Leaving Cert Physics Long Questions: 4. Circular Motion and SHM (all higher level)

Remember to photocopy 4 pages onto 1 sheet by going A3→A4 and using back to back on the photocopier

2015 - 2002

Solutions begin on page 7

Circular Motion

2012 Question 12 (a)
An Olympic hammer thrower swings a mass of 7.26 kg at the end of a light inextensible wire in a circular motion. In the final complete swing, the hammer moves at a constant speed and takes 0.8 s to complete a circle of radius 2.0 m. (i) What is the angular velocity of the hammer during its final swing? (ii) Even though the hammer moves at a constant speed, it accelerates. Explain. (iii) Calculate the acceleration of the hammer during its final swing (iv) Calculate the kinetic energy of the hammer as it is released.

2011 Question 6 (c)
A simple merry-go-round consists of a flat disc that is rotated horizontally.
A child of mass 32 kg stands at the edge of the merry-go-round, 2.2 metres from its centre.
The force of friction acting on the child is 50 N.
Draw a diagram showing the forces acting on the child as the merry-go-round rotates.

What is the maximum angular velocity of the merry-go-round so that the child will not fall from it, as it rotates?
If there was no force of friction between the child and the merry-go-round, in what direction would the child move as the merry-go-round starts to rotate?

2006 Question 6 (i) Define velocity. (ii) Define angular velocity. (iii) Derive the relationship between the velocity of a particle travelling in uniform circular motion and its angular velocity. (iv) A student swings a ball in a circle of radius 70 cm in the vertical plane as shown. The angular velocity of the ball is 10 rad s–1.
What is the velocity of the ball? (v) How long does the ball take to complete one revolution? (vi) Draw a diagram to show the forces acting on the ball when it is at position A. (vii) The student releases the ball when is it at A, which is 130 cm above the ground, and the ball travels vertically upwards. Calculate the maximum height, above the ground, the ball will reach. (viii) Calculate the time taken for the ball to hit the ground after its release from A.

Circular Motion and Gravity
2004 Question 12 (a) (i) State Newton’s universal law of gravitation. (ii) Centripetal force is required to keep the earth moving around the sun.
What provides this centripetal force? (iii) In what direction does this centripetal force act? (iv) Give an expression for centripetal force. (v) The earth has a speed of 3.0 × 104 m s–1 as it orbits the sun.
The distance between the earth and the sun is 1.5 × 1011 m
Calculate the mass of the sun.
(gravitational constant G = 6.7 × 10–11 m3 kg–1 s–2)

2013 Question 6 (i) State Newton’s law of universal gravitation. (ii) Explain what is meant by angular velocity. (iii) Derive an equation for the angular velocity of an object in terms of its linear velocity when the object moves in a circle.

The International Space Station (ISS), shown in the photograph, functions as a research laboratory and a location for testing of equipment required for trips to the moon and to Mars.
The ISS orbits the earth at an altitude of 4.13 × 105 m every 92 minutes 50 seconds.

(iv) Calculate (a) the angular velocity, (b) the linear velocity, of the ISS. (v) Name the type of acceleration that the ISS experiences as it travels in a circular orbit around the earth. (vi) What force provides this acceleration? (vii) Calculate the attractive force between the earth and the ISS. (viii) Hence or otherwise, calculate the mass of the earth.

(ix) If the value of the acceleration due to gravity on the ISS is 8.63 m s−2, why do occupants of the ISS experience apparent weightlessness? (x) A geostationary communications satellite orbits the earth at a much higher altitude than the ISS.
What is the period of a geostationary communications satellite?
(mass of ISS = 4.5 × 105 kg; radius of the earth = 6.37 × 106 m)

