Capacitors

Springs Past Questions

1

1.       (a)     The extension of a spring is directly proportional to the applied force          M1
as long as the elastic limit is not exceeded)                                                    A1

  (b)   (i)      Correct pair of values read from the graph
force constant = 12/0.080                                                                     C1
force constant = 150 (N m–1)                                                               A1

(ii)     extension, x = 20/12 × 80 (= 133.33) (mm)                                              C1
(E = ½ Fx)
energy = ½ × 20 × 133.33 × 10–3
energy = 1.33 (J)                                                                                  A1

  (iii)  The spring has not exceeded its elastic limit                                         B1

(iv)    (elastic potential energy = kinetic energy)
1/2kx2 = 1/2mv2                                                                                     M1
m and k are constant, therefore x µ v.                                                  M1

2

(a)     (i)      F = kx / k is the gradient of the graph                                                             C1

k = 2.0 / 250 ´ 10–3 = 8.0                                                                     A1

Correct unit for value given in (a)(i)

i.e. 0.008 or 8 ´ 10–3 requires N mm–1.

Allow N m–1 / kg s–2 if no working in (a)(i).

Do not allow unit mark if incorrect physics in part (a)(i)                      B1

(ii)     W = ½ (F ´ extension) / area under the graph                                       C1

          = ½ ´ 2.0 ´ 0.250

          = 0.25 (J)                                                                                    A1

  (b)   (i)      F = 8 ´ 0.15 = 1.2 (N)                                                                           A1

(ii)     Hooke’s law continues to be obeyed / graph continues as a straight

line / k is constant / elastic limit has not been reached                          B1

  (c)    (i)      1.     correct time marked on the graph with a V (t = 0.75 s or 1.75 s)   B1

2.       tangent in the correct place for downward velocity or implied
by values                                                                                     B1

          value between 0.95 to 1.1(m s–1)                                                A1

(ii)     1.     X marked in a correct place (maximum or minimum on graph)   M1

2.       relates the extension / compression to F = kx to explain why the

          force is a maximum or maximum extension gives max force or
maximum extension gives max acceleration                               A1

3

(a)     The extension of a spring is directly proportional to the applied force                   M1
as long as the elastic limit is not exceeded)                                                    A1

(b)     (i)      Correct pair of values read from the graph
force constant = 12/0.080                                                                     C1
force constant = 150 (N m–1)                                                               A1

(ii)     extension, x = 20/12 × 80 (= 133.33) (mm)                                              C1
(E = ½ Fx)
energy = ½ × 2/ × 133.33 × 10–3
energy = 1.33 (J)                                                                                  A1

  (iii)  The spring has not exceeded its elastic limit                                         B1

(iv)    (elastic potential energy = kinetic energy)
1/2kx2 = 1/2mv2                                                                                      M1
m and k are constant, therefore x µ v.                                                  M1

 

Y12 Power Answers

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

W= Fx = (10kg)(9.8Nkg^-1) (5m) = 490J

 

P = W/t = 490 J / 15s = 32.67 W

 

 

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.

 

W = Fx = (4N) (2m) = 8J

 

P = W/t =  8J/ 2s = 4 W

 

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?

W = Fx             F = 50kg x 9.8 = 490.5 N                     x = 4.5m

W = 490.5 x 4.5 = 2207.25 J

P = W/t = 2207.25 J / 5s = 442w

 

4        2 x 10⁶ kg of water per second flow over a large waterfall of height 50 m.

W = Fx                   F = 2 x 10⁶ kg x 9.8 = 1.96 x10⁺⁷N      x=50m

W= (1.96 x10⁺⁷N) (50m) = 9.8 x10⁺⁸ J

What power is available?

As 1W = I J per second the water falling over the fall transfers 9.8 x10⁺⁸ J of GPE per second the available power is 9.8 x10⁺⁸ W (ignoring losses due to friction)

 

5        The first-stage rocket motors of Saturn 5 burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers is the same as that necessary to lift the fuel through 1680 m.

W= Fx                    F = 13600kg x 9.8= 1.3328 x 10⁺⁵ N    x = 1680m

W = (1.3328x 10⁺⁵ ) (1680) = 2.24 x 10⁺⁸ J

What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

As this mass of fuel is burnt in 1 second then 2.24 x 10⁺⁸ J of work is done in 1 second so the power is 2.24 x 10⁺⁸ W

 

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?

At a steady velocity all work done by the engine is against friction or gravity.

