SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
1.(a)
(b)
The time taken for the ball to fall 0.7m is
(b)
The horizontal distance travelled in this time (range) is
Thus the ball will clear the net if it is hit 11.0m
from the net.
4.(a)
The horizontal distances travelled between each flash
is constant 4mm. The horizontal velocity is therefore
constant.
SOLUTIONS TO CHAPTER 1
2.
(b)
Measure the distances between the images vertically. The
distance between each image increases \ the vertical
velocity between each image increases.
\ A accelerates downwards.
(c)
There is no force in the horizontal direction \ it will not
accelerate horizontally.
(d)
Acceleration of B is also 9.8ms-2 downwards as this is
the direction of the force (gravity) on B. [ The vertical
movement of B is the same as A-draw horizontal lines
through the images of A and B to check this ].
(e)
B will hit the ground at greater speed. It has an initial
velocity that A hasn’t got, but also has horizontal
component of velocity \ it is always moving faster than A.
5.(a)
1
s = v0t +
at2
2
1
=0+
x 9.8 x 22
2
3.(a)
= 19.6m
266
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TO EXERCISES
MOTION
MOTION
5.(b)(i)
v = V0 + at
v = 0 + 9.8 (2)
v = 19.6ms-1 down
(ii)
Vector addition of velocities
7.(a)
Range = vH x t
= 900 x 0.64
= 576m
VR [ by Pythagoras ]
VR = 102ms-1 at 11.1˚ below the horizontal
(c)
The horizontal velocity stays constant as
there is no force horizontally.
The vertical velocity will increase as the
force of gravity acts vertically.
(d)
s = v 0t +
at2 (vertical movement)
120 x 2 = t2
9.8
\ t = 4.95s
(e)
Range = VH x t
= 100 x 4.95
= 495m
6.(a)
Final
By vector addition
vFinal = 900.02ms-1 at 0.399˚ below the horizontal (3sf)
8.(a)
Vertical velocity
= 100 sin80˚
= 98.5ms-1
Horizontal velocity
= 100cos 80˚
=17.4ms-1
(b)(i)
at 1s
vv = v0 + at
= 98.5 + (-9.8) x 1
= 88.7ms-1
(ii)
at 13s
vv = v0 + at
= 98.5 + (-9.8).13
= -28.9ms-1 (i.e. 28.9ms-1 down)
(c)
14ms-1
VR = 33.7ms-1 at 58.9˚
12ms-1
(b)
vH will not change as there is no force
horizontally.
vv will increase continuously due to the
downwards force of gravity.
28.9
(c)
At X, velocity is 14ms-1 as the vertical component of
velocity is zero at X.
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267
SOLUTIONS TO CHAPTER 1
120 = 0 (t) + (9.8) x t2
(b)
Vertical velocity after 0.64s
v = V0 + at
v = 0 + 9.8 (0.64)
v = 6.27ms-1 down
\ The resultant velocity on impact is:
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
8.(d)
vhorizontal only
= 17.4ms-1 horizontal
Energy at impact is all KE
K = 1 mv2
2
1
= 2 x 0.200 x (41.0)2
(e)
a = g = 9.8ms-2 down
= 168J Thus energy is conserved.
(f)
Maximum height
v2 = u2 +2as
\ 0 = (98.5)2 + 2(-9.8)s
\ s = 495m
9.(a)
s = v0t +
10.(a)
vhorizontal = 50 cos 20˚ = 47.0ms-1
vvertical = 50 sin 20˚ = 17.1ms-1
(b)
Range = vH x t
= 47.0 x 4.4
= 207m
at2
SOLUTIONS TO CHAPTER 1
\ 40 = 0(t) + 1 (9.8) x t2
2
40
x
2
\ 9.8 = t2
(c)
Vertical velocity on impact
v = v0 + at
v = 17.1 + (-9.8)(4.4)
v = 26.0ms-1 (down)
The resultant velocity
\ t = 2.86s
(b)
t = 2.86s
vH = 47.0ms-1
(c)
v = v0 + at
v = 0 + 9.8 (2.86)
v = 28.0ms-1
(d)
vv = 26.0ms-1
vH = 30.0ms-1
vv = 28.0ms-1
vR
VR = 41.0ms-1 at 43.0˚ below the horizontal
(e)
Range = vH x t
= 30 x 2.86
= 85.8m
(f)
Total Energy at top KE + PE
vR
VR = 53.7ms-1 at 29.0˚ below the horizontal
(d)
The ball gains the extra kinetic energy due to
the fact that the tee is 20m above the green. The extra
potential energy is converted to kinetic energy.
11.(a)(i)
Range = vH x t
= (2.0 x cos45˚) x 2.96
= 4.2m
(ii)
The range will be reduced (see text section 1.7.2)
(iii)
The range may increase if the change in the angle is small
enough. (see text section 1.7.2)
= 12 mv2 + mgh
= ( 12 x 0.200 x 30.02 ) + (0.200 x 9.8 x 40)
= 168J
268
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MOTION
MOTION
11.(b)(1)
Maximum height above the nozzle
v = v2 + 2as
\ = (2.0 x sin45˚)2 + 2 (-9.8)s
\ s = 0.1m
\ above ground level = 1.2 + 0.1 = 1.3m
(b)(ii)
The maximum height will be increased if the angle is
increased. This is because the vertical component of the
initial velocity will increase, hence a greater time of flight.
(c)(iii)
The maximum height will be decreased if the angle is
decreased. This is because the vertical component of the
initial velocity will decrease, hence a smaller time of flight.
12.(a)
Range = vH x t
290 = vH x 7.693
vH = 37.7ms-1
(ii)
The range will be reduced.
Range = vH x t
\ R α vH
\ reduced vH reduced range.
13.(a)
The time of flight t is given by
2v sinq
t=
g
\ if we increase q, sinq increases
\ the time of flight increases.
(b)
vH = v cosq
\ if we increase q , cosq decreases
\ vH decreases.
(c)
The range will decrease as vH decreases –
45˚ gives the maximum range.
v
vV
14.
Components of the initial velocity
45˚
vH = v cos45˚
v
vH
\ V = 37.7 = 53˙3ms-1 at 45˚ to the horizontal
cos45˚
(c)
37.7ms-1
(d)
Only vH = 37.7ms-1
(e)
Maximum height at half the time of flight
= 7.963 ÷ 2
= 3.98s
SOLUTIONS TO CHAPTER 1
(b)
vV
40˚
vH
VH = 13 cos40 = 9.96ms
Vv = 13 sin40 = 8.36ms
Range = VH x t
= 9.96 x 1.9
= ~19m
On impact, vertical velocity
v = v0 + at
v = 8.36 + (-9.8)(1.9)
v = -10.3ms-1
vH = 9.96ms-1
q
(f)(i)
No effect on the time of flight as it is dependent on the
vertical velocity.
(v = v0 + at)
vv = 10.3ms-1
vR = 14.3ms-1
VR = 14.3ms-1 at 46.0˚ i.e 14ms-1 at 46˚
below the horizontal
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16.(a)
vH = v0 cosq
(b)
vv = v0 sinq
(c)
A and E, B and D
15.(a)(i)
(d)
None
(ii)
(e)
C
(f)
None
(g)
Down
SOLUTIONS TO CHAPTER 1
(iii)
(h)
B , E (tangents to the curve)
(i)
Vertical velocity is zero
Acceleration is equal to g (i.e. 9.8ms-2)
17.
See text.
(iv)
(b)(i)
vH will not change – no force in the horizontal direction.
(ii)
Acceleration is constant \ graph will not change.
(iii)
vv will increase in the negative direction as it is
accelerating for a larger time.
(iv)
K will increase as the velocity increases.
270
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MOTION
MOTION
1.(a)
(e)
1.0ms-2
towards the centre of the circle.
2.(a)
(b)
Towards the centre of the circle.
towards the nucleus.
(b)
(c)
Towards the centre of the circle.
As
, the direction of the change of velocity is the
direction of the acceleration (and the force).
(c)
The nucleus and the electron experience the same force (in
magnitude) i.e. Newton’s Third Law.
, at constant r
a
(ii)
The acceleration is inversely proportional to the radius at
constant speed.
(d)
a
24
(e)
The electrostatic force of attraction between the charges.
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271
SOLUTIONS TO CHAPTER 2
(d) (i)
The acceleration is directly proportional to the square of
the speed.
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
2.(f)
Attraction described by Coulomb’s Law
5.(a)
Towards the centre of the track.
(b)
5
5
(This is the nuclear charge)
5
6.
60km / hr = 16.7ms-1
(a)
SOLUTIONS TO CHAPTER 2
3.(a)
(b)
(b)
(c)
Towards the centre of the track.
7.(a)
4.(a)
(b)
\ If the speed is doubled the force becomes
4 times larger.
(b)
(c)
(i)
8.(a)
(ii)
\New force =
272
(b)
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MOTION
MOTION
8.(c)
11.(a)
(4)
(b)
8
towards the centre of its orbit.
9.(a)
12.(a)
F
v
The force is at 90º to the velocity (a centripetal force) it
will no change the speed of the object, just its direction;
hence the velocity changes but not the speed.
As above, the speed will not change
will not change. The force does no work on
the moving particle.
10.(a)
Electrostatic force: - between an electron and nucleus in
an atom.
(b)
Gravitation force: - between the earth and moon.
(c)
Frictional force: - between car tyres and the road for a car
moving around a bend.
(d)
Tension force: - the force provided by the wire holding a
‘hammer’ being thrown on a sports day.
(b)
On each tyre
SOLUTIONS TO CHAPTER 2
(b)
\
2.723 x 10-3ms-2
(c)
Total force (weight)
(d)
Total average force (by Pythagoras)
= 5830N at 19.6º to the horizontal.
(e)
Magnetic force: - the force on a moving charge in a magnetic field (as in a cyclotron)
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13.(1)
(c)(i)
(ii)
(2)
6.0ms-2
SOLUTIONS TO CHAPTER 2
15.(a)
14.(a)
The frictional force between the car tyres and the road
provides the centripetal acceleration when a car moves
around a bend.
To improve road safety and / or enable the car to move
around the bend at a greater speed, a larger centripetal
force is often required.
This is often done by banking the curved part of the road.
The horizontal component of the normal force (see Q 13)
can provide an increased force that now also causes the
centripetal acceleration.
Thus the car can move around the corner with more safety
(as the force is now not solely dependent on the road
friction) and at a greater speed if required
i.e
[ \ increase F, allows an increase r. ]
(b)
Using the diagram in 13.(b)
FV is equal and opposite to FG
(b)
(c)
(d)
(e)
(f)
But the force providing the centripetal force
274
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MOTION
MOTION
16.
17.
(a)
SOLUTIONS TO CHAPTER 2
(b)
(c)
(d)
(e)
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275
SOLUTIONS
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ESSENTIALS STAGE
STAGE 22
1.
The gravitational force between two masses is
proportional to the product of the masses and inversely
proportional to the square of the distance between them.
2.(a)
Each 5.20 x 10-11N.
This is consistent with Newton’s third law.
(b)
\ double distance,
force
\ new force
(b)
,\
distance.
