y13 Circular Motion answers

I have left the answers for the question on Magnetic fileds as it will be relevant when you have studied unit 5

1.       (i)      (v = 2πr/t) t = 2π60/0.26 = 1450 s

Correct answer is 1449.96 hence allow 1.4 × 103 
Do not allow a bare 1.5 × 103

B1

  (ii)   correct substitution into F = mv2/r: eg F = (9.7 × 103 × 0.262)/60

C1

          F = 10.9 N

Allow 11 N

A1

[3]

 

  2.     (i)      THREE correct arrows at A, B and C all pointing towards
the centre (judged by eye)

Ignore starting point of arrow

B1

  (ii)   1.      Greatest reaction force is at C

This is a mandatory M mark. The second mark cannot be gained unless this is scored.

M1

         because it supports weight of sock AND provides the required
upward resultant (centripetal) force (WTTE)

Any indication that candidates think that the centripetal force is a third force loses this second and possibly the next mark. They must make correct reference to the resultant force that provides the required centripetal force/acceleration.

A1

2.      Least at A because sock’s weight provides part of the required
downward resultant (centripetal) force (WTTE)

Allow answers using the equation F = mv2/r
such as Nc – mg (at C) = centripetal force OR mv2/r
OR mg +NA (at A) = centripetal force OR mv2/r

B1

[4]

 

  3.     (i)      At top of loop, the centripetal force = mv2 / r (1)
= mg (1)
Thus speed at top v = √gr
= √9.81 × 9.17 / 2 (1)
= 6.7 m s–1                                                                                                       3

          or use an energy argument KE on entry = PE gained + KE at top

          (1 mark for idea, 1 mark for correct substitution in appropriate formula
and 1 mark for correct calculation of 6.7 m s–1)

          (NOTE The unexplained use of v2 = u2 + 2gs can only score a
maximum of 1 mark)

(ii)     Kinetic Energy at top = ½ × 86 × 6.72 (1)
= 1935 J (1)
Potential Energy at top = 86 × 9.81 × 9.17 (1)
= 7740 J (1)                                                                                                      4

  (iii)  Kinetic Energy on entry = ½ × 86 × 152
= 9675 J (1)
Sum of energies at top = 1935 + 7740
= 9675 J .......... QED (1)                                                                                  2

  (iv)  Any reference to loss of contact / centripetal force or wtte (1)
Comment on the consequences of taking off vertically or wtte (1)                  2

          Enacting the suggestion could result in disaster
At the point A in the loop, the velocity vector is purely vertical
Therefore there is no horizontal component of velocity
So no matter how fast the cyclist is travelling he will only be
projected vertically
And come (crashing?) down on the same point where he left off
The best that could happen ( with some skill ) is to return back along same path

[11]

 

  4.     (a)     B = F/Il with symbols explained or appropriate statement in words; (1)
explicit reference to I and B at right angles/define from F = BQv etc (1)          2

  (b)   (i)      arrow towards centre of circle                                                                 1

(ii)     field out of paper; Fleming’s L.H. rule/moving protons act as
conventional current                                                                               2

  (iii)  F = Bev allow BQv                                                                                  1

  (iv)  F = mv2/r; Bev = mv2/r; (2)
B = mv/er = 1.67 × 10–27 × 1.5 × 107/(1.6 × 10–19 × 60); = 0.0026; T (3) 5

allow Wb m–2

  (v)   the field must be doubled; (1)
B ∞ v (as m, e and r are fixed)/an increased force is required
to maintain the same radius (1)                                                               2

[13]

 

  5.     (a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                           2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown             2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

         Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                              5

  (b)   (i)      k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)                                                                               2

  (ii)   f = (1 /2π (k/m)0.5 (0)
= (1 /2π (2.0 × 106 / 9.5 × 105)0.5 (1)
= 0.23 Hz (1)                                                                                           2

  (iii)  If wind energy causes this frequency in the structure, the
amplitude increases / resonance occurs / or explanation of
resonance / ref. to natural frequency (1)
e.g. damping is necessary / mass change to shift resonant
frequency / change spring constant (1)                                                    2

[18]

 

  6.     the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                        3

MAXIMUM 3

[3]

 

 

Power Homework

Power

(Use    g = 9.8 Nkg^-1)

 

Examples

 

A         Find the work done in lifting a mass of 6 kg through a height of 10 m.

            W = Fx            W        work done in J

 F         force in N

x          distance moved in the direction of the force in m

W = 6 x 9.8 x 10

    = 588 J

B) An escalator carries 100 people of average mass 70 kg to a height of 6 m in one minute.

Find the power necessary to do this.