2008 Question 6 (i) State Newton’s law of universal gravitation. (ii) The international space station (ISS) moves in a circular orbit around the equator at a height of 400 km.
What type of force is required to keep the ISS in orbit? (iii) What is the direction of this force? (iv) Calculate the acceleration due to gravity at a point 400 km above the surface of the earth. (v) An astronaut in the ISS appears weightless. Explain why. (vi) Derive the relationship between the period of the ISS, the radius of its orbit and the mass of the earth. (vii) Calculate the period of an orbit of the ISS. (viii) After an orbit, the ISS will be above a different point on the earth’s surface. Explain why. (ix) How many times does an astronaut on the ISS see the sun rise in a 24 hour period?
(gravitational constant = 6.6 × 10–11 N m2 kg–2; mass of the earth = 6.0 × 1024 kg; radius of the earth = 6.4 × 106 m)

2005 Question 6 (i) Define angular velocity. (ii) Define centripetal force. (iii) State Newton’s Universal Law of Gravitation. (iv) A satellite is in a circular orbit around the planet Saturn.
Derive the relationship between the period of the satellite, the mass of Saturn and the radius of the orbit. (v) The period of the satellite is 380 hours. Calculate the radius of the satellite’s orbit around Saturn. (vi) The satellite transmits radio signals to earth. At a particular time the satellite is 1.2 × 1012 m from earth. How long does it take the signal to travel to earth? (vii) It is noticed that the frequency of the received radio signal changes as the satellite orbits Saturn. Explain why.
Gravitational constant = 6.7 × 10–11 N m2 kg–2 mass of Saturn = 5.7 × 1026 kg speed of light = 3.0 × 108 m s–1

2015 Question 6
In the circular orbit of a satellite around the Earth, the required centripetal force is the gravitational force between the satellite and the Earth.
The force can be determined using Newton’s law of universal gravitation. (i) Explain what is meant by centripetal force. (ii) State Newton’s law of universal gravitation. (iii) Derive the relationship between the period of a satellite, the radius of its orbit and the mass of the Earth.

A Global Positioning Systems (GPS) receiver can calculate its position on Earth to within a few metres.
It picks up radio-wave signals from several of the 32 GPS satellites orbiting the Earth.
GPS satellites orbit the Earth in Medium Earth Orbit (MEO) with a period of 12 hours.

(iv) Calculate the height of a GPS satellite above the Earth’s surface. (v) Calculate the speed of a GPS satellite. (vi) Calculate the minimum time it takes a GPS signal to travel from the satellite to a receiver on the surface of the Earth. (vii) Explain why GPS satellites are not classed as geostationary satellites. (viii) Radio-waves, such as those used by GPS satellites, have the lowest frequency of all electromagnetic radiation types.
What type of electromagnetic radiation has the next lowest frequency?
(mass of Earth = 5.97 × 1024 kg; radius of Earth = 6371 km)

Simple Harmonic Motion

2011 Question 12 (a)
State Hooke’s law.
A body of mass 250 g vibrates on a horizontal surface and its motion is described by the equation a = – 16 s, where s is the displacement of the body from its equilibrium position.
The amplitude of each vibration is 5 cm. (a) Why does the body vibrate with simple harmonic motion? (b) Calculate the frequency of vibration of the body? (c) What is the magnitude of (i) the maximum force, (ii) the minimum force, which causes the body’s motion?

2013 Question 11
Read the following passage and answer the accompanying questions.

A seismometer consists of a sensor that detects ground motion, attached to a recording system.
A seismometer that is sensitive to up-down motions of the ground, as caused by an earthquake, can be understood by visualising a mass hanging on a spring as shown in the diagram.
The frame and the drum move up and down as the seismic wave passes by, but the mass remains stationary.
If a recording system is installed, such as a rotating drum attached to the frame and a pen attached to the mass, this relative motion between the suspended mass and the ground can be recorded to produce a seismogram, as shown in the diagram.
Modern seismometers do not use a pen and drum.
The relative motion between a magnet that is attached to the mass, and the frame, generates a potential difference that is recorded by a computer.
(Adapted from www.iris.edu Education and Outreach Series No.7: How does a Seismometer Work?)

