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G484
The Newtonian World
MODULE 1:
Newton’s Laws and Momentum
Answer Booklet
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Lesson 1 questions – Newton’s First Law
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1
Without looking at your notes, define (a) Newton’s First Law, (b) Newton’s
Second Law and (c) Newton’s Third Law.
(a) Newton’s first law states that an object will remain stationary or continue at
a constant velocity unless acted on by a resultant force. (1) If it is acted on by
a resultant force it will either accelerate, decelerate or change direction (1)
depending on the direction of the force.(1) … (3)
(b)… Newton’s Second Law says that an object will accelerate quicker the
harder you push it. (1) But if you push two different objects with the same
force, (1) the heavier one won’t accelerate as much. (1)
OR
F=ma
(each symbol must be defined) (3)………………………………………………… (3)
(c)… When body A exerts a force on body B,(1) then body B exerts on body A
a force that is equal, (1) opposite in direction and of the same type (1)…… (3)
2
Define the SI unit of force.
…………………………………………………………………………………………..
The force needed to accelerate an object of 1kg by 1ms-2.……………………..
……………………………………………………………………………………… (2)
3
The figure below shows a ball of 50g resting on the strings of a racket held
horizontally.
Contact Force
Weight
a)
b)
Draw and label the 2 forces acting on it.
Calculate the weight of the ball and give the units.
(2)
Weight = mass x gravitational field strength
0.05 x 10
Weight = ……0.5……………… unit……N(ewtons)…………. (3)
Explain, with reference to Newton’s Laws what we know about the forces
acting on the ball if it is at rest.
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
c)
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Lesson 2 questions – Momentum
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1. Define linear momentum and state its SI unit.
……… Momentum=massvelocity [1]
SI unit of momentum is kgms–1. [1]
…………………………………………………………………………………………
…………………………………………………………………………………………(
2)
2. A bumper car collides at right-angles with a metal barrier and rebounds at the same
speed. A student suggests that the change in momentum of the car is zero.
Explain why the student is wrong.
Momentum is a vector quantity – it has both direction and magnitude. [1]
If the initial momentum of the car is p, then its final momentum must be –p
(see diagram):
Change in momentum, Δp= final momentum– initial momentum
Δp=–p–p= –2p (the change is not zero) [1]
…………………………………………………………………………………………(
2)
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3. Work out the momentum changes in the following cases:
(a) a mass of 3 kg slows from 3 m/s to 1.5 m/s;
Momentum change = mv – mu =
………- 4.5………………. Kgm/s (2)
(b) a mass of 500 g accelerates from 4 m/s to 8 m/s;
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………2 ………………. Kgm/s (2)
(c) a ball of 50 g moving at 25 m/s is caught;
………-1.25 ………………. Kgm/s (2)
(d) a force of 4 N acts on an object for 12s;
……48 …………………. Kgm/s (2)
(e) a ball hits a wall at 4 m/s and bounces back at 3.5 m/s. Mass of ball is 2 kg;
………26.25 ………………. Kgm/s (2)
(f) a stone of mass 750 g hits a wall at 20 m/s and bounces back at 15 m/s;
Momentum change = 0.75(15 – (-20)) = 15 kgm/s or 15 Ns
(notice that the direction change has been allowed for by the use of another minus
sign)
………15………………. Kgm/s (2)
(g) a ball of mass 1.5 kg hits the ground at 80cm/s and bounces back at 65 cm/s.
……2.175 …………………. Kgm/s (2)
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Lesson 3 questions – Newton’s Second Law
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1
Fig 1.1 shows a car of mass 1200kg, pulling a caravan of mass 400kg along a
horizontal road.
Fig 1.2
Calculate, for the first 20s of the journey
i)
The acceleration of the car
Gradient = 13/20
acceleration = …0.65…..ms-2 (2)
ii)
the resultant force acting on the car
Force = ma
1200x0.65
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resultant force on the car = 780…N (2)
iii)
the resultant force acting on the caravan.
Force = ma
400x0.65
resultant force acting on the caravan = ……260…. N (2)
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2. A boy running at 8 m/s is stopped by a friend in 2s. If the mass of the boy is 40kg
what is the force on him during stopping?
40x8 = Fx2
F = 160N
………160………………. N (2)
3. A ball of mass 20 g hits a vertical wall with a horizontal velocity of 12 m/s. If it
rebounds with a velocity of 10 m/s in the opposite direction and is in contact with the
wall for 0.05 s find:
(a) the change in momentum of the ball;
0.02x22 = 0.44 Ns
…………0.44 ……………. Kgm/s (2)
(b) the force of the wall on the ball.
