#### Transcript M1.4 Dynamics

```AS-Level Maths:
Mechanics 1
for Edexcel
M1.4 Dynamics 1
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Momentum
Contents
Newton’s laws of motion
Applying Newton’s laws
Momentum
Impulse
The principle of conservation of momentum
Examination-style questions
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Newton’s First Law of Motion
Newton’s First Law of Motion states that:
An object will remain at rest or move in a
straight line with a constant speed unless
acted on by an external force.
For example, the body in the following diagram is moving with
a constant velocity of 2 ms–1.
R
2 ms–1
F
D
W
There is no acceleration and so the forces are in equilibrium.
So,
F=D
and
R=W
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Newton’s Second Law of Motion
Newton’s Second Law of Motion states that:
The rate of change of momentum of a body is
proportional to the force applied to that body
and in the direction of the force.
Hence for a constant mass:
F = d  mv 
dt
F = m dv
dt
F = ma
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Newton’s Second Law example 1
Find the acceleration produced by a resultant force of 6 N
acting on a particle of mass 1.5 kg.
Applying Newton’s Second Law:
F = ma
6 = 1.5a
a=4
Therefore the acceleration produced is 4 ms–2.
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Newton’s Second Law example 2
A particle of mass 3 kg is accelerating at 4 ms–2. Find the
resultant force that produces this acceleration.
Applying Newton’s Second Law:
F = ma
F=3×4
 F = 12
Therefore the resultant force is 12 N.
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Newton’s Second Law example 3
Find the magnitude of the resultant force acting on a particle
of mass 2 kg needed to produce an acceleration of
(3i + 4j)ms–2.
Applying Newton’s Second Law:
F = ma
F = 2 × (3i + 4j)
= 6i + 8j
The magnitude of the force is 62 + 82 = 10 N.
Therefore the magnitude of the resultant force is 10 N.
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Newton’s Third Law of Motion
Newton’s Third Law of Motion states that:
For every action there is an equal and
opposite reaction.
For example when two surfaces are pressed against one
another then a normal reaction force acts between the two
surfaces.
This force acts perpendicular to the area of the surfaces in
contact.
Normal reaction force
Weight
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Momentum
Contents
Newton’s laws of motion
Applying Newton’s laws
Momentum
Impulse
The principle of conservation of momentum
Examination-style questions
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Question 1
A car is travelling on a straight horizontal road. It passes
point A with a speed of 15 ms–1, and 5 seconds later passes
point B with a speed of 20 ms–1.
The car has a mass of 1200 kg including the driver.
Assume acceleration is constant.
Find the driving force, D, if the resistance to motion is 500 N.
500 N
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1200 kg
D
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Question 1 Solution
Acceleration = 20 15 = 1 ms–2
5
-
Applying Newton’s Second Law:

5000 – R = 2050 × 2
D = 1200 + 500
D
= 1700
Therefore the driving force of the car is 1700 N.
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Question 2
A truck is travelling on a straight horizontal road. It passes
point A with speed of 10 ms–1, and 5 seconds later passes
point B with a speed of 20 ms–1.
The truck has a mass of 2050 kg including the driver.
Assume acceleration is constant.
Find the resistive force, R, to the motion if the truck produces
a driving force of 5000 N.
R
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2050 kg
5000 N
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Question 2 Solution
Acceleration = 20  10 = 2 ms–2
5
Applying Newton’s Second Law:

5000 – R = 2050 × 2
R = 5000 – 4100
R = 900
Therefore the resistive force is 900 N.
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Question 3
A car of mass 1450 kg including the driver is travelling along
a straight road which is inclined at an angle of 4° to the
horizontal.
The car is travelling at a constant speed uphill and is subject
to a resistive force of 850 N.
a) Draw a diagram to show all the forces acting on the car.
b) Calculate the driving force of the car.
a)
a=0
R
D
850 N
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4°
1450g
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Question 3 Solution
b) As the car is travelling at a constant speed, acceleration = 0.
The component of the car’s weight acting along the plane
is 1450g cos86°.
So the net force driving the car is D – 1450g cos86°.