(i) Seismic waves can be longitudinal or transverse.
What is the main difference between them? (ii) An earthquake generates a seismic wave that takes 27 seconds to reach a recording station. If the wave travels at 5 km s−1 along the earth’s surface, how far is the station from the centre of the earthquake? (iii) Draw a diagram to show the forces acting on the suspended mass when the seismometer is at rest. (iv) At rest, the tension in the spring is 49 N. What is the value, in kilograms, of the suspended mass? (v) What type of motion does the frame have when it moves relative to the mass? (vi) During an earthquake the ground was observed at the recording station to move up and down as the seismic wave generated by the earthquake passed. Give an equation for the acceleration of the ground in terms of the periodic time of the wave motion and the displacement of the ground. (vii) If the period of the ground motion was recorded as 17 seconds and its amplitude was recorded as 0.8 cm, calculate the maximum ground acceleration at the recording station. (viii) In some modern seismometers a magnet is attached to the mass and a coil of wire is attached to the frame. During an earthquake, there is relative motion between the magnet and the coil.
Explain why an emf is generated in the coil. (ix) (acceleration due to gravity, g = 9.8 m s−2)

Tricky maths questions - points to note * When using the F = k s expression for Hooke’s Law, s represents the extension, i.e. the distance between the new length and the original (natural) length.
However when using the expression for simple harmonic motion (a = -ω2 s) s represents the distance between the new length and the equilibrium position. ω= km

* Remember that the most common equation used here is the following:

2014 Question 12 (a) (i) State Hooke’s law. (ii) The elastic constant of a spring is 12 N m–1 and it has a length of 25 mm.
An object of mass 20 g is attached to the spring.
What is the new length of the spring? (iii) The object is then pulled down until the spring’s length is increased by a further 5 mm and is then released. The object oscillates with simple harmonic motion.
Sketch a velocity-time graph of the motion of the object. (iv) Calculate the period of oscillation of the object.
(acceleration due to gravity, g = 9.8 m s−2)

2009 Question 12 (a) (i) State Hooke’s law. (ii) When a sphere of mass 500 g is attached to a spring of length 300 mm, the length of the spring increases to 330 mm.
Calculate the spring constant. (iii) The sphere is then pulled down until the spring’s length has increased to 350 mm and is then released.
Describe the motion of the sphere when it is released. (iv) What is the maximum acceleration of the sphere? (acceleration due to gravity = 9.8 m s-2)

2007 Question 6 (i) State Hooke’s law. (ii) A stretched spring obeys Hooke’s law.
When a small sphere of mass 300 g is attached to a spring of length 200 mm, its length increases to 285 mm.
Calculate its spring constant. (iii) The sphere is pulled down until the length of the spring is 310 mm.
The sphere is then released and oscillates about a fixed point.
Derive the relationship between the acceleration of the sphere and its displacement from the fixed point. (iv) Why does the sphere oscillate with simple harmonic motion? (v) Calculate the period of oscillation of the sphere (vi) Calculate the maximum acceleration of the sphere (vii) Calculate the length of the spring when the acceleration of the sphere is zero.
(acceleration due to gravity = 9.8 m s–2)

2002 Question 6 (i) State Newton’s second law of motion. (ii) The equation F = – ks, where k is a constant, is an expression for a law that governs the motion of a body.
Name this law and give a statement of it. (iii) Give the name for this type of motion and describe the motion. (iv) A mass at the end of a spring is an example of a system that obeys this law.
Give two other examples of systems that obey this law. (v) The springs of a mountain bike are compressed vertically by 5 mm when a cyclist of mass 60 kg sits on it.
When the cyclist rides the bike over a bump on a track, the frame of the bike and the cyclist oscillate up and down.
Using the formula F = – ks, calculate the value of k, the constant for the springs of the bike. (vi) The total mass of the frame of the bike and the cyclist is 80 kg.
Calculate (i) the period of oscillation of the cyclist, (ii) the number of oscillations of the cyclist per second. (acceleration due to gravity, g = 9.8 m s-2)