0.44/0.05 = 8.8 N
…………8.8 ……………. N (2)
4. A parachutist lands on the ground. If his mass is 65 kg and he hits the ground at 4
m/s and stops in 0.5 s find:
(a) his change of momentum;
65x4 = 260 Ns
………260 ………………. Kgm/s (2)
(b) the force of the ground on him.
260/0.5 = 520 N
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……… …520 ……………. N (2)
5. A second parachutist with the same mass lands with the same velocity but forgets
to bend his knees and stops in 0.01 s. Find his change of momentum and the force
of the ground on him.
260/0.01 = 26000 N!
…………260 ……………. Kgm/s (2)
………26000 ………………. N (2)
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6. F=ma is a special case of Newton’s Second Law for a body that has a constant
mass but changing velocity.
a. Define a change in linear momentum, Δp in terms of Δt and F.
…………………………………………………………………………………………
…………………… Δp = F Δt
…………………………………………………………………………………………
………………………………………………………………………… (1)
b. Define acceleration.
………………a= Δv/ Δt
………………………………………………………………………………… (1)
Show that F=ma is true for a body that has a constant mass but changing
velocity using your definitions from (a) and (b).
Since FΔt = Δp = Δ(mv);
And acceleration is the change in velocity (Δv) in a change in time (Δt)
or a = Δv/Δt;
If an object of constant mass has a changing velocity we can write that:
FΔt = m Δv
If we rearrange this to make F the focus of the equation we can get:
F = m Δv/Δt
We have said that Δv/Δt = a and so substituting this in we now get
F = ma which is a special case of Newton’s Law when mass remains constant and
velocity changes.
QED
(2)
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Lesson 4 questions - Newton’s third law
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1.
State Newton’s 3rd Law.
When body A exerts a force on body B, then body B exerts on body A a force
that is equal, opposite in direction and of the same type. (2)
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2.
When a golf ball is dropped to the pavement, it bounces back up. (a) Is a
force needed to make it bounce back up? (b) If so, what exerts the force?
a.
……………Yes…………………………………………………
……………………………………………………………………………………… (1)
b.
………The pavement’s contact force must be greater than the
weight of the ball in order for the ball to accelerate
vertically.………………………………………………………………..
……………………………………………………………………………………… (2)
3.
If you walk along a log floating on a lake, why does the log move in
the opposite direction?
……… Newton’s 3rd Law says that if I push the log away with a force it will push
back the same amount. The frictional force between your foot and the log is greater
than the frictional force between the water and the log and so the log moves in the
opposite direction to the dircetion I walk.………………………………… (2)
4.
When an object falls freely under the influence of gravity there is a net force
mg exerted on it by the Earth. Why doesn’t the Earth move?
…………The weight exeerted on me by the Earth is the same as the weight exerted
by me on the Earth. My inertia is less than the Earth’s and so I accelerate much more
than the Earth.……………………………………………………………..
……………………………………………………………………………………… (2)
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5.
A person exerts an upward force of 40 N to hold a bag of shopping. Describe the
“reaction” force (Newton’s third law) by stating (a) its magnitude, (b) its
direction, (c) on what object it is exerted, and (d) by what object it is exerted.
a) …………40N………………………………………………………
……………………………………………………………………………………… (1)
b) ……upwards……………………………………………………
…………………………………………………………………………………………..
……………………………………………………………………………………… (1)
c) ………Bag of shopping………………………………
……………………………………………………………………………………… (1)
d) …The Earth…………………………………………………………
……………………………………………………………………………………… (1)
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9. According to legend, the Russian mystic Nostradamus's horse once refused to
pull a cart after being hit on the head by an apple. The horse reasoned that,
according to Newton's third law, any force that he exerted would be balanced
by an equal and opposite force. The cart would thus never move, no matter
how hard the horse tried.
Explain why this horse is wrong.
…The cart feels the
unbalanced force applied by
the horse and therefore
accelerates if this force is
greater than the frictional force
of the ground on the cart. On
the other hand, the horse feels
an equal and opposite force applied by the cart but is able to accelerate
forward if the frictional force of the ground on the horse is greater… (3)
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Lesson 5 questions – Impulse
1
a)
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This question is about kicking a football.
The graph below shows how the force F applied to a ball varies with time t
when it is kicked horizontally. The ball is initially at rest.
i)
Use the graph to find
1
the maximum force applied to the ball
maximum force = ……48…………… N
2
ii)
the time the boot is in contact with the ball.
Time = ……0.25………….s (1) Both answers correct for 1 mark
What is the quantity called that the mean force multiplied by the time
of contact describes?
……………………Impulse (accept change in momentum)………………… (1)
iii)
How would you use the graph to estimate this quantity delivered to the
football?
…………… Estimating area under graph (1)…..
……………………………………………………………………………………… (1)
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iv)
b)
Use the graph to estimate this quantity delivered to the football.