Work done = F x = 100 x (70 x 9.8) x 6 J

Power = rate of doing work = work done per second

           = (100 x 70 x 9.8 x 6) / 60 s = 6860 W or 6.86 kW

 

1. Find the work done to lift a mass of 10 kg through a height of 5 m. If this takes 15 s what is the power?

 

2. A mass of 5 kg is pulled along a surface at a constant speed. Frictional force between the mass and the surface is 4N what work is done in moving the mass 2 m along the surface? Find the power if this takes 2 s.

 

3. A girl of mass 50 kg runs up a flight of steps 4.5 m high in 5 s. What power does she develop?

 

4. 2000000 kg of water per second flow over a large waterfall of height 50 m. What power is available?

 

5. The first-stage rocket motors of the Saturn 5 moon rocket burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers was the same as that necessary to lift the fuel through 1680 m. What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

 

6. A gas turbine locomotive developing 6330 kW pulls a train at a steady velocity of 108 kmh^{- 1}. What is the total force in Newtons resisting the motion?

 

7. A mechanical loader driven by an engine developing 3 kW lifts a total load of 200000 kg to a height of 3 m in one hour. What is the work done on the load? How much power is developed lifting the load? How does this compare with the work done by the motor in the same time?

 

 

KE and GPE

Gravitational Potential Energy & Kinetic Energy

acceleration due to gravity = 9.8 m/s2

1. A car has a mass of 750 kg. Calculate its kinetic energy at the following velocities
(a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 30 m/s (e) 35 m/s {NB 1 m/s =2 m p h}

2. (a) A car has a mass of 1000 kg. Calculate its KE at a velocity of 25 m/s.
(b) A train has a mass of 37 tonnes {1 tonne = 1000 kg}. If it also has a velocity of 25 m/s what is its kinetic energy ?

3. A woman has a mass of 65 kg.What is her GPE at the top of a 12 metre diving board ?

4. A cat has a mass of 6 kg. What is its GPE at the top of a tree that is 4.2 m above the ground ?

5. The Eiffel Tower is 300 m high. What is the GPE of a 100 g bird perched on the top of it ?

6. The Great Pyramid of Khufu (Cheops to the Greeks) is 146 m high. It is made of stones of mass 250 t. What is the GPE of the topmost stone ?

7. The Empire State Building is 449 m high(but this includes a 68 m TV mast).What is the GPE of a ball of mass 400 g at the top of the building (not the TV mast) ? If the ball is dropped over the side what will happen to the GPE ? What will be its speed at the instant before it hits the ground ? (Ignoring air resistance)

8. Olympus Mons is a volcano on Mars. It is 25 km high. What would be the GPE of a human of mass 50 kg on its summit ? (surface gravity = 0.38 that of Earth) Compare this with the same human standing on the summit of Everest 8534 m high.

9. The space shuttle has a mass of 80 t. Escape velocity (the velocity an object has to travel at in order to escape the earth's gravitational field) is 11 km/s. What is its KE at this speed? If all this KE is turned into GPE how high would it be ? What assumption do you have to make in your calculation.

How much energy would it take to accelerate the space shuttle to light speed (300 000 km/s). {This is not possible, also the equation you will use to calculate KE is not valid near the speed of light but it will do for this purpose} A nuclear power station produces around 500 MW
(500 000 000 J per second). How long would 100 power stations take to produce this amount of energy?
How feasible is travelling close to the speed of light?
The nearest star is about 4 light years away. How long would it take the space shuttle to get there at 11 km/s (22 000 mph)? {1 light year is the distance light will travel in 1 year}
How feasible is interstellar travel?

Markschemes for questions set today

Some notes on electro mag can be found here

http://twyfordphysics.blogspot.co.uk/2011/02/y13-magenetic-fields-due-to-current.html

http://twyfordphysics.blogspot.co.uk/2010/04/y13_26.html

Markscheme for questions on Cosmology

1.       any 4 from:
end of H burning/red giant/supergiant (1)
onset of He fusion/fusion of heavier nuclei (1)
gravitational collapse of core (1)
supernova explosion/ star explodes (1)
suitable mass limit (chanderasekha limit 1.4M) (1)
supported against gavity by neutron gas pressure/ ref to
Fermi pressure (1)
internal structure protons and electrons combined/ very
thin atmosphere/ metallic crust (1)                                                                             4

[4]

  2.     Any 5 from
red shift data for galaxies (accept stars)                                                                     1
calculate velocity from red shift                                                                                1
galaxies/ stars receding from Earth                                                                             1
distance data for galaxies/ stars                                                                                  1
velocity α distance / v/r = constant / v-r graph straight line                                         1
universe began at a single point                                                                                 1

[5]