Solutions
2015 Question 6 (i) Explain what is meant by centripetal force.
The force - acting in towards the centre - required to keep an object moving in a circle is called centripetal force. (ii) State Newton’s law of universal gravitation.
Newton’s law of gravitation states that any two point masses in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. (iii) Derive the relationship between the period of a satellite, the radius of its orbit and the mass of the Earth.
Centripetal force = gravitational force mv2/r = Gm1m2/r2 v2 = Gm/r

Also T = 2πr/v
T2 = 4π2r3/GM (iv) Calculate the height of a GPS satellite above the Earth’s surface.
T2 = 4π2r3/GM r = 2.66 × 107 m h = (2.66 × 107 – radius of the earth) = 2.023 × 107 m (v) Calculate the speed of a GPS satellite. v2 = GM/r v = 3.869 × 103 m s–1 (vi) Calculate the minimum time it takes a GPS signal to travel from the satellite to a receiver on the surface of the Earth. v = s/t t = 0.067 s (vii) Explain why GPS satellites are not classed as geostationary satellites. Periodic time is not 24 hours // Not in the same place above the Earth (viii) Radio-waves, such as those used by GPS satellites, have the lowest frequency of all electromagnetic radiation types. What type of electromagnetic radiation has the next lowest frequency?
Microwaves

2014 Question 12 (a) (i) State Hooke’s law.
Hooke’s Law states that when an object is stretched the restoring force is directly proportional to the displacement, provided the elastic limit is not exceeded. (ii) What is the new length of the spring?
F = k (extension). extension = F/k = (0.02 × 9.8)/12 = 0.0163 m
New length = 41.3 mm (iii) Sketch a velocity-time graph of the motion of the object. axes labelled periodic graph correct sinusoidal shape, beginning with v = 0 (iv) Calculate the period of oscillation of the object. ω2 = k/m (= 600)
T = 2π/ω
T = 0.256 s

2013 Question 11 (a) Seismic waves can be longitudinal or transverse. What is the main difference between them?
A Longitudinal Wave is a wave where the direction of vibration is parallel to the direction in which the wave travels.
A Transverse wave is a wave where the direction of vibration is perpendicular to the direction in which the wave travels. (b) How far is the station from the centre of the earthquake? s = vt = 5000 × 27 = 135000 m = 135 km (c) Draw a diagram to show the forces acting on the suspended mass when the seismometer is at rest.
Weight acting downwards and tension acting upwards (d) At rest, the tension in the spring is 49 N. What is the value, in kilograms, of the suspended mass?
W = mg m = 5 kg (e) What type of motion does the frame have when it moves relative to the mass?
Simple Harmonic Motion (f) Give an equation for the acceleration of the ground in terms of the periodic time of the wave motion and the displacement of the ground. a = 2 s

T = 2πω a=4π2sT2

(g) If the period of the ground motion was recorded as 17 seconds and its amplitude was recorded as 0.8 cm, calculate the maximum ground acceleration at the recording station. amax=4π2(0.008)172 amax = 0.0011 m s-2 (h) Explain why an emf is generated in the coil.
Magnetic field passing through the coil is changing OR The coil cuts the magnetic flux

2013 Question 6 (i) State Newton’s law of universal gravitation.
Newton’s Law of Gravitation states that any two point masses in the universe attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. (ii) Explain what is meant by angular velocity.
Angular Velocity is the rate of change of angle with respect to time. (iii) Derive an equation for the angular velocity of an object in terms of its linear velocity when the object moves in a circle. θ= sr θt= str

But = /t and v = s/t: = vr
Cross-multiply to get: v= r (iv) Calculate (a) the angular velocity, (b) the linear velocity, of the ISS. ω= 2πT= 2π5570 = 1.1 × 10-3 s-1 v =rω v = (6.783 × 106) × (1.1 × 10−3) = 7651.5 m s−1 (v) Name the type of acceleration that the ISS experiences as it travels in a circular orbit around the earth.
Centripetal / Gravitational ?? (vi) What force provides this acceleration?
Gravitational
(vii) Calculate the attractive force between the earth and the ISS.