Impulse = ………6.5 +/- 1 (1)……………. Ns (1)
The mass of the ball is 0.5kg. Use your answer to (a) to calculate
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i)
a=F/m (1)
the maximum acceleration of the ball
acceleration = ……96 (1)……………….ms-2 (2)
ii)
the final speed of the ball
Ft=mv (1)
v=a(ii)/0.5
=2a(ii) (1) ecf
Speed = ……13…………….. ms-1 (2)
iii)
the kinetic energy of the ball after the kick.
Ke=1/2 mv2
ke=1/2 x 0.5 x b(ii)
Kinetic energy = ……42………………….. J (2)
c)
The ball hits the wall with a speed of 14 ms-1. It rebounds from the wall along
its initial path with a speed of 8.0 ms-1. The impact lasts for 0.18s. Calculate the mean
force exerted by the ball on the wall.
Ft=mv-mu
Ft=0.5 (8-(-14))
F=11/0.18
Force = ……61.1……………. N (3)
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2
A 0.30 g fly moving at 1.5ms–1 is trapped by a spider’s web. The fly comes to
rest in a time of 0.40 s. Calculate the magnitude of:
a) the change in momentum of the fly, (give the units);
Δp= final momentum – initial momentum
Δp=0 – (0.30×10–3×1.5) (the final momentum is zero) [1]
Δp= 4.5×10–4 kgms–1 (magnitude only) [1]
change in momentum = ……4.5×10–4 …………….. units …… kgms–1 …[1]… (3)
b) the average force exerted by the web on the fly.
Newton’s second law: F = Δp / Δt [1]
F = 4.5×10–4/0.4 [1]
F = 1.13x10–3N [1]
Average force = ……1.1x10–3……………… N (3)
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3
a) State Newton’s second law.
…………………The rate of change of momentum of an object is directly proportional to the
net force acting on the object and takes place in the direction of this force. ………….
…………………………………………………………………………………… (1)
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b) A ball of mass 320 g hits the ground at right-angles at a speed of 15ms–1and
rebounds vertically at a speed of 7.0ms–1. The ball is in contact with the ground for a
time of 0.16 s.
i) Show that the change in momentum of the ball is about 7.0kgms–1.
Momentum, p = mass × velocity [1]
Δp = final momentum – initial momentum
Δp = (0.320 × –7.0) – (0.320 × 15) [1]
Δp = 7.04kgms–1  7.0kgms–1 (magnitude only)
(2)
ii) Calculate the average force on the ball during impact with the
ground.
Δp = 7.0kgms–1, Δt = 0.16 s
F = Δp/ Δt [1]; F=7.0/0.16 [1]
F  44N [1]
Average force = ………44………..N (3)
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Lesson 6 – 8 questions – Conservation of Momentum
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1. State the principle of conservation of momentum and explain what is meant by
elastic collisions and inelastic collisions.
……The momentum before a collision or explosion = the momentum after that
collision or explosion………………………………………………………………..
…………………………………………………………………………………………..
……In elastic collisions momentum and KE are conserved.………..
…………………………………………………………………………………………..
……In inelastic collisions momentum is conserved but the KE before the collision is
greater than afterwards.…………..
……………………………………………………………………………………… (3)
2 In the fission of a uranium 235 nucleus a neutron collides with the nucleus causing
it to break up. The particles formed are two smaller nuclei and three neutrons. Is the
linear momentum of the system conserved during the collision? Explain your answer.
…………………………………………………………………………………………..
Yes – momentum is conserved in all explosions and collisions as long
as all particles involved are taken into account
…………………………………………………………………………………………..
…………………………………………………………………………………………..
…………………………………………………………………………………………..
……………………………………………………………………………………… (2)
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3A ball of mass 210 g moving at a speed of 23ms–1 hits a wall at right-angles and
rebounds at the same speed. The ball is in contact with the wall for 0.31 s.
a) Calculate the change in momentum of the ball.
Δp=mΔv = 0.210 × (–23 – 23) (original direction taken as ‘positive’) [1]
Δp= –9.66kgms–1 ≈ –9.7kgms–1 [1]
(The minus implies that the force exerted by the wall on the ball is in the opposite
direction to its initial direction of travel.)
change in momentum = ……–9.7kgms–1 …………..(2)
b) Is the momentum of the ball conserved? Explain your answer.
The momentum of the ball itself is not conserved. [1]
The total momentum of the wall and the ball is conserved. The wall gains momentum
equal to 9.7kgms–1 but because it is massive its velocity is negligible. [1] ……… (2)
c) Calculate the magnitude of the average force acting on the ball.