  3.     (a)     Any two
stars rotate around galactic centre                                                                     1
star with velocity component towards Earth                                                     1
reference to motion/shape of galaxy                                                                1
or other valid points eg blue shift

  (b)   Ho = 75/ 3 × 1019 s–1                                                                                        1
t ≈ 1/ 2.5 × 10–18                                                                                              1
t ≈ 4 × 1017 s                                                                                                    1

[5]

  4.     Hydrogen atoms/particles (1)
Collapse under gravity/ decrease of gpe (1)
Increase in kinetic energy/ temperature (1)
Fusion of protons (1)
Energy released/ ref. to E = ∆mc2 (1)

[5]

5.       Any 6 from
Nuclear/hydrogen burning ends (1)

          Mass > Chandrasekhar limit (1)

          Expanding gas/planetary nebular/red giant (1)

          Gravitational collapse /ref. to burning He or higher metals (1)

          Correct ref. to (Fermi) pressure/ radiation pressure (1)
(must have ref. to pressure or force from radiation.)

          Neutron star (neutron by itself, not enough) (1)

          Correct reference to Schwarzschild radius/
allow mass> 3M/ allow ref. critical radius (1)

          Black Hole (1)                                                                                                           6

[6]

  6.     Ho2 = (1 × 10–26 × 8 × π × 6.67 × 10–11) / 3                                                             C1
Ho = 2.36 × 10–18 s–1                                                                                              A1

[2]

  7.     (i)      v/c = ∆λ / λ (1)
∆λ = 656.3 × 10–9 × 6.1 / 3 × 108 (ignore minus sign) (1)
∆λ = 1.33 × 10–14 m (1)                                                                                    3

  (ii)   Graph: any 4 points plotted correctly (1)
all correct (1)                                                                                                    2

  (iii)  graph: draw curve, reasonable attempt (1)                                                        1

(iv)    Either point where star moves perpendicular to line of sight (1)                       1

  (v)   time = 72 h ± (1)h (ecf read value from their graph ± 1 h) (1)                          1

  (vi)  r = 3√(6.7 × 10–11 × 4 × 1030 × [72 × 3600]2/ 4π2) ecf (1)
r = 7.70 × 109 m ecf . (1)                                                                                  2
(use of t = 72h 1/2)

[10]

Doppler Shift

8

a Which transition

Use of (

Δ)E = hc/λ OR (Δ)E = hf and f = c/λ 1

Use of 1.6 x 10

-19 1

Correct answer [1.9 eV]

1

C to B / -1.5 to – 3.4

1 4

[Accept reverse calculations to find wavelengths]

e.g.

(6.63 x 10

-34 J s)(3.00 x 108 m s-1)/

(656 x 10

-9 m)(1.6 x 10-19 J eV-1)

= 1.9 eV

b Explanation of absorption line

QOWC

1

Light of this wavelength is absorbed by hydrogen

1

In the outer part of the Sun (OR Sun’s atmosphere)

1

Absorbed radiation is reemitted in all directions

1

Transition from B to C (OR -3.4 to -1.5)

1 Max 4

c i Why galaxy receding

Wavelength increased (OR stretched) / red shift /

frequency decreased

1

c ii Calculation of speed

Use of

v/c = Δλ/λ 1

Correct answer [1.78 x 10

7 m s-1] 1 3

e.g.

(3.00 x 10

8 m s-1)(695 x 10-9 m – 656 x 10-9 m)/

(656 x 10

-9 m)

= 1.78 x 10

7 m s-1

[N.B. Use of wrong wavelength in denominator gives

1.68 x 10

7]

OR

Use of

v = fλ and v/c = Δf/f 1

Correct answer [1.67 x 10

7 m s-1] 1

[Look for

(4.573 x 10

14 Hz – 4.317 x 1014 Hz) / (4.537 x 1014 Hz)

N.B. Use of wrong frequency in denominator gives 1.78

x 10

7]

Total 11

Questions on resistivity

Resistivity

1. A coil of constantan wire of length 2 m has an area of cross-section of

7 x 10^-7 m². If its resistivity is 49 x 10^-8 Ω m, find its resistance.

2. A coil of wire 4 m long has a resistance of 20 Ω. If its resistivity is

108 x 10^-8 Ω m, find its area of cross-section.

3. A 1.75 m length of tin wire with a diameter of 0.70 mm is found to have a resistance of 0.5 Ω Find the resistivity of tin.

4. A resistor of 2 Ω is to be made from manganin wire of resistivity

42 x 10^-8 Ω m having a diameter of 0.42 mm. What length is required?

5. Find the resistance of a wire 2.8 m long with a resistivity of 22 x 10^-8 Ω m and a diameter of 0.56 mm.
6. A wire has a resistance of 2 Ω Find the resistance of a wire of the same material which is three times as long, but with half the diameter.