Force is now 16 times the size
\ new force
r2
r2
r
3.(a)
r2
1.983 x 1020N attraction
SOLUTIONS TO CHAPTER 3
5.(a)
(b)
Both accelerate. They both experience the
same force \ they accelerate. But the moon
accelerates more because it has a
smaller mass.
4.(a)
r2
(c)
(d)
Same force – the distance is taken from the centre of each
mass no change.
6.
The gravitational force is a mutual force. That is each body
exerts a force on the other body. The forces are equal in
magnitude and opposite in direction. Thus the law of
universal gravitation is consistent with Newton’s third law
which states “ if body a exerts a force on body B, then body
Bexerts an equal and opposite force on body A”
7.(a)
The force on a mass(m1) in the earth’s
gravitational field is F = m1 g (i.e. weight). But
also the force is given by
(b)
r2
r2
r2
(b)
(c)
Double distance will
r2
the force, because
r2
(c)(i)
r2
if double radius then
g
\ new g is
= 2.46ms-2
276
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MOTION
MOTION
7.(c)(ii)
\ g increases by 20% = 11.8ms-2
(c)
Period squared on the Y ( vertical axis ) and radius cubed
on the X ( horizontal )axis.
(iii)
\ half mass, half g; but half radius, 4 x g
\ ”gnew” = 2g = 19.7ms-2
8.
Different types of materials beneath the surface will
change the value
Eg. - iron ore bodies compared to sand etc.
Also slight differences in the Earth’s
radius will change “g” (mountains etc)
r3
11.(a)
The gravitational force of attraction between the satellite
and the earth provides the centripetal force needed to
keep the satellite in it’s orbit.
\
=
\
, hence
(b)(i)
No effect,
9.(a)
- The mass of the satellite has no effect
As an object O moves below the surface,
the mass above it attracts it in the opposite
direction to the mass below it the resultant
force is less less g.
(b)
At the centre O is attracted equally in all
directions the resultant force is zero
\g=0
10.(a)
Gravitational attraction between the satellite and the
Earth.
SOLUTIONS TO CHAPTER 3
(ii)
reduce the radius , velocity will increase.
If r is halved, v is 2 time bigger.
\ 20,130 kmh-1
(iii)
less mass (mars)
\ lower velocity
\
\
(b)
The force between the satellite and the earth is given by
\
Provides the centripetal force and hence the centripetal
acceleration.
\
but
\
4
12.
If they move at different speeds at the same radius the
periods would be different; cannot be \ cannot have
different speeds.
13.(a)
periods =
= 96 mins
= 5760s
days
(b)
\
E
hence
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17.
Advantages
- Communication “dishes” and antennae can be fixed
in one direction – less complicated.
- Can maintain continuous communication
Disadvantages
- Too high signal strengths need to be high and
time delays are a problem, especially with TV
interviews.
13.(c)
14.(a)
18.
The gases in the atmosphere will slow the satellite down
\ it would spiral inwards.
19.(a)
7550
SOLUTIONS TO CHAPTER 3
(b)
ms-1
(b)
7550
5830s
15.(a)
It has the same period as the Earth
i.e. T = 24 hours.
\ The satellite it stays above the same point on the Earth
at all times.
7.28 x 103
= 6.49 x 103s
(c)
7
(d)
(b)
T = 24 hours
(c)
The centre of the orbit must coincide with the centre of
the Earth, and it must therefore be an equatorial orbit,
and Adelaide is not on the equator.
(d)
Same direction as the Earth \ West to East, to ensure that
it stays above the same place on the equator at all times.
(e)
6
(e)
T = 24 hours
= 8.64 x 104 s
(f)
Height above surface
16.
The gravitational force between the satellite and the earth
acts along a line centre to centre of each mass. Therefore
the centre of the orbit of the satellite must be the centre of
the earth.
278
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MOTION
MOTION
20.(a)(i)
Orbit Radius
(ii)
1
(iii)
SOLUTIONS TO CHAPTER 3
(iv)
m
m
(b)
If the satellite loses kinetic energy it slows down and it’s
velocity gets smaller. Thus the centripetal force at this
height will reduce. Therefore the now larger gravitational
force will produce a resultant force inwards towards the
centre of the earth. The satellite will begin to
spiral inwards.
Consequently, as the radius gets smaller the period will
get smaller. Therefore the number of revolutions
per day will increase.
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1.(a)
f
(e)
Away from the table at 90º.
(4)
Into the table at 90º. - The table experience on equal and
opposite force to the ball (Newton’s Third Law).
c
90˚ away from floor.
3.(a)
(b)
(c)
Away from the floor
(b)
N away from the floor at 90˚
(d)
(c)
Force on floor = 12N in the initial direction of the
ice-cream (Newton’s Third Law)
SOLUTIONS TO CHAPTER 4
ms-2 away from the floor
(e)
60N In the initial direction of the ball– Newton’s Third
Law – equal and opposite forces.
(f)
Work W = F.s
Force acts while ball is in contact with wall. While in
contact s = 0
4.(a)
g
(b)
(g)
Δv will be smaller
Δv = -5 - 6 = -11.0ms-1
and as a =
, Then acceleration will be smaller.
2.(a)
ms-1
(c)
5.(a)
+
away from the table at 90º.
(b)
5.0sN towards the hand
(b)
a = 140ms-2 away from the table.
280
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MOTION
MOTION
5.(c)
7.(a)
Use 3 images to represent the velocity
375N away from the hand.
(d)
375N is the force on the cricketer’s hand.
(e)
The time of “collision” would be increased
\ The acceleration would decrease.
\ The force would be less.
6.(a)
(i) Golf Club
direction as shown
(b)
Because
The direction of
will be the direction of F
force will be in the same direction.
8.
Law of conservation of momentum
SOLUTIONS TO CHAPTER 4
(ii) Racquet
9.(a)
(b)(i) Golf Club
(b)
(c)
(ii) Racquet
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9.(d)
Final Velocity
(b)
new velocity = 1.5ms-1 in the direction of B. ( west )
13.
10.(a)
The initial total momentum of the system is zero.
\ The final total momentum of the system is zero.
SOLUTIONS TO CHAPTER 4
(b)
i.e. 2.2sN in the direction of B.
but the vector addition triangle
11.(a)
V
V
is equilateral
V
14.
(as shown on diagram)
(b)
3900N
(c)
The total momentum of A and B is 7.1m as shown.
[ use pythigras ] But because
the
momentum of the larger mass is 7.1m in the opposite
direction to that of the total momentum of A + B.
3900N
12.(a)
i.e. 6.0m in the direction of B (west)
282
15.
Possible answer could be based on the fact that the
plasticine compresses a little on collision
increasing the time of the collision reducing the collision
force.
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MOTION
MOTION
16.(a)
zero (
momentum
18.(a)
) - Because of the law of conservation of
(b)
(b)
24sN to the right \p of 2.5kg trolley 24sN left.
(c)
(c)
i.e velocity of 2.5kg trolley = 9.6ms-1 left
(d)
(d)
SOLUTIONS TO CHAPTER 4
The change in momentum of trolley mass 6.0kg
-2
19.
(force on smaller trolley is equal and opposite)
\ The force pushing the trolley apart is 60N
17.(a)
(b)
\ both have the same recoiling momentum
(c)
20.(a)
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23.
(b)
(c)
a solution can be found using a scale diagram or using the
cosine rule.
SOLUTIONS TO CHAPTER 4
21.
24.(a)
No. It is moving with constant speed indicated by the
equal spacing between the images in a straight line \ not
accelerating \ no force.
(b)
(c)
22.
(d)
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MOTION
25.(a)
(b)
By the law of conservation of momentum
The change in momentum of the rocket will be
Δp = 8.1 x 107 sN in opposite direction to the fuel.
(c)
Using Δp = mΔv the change in velocity of the
rocket will be
-1
acceleration a =
4
in the direction of movement of the rocket.
27.(a)(i)
-16
SOLUTIONS TO CHAPTER 4
(ii)
(b)
(b)(i)
(c)
(ii)
(d)
26.(a)
(iii)
(c)(i)
No effect – acceleration would be constant.
(ii)
Acceleration is directly proportional to the mass of xenon
gas emitted each second.
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28.(a)(i)
Photon is absorbed by the black sail
final momentum = 0
(d)(i)
Intensity refers to the number of photons per area.
(ii)
Less photons per area will mean a decreasing force on the
sail, hence a decreasing acceleration. Therefore the speed
of the space craft will increase at a decreasing rate the
time will increase.
(ii)
Total number of photons hitting the sail per second is
SOLUTIONS TO CHAPTER 4
\ total change of momentum/second
(iii)
Change in momentum of the space craft and sail
= 8.08 x 10-3 sN (Law of conservation of momentum)
Change in velocity of the space craft per second
-4
2
(b)(i)
(ii)
(c)
The change in the momentum of the photons will be double the force on the sail will double the acceleration will
double. Hence the time taken will be less.
286
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- SOLUTIONS
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TO EXERCISES
MOTION
MOTION
3.(a)
1.(a)
Coulombic Force
9
r2
r2
1.80 x 10-2N
(b)
(b)(i)
-6
r2
4.50 x 10-3 N attraction
(ii)
(c)
(iii)
Same magnitude force now repulsion.
(c)
Equal and opposite forces as per Newton’s Third Law.
new
4.
r2
(d)
as F = ma, need the mass of q2.
2.(a)
SOLUTIONS TO CHAPTER 5
= 0.29N (2sf)
-6
r2
r
r = 328m ∼ 330m
r2
( 2sf )
F = 8.14 x 10-8N
(b)
The attraction force is a centripetal force.
8.14
-11
v = 2.18 x 106ms-1
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287
SOLUTIONS
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ESSENTIALS STAGE
STAGE 22
5.
Force between A and C
7.(a)
Force between B and C
r2
9
\ resultant force on C
(b)
90cm is 3 times the distance as
SOLUTIONS TO CHAPTER 5
at 90cm will be
6.
Force between A and B
6
r2
, the field
=
9
9 NC-1 outwards from the charge
(c)
2
8.(a)
(b)
2.5 x 103 x 1.6 x 10-19
(c)
288
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TO EXERCISES
MOTION
MOTION
9.(a)
C
(b)
r2
-7
11.
The strength of the field is shown by the number of
field lines crossing a unit area. More field lines implies a
stronger field. The direction of the field is shown by the
arrows on the lines.
12.(a)
-19
(b)
7.7
10.(a)
Magnitude of the electric field at X due to A
r2
5
-1
(c)
The charge is evenly distributed along the parallel plates,
because the charge density along the plates is constant.
The equally spaced field lines are therefore representative
of this.
In the pear-shaped conductor the charge is not evenly distributed. Therefore the lines are closer together where the
charge density is higher, and less closer where the charge
density is lower.
13.
The electric fields are stronger near sharp points on
conductors - such as pear-shaped object.
If the resultant electric field at the points is strong enough
it can attract molecules in the air? towards the point especially polar molecules. When these molecules touch
the conductor they have some of their charge nuetralized.
This results in the molecules now carrying a patial charge,
become partially ionised are then repel away
from the surface.
Consequently the conductive surface has its charge
gradually reduced. This is called a “corona discharge”.