F = Δp / Δt [1]
Δp= –9.66kgms–1, Δt = 0.31 s
F = –9.66 /0.31 (magnitude only) [1]; F ≈ 31N [1]
Average Force = ……31N …………….(3)
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4 A ball mass 2 kg moving left to right at 1.5 ms -1 collides with a ball of mass 3 kg
moving from right to left at 0.2 ms-1. After the collision the 2 kg ball moves from left
to right at 1 ms-1.
Draw a diagram.
Appropriate diagram drawn
Calculate:
(a) the change of momentum of the 2 kg ball
= m[v – u]
= 2[1 - 1.5]
= -1 Ns
Change in momentum = ………-1 Ns…………… (2)
(b) the momentum of the 3 kg ball after the collision
Momentum change
Momentum of 3 kg ball after collision
= -3x0.2 + 1
= 0.4 Ns
momentum = …………0.4 Ns ………… (2)
(c) the velocity of the 3 kg ball after the collision
Velocity of 3 kg ball = 0.4/3
= 0.13 ms-1
velocity = ………0.13 ms-1…………… (2)
5 A 1 kg ball of plasticene moving at 4 ms-1 collides with another ball of mass 2 kg
initially at rest. They stick together and move off.
Draw a diagram
Appropriate diagram drawn
Calculate:
(a) the total momentum of the system before the collision
Momentum
= 1x4 + 2x0
= 4 kgms-1
total momentum =………4 Ns ……………… (2)
(b) the velocity of the combined balls after impact
Velocity
= Momentum/mass
= 4/3
= 1.33 ms-1 in the direction that the 1 kg was moving before the
collision
velocity = ……1.33 ms-1 ………… (2)
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(c) the kinetic energy converted to other forms
Kinetic energy converted
= 0.5x1x16 – 0.5x3x1.77
= 8 – 2.65
kinetic energy = …5.35 J …………… (2)
6 A radium 226 nucleus decays by the emission of an alpha particle into a radon 222
nucleus. If the velocity of the alpha particle is 2x106 ms-1 what is the recoil velocity of
the nucleus?
222xv = 4x2x106
Therefore: v = 8x106/222
Recoil velocity = …3.6x104 ms-1……………… (2)
7The diagram shows two toy trains T and R held in place on a level track against the
force exerted by the compressed spring.
When the trains are released, R moves to the right at a speed of 3.8ms–1. The spring
takes 0.25 s to uncoil to its natural length. Calculate:
a) the velocity of train T;
Initial momentum = final momentum [1]
Moving towards the right is taken as the ‘positive’ direction.
0=(0.500×3.8)+(0.310×v) (v is the velocity of T) [1]
v = – 0.500×3.8 /0.310 (the minus sign means that T moves to the left) [1]
v = –6.13ms–1 ≈ –6.1ms–1 [1]
velocity = ……–6.1ms–1……… (4)
b) the average force exerted by the spring on each train.
F = Δp / Δt [1]
Δp= 0.500×3.8=1.9kgms–1, Δt =0.25 s [1]
F =1.9/0.25 [1];
average force = …7.6N …[1]………. (4)
8A 850kg cannon fires a 20kg shell at a velocity of 180ms–1.
a) Calculate the final momentum of the shell.
p=mv = 20×180 [1]
p= 3.6×103 kgms–1 [1]
final momentum = ……3.6×103 kgms–1 …………….(2)
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b) What is the magnitude of the momentum of the cannon immediately after
the shell is fired? (You may assume that the cannon is initially at rest.)
The momentum is conserved in this explosion. The momentum of the cannon is equal
in magnitude but opposite in direction to that of the shell. [1]
Momentum of the cannon= 3.6×103 kgms–1 [1]
magnitude of the momentum = ……3.6×103 kgms–1 ………………(2)
c) Calculate the recoil velocity V of the cannon.
Using the answer from b, we have:
850×V= 3.6×103 [1]
V = 3.6×103 / 850[1]; V≈4.2ms–1 [1]
recoil velocity = ……4.2ms–1 ………………[3]
9The diagram shows flour falling onto a horizontally moving conveyor.
The flour falls vertically onto the conveyor belt at a constant rate of 3.2kg s–1. The
conveyor belt is moving at a constant speed of 1.5ms–1. Calculate the horizontal force
required to keep the belt moving.
In a time interval of 1 s we have:
change in the horizontal momentum of the flour Δp= 3.2 × 1.5= 4.8kgms–1 [2]
Δt =1.0 s
Using Newton’s second law F Δp / Δt = [1]; so we have F =4.8/1.0 = 4.8 N [1]
Horizontal force = ……4.8 N …………(4)
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10 stationary radioactive nucleus of mass M ejects an alpha particle of mass m at a
speed of 2.0x107ms–1. Given M = 55m, calculate the kinetic energy of the alpha
particle as a percentage of the final total kinetic energy.
kinetic energy of the alpha particle = ……98………..% [6]
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