F = 3.884 × 106 N (viii) Hence or otherwise, calculate the mass of the earth.
F= GM1M2d2

M1= Fd2GM2

M1 = 5.95 × 1024 kg (other methods also acceptable – can you think of two others?) (ix) Why do occupants of the ISS experience apparent weightlessness?
They are in freefall // ISS accelerating at the same rate as occupants
What is the period of a geostationary communications satellite?
One day

2012 Question 12 (a) (i) What is the angular velocity of the hammer during its final swing?

T= 2π
= 2π0.8

= 7.85 rad s-1 (ii) Even though the hammer moves at a constant speed, it accelerates. Explain.
The direction changes (continuously) (iii) Calculate the acceleration of the hammer during its final swing. a = ω2r a = (7.85)2(2) a = 123.37 m s-2 towards the centre of orbit (iv) Calculate the kinetic energy of the hammer as it is released.
K.E. = ½ mv2
K.E. = ½ m(r ω) 2
K.E. = 896 J

2011 Question 6 (c) (i) Draw a diagram showing the forces acting on the child as the merry-go-round rotates.

(ii) What is the maximum angular velocity of the merry-go-round so that the child will not fall from it, as it rotates?
F = mω2r
50 = 30 ω2(2.2) ω = 0.842 rad s-1 (iii) If there was no force of friction between the child and the merry-go-round, in what direction would the child move as the merry-go-round starts to rotate?
The child would remain stationary / any appropriate answer.

2011 Question 12 (a) (i) State Hooke’s law.
For a stretched string the restoring force is proportional to displacement (ii) Why does the body vibrate with simple harmonic motion?
The acceleration is proportional to the displacement (iii) Calculate the frequency of vibration of the body? ω2 = 16 ω = 4 f = 1/T and T = 2π/ω so f = ω/2π f = 0.64 Hz (iv) What is the magnitude of (i) the maximum force, (ii) the minimum force, which causes the body’s motion? a max = (–)16(0.05) = 0.80 (Fmax occurs when acceleration / displacement is a maximum)
Fmax = (0.250)(0.80) = 0.20 N
Fmin = 0

2010 Question 6 (i) State Newton’s law of universal gravitation.
Force between any two point masses is proportional to product of masses and inversely/indirectly proportional to square of the distance between them.

(ii) Use this law to calculate the acceleration due to gravity at a height above the surface of the earth, which is twice the radius of the earth.
Note that 2d above surface is 3d from earth’s centre

where d = 6.36 × 106 m

gnew = 1.09 m s-2

(iii) Explain why the spacecraft continues on its journey to the moon, even though the engines are turned off.
There are no external forces acting on the spacecraft so from Newton’s 1st law of motion the object will maintain its velocity. (iv) Describe the variation in the weight of the astronauts as they travel to the moon.
Weight decreases as the astronaut moves away from the earth and gains (a lesser than normal) weight as she/he approaches the moon (v) At what height above the earth’s surface will the astronauts experience weightlessness?
Gravitational pull of earth = gravitational pull of moon =

dE = 9 dm and dE + dm = 3.84 × 108 m

10 dm = 3.84 × 108 dm = 3.84 × 107 dE = 3.356 × 108

Height above the earth = (3.356 × 108) – (6.36 × 106) = 3.39 × 108 m

(vi) The moon orbits the earth every 27.3 days. What is its velocity, expressed in metres per second? v= 2πrT v= 2π(3.84 ×108)27.3 ×24 ×24×60

v = 1022.9 m s-1 (vii) Why is there no atmosphere on the moon?
The gravitational force is too weak to sustain an atmosphere.