Applying Newton’s Second Law along the plane,
D – 850 – 1450g cos86° = 0
D = 850 + 1450g cos86°
D = 1841 (to 4 s.f.)
Therefore the driving force of the car is 1841 N.
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Question 4
A truck of mass 2000 kg is travelling along a straight road
inclined at an angle θ to the horizontal. The truck is travelling
at a constant speed and is experiencing a resistive force of
1000 N. The driving force of the truck is 4750 N.
Calculate the value of θ.
R
a=0
4750 N
1000 N
θ
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2000g
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Question 4 Solution
Applying Newton’s Second Law up the plane:
4750 – 1000 – 2000g cos(90 – θ)° = 0
3750 = 19600 sinθ
sinθ = 3750 ÷ 19600
 θ = 11.0° (to 3 s.f.)
Therefore the road is inclined at an angle of 11.0° to the
horizontal.
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Question 5
A lift is accelerating upwards at 2 ms–2. A woman of mass 60
kg is standing in the lift.
Find the normal contact force between the woman and the
floor of the lift.
Applying Newton’s Second Law: 
R – 60g = 60 × 2
R
60g
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a = 2 ms–2
R = 60g + 120
R = 708
Therefore the normal contact force
is 708 N.
© Boardworks Ltd 2005
Question 6
A block of mass 3 kg is being pulled up a rough plane
inclined at an angle of 20° to the horizontal by a force of
20 N parallel to the plane.
The block is accelerating at a rate of 2 ms–2. Friction acts as
a resistive force.
Find the frictional force.
a = 2 ms–2
R
20 N
F
20°
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3g
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Question 6 Solution
Applying Newton’s Second Law up the plane,
20 – F – 3
F
g
c
o
s
7
0
°
=3
= 20
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Momentum
When an object is in motion it is said to have momentum.
The momentum of an object is given by the product of its
mass and its velocity.
momentum = mass × velocity
Momentum is a vector quantity since it depends upon the
velocity of the particle. We can therefore write the above
formula as
p = mv
A stationary particle has no velocity and so its momentum
is zero.
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Momentum
Multiplying mass and velocity gives the units for momentum
as kilogram metres per second (kg ms–1).
Momentum can also be given in Newton seconds (Ns).
Newtons, the basic units of force, are given by N = kg ms–2,
so
Newton seconds = N × s
= kg ms–2 × s
= kg ms–1
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Calculating momentum 1
A truck of mass 5 kg is travelling at 10 ms–1. Calculate the
momentum of the truck and the direction in which it is acting.
10 ms–1
Momentum = mass × velocity
= 5 × 10 Ns
5 kg
= 50 Ns
The momentum of the truck is acting in the direction of
motion.
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Calculating momentum 2
Calculate the momentum of a bullet of mass 4 g travelling at
750 ms–1.
Mass of the bullet = 4 g = 0.004 kg
Momentum = mass × velocity
= 0.004 × 750 Ns
= 3 Ns
Again, the momentum is acting in the direction of the
motion.
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Momentum
Contents
Newton’s laws of motion
Applying Newton’s laws
Momentum
Impulse
The principle of conservation of momentum
Examination-style questions
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Impulse
The impulse on a body is defined as its change in momentum.
Impulse = change in momentum
= mv – mu
where u is the initial velocity and v is the final velocity.
Impulse is denoted by the vector I.
Note that the impulse I that a body A exerts on a body B
is equal to the magnitude of the impulse that B exerts on
A but in the opposite direction.
Since impulse is change in momentum it is also measured in
Newton seconds.
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Calculating impulse 1
A ball of mass 0.5 kg hits the floor with a speed of 8 ms–1. It
rebounds with a speed of 6ms–1. Find the impulse exerted by
the floor on the ball.
Taking the upwards direction to be
positive:
Impulse (in Newton seconds)
= (0.5 × 6) – (0.5 × –8)
=3+4
=7
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Calculating impulse 2
A ball of mass 0.2 kg falls into a lake. Its velocity before
impact with the water was 15 ms–1 and after impact with the
water 10 ms–1.
Calculate the impulse exerted on the ball by the water.
Taking the downwards direction to be positive:
Impulse = 0.2 × 10 – 0.2 × 15
=2–3
= –1 Ns
Therefore the impulse exerted on the ball by the water is 1 Ns
in an upwards direction.