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SOLUTIONS TO CHAPTER 5
(b)
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
14.(a)
Towards the right.
(b)
Field lines radiate from positive to negative.
\ if the small charge is attracted to the right it
must be negative.
SOLUTIONS TO CHAPTER 5
(c)
(d)
(e)
-6
-6
290
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TO EXERCISES
MOTION
MOTION
1.(a)
See text.
(b)
W
2.(a)
W
5.(a)
towards the negative plate
(b)
(c)
-16
(b)
-27
(d)
SOLUTIONS TO CHAPTER 6
(c)
No. Energy = W = ∆Vq which is independent of mass.
3.(a)
3.8 x 1011
(b)
∆V changes to 300V
(e)
(i)
(ii)
4.(a)
6.
1eV = 1.60 x 10-19 J
1KeV = 1.60 x 10-16 J
1MeV = 1.60 x 10-13 J
7.(a)
-17 towards plate B
(b)
(b)
(c)
Initial K = 2.00eV. As the electron enters the field its
kinetic energy converts to potential energy. PE = 2.0eV
when K = 0. But the electron needs 4.00eV to comletely
cross between the plates therefore it will stop halfway
between the plates
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ESSENTIALS STAGE
STAGE 22
7.(d)
The electron will experience a constant acceleration
towards plate A. Therefore the electron will slow down,
stop, and then accelerate back to the hole in the plate.
(e)
(d)
No force in the horizontal direction
\ no acceleration \ no change of velocity.
(e)
(f)
-19
2
SOLUTIONS TO CHAPTER 6
2
11.(a)
( Which of course, coincides with the answer in part (c)
8.
(b)
(c)(i)
2
9.
A. – electrons are negative \ are attracted to the upper
plate. \parabolic path curving upwards.
10.(a)
(ii)
(b)
(iii)
If not evacuated protons will collide with air molecules –
losing energy and \ slowing down \ changing the radius
of curvature. This will affect the period of motion of the
protons, which will in turn eventually disrupt the supply
of energy to the protons which will result in the protons
not escaping the cyclotron.
(c)
292
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TO EXERCISES
MOTION
MOTION
12.(a)
W
(b)
13.
14.(a)
A – negative
B – positive
(b)
B has the larger mass.
The displacement vertical is given by
Now this is smaller for B. Because the time of
SOLUTIONS TO CHAPTER 6
flight in the plates is the same for A and B,
the acceleration (a) for B is less because its downwards
displacement (s) is less. But each experiences the same
force ( F = Eq )
\ B has the bigger mass
(c)
Same time for each. No force in the horizontal direction
for each particle. \ both move at constant speed
horizontally \ t is the same. For both A & B.
(d)
As the magnitude of the charge is the same for both
A & B, and the electric field strength is constant.
F = Eq implies that the force on A & B is the
same in magnitude.
(e)
A has the greater acceleration.
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293
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
2.
The magnitude of the vector field is shown by the number
of lines per unit area. i.e. stronger field is shown by more
lines per unit area.
1.(a)
(b)
Direction by the arrows on the lines at a given point shows
the directional nature of the field at the point.
3.
(c)
SOLUTIONS TO CHAPTER 7
(iii)
Uniform field in the middle of the solenoid.
(lines nearly parallel)
(iv)
at A
at B
at C
(d)
4.
(a)
∼ 0.1N
(b)
I
(e)
∼ 0.7N
(c)
By right hand rule the force is perpendicularly out of the
page.
5.
294
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TO EXERCISES
MOTION
MOTION
6.(a)
9.(a)
F = BIΔ sinq
= 8 x 10-4 x 1.5 x Δ sin30
\ F = 8 x 10-4 x 1.5 x sin30 = 6.0 x 10-4Nm-1 Δe
(b)
On a 20cm section of wire
F = 6 x 10-4 x 0.2 = 1.2 x 10-4N
(b)
No force.
10.(a)
No force ( q = 0˚ )
(c)
(b)
No force ( q = 0˚ )
(c)
(d)
Force out of the page
11.
Vector addition of the permanent magnetic field B and
that generated by the current in the wire.
(b)
2
at A
F = 0.45 x 0.9 x 0.8 = 0.32N
(c)
2
at B
8.(a)
Out of the page
(b)
Into the page
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SOLUTIONS TO CHAPTER 7
7.(a)
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
13.(a)
at C
(b)
at D
(c)
SOLUTIONS TO CHAPTER 7
12.(a)
The cone is caused to vibrate at the same frequency as
the original sound being reproduced, and because the
cone is large it sets up sound waves of large amplitude
i.e. louder.
(d)
(b)
The initial current would cause the cone to move out/
in once producing a single pulse. Then no further
movement \ no further sound.
(c)
The cone would move in and out producing a
continuous wave of pulse \ sound of the frequency of
the switching.
(e)
(d)
A stronger magnet produces a stronger magnetic field
and as the force is proportional to the field strength a
larger force is produced.
(e)
In this way the coil current is always at 90º to the
magnetic field lines \ maximum force, at all times.
296
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TO EXERCISES
MOTION
MOTION
1.(a)
3.(a)
by a right hand rule the particle is negative
(b)
v
(d)
v
(e)
(d)
The magnetic force is always at 90º to the velocity of the
proton \ the proton moves in a circular path \ the force
acts as a centripetal force.
(f)
2.(a)
(b)
(c)
No force
(d)
No force
4.(a)
By right hand rule:
A is positive
C is negative
(b)
• Charge on C is bigger than charge on A
• Mass A is bigger than the mass of C
• Velocity of A could be bigger than C’s velocity
5.
Magnetic force F = Bqv sinq
If q = 90º then F = Bqv
But this force is a centripetal force because it is at 90º to
the charge’s velocity.
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SOLUTIONS TO CHAPTER 8
(c)
v
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
(e)
Moving at 90º to the field ensures the maximum force.
F = Bqv sinq\ if q = 90˚, F is a maximum.
6.(a)
5
8.(a)
(b)
(b)
(c)
Weight of alpha particle
(c)
SOLUTIONS TO CHAPTER 8
i.e. The magnetic force is about 1011 times the size of its
weight.
7.(a)
(d)(i)
i.e. T does not depend on diameter
\ no effect.
(b)
(ii)
(c)
9.(a)
Radius of the circular path in the magnetic field is
=
=
(d)
The time to traverse a semicircle
The electrons now move in a straight line – no force
outside the field \ obeys Newton’s First Law.
298
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TO EXERCISES
MOTION
MOTION
(b)
10.(a)(i)
(ii)
SOLUTIONS TO CHAPTER 8
(iii)
(b)
The proton gains kinetic energy as it moves across
the gap between the Dees. Its speed increases each time,
but in the magnetic
field the radius is
\ a bigger speed means a bigger radius.
(c)(i)
the energy gained by the proton is directly proportional
to the accelerating voltage.
ie. Energy
\ double ∆V, double the energy of the proton.
(ii)
No effect
(iii)
The number of times that the proton crosses the
electric field descreases. More energy per crossing,
results in bigger radius valves, which results in less
rotations before the proton emerges.
(iv)
No effect using the equation
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SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
1.
Using the wave equation
(c)
The rods of the receiving antenna must be orientated
horizontally. The electric fields of the E/M waves are
oscillating in a horizontal direction. \ they will cause
the electrons in the antenna to oscillate in a horizontal
direction.
(d)
The electrons in a receiving antenna will oscillate with the
same frequency as the electric field in the
E/M wave. \ The alternating potential difference
generated in the antenna will have the same frequency as
the wave \ f = 1.0MHz.
2.
8
8
SOLUTIONS TO CHAPTER 9
3.(a)
A wave is ‘plane polarised’ if the oscillations are all
confined to one plane.
(b)
Only transverse waves can be polarised.
(c)
The plane of polarisation of an electromagnetic wave is
the plane defined by the oscillation of the electric field
vector and the direction of travel of the wave.
(d)
When an electromagnetic wave is emitted by a dipole
antenna all the electric field vectors are oscillating in
the same direction as the rods of the antenna (as this is
the direction in which the electrons in the antenna are
oscillating). Thus all the oscillations of the vectors are
confined to one plane. The wave is polarised.
4.(a)
The plane of polarisation is horizontal. Electrons in
the rod of the antenna are oscillating in a horizontal
direction. Vectors are oscillating in a horizontal
direction.
(b)
f = 1.0MHz .The frequency of oscillation of the E vector
is equal to the frequency of oscillation of charges in the
rods, and equal to the frequency of oscillation of the
alternating potential difference applied to the rods.
300
(e)
The signal strength in the receiving antenna is greatest
when the rods are aligned in the direction of polarisation
of the E/M wave. If the country channels are polarised
perpendicularly to city channels, country viewers can
pick up their local channel (by orientating their antennae
appropriately) with minimal interference from the city
broadcasts.
5.(a)
-6
-6
(i)
The first beam travels a distance s = c x t
(ii)
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TO EXERCISES
MOTION
MOTION
5.(b)(i)
First beam travels a distance s = c x t
Extra distance travelled by the ‘second’ beam
∼ 51.3m
(ii)
Distance = speed x time
= 75 x 2.896 x 10-6
= 2.172 x 10-4m
= 0.22mm
SOLUTIONS TO CHAPTER 9
(c)(i)
(ii)
Distance travelled by ‘beam 1’, s = c x t1
height above the sea
6.
See notes and text.
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ESSENTIALS STAGE
STAGE 22
1.
AT Q
The path difference from sources
A and B to Q
= 7.05m − 3.70m
= 3.35m
= 2.5λ
∴waves will destructively interfere at Q
∴no sound is heard.
4.
S1
S2
f = 891 kHz = 8.91× 105 Hz
SOLUTIONS TO CHAPTER 10
Using c = f λ, λ =
∴λ=
c
f
3.00×108
= 336.7m
8.91× 105
Maximum amplitude
Path difference = mλ where m = 0,1, 2, 3, ....
The path difference between the two waves to the radio
= 2 × 252.5 = 505m
Minimum amplitude
Path difference = (m + 1⁄2) λ where m = 0,1, 2, 3,
Now,
2.(a)
The path difference to B from sources A and C
= 38m − 22m
=16m
= 4λ
∴ constructive interference at B.
(b)
The path difference to C from sources A and B.
= 60m − 22m
= 38m
=9
λ
∴ destructive interference at C.
3.
At P
The path difference from sources A and B
= 6.49m − 2.47m
= 4.02m
= 3λ
∴ sound will constructively interfere at P and a loud
sound will be heard.
302
505
=1
336.7
λ
∴ The direct wave, and the reflected wave will
destructively interfere.
∴ Poor reception.
5.
d =0.15mm=1.5×10-4m
λ = 540nm = 5.40×10-7m
L = 20cm = 0.20m
(a)
For the 3rd minimum
PD = d sinθ =
∴ sinθ =
∴ sinθ =
λ
λ
d
x 5.40 x 10-7
1.5 x 10-4
∴ θ = 0.52˚
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MOTION
MOTION
5.(b)
For the 5th minimum
d sinθ =5λ
8.