 

Mark Scheme February assessment

1.       The sum of the currents entering a point / junction is equal to the sum of the currents

leaving (the same point) Or ‘Algebraic sum of currents at a point = 0’                    B2

(–1 for the omission of ‘sum’ and -1 for omission of ‘point’/ ‘junction’)

(Do not allow I1 + I2 = I3 + I4 unless fully explained)

[2]

  2.     A:   gamma / γ (ray/radiation/wave)                                                                        B1
       λ = 10–16 to 10–10 (m)                                                                                       B1
B:    infrared / IR / i.r.                                                                                              B1
       λ = 7 × 10–7 to 10–3 (m)                                                                                    B1

[4]

  3.     (i)      v   = I / nAe = 0.0025 / (8.5 × 1028 × 1.1 × 10–7 × 1.6 × 10–19) (1)
     = 1.67 × 10–6 m s–1 (1)                                                                                2

  (ii)   Free electron concentration (or wtte) is much smaller in the thermistor than
in the wire.                                                                                                       1

[3]

  4.     (a)     R = R1 + R2 / R = 200 + 120 / R = 320                                                            C1
current =                                                                                                 C1
current = 2.5 ´ 10–2 (A)                                                                                 A0

  (b)   V = 25 ´ 10–3 ´ 120 / V =
V = 3.0 (V)                (Possible ecf)                                                                B1

  (c)   p.d. across the 360 (W) resistor = p.d. across the 120 (W) resistor /
There is no current between A and B / in the voltmeter                                 B1
(Allow ‘A & B have same voltage’ - BOD)

          The p.d. calculated across 360 W resistor is shown to be 3.0 V /
The ratio of the resistances of the resistors is shown to be the same.             B1

[5]

  5.     (i)      plane polarised light vibrates (travels) in one plane only                                B1
(look for reference to one plane of oscillation)

(ii)     only transverse waves can be polarised/AW                                                  B1
sound waves are longitudinal/not transverse/AW                                           B1

[3]

6.       (a)     (i)      amplitude = 3.75 cm {allow 3.7 to 3.8)                                                B1

(ii)     when t = 1.8 ms displacement = ANY negative value                           B1

(–)3.35 cm (ALLOW 3.3 TO 3.4)                                                         B1

(iii)    period = 2.64 ms (allow 2.64 to 2.68)                                                   B1

(iv)    frequency = 1/period                                                                            C1

= 1/(2.64 × 10 – 3) = 379 Hz (379 to 373 or 380) {ecf for T}              A1

  (b)   recall of v = fl                                                                                               C1

l = v/f = 300/379 = 0.79 m (or 0.8 m) {allow ecf from (iv)}                         A1

[8]

  7.     Any three properties from:                  (–1 for each error or contradiction)        B1 ´ 3

1.      Travel at the speed of light / c / 3 ´ 108 m s–1 (NOT ‘same speed’)

2.      Travel through vacuum / ‘free space’

3.      Have oscillating electric and magnetic fields

4.      They are (all) transverse waves / can be polarised

5.      Allow: ‘They show diffraction / reflection / refraction / interference’

6.      Allow: ‘Consist of photons’

Any three regions from the list below:                                                               B1 ´ 3

Gamma (rays / radiation) / g (rays) ; X-rays ; u.v ; ir ; microwaves ; radio waves

(NOT ‘radio’)

One suitable application for the opted region.                                                         B1

(E.g.: Gamma rays for radiotherapy / sterilisation;

X-rays for taking pictures of skeleton / bones; u.v for tanning; ir for TV remote
control; microwaves for cooking / mobile phones; radio waves for communication)

(Note: Reference to alpha, beta and gamma can only score the last marking point)

[7]

  8.     (i)      evidence of knowledge of:
full/max transmission when the (transmission axis of) polarising sheet is
parallel to the light’s plane of polarisation/vibrations                                     B1
no transmission when the (transmission axis of) polarising sheet is at right
angles to light’s plane of polarisation/vibrations                                            B1

(ii)     reflected light from surface is partially plane polarised                                  B1
polarising sheet is placed at right angles to reflected light’s polarisation        B1
plane/AW

[4]

  9.       (b)                                                    (Allow other subject, with or without D)     C1
(charge =) 7800 ´ 0.23                                                                                            C1
1.794 ´ 103 » 1.8 ´ 103 (C)                                                  (Ignore minus sign)     A1
(1.8 ´ 106 (C) scores 2/3)

(c)     (number =)           (Possible ecf)                                                    C1
(number =) 1.12 ´ 1022 » 1.1 ´ 1022                                                              A1

[6]

Answers to questions on probing the nucelus