3)
Total (7)

Q6 a) Stopping distance = distance trvelled from when see obstacle to when carcomes to rest (1)
b) Velocity. The faster you go, the bigger the braking dist (prop to v^2)
Friction between tyres and road -> more friction, shorter braking dist
How hard you press the brake pedal -> more force, shorter braking dist
(any two) (4)
c) GPS: Satellites send out coded timing radio pulses
Receiver calculates time difference.
Receiver calculates dist from each satellite by d = ct
Receiver has ephemera of satellite positions at each time
Receiver uses trilateration / triangulation to work out position on surface of earth
Receiver gives result as latitude and longitude. (4)
Total (9)

Q7 a) Elastic or strain energy (1)
b) i) strain = ext / original length = 0.35E-3 / 1.2 (convert mm to m) = 2.9E-4 (1)
ii) stress = strain x E = 2.9E-4 x 1.9E11 = 5.51E7 Pa
F = stress x cross sectarea = 5.51E7x1.4E-7 = 7.71N (2)
c) i) 1) 1nm = 1E-9m (1)
2) Plastic - does not return to original size and shape when deforming forces are removed (1)
ii) ratio = 60 / 1.2 = 50x
iii) stronger / lighter / stiffer (2)
Total 9

Col

Thanks for the virtual markscheme man. My answers are alot like these, except I used rounded values for some, which the actual markscheme will allow.

Reply 12

teachercol

OP

9

2sf or 3sf answers are always OK. Mark schemes give 3sf normally to help markers.

Reply 13

teachercol

OP

9

car Mass and road conditions are Ok as long as you explain effect on braking distance.

Reply 14

DeanK2

teachercol

Here are my answers

OCR Physics A - Mechanics 13/1/2009

Q1 a) vector = physical quantitiy with magnitude and direction (1)
b) vectors: acceleration, displacment, weight (1)
c) i) A constant acceleleration from t=0 to t=4s
constant velocity from t=4s to t=10s (2)
ii) distance = area under graph or s=1/2(u+v)t
= 1/2 x 4 x (10+24)
= 68m (2)
iii) 1) same velocity when lines cross t = 1.2s (1)
2) acc of A = (v-u)/t = (24-10)/4 = 3.5 ms-2
so Sa= 10t +1/2x3.5xt^2
sb = 14t
sa=sb when overtakes
10t + 1/2 x 3.5 x t^2 = 14t
1/2 x 3.5 t = 14
t = 2.3 s (2) (a lot of work for 2 marks)
Total (9)

Q2 a) arrow is vertically down (1)
b) Gravitational potential energy is converted to kinetic energy (2)
c) There are no forces acting horizontally if we ignore air resistance (1)
d) vert motion ; s = ut + 1/2 at^2 with u=0
t^2 = 1.3/(1/2x9,81)
t = 0.51s (1)
e) vert v = u + at = 0 + 9.81 x 0.5 = 5.0ms-1
v = sqrt( vvert^2 + vhoriz^2)
= sqrt(5.0^2 + 7.0^2)
= 8.6ms-1 (3)
Total (8)

Q3 a) 1N = 1 kg x 1ms-2 or IN is the force needed to give a mass of 1kg an acceleration of 1ms-2 (1)
b) m increases due to relativistic effects (1)
c) i) Resultant force = 200 - 80 = 120N
a = F/m = 120/900 = 0.13ms-2 (2)
ii) Air resistance increases with speed so resultant force and acc are not constant.(1)
d) T - mg = ma
T = 72( 9,81+1.4)
= 807N (3)
Total (8)

Q4 a) Torque of a couple = one of forces x perpendicular distance between them (1)
b) Both are force x distance
c) i) Moment = Fd
= 6.0x 0.40 = 2.4Nm (1)

ii) The perpendicular distance becomes zero at bottom of swing (1)
d) Let plate hang vertically. Attach plumbline to P. Draw a vertical line through P
Repeat from a different point. G is where lines cross ( or use balance method) (3)
e) F x 3.5E-2 = 18 x14E-2 + 60 x 32E-2 ( can work in N cm and leave out conversion factor E-2)
F = 621N (3)
Total (10)

Q5 a) Vh = 20 cos 38 = 15.8N (1)
Vv = 20 sin 38 = 12.3N (1)
b) i) Resolve vertically W = 2 x 20 sin 38 = 24.6N (2)
ii) Mass = weight / g = 24.6 / 9,81 = 2.51kg
Density = mass/volume = 2.51/2.9E-4 = 8650 kgm-3 ( looks like steel) (3)
Total (7)

Q6 a) Stopping distance = distance trvelled from when see obstacle to when carcomes to rest (1)
b) Velocity. The faster you go, the bigger the braking dist (prop to v^2)
Friction between tyres and road -> more friction, shorter braking dist
How hard you press the brake pedal -> more force, shorter braking dist
(any two) (4)
c) GPS: Satellites send out coded timing radio pulses
Receiver calculates time difference.
Receiver calculates dist from each satellite by d = ct
Receiver has ephemera of satellite positions at each time
Receiver uses trilateration / triangulation to work out position on surface of earth
Receiver gives result as latitude and longitude. (4)
Total (9)

Q7 a) Elastic or strain energy (1)
b) i) strain = ext / original length = 0.35E-3 / 1.2 (convert mm to m) = 2.9E-4 (1)
ii) stress = strain x E = 2.9E-4 x 1.9E11 = 5.51E7 Pa
F = stress x cross sectarea = 5.51E7x1.4E-7 = 7.71N (2)
c) i) 1) 1nm = 1E-9m (1)
2) Plastic - does not return to original size and shape when deforming forces are removed (1)
ii) ratio = 60 / 1.2 = 50x
iii) stronger / lighter / stiffer (2)
Total 9

Col