The engine does 6.33 x10⁶ J of work every second

As W = Fx the force exerted by the engine is equal Work done per second divided by the distance travelled per second

F = W/x            108 kmh^-1 = (108000m)/(60 min x 60 sec) = 30 ms^-1

F = 6.33 x10⁶ J/ 30m = 211 000N = 2.11 x 10⁵ N

Nov Assesment Mark Scheme

SHM test

1.       The acceleration of the oscillator is directly proportional to the                              B1
displacement (the term displacement to be included and spelled correctly
to gain the mark).
with the acceleration always directed to a fixed / equilibrium point                        B1

[2]

 

  2.     (a)     arrow towards centre of planet                                                                         1

(b)     (i)      g = GM/R2                                                                                               1

(ii)     gS/gO = R2/25R2; gS = 40/25 (= 1.6 N kg–1)                                             2

(iii)    gC/gO = R2/16R2 giving gS = 40/16 (= 2.5 N kg–1)                                    1

(iv)    average g = (2.5 + 1.6)/2 = 2.(05) (1)
Δp.e. (= mgavR) = 3.0 × 103 × 2.05 × 2.0 × 107; = 1.2 × 1011 (J) (2)       3

  (c)   g = v2/r; = 4π2(5R)/T2 (2)
1.6 = 4 × 9.87 × 1.0 × 108/T2 giving T2 = 24.7 × 108 and T = 5.0 × 104 (s) (2) 4

[12]

 

  3.     (a)     (i)      Fig. 1 : x and a in opposite directions/acceleration towards
equilibrium point/AW; (1)
Fig. 2 : proportional graph between x and a/AW (1)                                2
Figures not identified max. of 1 mark

  (ii)   a = 4π2f2x; 50 = 4π2f2.50 × 10–3; giving f2 = 25 and f = 5.0 Hz              3

  (iii)  cosine wave with initial amplitude 25 mm; decreasing amplitude; (2)
correct period of 0.2 s (for minimum of 2.5 periods); (1)                       3

  (b)   (i)      the acceleration towards A/centripetal acceleration or force;
is constant                                                                                               2

  (ii)   a = v2/r; so 50 = v2/10; v2 = 500 giving v = 22.4 m s–1                            3

[13]

 

  4.     (a)     (i)      v = 2πrf = 2π × 0.015 × 50; = 4.7 (m s–1)                                                2

  (ii)   a = v2/r = 4.72/0.015; = 1.5 × 103 (m s–2) ecf(a)(i)                                   2

  (iii)  the belt tension is insufficient to provide the centripetal force; (1)
so the belt does not ‘grip’ the pulley/does not hold the belt against
the pulley/there is insufficient friction to pull/push/move the belt. (1)    2
alternative argument the belt does not ‘grip’ the pulley/there is
insufficient friction to pull/push/move the belt; because of its
inertia/insufficient to provide force for acceleration of (belt)-drum

  (b)   resonance occurs; when the natural frequency of vibration of the (1)
panel = rotational frequency of the motor (1)                                                   2

 [8]

 

 

shm assessed hw markscheme

1.       (a)     The resultant force is zero (WTTE)

For the first mark allow
-      sum of forces is zero,
-      upward force = downward force,
-      forces cancel each other
BUT do not allow forces are balanced

B1

          Forces are weight and force from the spring (allow tension)

Allow force of gravity for weight

B1

  (b)   (i)      acceleration is (directly) proportional to displacement

M1

          and is directed in the opposite direction to the displacement. (WTTE)

allow a = –(2πf)2 x, provided a and x are identified and –ve sign must be explained.
Do not allow “acceleration is prop to negative displacement for second mark.
Allow always towards the equilibrium position

A1

  (ii)   x = acos2πft  2πf = 7.85 (expressed in any form)

M1

          f = (7.85/2π) = 1.25 (1.249Hz)

Do not allow use of the fig to show T = 0.8s and hence
f = 1.25 Hz. This scores 0.

A1

  (iii)  correct substn in Vmax = (2πf)A  Vmax = 2π × 1.25 × 0.012

Many will forget to change 12 mm into 0.012m and have
v = 94 m s–1 this scores 1 mark.