2009 Question 12 (a) (i) State Hooke’s law.
When a string is stretched the restoring force is proportional to the displacement. (ii) When a sphere of mass 500 g is attached to a spring of length 300 mm, the length of the spring increases to 330 mm. Calculate the spring constant.
When the mass of 500 g is attached the new force down = mg = (0.5)(g).
Because the spring is in equiblibrium this must be equal to the force up (which is the restoring force).
Hooke’s law in symbols: F = k x
 (0.5)(g) = kx  k = F/x = 0.5g/0.030  k = 163.3 N m-1 (iii) The sphere is then pulled down until the spring’s length has increased to 350 mm and is then released.
Describe the motion of the sphere when it is released.
It executes simple harmonic motion because the displacement is proportional to t he acceleration. (iv) What is the maximum acceleration of the sphere?
F = ma = kx a = kx/m = (163.3)(0.02)/(0.5) = 6.532 m s-2 OR a = 2 x  2 = k/m = 163.3/0.5 a = 6.532 m s-2

2008 Question 6 (i) State Newton’s law of universal gravitation.
Newton’s Law of Gravitation states that any two point masses in the universe attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. (ii) What type of force is required to keep the ISS in orbit?
Gravity
(iii) What is the direction of this force?
Towards the centre of the orbit / inwards / towards the earth (iv) Calculate the acceleration due to gravity at a point 400 km above the surface of the earth. = mg   g = (6.6 × 10–11)( 6.0 × 1024) / (400 000 + 6.4 × 106)2  g = 8.6 m s-2 (v) An astronaut in the ISS appears weightless. Explain why.
He is in a state of free-fall (the force of gravity cannot be felt). (vi) Derive the relationship between the period of the ISS, the radius of its orbit and the mass of the earth.
See notes Circular Motion chapter for a derivation.

(vii) Calculate the period of an orbit of the ISS.

 T2 = 3.1347 × 107 T = 5.6 × 103 s

(viii) After an orbit, the ISS will be above a different point on the earth’s surface. Explain why.
The ISS has a different period to that of the earth’s rotation (it is not in geostationary orbit). (ix) How many times does an astronaut on the ISS see the sun rise in a 24 hour period?
(24 ÷ 1.56 + 1) = 16 ( sunrises).

2007 Question 6 (i) State Hooke’s law.
For a stretched string the restoring force is proportional to the extension. (ii) A stretched spring obeys Hooke’s law.
When a small sphere of mass 300 g is attached to a spring of length 200 mm, its length increases to 285 mm.
Calculate its spring constant.
F = mg = ks
(0.30)(9.8) = (k)(0.085) k = 34.6 N m-1 (iii) The sphere is pulled down until the length of the spring is 310 mm.
The sphere is then released and oscillates about a fixed point.
Derive the relationship between the acceleration of the sphere and its displacement from the fixed point.
F = - ks  ma = - ks  a = - (k/m)s  a α -s  a = - k s (iv) Why does the sphere oscillate with simple harmonic motion?
Its acceleration is proportional to its displacement from a fixed point. (v) Calculate the period of oscillation of the sphere.
From above:ω2 = k/m  ω2 = 34.6 / 0.3 ω = 10.7 T = 2π/ω = 2π/10.7 = 0.58 ≈ 0.6  T = 0.6 s (vi) Calculate the maximum acceleration of the sphere.
This occurs when s is a maximum, i.e. when s = amplitude = 0.310 – 0.285 = 0.025 m. a = -ω2s  a = - (10.7)2 (0.025)  a = (-) 2.89 m s-2 (vii) Calculate the length of the spring when the acceleration of the sphere is zero.
This occurs at the fixed point when l = 0.285 m