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Momentum
Contents
Newton’s laws of motion
Applying Newton’s laws
Momentum
Impulse
The principle of conservation of momentum
Examination-style questions
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The principle of conservation of momentum
The principle of conservation of linear momentum states
that if two objects moving in a straight line collide, the total
momentum before the collision is equal to the total momentum
after the collision.
We can prove this as follows:
Let the forces acting on two particles of masses m1 and m2
in an isolated system be F1(t) and F2(t) respectively, where
the forces are variable and functions of time.
Note that in an isolated system there are no external forces
such as air resistance and friction.
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The principle of conservation of momentum
By Newton’s Third Law:
F1 = –F2
 F1 + F2 = 0
From Newton’s second Law:
m1a1 + m2a2= 0
Integrating with respect to time we get:
m1v1 + m2v2= constant
Therefore, the total momentum of the system is conserved.
In general, for two particles of mass m1 and m2 we have,
m1u1 + m2u2 = m1v1 + m2v2
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Modelling collisions
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Conservation of momentum 1
Two smooth spheres A and B are projected towards each
other with speeds of 5 ms–1 and 7 ms–1 respectively. Sphere
A has mass 0.3 kg and sphere B has mass 0.4 kg. After they
collide sphere A rebounds with a speed of 6 ms–1. Calculate
the speed with which sphere B rebounds.
Taking the positive direction of motion to be from left to right:
(5 × 0.3) + (–7 × 0.4) = (–6 × 0.3) + (v × 0.4)
1.5 – 2.8 = –1.8 + 4v
5 ms–1
7 ms–1
4v = 0.5
Before impact
0.3 kg
0.4 kg
v = 0.125
A
B
Therefore, sphere B
v ms–1
6 ms–1
moves to the right with a
speed of 0.125 ms–1.
After impact
0.3 kg
0.4 kg
A
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B
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Conservation of momentum 2
A particle A of mass 0.3 kg collides directly with a particle B
of mass 0.1 kg. Immediately before the collision A has a
velocity of 5 ms–1 and B is at rest. After the collision the
particles coalesce into a single particle C and it continues to
move in the same direction as A.
Find the velocity of C immediately after the collision.
Total momentum before collision = total momentum after
0.3 × 5 = 0.4 × v
0.4v = 1.5

v = 3.75
Therefore C moves with velocity 3.75 ms–1 immediately after
the collision.
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Impulsive tension (jerk) in a string
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Jerk in a string
Two smooth spheres A and B of mass 2 kg and 3 kg
respectively are at rest on a smooth, horizontal surface. They
are connected by a light, inextensible string which is slack.
A is projected directly away from B with a speed of 5 ms–1.
Find their speed when the string becomes taut.
Before Jerk
5 ms–1
0 ms–1
2 kg
3 kg
A
Let the speed when the
string is taut be v ms–1
(2 × 5) + (3 × 0) = 5v
B
5v = 10
After Jerk
A
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v ms–1
v ms–1
2 kg
3 kg
B
v=2
The spheres move to the
left with a speed of 2 ms–1.
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Examination-style questions
Contents
Newton’s laws of motion
Applying Newton’s laws
Momentum
Impulse
The principle of conservation of momentum
Examination-style questions
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Examination-style question 1
A car is travelling along a straight road. Initially it is travelling
at a constant speed on a horizontal section of the road.
The mass of the car and the driver is 850 kg and the car is
exerting a driving force of 1100 N.
During the second stage of the journey the road inclines at
an angle of 3° to the horizontal. Find the driving force that the
car must exert if it is to accelerate at a rate of 0.5 ms–2.
Assume that the resistive force remains constant through the
whole motion.
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Solution 1
By Newton’s First Law the resistive force acting on the car
must be 1100N.
a = 0.5 ms–2
R
D
1100 N
3°
850g
Applying Newton’s Second Law up the plane,
D – 1100 – 850g cos87° = 850 × 0.5
D = 1960 (to 3 s.f.)
Therefore the driving force required to generate an
acceleration of 0.5 ms–2 is 1960 N.