λ = 693nm = 6.93 x 10-7m
L=3.5m
B3 B2 B1 0 B1 B2 B3
∴ sinθ = 5×5.40 × 10-7
1.5×10-4
6∆y
∴ θ = 1.0˚
∴ 6∆y = 4.0cm
∴ ∆y = 0.667cm
\ ∆y = 6.67×10-3m, ∆y = 6.7 x 10-3m ( 2 sf )
(c)
λL
∆y =
d
5.40×10-7 × 0.2
=
1.5 × 10-4
(b)
∆y = λL
d
∴ d = λL
∆y
= 7.2 x 10-4m = 0.72mm
(d)
Distance will equal 4
fringe separations.
∴ d = 6.93 x 10-7 x 3.5
6.67 x 10-3
∴ d = 3.6 x 10-4m = 0.36m
O B1 B2 B3 B4 B5
∴d=4
× 0.72
6.
Blue, Green, Red
as λB < λG < λR.
λL
\ ∆y α λ \ The smallest λ will produce the
d
smallest fringe separation ]
[ ∆y =
7.
d = 3.4 x 10-4m
L = 0.30m
(a)
8∆y = 3.7mm
3.7x10-2
∴ ∆y =
= 4.6 x 10-4m
8
∴ ∆y α
1 ( a constant λ and L )
d
∴ as the separation d increases the distance between
adjacent fringes decreases.
i.e. the pattern “contracts”.
10.
λB = 440nm = 4.40 x 10-7m
λG =540nm = 5.40 x 10-7m
(a)
mλ = d sinq
( m = 1 )
1λ = d
sinq
5.40 x 10-7
∴d=
sin4.2°
∴
∴ d =7.37 x 10-6m
∴ d =7.4 x 10-6m
(b)
λL
∆y =
d
\λ=
d∆y
L
= 3.4×10-4 x 4.6 x 10-4
0.30
= 5.2×10-7m
= 520nm
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SOLUTIONS TO CHAPTER 10
∴ d = 3.2mm
9.
∆y = λL
d
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
10.(b)
For the second order green ( m = 2 )
2λ
d
2×5.4×10-7
=
7.37×10-6
\ sinq =
λ = 523nm=5.23 x 10-7m
\ θ = 8.43˚
(a)
For the 2nd order maximum
d sinθ = 2λ
For the second order blue ( m = 2 )
2λ
d
2 × 4.4 × 10-7
=
7.37 × 10-6
\ sinq =
11.
λ1 = 460nm = 4.60 x 10-7m
λ2 = ?
The 2 fringes in question have the same angular
position.
\ for λ1 d sinθ = 6λ1
\ for λ2 d sinθ = 3
7
λ
2 2
12
\ λ2 =
λ
7 1
\ 6λ1 =
\ sinθ = 2λ
d
∴ sinθ =
\ q = 6.86˚
\ angular separation = 1.6˚ ( 2.s.f )
SOLUTIONS TO CHAPTER 10
13.
d = 1 cm = 2.22 x10-6m
4500
λ2
as d sin q is
the same for both
\ λ2 = 7.89 x 10-9nm
\ l2 = 7.89nm
12.
If one slit is blocked we no longer see two- slit
interference. We will now see a single slit diffraction
pattern centred on the open slit.
2 x 5.23 x 10-7
2.22 x 10-6
∴ θ = 28.1˚
(b)
Maximum angular displacement = 90˚
∴ sin90 = mλ
d
1×d
\
=m
λ
\ m = 1 x 2.22 x 10-6
5.23×10-7
∴ m = 4.2
∴ see 4 orders on each side of the central max.
∴ 9 fringes
i.e. 9 beams.
14.(a)
d = 1 cm = 1.6667 x 10-6 m
6000
λ1 = 5.89×10-7m
λ2 = 5.896×10-7m
for
λ1 = 589.0nm
d sinq = 2λ
\ sinq = 2 x 5.89 x 10-7
1.6667 x 10-6
q1 = 44.97˚
for
λ2 = 589.6nm
q2 = 45.03˚
\ angular separation ∆Q = 0.06˚
(b)
Using the smallest λ = 589nm
Highest order maximum is
m = d = 1.6667 x 10-6 = 2.8
λ
5.89 x 10-7
\ 2nd order only.
304
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TO EXERCISES
MOTION
MOTION
15.
d=
1
cm=2.0 x 10-6m
5000
For the m = 3, θ = 40.6˚
\ d sinθ =3λ
d sinθ
3
2.0×10-6 x sin40.6˚
∴λ=
3
λ=
d sinθ = 3λ
d
3×4.75×10-7
∴ sinθ =
d
1.425×10-6
∴ sinθ =
d
∴ The 3rd order blue constructively interferes at a larger
angle than the 2nd Red.
∴ The 2nd and 3rd orders do not overlap.
∴ λ = 4.34×10-7m
∴ λ = 434nm
18.
λ = 6.328 x 10-7m
16.
d=
For the 3rd order Blue
d sinθ = 3λ
1
cm = 2.0 x 10-6m
5000
(a)
1.672
tanθ =
3.705
(a)
1.085
4.0
(b)
\ q = 15.2˚
(b)
mλ = d sinθ
\ λ = d sinθ
\ λ = 2 x 10-6 x sin 15.2
∴ λ = 5.24 x 10-7m
∴ λ = 524nm
SOLUTIONS TO CHAPTER 10
tanθ =
∴ θ = 24.29˚
(m=1)
17.
λ blue = 4.75 x 10-7m
λ red = 6.95 x 10-7m
1λ = d sinθ
1λ
sinθ
∴d=
6.328×10-7
∴ d = Sin 24.29
∴ d =1.5384×10-6m
1
1
d
1.5384×10-6
∴ number of lines =
=
∴ number of lines 6.50 x 105 /m
∴ number of lines 6500/cm
For the 2nd order red, d sinq = 2λ red
2λ
d
2×6.95×10-7
∴ sinθ =
d
1.39×10-6
∴ sinθ =
d
∴ sinθ =
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ESSENTIALS STAGE
STAGE 22
19.
d=
(d)(i)
1
= 3.175 x 10-6m
3.150 x 105m
For the 5th order
mλ = d sinθ (m=5)
∴ sinθ = 5λ
d
Maximum value of sinθ = 1
5λ
<1
d
d
∴λ<
5
∴ λ < 3.175 x 10-6
5
The distance measured is 22mm. This 22mm encompasses
8 fringe separations, ∆y.
∴ 22 = 8∆y
∴ ∆y =
22
= 2.75mm
8
* Note: it is more precise to measure the whole pattern
and divide, by 8 in this case, than to measure one∆y value.
∴
(ii)
∴ λ < 6.35 x 10-7m
∴ λ < 635nm
SOLUTIONS TO CHAPTER 10
ie. for all wavelengths less than 635nm, a 5th order will
be observed.
20.
The path difference for the 4th order is 4λ
• The incondescent globes are not
22.
Use the ruler on the photo to measure the distance
between a large number of constructuve interferes –
say 5 or 10.
• Multiple reflections occur off walls ∴ not a
The distance between 6 constructuve interferes is about
10mm.
• Incondescent globes are not coherent
monochromatic
double slit interference but a many slit pattern.
ie. not two sources but multiple sources.
• There is an infinite mix of patterns
• The fringe separation are too small
∴ ∆y =
10
= 2mm
5
(remember, 6 construcive interferes encompass 5 fringe
separations)
to distinguish.
21.(a)
Diffracted beams must be able to overlap to produce an
interference pattern. ∴ ‘L’ needs to be large.
Also ∆y =
λL
, then
d
∆y α L. ∴ too see a pattern and to measure a fringe
separation, the L needs to
be relatively large.
(b)
∆y α 1
d
∴ To observe a ‘fringe’ d needs to be small. Also, if ‘d’ is
small the coherent beams will diffract more and ∴ will
overlap over a bigger area ∴ more observable.
(c)
∆y α L ∴ if the screen is moved closer the
fringe separation will be smaller ∴ more difficult
measure.
306
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SOLUTIONS
PHYSICS
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PHYSICS
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- SOLUTIONS
OF PROJECTILE
CIRCULAR
TO EXERCISES
MOTION
MOTION
1.
E = hc = 6.63 x 10-34 x 3.00 x 108
5.90 x 10-7
λ
\ E = 3.37 x 10-19J
2.
E=
hc 6.63 x 10-34 x 3.0 x 108
=
0.91 x 10-7
λ
\ E = 2.2 x 10-18J
P= h
λ
6.63
x 10-34
=
0.91 x 10-7
P = 7.3 x 10-27sN
3.
P = E = 13.6 x 1.6 x 10-19
3 x 108
c
P = 7.25 x 10-27sN
P= h \λ= h
λ
P
6.63
x
10-34
\λ=
= 2.9 x 10-13m
2.3 x 10-21
c
c=fλ\f=
λ
3.00
x
108
\f=
2.9 x 10-13
f = 1.03 x 1021Hz = 1.0 x 1021Hz
E = hf = 6.63 x 10-34 x 1.03 x 1021
E = 6.9 x 10-13 J = 4.3MeV
5.
P = h = 6.63 x 10-34
λ
4.5 x 10-7
= 1.47 x 10-26sN
violet
λ = 750nm
λ = 400nm
Red p = h = 6.63 x 10-34
7.50 x 10-7
λ
p = 8.84 x 10-28sN
Violet p = h = 6.63 x 10-34
λ 4.00 x 10-7
p = 1.66 x 10-27sN
7.(a)
c=fλ\λ=
c
f
λ = 3.00 x 108 = 4.29 x 10-7m
7.00 x 1014
(b)
6.63 x 10-34
p= h =
4.29 x 10-7
λ
p = 1.55 x 10-27sN
(c)
Pi = Pf ( Law of conservation of linear momentum )
\ Pphoton = Pelectron
\ Pe = 1.55 x 10-27 = meVe
\ Ve = 1.55 x 10-27
9.11 x 10-31
\ Ve = 1.70 x 103 ms-1
8.
Intensity is proportional to the number of
photons passing through a given area.
i.e. higher intensity means more photons.
9.
60 watts = 60 Joules per second
\E = hf of red photon = hc
λ
6.63
x
10-34
x
3.00
x
108
\E=
7.00 x 10-7
\ E = 2.84 x 10-19 J
\ number of photons = total energy ÷ energy of photon
\ Δp = -2pi = 2 x 1.47 x 10-26
= 2.94 x 10-26 = 2.9 x 10-26sN ( 2sf )
away from the wall.
=
60
= 2.11 x 1020 per second
2.84 x 10-19
\ 2.1 x 1020 per second
≡ 2.1 x 1020 x 60 x 60 photon / hour
≡ 7.60 x 1023 photon / hour
10.
The photoelectric effect is the ejection of
electrons from a metal surface by light photons.
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307
SOLUTIONS TO CHAPTER 11
4.
6.
Red
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
11.
Minimum frequency photons that will eject
photo- electrons from a metal surface.
No – different metals have different f0 values.
(c)
Some ejected electrons lose kinetic energy due to
collisions with other atoms and electrons after they are
released by the incoming photons.
12.