2006 Question 6 (i) Define velocity.
Velocity is the rate of change of displacement with respect to time. (ii) Define angular velocity.
Angular velocity is the rate of change of angle with respect to time. (iii) Derive the relationship between the velocity of a particle travelling in uniform circular motion and its angular velocity. θ = s /r θ /t = s/rt ω = v /r v = ω r (iv) A student swings a ball in a circle of radius 70 cm in the vertical plane as shown. The angular velocity of the ball is 10 rad s–1. What is the velocity of the ball? v = ω r = (10)(0.70) = 7.0 m s-1 (v) How long does the ball take to complete one revolution?
T= 2πr/v = 2π(0.70)/v = 0.63 s (vi) Draw a diagram to show the forces acting on the ball when it is at position A.
Weight (W) downwards; reaction (R) upwards; force to left (due to friction or curled fingers) (vii) The student releases the ball when is it at A, which is 130 cm above the ground, and the ball travels vertically upwards. Calculate the maximum height, above the ground, the ball will reach. v2 = u2+ 2as  0 = (7)2 + 2(-9.8) s / s = 2.5(0) m  max. height = 2.5 + 1.30 / 3.8 m (viii) Calculate the time taken for the ball to hit the ground after its release from A. s = ut + ½ at2
-1.30 = 7t – ½ (9.8)t2 t = 1.59 s

2005 Question 6 (i) Define angular velocity.
Angular velocity is the rate of change of displacement with respect to time. (ii) Define centripetal force.
The force - acting in towards the centre - required to keep an object moving in a circle is called Centripetal Force. (iii) State Newton’s Universal Law of Gravitation.
Newton’s Law of Gravitation states that any two point masses in the universe attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. (iv) Derive the relationship between the period of the satellite, the mass of Saturn and the radius of the orbit.
See notes on the Circular Motion chapter for the derivation. (v) The period of the satellite is 380 hours. Calculate the radius of the satellite’s orbit around Saturn. T = 380 × 60 × 60 = 1.37 × 106 s r3 = T2GM/4π2  r3 = (1.37 × 106)2(6.7 × 10–11)( 5.7 × 1026)/ 4π2  r = 1.2 × 109 m (vi) The satellite transmits radio signals to earth. At a particular time the satellite is 1.2 × 1012 m from earth. How long does it take the signal to travel to earth? v = s/t
(3.0 × 108) = (1.2 × 1012)/t t = 4000 s (vii) It is noticed that the frequency of the received radio signal changes as the satellite orbits Saturn. Explain why.
Doppler Effect due to relative motion between source of signal and the detector

2004 Question 12 (a) (i) State Newton’s universal law of gravitation.
Newton’s Law of Gravitation states that any two point masses in the universe attract each other with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. (ii) Centripetal force is required to keep the earth moving around the sun.
What provides this centripetal force?
Gravitational pull of the sun. (iii) In what direction does this centripetal force act?
Towards the centre. (iv) Give an expression for centripetal force.

(v) Calculate the mass of the sun. and Equating gives v2 Ms = v2R/G
 Ms = (3.0 × 104)2 ( 1.5 × 1011)/ 6.7 × 10–11 = 2.0 × 1030 kg.

2002 Question 6 (i) State Newton’s second law of motion.
Newton’s Second Law of Motion states that the rate of change of an object’s momentum is directly proportional to the force which caused it, and takes place in the direction of the force. (ii) Name this law and give a statement of it.
Hooke’s Law states that when an object is stretched the restoring force is directly proportional to the displacement, provided the elastic limit is not exceeded. (iii) Give the name for this type of motion and describe the motion.
Simple harmonic motion; an object is said to be moving with Simple Harmonic Motion if its acceleration is directly proportional to its distance from a fixed point in its path, and its acceleration is directed towards that point. (iv) Give two other examples of systems that obey this law.
Stretched elastic, pendulum, oscillating magnet, springs of car, vibrating tuning fork, object bobbing in water waves, ball in saucer, etc. (v) Using the formula F = – ks, calculate the value of k, the constant for the springs of the bike.
F = – ks  mg = – ks  60 × 9.8 = -k (.005)  588 = -k (.005)  k = 1.2 × 105 N m-1 (vi) Calculate the period of oscillation of the cyclist. k/m = ω2 ω = 38 s-1
T = 2π/ω = 0.16 s (vii) Calculate the number of oscillations of the cyclist per second. f = 1/T approximately = 6