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Examination-style question 2
A particle A collides with a particle B.
A has a mass of 0.3 kg and is moving with a velocity of
 1
 3 
–1
–1
ms
immediately
before
the
collision
and
ms
 2
 4 
 
 
immediately after the collision.
a) Find the change in the momentum of A due to this
collision and state the change in the momentum of B.
The mass of B is 0.2 kg and immediately before the collision
 1  –1
it has a velocity of   ms .
6
b) Find the velocity of B immediately after the collision.
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Solution 2
 1
3
a) Change in momentum of A = 0.3    0.3  
 2
 4 
 0.6 
=

1
.
8


 0.6 
Therefore the change in momentum of B is 
Ns.

 1.8 
b) For B,
 v1 
 1   0.6 
0.2    0.2   = 

v
6

1
.
8




 2
 v1 = 4 and v2 = 3
Therefore the velocity of B immediately after the collision is
 4
–1
 3  ms .
 
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Examination-style question 3
Two particles A and B are moving directly towards each other
when they collide. A is of mass 0.4 kg and B is of mass 0.8
kg.
Immediately before the collision A is moving with a speed of
5 ms–1 and B is moving with a speed of 4 ms –1.
Immediately after the collision the particles move away from
each other with speeds of a ms–1 and b ms–1 respectively.
a) Show that a – 2b = 3
b) After the collision A moves 7 m in 2 seconds. Find the
values of a and b.
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Solution 3
A
A
5 ms–1
4 ms–1
0.4 kg
0.8 kg
B
a ms–1
b ms–1
0.4 kg
0.8 kg
B
a) Taking left to right the be the
positive direction, by
conservation of momentum,
0.4 × 5 – 0.8 × 4 = –0.4 × a + 0.8 × b
2 – 3.2 = –0.4a + 0.8b
4a – 8b = 12
a – 2b = 3
b) A moves 7 m in 2 seconds  a = 3.5
 3.5 – 2b = 3
 2b = 0.5
 b = 0.25
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Examination-style question 4
Two trucks A and B, moving in opposite directions on the
same horizontal track, collide. The mass of A is 800 kg and
the mass of B is m kg. Immediately before the collision the
speed of A is 5 ms–1 and the speed of B is 4 ms–1.
Immediately after the collision the trucks are joined together
and move with the same speed, 1 ms–1. The direction of
motion of A is unchanged by the collision. Find
a) the mass of truck B
b) the magnitude of the
impulse exerted on A in the
collision.
c) State any assumptions that
you have made.
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Before
A
5 ms–1
4 ms–1
800 kg
m kg
B
1 ms–1
After
(800 + m) kg
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Solution 4
a) Using Conservation of Momentum,
5 × 800 – 4m = 800 + m
4000 – 4m = 800 + m
3200 = 5m
m = 640
 mass of truck B is 640 kg
b) Impulse = change in momentum,
I = 800 × 1 – 800 × 5
= 800 – 4000
= –3200
 the magnitude of the impulse exerted on A is 3200 Ns.
c) The trucks are modelled as particles and there are no
external forces, such as friction, acting.
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Examination-style question 5
Two particles A and B have masses m kg and 2 kg
respectively. The particles are moving directly towards each
other on a smooth horizontal surface when they collide.
Immediately before the collision the speed of A is 6 ms–1 and
the speed of B is 2 ms–1.
Given that the impulse exerted on B by A is 10 Ns,
a) find the speed of B immediately after the collision.
b) If the speed of A immediately after the collision is 2 ms–1 in
the same direction as before the collision, calculate the
value of m.
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Solution 5
6 ms–1
A
m kg
2 ms–1
B
2 kg
2v + 4 = 10
v ms–1
B
6 ms–1
A
A
m kg
2 kg
2 ms–1
B
2 kg
2 ms–1
3 ms–1
m kg
2 kg
B
a) Taking left to right to be the positive
direction, change in momentum of B is
 2v = 60

v = 30
Therefore the speed of B after the
collision is 3 ms–1.
b) Taking left to right to be the positive
direction, by Conservation of Momentum
6m – 4 = 2m + 6
 4m = 10
Therefore the mass of A is 2.5kg.
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