Photo - electric equation
(d)
K = ΔVs q
K = hf - W = hc - W
λ
6.63
x
10-34 x 3 x 108 -W
\K=
3.81 x 10-7
\ ΔVs = 1.538 x 10-18 = 9.6 volts
1.6 x 10-19
[ W = 2. 1 x 1.6 x 10-14 = 3.36 x 10-19 J ]
\ K = 1.9 x 10-19J
\ electrons will be ejected ( i.e. K is positive )
13.(a)
\ f0 = W = 2.32 x 1.6 x 10-19
h
6.63 x 10-34
SOLUTIONS TO CHAPTER 11
\ f0 = 5.60 x 1014Hz
( 3sf )
(b)
K = hf - W = hc - W
λ
K = 5.1 x 1019 - 3.71 x 10-19
\ K = 1.4 x 10-19 J
14.
Maximum speed electrons will be produced by the
highest energy photons
i.e. violet ( l = 400nm )
\ K = hc - W
λ
= 4.97 x 10-19 - 2.32 x 1.6 x 10-19
= 1.26 x 10-19 J
\ 1.26 x 10-19 = 1/2mv2
\ 1.26 x 10-19 = 1/2 x 9.11 x 10-31v2
\ v = 5.26 x 105ms-1 ( 3sf )
16.
see text
17.
W = 2.3eV
\ W = 2.3 x 1.6 x 10-19
= 3.68 x 10-19 J
En of photon = hc
λ
6.63
x
10-34
x 3 x 108 = 3.43 x 10-19J
\ En =
5.8 x 10-7
Thus the photon energy is not enough to
overcome the work function \ electrons will
not be emitted.
18.(a)
Larger intensity increases the number of photons, all still
having the same energy; and if they are not energetic
enough to overcome the work function at low intensity,
they still will not be at high intensity.
(b)
Photo-electric effect is a one photon-one
electron interaction. Therefore if the intensity of the light
is greater, this will mean more photons \ more chance of
collisions with electrons \ assuming the frequency, more
photon- electron collisions will produce more released
electrons \ a greater photo-current.
19.(a)
Kmax
(J)
15.(a)
K = hf0 \ f0 = W
h
4.2
x
1.6
x 10-19
\ f0 =
6.63 x 10-34
\ f0 = 1.0 x 1015Hz
(b)
K = hc - W
λ
\ K = 6.63 x 10-19 x 3 x 108 - 4.2 x 1.6 x 10-19
0.9 x 10-7
(Threshold frequency).
f0
(work function)
K = 1.538 x 10-18 J
K = 1.5 x 10-18 J ( 2sf )
308
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TO EXERCISES
MOTION
MOTION
(b)
Slope = h = rise
run
[ 3.4 x 1.6 x 1019 ]
\h=
[ 12.2 - 6.5 ] x 1014
h = 9.5 x 10-34 Js
20.(a)
see text
(b)
\v=
\v=
2ΔVq
m
2 x 50,000 x 1.6 x 10-19
9.11 x 10-31
v = 1.3 x 108ms-1
(b)
• The density needs to be high so that the
incoming electrons are “slowed down” more
efficiently, which will result in a greater
number of photons in the X-ray region.
• The melting point needs to be high. This is
because most of the electron energy is lost as
heat which could then melt the target.
• The target should have a high atomic number.
(c)
Most of the electron energy is converted to heat by
target collisions \ target might melt if not cooled.
21.(a)
See text.
(b)
See text.
22.(a)
K = ΔVq
K = 60,000 x 1.6 x 10-19
K = 9.6 x 10-15 J
(b)
K = hfmax
\ f max = K = 9.6 x 10-15 = 1.447 x 1019Hz
h
6.63 x 10-34
\ f max = 1.4 x 1019Hz ( 2sf )
(c)
h
6.63 x 10-34
=
λ
( 3 x 108 ÷ 1.4 x 10-19
p = 3.1 x 10-23 SN
1015 x ΔVq 98
En
=
x
time
1
100
1015 x 8 x 10-15
98
P=
x
100
1
Power =
P = 7.84 Watts = 8W
(c)
Increase the current in the filament \ more electrons
emitted \ more target hits \ more X-rays photons.
24.
Energy of electron
K = ΔVq = ΔVe.
Now if all the electron’s energy is transferred
as a photon then
ΔVe = hfmax
\ fmax =
ΔVe
h
(b)
Straight lines as fmax a ΔV
(c)
Slope =
e
h
From the graph fmax a DV
\ as it is a straight line through the origin it is of the form
y = mx + c
\ fmax = slope x DV
e
but fmv = DV
h
e
\ slope =
h
25.(a)
‘Hard’ X-rays are X-rays with high penetrating power and
hence high photon energies and frequencies.
(b)
Hard X-rays are produced in X-ray tubes by high
accelerating voltages, i.e. 100,000V.
(c)
The degree of absorption of X-rays by body tissue is called
the attenuation of the X-rays.
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309
SOLUTIONS TO CHAPTER 11
This ensures the electrons knocked out of
the inner energy levels of the target atoms by
the incoming electrons will give out X-ray
frequency photons when they revert back to
the ground state. \ increasing the intensity of
the emitted X-rays.
p=
23.(a)
ΔVq = 1/2mv2
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
1.(a)
(b)
λ = h = 6.63 x 10-34
60 x 20
mv
λ= h
mv
λ = 6.63 x 10-34
9.11 x 10-31 x v
λ = 5.5 x 10-37m
But K = 1/2mv2
(b)
λ= h =
mv
6.63 x 10-34
1.675 x 10-27 x 1 x 106
λ = 4.0 x 10-13 m
(c)
6.0eV = 9.6 x 10-19J
\ K = 1/2 mv2 = 9.6 x 10-19
\v=
9.6 x 10-19 x 2
9.11 x 10-31
v = 1.45 x 106ms-1
\λ=
6.63 x 10-34
9.11 x 10-31 x 1.45 x 106
SOLUTIONS TO CHAPTER 12
λ = 5.0 x 10-10m
\v=
2K
m
\v=
2 x 3.2 x 10-15
9.11 x 10-31
\ λ = 8.7 x 10-12m
( 2sf )
(c)
λ= h
mv
\λa 1
v
\ as electrons accelerate v increases
\ λ decreases
4. (a)
You do! However, as your de Broglie wavelength is very
small your diffraction is difficult to detect.
2.(a)
a = Δv
Δt
\ a = 17.1 x 106
1 x 10-6
a = 1.7 x 1013ms-2
\ F = ma
= 9.11 x 10-31 x 1.7 x 1013
F = 1.6 x 10-17N
( 2sf )
(b)
λ1 = h
mv1
λ2 = h
mv2
λ1 = 3.83 x 10-11m
λ2 = 3.83 x 10-10m
\ Δλ = 3.4 x 10-10
( 2sf )
i.e. de broglie wavelength increases as it slows down
3.(a)
K = ΔVq
K = 20,000 x 1.6 x 10-19
K = 3.2 x 10-15J
(b) In a diffraction grating the slit width is smaller than in
a 2-slit setup. The ‘d’ value is closer to the wavelength of
light \ light diffracts more.
m λ = dsinq
\ sinq = λ for m = 1
d
\ as d decreases, sinq increases
\ q increases
\ The angular deviation of the orders increases
(c) Electrons have a similar wavelength to the crystal
spacings \ they show ‘large’ angular diffraction.
5.
Low energy electrons do not penetrate deeply into the
nickel surface \ only the interference of the beams
reflected off the top layer of the crystal
need be considered.
6.(a)
En = 54 x 1.6 x 10-19
= 8.64 x 10-18
= 8.6 x 10-18J
(b)
En = K = 1/2mv2
2K =
2 x 8.64 x 10-18
m
9.11 x 10-31
v = 4.4 x 106ms-1 (2sf )
\v=
310
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MOTION
MOTION
(c)
p = mv
= 9.11 x 10-31 x 4.3 x 106
p = 3.97 x 10-24sN
p = 4.0 x 10-24sN ( 2sf )
(d)
λ= h
mv
λ = 6.63 x 10-34
3.97 x 10-24
λ = 1.67 x 10-10m
λ = 1.7 x 10-10m
(e)
mλ
= d sin q ( m = 1 )
\1.67 x 10-10 = 2.2 x 10-10 sin q
\ q = 49.4˚
q = 49˚ ( 2sf )
(b)
Take electrons of
λ = 0.05nm
= 0.05 x 10-9
= 5 x 10-11m
\λ= h
p
h
\p= λ
\ p = 6.63 x 10-34
5 x 10-11
= 1.3 x 10-23sN
( 2sf )
(c)
p
p = mv \ v =
m
\ v = 1.3 x 10-23
9.11 x 10-31
v = 1.43 x 107ms-1
\ K = 1/2mv2 = ΔVq
(b)
λ= h
mv
at 100,000V
Velocity of electrons is found using
2K
m
2 x 1.6 x 10-14
\v=
9.11 x 10-31
v = 1.87 x 108ms-1
v=
\λ=
10.(a)
Electron wavelengths about 10 times smaller than this, i.e.
0.05nm.
SOLUTIONS TO CHAPTER 12
7.(a)
Range. En = ΔVq
\ En = 50,000 x 1.6 x 10-19
= 8.0 x 10-15J
\ En = 1.6 x 10-14J at 100,000V
\ The energies vary from
8.0 x 10-15J to 1.6 x 10-14J
(b)
enables smaller objects, less than 400nm, to be studied.
optical microscopes are limited to observing objects no
smaller than 10-7m i.e. in the visible light range. Using
electrons, objects similar to the wavelength of eletrons
(= 10-10 ) can be observed.
6.63 x 10-34
9.11 x 10-31 x 1.87 x 108
λ1 = 3.9 x 10-12m
at 50,000V, Velocity = 1.33 x 108ms-1
6.63 x 10-34
9.11 x 10-31 x 1.33 x 108
= 5.5 x 10-12m
\λ=
8.
The electrons are moving charges.
\ in electrics fields they will experience forces
F = Eq
In magnetic fields they will experience forces
F = Bqv sinq
\ velocities, directions, etc. can be altered.
1/2mv2
q
1/2 x 9.11 x 10-31 x (1.43 x 107 )2
\ ΔV =
1.6 x 10-19
\ ΔV =
\ ΔV = 582 volts 600v
11.
1λ = d sin q
\ λ = 2.0 x 10-10 sin 9˚
\ l = 3.13 x 10-11m = 3.1 x 10-11m (2sf)
Energy use l = h
p
\p= h
l
\ p = 6.63 x 10-34 ÷ 3.13 x 10-11
\ p = 2.12 x 10-23sN
\ velocity =
p
= 2.12 x
m
1.673 x 10-27
velocity = 1.27 x 104ms-1
\ K = 1/2mv2
= 1/2 x 1.673 x 10-27 x ( 1.27 x 104 )2
\ energy = 1.3 x 10-19 J ( 2sf )
9.(a)
better resolution
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311
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
(b)
neutron velocity would be
12.