C1

          Vmax = 0.094 m s–1

A1

  (c)   roughly sinusoidal graph of correct period ie 0.8s

B1

          90° out of phase with displacement graph (i.e. starts at origin
with -ve initial gradient)

B1

          maximum velocity correctly shown as 0.094 {allow ecf from (iii)}

B1

[11]

 

  2.     (i)      1.      Measure the time t for N oscillations.                                                   M1
frequency f = N/t                                                                                  A1

2.       Measure the amplitude A of the oscillations using the ruler.                 M1
maximum speed is calculated using: vmax = (2πf)A                               A1

  (ii)   The maximum speed is doubled                                                                    B1
because the frequency is the same and vmax =  2 pi fA

(iii)    F = (–) kx and F = ma
Therefore ma = (–) kx                                                                                   M2
ω2 = k / m
T = 2Pi root k over m                                                                                              M1

[9]

 

  3.     (a)     (i)      A motion in which the acceleration/force is proportional to the
displacement; (1)
directed towards the centre of oscillation/equilibrium position/AW
or a α -x or a = –ω2x or a = –4π2f2x; symbols must be identified (1)     2

  (ii)   T = 0.25 s or f = 1/T; f = 4 (Hz) (2)                                                         2

(iii)    a = –4π2f2A; = 4 × 9.87 × 16 × 0.005; = 3.2 (m s–2) ecf a(ii) (3)             3

  (b)   (i)      Resonance occurs at /close to the natural frequency of an oscillating (1)
object/system; caused by driving force (at this frequency); when (1)
maximum energy transfer between driver and driven/maximum
amplitude achieved (1)                                                                            3

3 marking points in any sensible order

  (ii)   1   reduced amplitude; as resonance frequency lower
     or resonance will occur at lower frequency; as greater
     inertia/reduced natural frequency/AW in terms of amplitude change (2)

          2   reduced amplitude; as resonance frequency higher
     or resonance will occur at a higher frequency; as larger restoring
     force/increased natural frequency/AW in terms of amplitude change (2)           4

[14]

 

  4.     (a)     (i)      acceleration ∞ displacement; indication of restoring force by negative
sign/acc. in opp. direction to displacement/acc. towards origin/AW        2

  (ii)   linear graph through origin; negative gradient                                          2

  (b)   (i)      0.05 (m)                                                                                                  1

  (ii)   4π2f2 = a/A; = 12.5/0.05 = 250 so f = 2.5(1) Hz; T = 1/f (= 0.4 s)           3

  (c)    (i)      cosine wave; correct period of 0.4 s; correct amplitude of 0.05 m          3

  (ii)   0; 0.1/0.3/0.5/0.7/0.9 (s)                                                                          2

[13]

 

 

Density Calculations

Kepler found three laws which governed planetary motion. One linked their orbital periods and their distance from the sun. The orbital periods can be found by painstaking observation of the motions of the planets in the sky.  Kepler’s 2^nd Law can then be used to find the radius of orbit. Once this is known Newton’s Law of gravitation can be used to calculate the mass of a planet. The diameter of a planet can then be calculated form the angle it subtends when viewed from Earth. 

Name	Radius km	Mass kg
Sun	695000	1.99x1030
Mercury	2440	3.3x1023
Venus	6052	4.87x1024
Earth	6378	5.97x1024
Mars	3397	6.42x1023
Jupiter	71492	1.90x1027
Saturn	60268	5.68x1026
Uranus	25559	8.68x1025
Neptune	24766	1.02x1026
Pluto	1137	1.027x1022
Moon	1738	7.35x1022
Ganymede	2634	1.48x1023

1.     Using the table opposite calculate the average densities of the objects listed.

2.     The planets can be divided into two groups called Jovian and Terrestrial. Identify which belongs to each group, and why they are so named.

3.     The average density of the surface rocks of the earth is 2 800kg m^-3.  How does this compare with your value. The earth is thought to have a dense iron core. Why do you think this conclusion has been drawn?

4.       Comment upon my use of the term average density in question 1.

5.     Compare your values for Mars and the Earth.  The surface rocks on Mars are similar to igneous rocks on Earth. Mars is thought to have a core rich in sulphur as well as iron, explain why.

6.     The Moon and the Earth are the same age. This has been verified from rocks returned by American and Soviet missions. Compare the densities of the two bodies.

7.     There are four theories of the formation of the Earth Moon system

a)           That the two bodies formed together at the same time.

b)           That a proto-planet of the size of Mars collided with the proto-earth. Both planets were destroyed. The bulk of the material accreted to form the Earth but a small amount of lighter material spun out and formed the Moon.

c)            The teardrop theory where the proto moon was torn away from the Earth leaving a hole now filled by the Pacific Ocean. (Continental drift?)

d)           The moon was a stray planet that the earth captured. (compare to other small objects)

 

 

 

8.     The planet mercury superficially resembles the Moon. Do you think its internal structure is similar? Give reasons for your answer.

9.     The Jovan planets are sometimes called "Gas Giants". With reference to their densities give reasons why. Neptune and Uranus are both blue gas giants. Some may call them sister planets. Comment upon this.

10.  Should Pluto be classed as a terrestrial planet?