90 eV = 90 x 1.6 x 10-19J
= 1.44 x 10-17 J
\ 1/2mv2 = 1.44 x 10-17
2 x 1.44 x 10-17
9.11 x 10-31
\v =
v = 5.6 x 106ms-1
p = mv
= 9.11 x 10-31 x 5.6 x 106
p = 5.1 x 10-24sN
momentum of neutron
p = mv
= 1.675 x 10-27 x 1.35 x 105
p = 2.26 x 10-22sN
\l= h
mv
6.63 x 10-34
l=
5.1 x 10-24
6.63 x 10-34
\l= h =
p
2.26 x 10-22
l = 1.3 x 10-10m
\ for the first order
1l = d sin q
1.3 x 10-10 = 0.306 x 10-9 x sin q
For the first order
1l = d sin q
\ = sin q =
SOLUTIONS TO CHAPTER 12
1.52 x 10-17 x 2
1.675 x 10-27
v = 1.35 x 105ms-1
v=
l = 2.93 x 10-12m
1.3 x 10-10
3.06 x 10-10
2.93 x 10-12
\ sin q = l =
3.0 x 10-10
d
q = 0.56˚
q = 25˚
13.(a)
95eV = 1.52 x 10-17 J
\ 1/2mv2 = 1.52 x 10-17
\v=
1.52 x 10-17 x 2
9.11 x 10-31
v = 5.77 x 106ms-1
\ p = mv
= 9.11 x 10-31 x 5.77 x 106
\ p = 5.26 x 10-24sN
\l= h
p
6.63 x 10-34
=
5.26 x 10-24
l = 1.26 x 10-10m
for the first order diffraction
1l = d sin q
\d=
1.26 x 10-10
sin 25˚
\ d = 3.0 x 10-10m
312
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CIRCULAR
TO EXERCISES
MOTION
MOTION
1.(a)
3.(a)
E1 = 3.61 - 3.19 = 0.42eV
E2 = 3.19 - 2.10 = 1.09eV
E3 = 3.61 - 2.10 = 1.51eV
hf = hc = E
l
\ l = hc
E
6.63 x 10-34 x 3 x 108
\ l1 =
0.42 x 1.6 x 10-19
= 2.96 x 10-6m
(b)(i)
The visible spectrum terminates at the first excited
state the smallest energy transition E3 → E2 will give the
largest l photon
\ lmax implies fmin implies Emin
Emin = 1·9eV
= 1.9 x 1.6 x 10-19J
= 3.04 x 10-19J
lmax = 6.54 x 10-7m
(ii)
ΔE = E∞ - E2
ΔE = 3.4eV
\ hc ΔE
l
\ l = hc
ΔE
6.63
x 10-34 x 3.0 x 108
\l=
3.4 x 1.6 x 10-19
= 1.14 x 10-6m
\ l3 = 0.42 x 2.96 x 10-6
1.51
= 8.23 x 10-7m
(b)(i)
Atoms may be raised to first or second excited states.
\ Emitted photons will have energy
3.19eV, 2.10eV, 1.09eV
(ii)
Scattered electrons will have energies of
3.60eV, 1.50eV, 0.41eV
\ l = 3.66 x 10-19m
\ l = 3.7 x 10-7m( 2sf )
2.(a)
A: 10eV
B: 12eV
C: 12.8eV
D: 2eV
(b)
ΔE = 2eV = 2 x 1.6 x 10-19J
\ ΔE = 3.2 x 10-19J
\ hf = 3.2 x 10-19
3.2 x 10-19
= 4.8 x 1014Hz
6.63 x 10-34
3.0 x 108
c
l=
=
= 6.3 x 10-7m
f
4.8 x 1014
\f=
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313
SOLUTIONS TO CHAPTER 13
\ hf = hc = 3.04 x 10-19J
l
hc
\l=
3.04 x 10-19
= 6.63 x 10-34 x 3 x 108
3.04 x 10-19
la 1
E
l2 E2
\
=
l1 E1
E
\ l2 = 1 x l1
E2
= 0.42 x 2.96 x 10-6
1.09
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
4.
The single electron of hydrogen may be raised to any
one of a large number of higher energy excited states.
And, when the hydrogen spectrum is observed even a
small sample of gas would contain a very large number
of atoms being excited at any time.
As these excited electrons revert to the ground state
a large number of different energy transitions are
possible, giving rise to a large number of spectral lines.
But remember, It is the number of possible energy
states that determines the number of lines of any atom
not the number of electrons.
5.(a)
E4 → E0, E4 → E1, E4 → E2, E4 → E3,
E3 → E0, E3 → E1, E3 → E2,
E2 → E0, E2 → E1,
E1 → E0
SOLUTIONS TO CHAPTER 13
\ 10 transitions are possible
\ 10 different photons could be emitted.
(b)
lmax implies fmin implies Emin = 0.31eV
[ E4 → E3 ]
hf = hc = E
l
\ l = hc
E
= 6.63 x 10-34 x 3 x 108
0.31 x 1.6 x 10-19
lmax = 4.01 x 10-6m
lmin = implies fmax implies Emax = 13.06eV
[ E4 → E0 ]
\ l = hc
E
= 6.63 x 10-34 x 3 x 108
13.06 x 1.6 x 10-19
lmin = 9.52 x 10-8m
6.(a)
Ionisation energy = 10.4eV
(d)
Fluorescence is the conversion of high energy photons
(absorbed) by atoms into low energy photons, when
re-emitted by the atoms.
Referring to the diagram, if an 8.8eV photon is incident on
the mercury atom it can be absorbed by the atom which is
raised to the third excited energy state.
When the atom reverts to the ground state, it may do so
directly emitting a photon of energy 8.8eV, or it may do so
in stages emitting a subset of the following photons.
2.1eV, 3.9eV, 4.9eV, 1.8eV, 6.7eV
(e) See above (d)
7.
As the energy levels of the hydrogen atom increases, the
difference in energy between energy levels decreases.
The Paschen series of lines consists of all transitions that
end in the second excited state. The maximum possible
energy of such a transition is 1.51eV
(this is the series limit).
The Balmer series of lines consists of all transitions that
finish in the first excited state. The minimum energy of
such a transition is 1.9eV and the maximum is 3.4eV.
The Balmer series consists of four lines in the visible and
the rest in the UV part of the spectrum.
The Paschen lines are all of less energy than the Balmer,
and are all in the infra red region of spectrum.
8.
Continuous range of frequencies is emitted by vibrating
atoms/molecules (i.e. charges) which emit electromagnetic
radiation of the same frequency as the frequency of
oscillation of the charge.
Typically, a continuous range of frequencies is emitted by
solids and liquids, when heated.
In a gas (vapour), the atoms are not vibrating along bonds.
They are moving in straight lines, colliding with each
other and with the sides of the container.
When these atoms are excited (e.g. by heating), they
can be raised to higher energy states. They emit their
radiation when the atom reverts to a lower energy state,
losing energy and emitting a photon of energy equal to the
energy difference between the states.
Because only a finite set of discrete energy states exist,
there can only be a discrete set of transitions, therefore a
discrete set of frequencies.
(b)
E3 → E2: ΔE = 1.8eV
(c)
Assuming mercury atoms are in the ground state,
photons removed will be
4.9eV, 6.7eV, 8.8eV
314
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CIRCULAR
TO EXERCISES
MOTION
MOTION
9.
Metallie atoms in a filament vibrate along the metalmetal bonds with a continuous range of frequencies.
A typical graph of the emitted photons at a given
temperature is shown below.
10.
A 200W globe uses more energy per second
than a 25W globe.
\ in 200W globes, atoms are vibrating with frequencies
in a higher range.
\ a 200W globe has a greater proportion of
the blue-violet frequencies of the visible spectrum than
the 25W.
\ it appears ‘whiter’.
Also, as the 200W globe uses wave energy it will appear
to be brighter than the 25W globe.
11.
The line emission spectrum is a set of bright coloured
lines on a dark background.
The line absorption spectrum is a continuous spectrum
of white light (ROYGBIV) with dark lines on this
coloured background.
The position (frequency) of the dark lines in the
absorption spectrum is the same as the position of the
bright lines in the emission spectrum.
13.
In a spectroscope we are actually observing reinforcement
images of the slit through which light enters the apparatus.
A fine slit therefore appears as a fine rectangle.
i.e. a ‘line’.
14.
l1 = 259nm
= 2.59 x 10-7m
f1 = c = 3.0 x 108
l1
2.59 x 10-7
SOLUTIONS TO CHAPTER 13
As the temperature increases, the graph shifts to
the right as the atoms vibrate with a higher range
of freqencies. As the filament globe heats up the
frequencies of vibration increase and so the mean
electromagnetic emission frequency increases and the
waves change from infra red → red (glows red hot).
As the temperature increases further the filament will
eventually emitting wave from red → violet.
i.e. it glows white hot.
12.
Absorption lines are caused by photons being absorbed
by the atoms of an element. The atom is raised to a higher
energy state.
The energy of the absorbed photon must be exactly the
same as the energy difference between the two states.
Balmer absorption lines require that the atom was being
raised from the first excited state to a higher state.
(as very nearly all)
But at room temperature the atoms are not in the first
excited state – all atoms are in the ground state.
\ no Balmer absorption lines are observed at room
temperature.
= 1.16 x 1015Hz
fa 1
l1
l
f2
\
= 1
l2
f1
i.e. f2 =
l1
x f1
l2
l2 = 254nm
\ f2 =
259 x 1.16 x 105
= 1.16 x 1015
254
= 1.18 x 1015Hz
l3 = 251nm
\ f3 =
259
x 1.16 x 1015
251
= 1.20 x 1015Hz
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315
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
15.
If an atom is in a meta-stable excited state, it can be
stimulated to revert to a lower energy state if a photon,
of energy equal to the energy difference between the
two states, is incident on it.
When the atom ‘drops’ to the lower energy state it emits
a photon with the same frequency, direction of travel
and phase as the intial incident photon.
For stimulated emission to occur the atom must have a
meta-stable state, for the electron must be in the excited
state to be stimulated by the photon.
SOLUTIONS TO CHAPTER 13
16.
Population inversion occurs when there are more
atoms, in a sample of material, in an excited state than
the ground state.
For this to occur the atom must have a meta- stable
excited state, so that the electrons can resicle in this
state for a time.
If it did not, then when individual atoms are excited
they immediately revert to the ground state.
- spontaneous emission
\ a population inversion could not occur.
17.(a)
The higher state is meta-stable
(b)
E photon = 1.96eV
(c)
Zero. When stimulated emission occurs the photons
are emitted with the same phase as the stimulating
photon.
(d)
hf = hc = ΔE
l
\ l = hc
ΔE
= 6.63 x 10-34 x 3 x 108
1.96 x 1.6 x 10-19
= 6.34 x 10-7m
316
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SOLUTIONS
PHYSICS
PHYSICS
PHYSICS
OF UNIFORM
- SOLUTIONS
OF PROJECTILE
CIRCULAR
TO EXERCISES
MOTION
MOTION
1.(a)
Nucleon: a generic term used to identify a nuclear
particle, i.e. either a proton or a neutron.
(b)
The atomic number (Z) of a nucleus is the number of
protons in that nucleus.
(c)
Mass number (A) of a nucleus is the total number of
nucleons in that nucleus (protons and neutrons).
2.(a)
Pb214
82
Atomic no. Z = 82
Mass no. A = 214
(b)
Pb215
82
Identical to Pb214 in all respects expect that it has one
more neutron in its nucleus.
Pb215 has the same chemical properties as Pb214, but
different physical properties (e.g. density).
Symbol
#proton
#neutron
#nucleon
Fe56
26
Ba141
56
O16
8
Np239
93
Li6
3
26
30
56
56
85
141
8
8
16
93
146
239
3
3
6
4.
An element is defined by the number of protons in its
nucleus (atomic number).
An isotope of an element has the same atomic number
but a different mass number.
i.e. different isotopes of the same element have the
same number of protons but have a different number of
neutrons.
(b)
O16
8 → 8p, 8n
O17 → 8p, 9n
8
(c)
Too many neutrons or too few neutrons.
There are certain combinations of protons and neutrons
that are stable.
If an isotope of an element has more neutrons than a
stable combination (or less) it will decay radioactively.
6.(a)
There is a stronger nucleon force holding
the nucleons together.
(b)
It is a very strong force (the strongest known force).
It is of very short range (≈ 10-15m). It does
not act over a distance greater than approximately the
diameter of a few nucleons.
It is generally attractive but over extremely small distances
it is repulsive.
It is charge independent. The force between two nucleons
is the same, no matter whether they are proton or neutron.
(c)
If there are greater than 83 protons the cummulative
repulsive force of these protons overcomes the short range
attractive force between nucleons.
7.
Mass H2 = 3.344 x 10-27kg
mp = 1.673 x 10-27kg
mn =1.675 x 10-27kg
\ mp + mn = 3.348 x 10-27kg
\ mass defect Δm = 0.004 x 10-27kg
= 4 x 10-30kg
E = mc2
= 4 x 10-30 x 9 x 1016
= 3.6 x 10-13J
= 3.6 x 10-13
1.6 x 10-19
= 2.25 x 106eV
= 2.25 MeV
8.
Mass Fe56 = 9.2860 x 10-26kg
26
(a)
mass of 26p = 26 x 1.673 x 10-27
= 4.3498 x 10-26kg
mass of 30n = 30 x 1.675 x 10-27
= 5.025 x 10-26kg
\ 26mp + 30mn = 9.3748 x 10-26kg
\ Δm = ( 9.3748 - 9.2860 ) x 10-26kg
= 8.88 x 10-28kg
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317
SOLUTIONS TO CHAPTER 14
3.
5.
Isotopes of an element are identical in their chemical
behaviour.
\ if radioactive isotopes of an element occur in
nature they will follow the same ecological/ chemical
environmental/foodchain path as their stable
counterparts.
\ they will remain together.
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
(b)
E = Δmc2
= 8.88 x 10-28 x 9 x 1016
= 7.992 x 10-11
(b)
Y9 mass no. = 9
4
atomic no. = 4
= 7.992 x 10-11 eV
1.6 x 10-19
= 4.995 x 108eV
= 500MeV
9.
H2 = Δm = 4 x 10kg (see question 7.)
1
H3 = 2mn + mp = 5.023 x 10-27kg
1
\ Δm = (5.023 - 5.0089) x 10-27
= 1.41 x 10-29kg
\ BE of H3 > BE of H2
1
1
as mass defect is greater.
SOLUTIONS TO CHAPTER 14
10.
The H2 nucleus loses mass as it forms from the proton
1
and the neutron. This loss of mass ( mass defect ) is
converted to energy - specifically a gamma photon.
11.(a)
Blinding energy of U238 = 1800MeV
92
= 1.80 x 109 x 1.6 x 10-19
= 2.88 x 10-10J
(b)
E = Δmc2
\Δm = E
c2
2.88
x 10-10
=
9 x 1016
(b)
Δm = ( 8.367 - 8.319 ) x 10-27kg
= 4.8 x 10-29kg
\ E = Δmc2
= 4.8 x 10-29 x 9 x 1016
= 4.32 x 10-12J
=
4.32 x 10-12J
1.6 x 10-19
= 2.7 x 109eV
= 27MeV
15.
Conservation of Mass Number ( or Nucleon Number )
Conservation of Charge ( i.e. Atomic Number )
Conservation of Total Mass/Energy
Conservation of Momentum
16.
Po210 → Pb206 + He4
84
82
2
= 3.2 x 10-27kg
12.(a)
He4 + N14 → O17 + H1
2
7
8
1
(a)
Before decay total momentum ( Pt )
pt = 0
\ By conservation of momentum
pt = 0 after the decay
\ pPb + pα = 0
\ pPb = -pα
\ Pb nucleus and the α particle must move off in
opposite directions.
(b)
H2 + H3 → He4 + n1
1
1
2
0
(c)
Ra226 → Rn222 + He4
88
86
2
(d)
Po210 → Pb206 + He4
84
82
2
(b) V
Pb
Vα
(e)
Al27 + He4 → P30 + n1
13
2
15
0
ISO 2.5
14.(a)
H2 + H3 → He4 + n1
1
1
2
0
mass of reactants on left hand side
= ( 3.344 + 5.023 ) x 10-27
= 8.367 x 10-27kg
mass of products on right hand side
= ( 6.644 + 1.675 ) x 10-27
= 8.319 x 10-27kg
\ Energy is released as the mass descreases.
=
mα
1
=
mPb 51.5
\ Vα : VPb = 51.5 : 1
(f)
n1 + N14 → C14 + H1
0
7
6
1
(c)
(g)
C12 + H1 → N13 + γ0
6
1
7
0
\ Kα : KPb = 51.5 : 1
13.(a)
X3 mass no. = 3
1
atomic no. = 1
318
Kα
m
51.5
= Pb =
1
KPb
mα
proton
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SOLUTIONS
PHYSICS
PHYSICS
PHYSICS
OF UNIFORM
- SOLUTIONS
OF PROJECTILE
CIRCULAR
TO EXERCISES
MOTION
MOTION
17.
In all neclear reactions, energy is either taken in or
given out therefore in all nuclear reactions, either mass
is converted to energy or energy is converted to matter.
( E = Δmc2 )
i.e mass is not conserved.
18.
n1 + N14 → C14 + H1
0
7
6
1
(a)
Mass of reactants on left hand side
= 2.3252 x 10-26 + 1.675 x 10-27
= 2.4927 x 10-26kg
Mass of products on right hand side
= 2.3252 x 10-26 + 1.673 x 10-27
= 2.4926 x 10-26kg
\ mass lost = 0.0001 x 10-26
= 1.0 x 10-30kg
\ energy is released.
=
9 x 10-14
1.6 x 10-19eV
= 5.63 x 105eV
19.
H2 + C12 → N13 + N1
1
6
7
0
(a)
The cyclotron can accelerate the deutrons ( H2 )
to energies in excess of 0.28MeV. This kinetic energy of
these particles is the energy input needed to cause this
reaction to proceed.
(b)
0.28MeV is just the energy needed to create the extra
mass of the products ( E = Δmc2 ).
But the deuteron is moving and momentum must be
conserved.
\ the products are moving and hence have
kinetic energy.
\ the extra 0.05MeV of energy must be added to
provide for the kinetic energy of the reactants.
E
c2
4.48 x 10-14
=
9 x 1016
But E = Δm =
Δm = 5.0 x 10-31kg
20.
H1 + O16 → N13 + He4
1
8
7
2
(a)
Mass of reactants on left hand side
= ( 1.673 x 10-27 ) + ( 2.65527 x 10-26 )
= 2.82346 x 10-26kg
Mass of products on right hand side
= ( 2.15900 x 10-26 ) + ( 6.64462 x 10-27 )
= 2.82346 x 10-26kg
\ energy absorbed
( mass products > mass reactants ).
(b)
\ Δm = 0.00089 x 10-26kg
= 8.9 x 10-30kg
\ E = Δmc2
= 8.9 x 10-30 x 9 x 1016
= 8.0 x 10-13J
=
8.0 x 10-13
1.6 x 1019
eV = 5.0 x 106eV
= 5MeV
21.
32P is produced in a nuclear reactor by placing
32S into the reactor where it is exposed to a very larger
number (flux) of neutrons.
Randomly, same neutrons will be captured by the 32S
nuclei and the following process produces the
radioactive isotope 32P .
15
1n + 32S → 32P + 1H
0
16
15
1
22.
Cyclotrons are used to accelerate protons to
very high speeds.
These high speed protons have enough energy to
overcome the Coulombic repulsion when they collide with
18O nuclei. The nuclear reaction that results produces the
8
radio isotope 18F.
9
1H + 18O → 18F + 1n
1
8
9
0
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319
SOLUTIONS TO CHAPTER 14
(b)
E = Δmc2
= 1 x 10 x 9 x 1016
= 9 x 10-14J
(c)
E = 0.28MeV
E = ( 2.8 x 105 ) x ( 1.6 x 10-19J )
= 4.48 x 10-14J
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
1.
As the number of protons increases the Coulumbic
repulsion between the protons increases, therefore
more neutrons, that can supply a nuclear force to
overcome the Coulumbic force, are required. The ratio
of neutrons to protons increases.
2.
•
•
•
•
Alpha decay – emits alpha particles.
Beta minus decay – emits an electron.
Beta plus decay – emits a positron.
Spontaneous fission.
3. (a), (b), (c)
(c)
The ratio of the energies is inverse to the ratio
of their masses.
mass ( Th ) > mass α
\ K α > KTh
Kα
M
378.5277 x 10
= Th =
Mα 6.644889 x 10-27
KTh
57.0
=
1
7.
The electrons come from a breakdown of a nucleon (i.e. a
proton or neutron).
(See b– and b+ decay).
SOLUTIONS TO CHAPTER 15
8.
When an unstable nucleus decays by a decay, the resulting
nucleus could be left in the ground state or excited states,
e.g.: [X → a + Y]
Y* are different excited states of Y
(d)
There are no stable nuclei above Z = 83
4.
Ra226 → He4 + Rn222
88
2
86
9.(a)
U → Pb206 each α decay reduces the mass number by 4
\ 238 - 206 = 32
\ 8 α decays.
5.
(a) X
(b) Z
(c) Y
6.(a)
mass U = 3.851816 x 10-25 kg
[ mass Th = 3.785277 x 10-25
+ mass He = 6.64489 x 10-25 ]
= 3.8517259 x 10-25 kg = mass products
\ loss of mass = Mu - M products
Δm = 0.00009 x 10-25
= 9.01 x 10-30 kg
(b)
E = Δmc2
= 9.01 x 10-30 ( 3.0 x 108 )2
= 8.11 x 10-13J
320
Thus the energy of the emitted alpha particles will be
different.
Therefore they will travel different distances in a cloud
chamber before they stop, therefore different length
tracks.
(b)
Atomic number drops by 10 - but 8α decays will
reduce it by 16. Now each b- decay increases the atomic
number by 1.
\ 6 b- decays must occur.
10.
Po216 → Pb212 + He4
84
82
2
\ α particle emitted.
11.
(a) Z reduces by 2
A reduces by 4
(b) Z increases by 1
A constant
(c) Z decreases by 1
A constant
(d) A and Z constant.
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SOLUTIONS
PHYSICS
PHYSICS
PHYSICS
OF UNIFORM
- SOLUTIONS
OF PROJECTILE
CIRCULAR
TO EXERCISES
MOTION
MOTION
12.
n1 + Th232 → Th233
0
90
90
bTh233 → Pa233 + e- + v
90
91
Pa233 → U233 + e- + v
91
92
13.
• Increasing your distance from the source.
• Reducing your exposure time.
• Shielding yourself from the source.
14.(a)
b-decay - it has an excess number of neutrons
O18 → F18 + e0 + v + γ
8
9
-1
(b)
+
b decay - too few neutons \ proton decay.
C10 → B10 + e0+ + v + γ
6
5
1
15.(a)
19.
A free proton cannot spontaneously convert into
a neutron.
i.e. P1 → n1 + e+ is not possible.
However in b+ decay the proton is bound within the
nucleus and it is the total mass of the reactants and products that we need to consider - not the mass of an individual nucleon.
20.
At start..........................12Bq
After 1 t1/2....................6Bq
After 2 t1/2....................3Bq
After 3 t1/2....................1.5Bq
\ 3 equal half life in 3 days implies t1/2 = 1 day
21.
300 decays per 2 minutes
\ activity = 300
120
= 2.5Bq
( i.e. 2.5 decays per second )
(b)
15.2 = 4 half-lives
\ amount remaining
= 20 = 20 = 1.25g
24
16
16.
11 hours
i.e. 11 hours ⇒ 1/2 left
22 hours ⇒ 1/4 left
\ 11 hours
17.
Half remaining after 1.5 seconds
\ t1/2 = 1.5 seconds
18.(a)
72 days = 3 half-lives
\ amount left
(b)
The activity 100, to the activity 50 takes about 60 seconds
\ t1/2 approx. = 60 seconds you would then need to take
other points from the graph and averages the results to
reduce the effect of random errors.
23.(a)
Activity 200Bq → 100Bq takes about 4 minutes
\ t1/2 = 4 minutes
(b)
The half-life is the time taken for half the remaining atoms
in a sample to decay.
= 3.2 x 1020
23
= 0.4 x 1020
= 4.0 x 1019 atoms
(b) An infinite time.
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321
SOLUTIONS TO CHAPTER 15
22.(a)
Take 2 Bq off each activity before plotting to account for
the background.
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
24.
Gamma photons have no mass or charge, therefore they
are not readily stopped by mechanical and electrostatic
collisions with other matter – Alpha particles are
heavier and have a charge \ interact with matter and
do not penetrate very far.
28.
SOLUTIONS TO CHAPTER 15
25.(a)
Positrons for the b+decay of a nucleus such as 18F travel
9
only a small distance in body material and they are
effectively stationary when they collide with electrons
in the body.
Hence, the intial total momentum of the e–/e+collision
is neary zero.
Therefore by the law of conservation of momentum
two photons of equal energy must result from the
annihilation of the mass to ensure that the final
momentum of the system is also zero. These two
photons must move off in opposite directions such that
the vector additon of the momentum of each photon
adds to zero.
(b)
The total mass of the e– and e+is twice the mass
of the e–.
\ Δm = 2me = 2 x 9.11 x 10-31
= 1.822 x 10-30kg
but the mass is converted to energy
\ E = Δmc2
= 1.822 x 10-30 x ( 3.00 x 108 )2
= 1.6398 x 10-13J
= 1.025 x 106eV
= 1025KeV
\ each photon has energy
1025 ÷ 2 = 512KeV
26.
• destroy dense areas of a body and bone
marrow, intestine.
• ionise atoms.
• cause acids to form that attack cells.
• produce cancer.
27.(a)
Activity = 10 x 0.226
= 2.26Bq
(b)
10g of old bone
Activity = 0.28Bq
this is about 3 half-life values.
\ Bones are about
3 x 5730 = 17,200 years old
322
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SOLUTIONS
PHYSICS
PHYSICS
PHYSICS
OF UNIFORM
- SOLUTIONS
OF PROJECTILE
CIRCULAR
TO EXERCISES
MOTION
MOTION
1.
Spontaneous nuclear fission is the process in which a
very large nucleus splits into two smaller nuclei.
6.
• Fission gives out huge amounts of energy for small
Induced fission is caused by the capture of a nucleon
(neutron) that causes instability and the eventual split
of the large nucleus (i.e. forced fission).
•
2.
See text.
•
3.
mass reactants:
n = 1.675 x 10-27kg
U = 3.9017 x 10-25kg
\ mass = 3.91845 x 10-25kg
\ E = Δm2
= 3.64 x 10-28 x ( 3 x 108 )2
= 3.276 x 10-11J
(b)
En of gamma photon 10% of this energy
\ En = 3.276 x 10-12J
\ E = hf
\ f = E = 3.276 x 10-12
h
6.63 x 10-34
= 4.9 x 1021Hz
4.
n1 + U235 → Sn132 + M101 + 3n1
0
92
50
42
0
Conserved:
(i) mass/energy
(ii) nucleons
(iii) charge
(iv) momentum.
5.
The fission of a U235 nucleus produces about 230MeV.
230MeV is equivalent to 3.7 × 10-11 joules.
This is about 107 times the amount of chemical energy
given out by the combustion of a molecule of methane.
•
7.
n1 + U235 → Xe140 + Sr94 + 2n + γ
0
92
54
38
n1 + U235 → Sn132 + Mo101 + 3n + γ
0
92
50
42
Could be different because of the energy of the neutrons.
If the neutron energies are different, different fissions are
likely.
\ it is important to control the energy of neutrons
in reactors.
8.(a)
All the nuclei have an excess of neutrons compared to
their stable isotopes.
\ they will decay by b– emission which is neutron decay.
(b)
[ 1n → 1p + e- + ϑ ]
0
1
\ 141Ba → 141La + e0 + V
56
57
-1
92Kr → 92Rb + e0 + V
36
37
-1
etc, etc.
9.(a)
If they have low energy neutrons are more likely
to be captured, therefore upsetting the n
p ratio and
promoting fission.
High energy neutrons are likely to collide and shatter
nuclei or reflect off.
(b)
The moderator slows the neutrons down to thermal
energies so that effective neutron capture can occur.
10.
When high energy neutrons collide with D2O molecules
they transfer more of their energy and momentum to the
D2O.
\ it only takes a few collisions between a neutron and
D2O molecules to reduce the neutron energy to a level
that will ensure capture.
Larger moderator molecules will cause the neutrons to
collide and ‘bounce off ’ with most of their energy and
momentum retained.
\ many more collisions are necessary to ensure
capture.
\ D2O is more effective as a moderator.
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323
SOLUTIONS TO CHAPTER 15
mass products:
Ba = 2.28922 x 10-25kg
Kr = 1.57534 x 10-25kg
3n = 5.025 x 10-27kg
\ mass = 3.91481 x 10-25kg
\ mass defect = mr - mp
= 3.64 x 10-28kg
•
amounts of mass compound with methane burnt
in oxygen for example.
Fossil fuels use too much mass and produce too
much waste (CO/CO2).
The amount of waste produced by fission is very
much less.
Fission does not produce harmful green-house
gases.
Fisson does have a problem caused by the radioactive waste products.
SOLUTIONS
PHYSICS
PHYSICS ESSENTIALS
ESSENTIALS STAGE
STAGE 22
11.
Naturally occurring uranium ore does not have enough
U235 to maintain a chain reaction.
\ the fuel is enriched to ensure that enough neutrons
can collide with U235 nuclei to promote fission and the
chain reation.
SOLUTIONS TO CHAPTER 16
12.
Mass reactants;
2m H21 = 3.34357 x 10-27
\ mass total = 6.68714 x 10-27kg
Mass products:
m He = 5.00683 x 10-27
m n = 1.675 x 10-27
\ mass defect Δm
= 0.00531 x 10
\ E = Δmc2
= 5.31 x 10 x ( 3 x 10 )
= 4.78 x 10-13J
13.
To fuse the nuclei we need to overcome the coulumbic
repulsion of the nuclei (positive charges). This is
usually done by giving the nuclei large kinetic energies
by subjecting them to very high temperature.
14.
E = ΔVq
\ E = 100,000 x 1.6 x 10-19
= 1.6 x 10-14J.
(b)
En = K = 1/2mv2
\ 1.6 x 10-14 = 1/2 x 3.34357 x 10-27 x v2
\ v2 = 1.6 x 10-14 x 2
3.34357 x 10-27
= 3.1 x 106ms-1
15.
Mass reactants:
m H2 = 3.34357 x 10-27
1
m H1 = 1.673 x 10-27
1
mass = 5.01657 x 10-27kg
m He3 = 5.00683 x 10-27kg
2
\ mass defect Δm
= 0.00974 x 10-27
= 9.74 x 10-30kg
\ maximum energy of the gamma photon
E = Δmc2
= 9.74 x 10-30 ( 3.0 x 108 )2
En = 8.766 x 10-13J
\ using E = Δmc2
3.456 x 1013 = Δm x ( 3 x 108 )2
\ Δm = 3.456 x 1013
( 3 x 108 )2
= 3.84 x 10-4kg
= 0.38kg
17.
(i) Energy generation
(ii) Manufacture of radio-isotopes
(iii) Nuclear research
18.(a)
Core: Contains the fuel rods and control rods where
fission occurs. Basically a group of fuel rods in a high flux
of neutrons where energy is produced.
(b)
Rods of enriched uranium: The rods contain about 4%
U235 - the rest being U238 . The U235 is the fuel that under
goes fission.
(c)
The moderator: D2O (heavy water) – reduces the energy
level and momentum of the neutrons so that capture can
occur to promote fission.
(d)
Control rods: Typically Boron and Cadmium. They
‘soak up’ neutrons so that the number of neutrons (slow)
colliding with the fuel can be controlled, therefore
controlling the chain reaction and hence the energy ouput.
19.
The energy output is controlled by moving the control
rods in and out of the core. The further the rods are
inserted the more neutrons are absorbed. Therefore, the
reaction chain is reduced and so is the energy. The reverse
gives more energy.
20.
n10 + Li63 → H31 + H31 + H11
He32 + He32 → He42 + H11 + H11
21.(a)
E = Δmc2
\ 4.3 x 10-12 = Δmc2
\ Δm = 4.3 x 1012
( 3 x 108 )2
\ Δm = 4.8 x 10-29kg
16.
400 Mega watts
= 400 x 106 watts
= 400 x 106 joules per sec.
= 400 x 106 x 60 x 60 x 24 joules per day
= 3.456 x 1013 joules per day
324
Essentials Workbook © Adelaide Tuition 2013. All rights reserved, copying of any pages is strictly prohibited by law.
SOLUTIONS
PHYSICS
PHYSICS
PHYSICS
OF UNIFORM
- SOLUTIONS
OF PROJECTILE
CIRCULAR
TO EXERCISES
MOTION
MOTION
(b)(i)
E = Δmc2
= 4 x 109 x ( 3 x 108 )2
\ E = 3.6 x 1026 joules per second
(ii)
Number of reactions
= 3.6 x 1026
4.3 x 10-12
= 8.4 x 1037 reactions per second
(iii)
total mass
mass lost per second
T = 2.0 x 1030
4 x 109
Time =
T = 5 x 1020 seconds
= 1.6 x 1013 years.
SOLUTIONS TO CHAPTER 16
Essentials Workbook © Adelaide Tuition 2010. All rights reserved, copying of any pages is strictly prohibited by law.
325