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By: Bryon Long
Table of Contents
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Introduction to Dynamics
Newton’s First Law of Motion
Newton’s Second Law of Motion
Law of Universal Gravitation
Gravitational Field Strength
Normal Force
Frictional Force
Elastic Force – Hooke’s Law
Newton’s Third Law of Motion
Momentum
Impulse
Elastic Collision
Inelastic Collision
Explosions
Conclusion
References
Formulae
Introduction to Dynamics
Dynamics is the study of forces.
We will be focusing on the forces that explains
why objects move.
Sir Isaac Newton formulated three laws of
motion and the universal law of gravitation in
1665.
These laws are the foundation of Dynamics.
Newton’s First Law of Motion
States that: an object will remain at
constant velocity unless acted on by an
unbalanced force.
Also known as the Law of Inertia.
Inertia is mass dependent.
The larger the mass, the more resistant the
object is to the change in motion.
Newton’s First Law of Motion
Forces are not needed for an object to
maintain constant velocity.
Unbalanced forces cause the object to slow
down and come to a rest.
For example - when the platform an object
is traveling on suddenly stops, the object
will continue to travel in the direction of
motion until an unbalanced force such as a
brick wall stops it.
Newton’s First Law of Motion
Object is traveling 100 m/s
Newton’s First Law of Motion
Object is traveling 100 m/s
Newton’s First Law of Motion
Object is traveling 100 m/s
Newton’s First Law of Motion
Object is traveling 100 m/s
Newton’s First Law of Motion
Object is traveling 100 m/s
Newton’s First Law of Motion
Object is traveling 100 m/s
Newton’s First Law of Motion
Object is traveling 0 m/s
Newton’s Second Law of Motion
States that: the rate of change in an object’s
velocity is directly proportional to the net
force, and inversely proportional to the
object’s mass.
The net force is dependent on the change
of velocity (acceleration) and mass of the
object.
Newton’s Second Law of Motion
F = ma
F = Force
m = mass
a = acceleration
The unit of force is newton (N)
Newton’s Second Law of Motion
What is the force of a 15 kg object
travelling with an acceleration of 2.5 m/s².
F=?
F
= ma
m = 15 kg
= (15 kg)(2.5 m/s²)
a = 2.5 m/s²
= 37.5 N
Newton’s Law of Universal Gravitation
States that: the gravitational force between two
masses is directly proportional to the product of
their masses.
Gm m
Fg


r
Unit is in Newton (N)
1
2
2
Newton’s Law of Universal Gravitation
Two objects are 2.20 m apart. One object has a
mass of 20.0 kg and the other has a mass of 12.0
kg. What is the gravitational force between them?
Gm1 m2
Fg 
r2
2


N

m
11
 6.67  10
(20.0kg)(12.0kg)
2
kg 


(2.20m) 2
 3.31  10 9 N
Gravitational Field Strength
Invisible fields that surround masses.
Can be scalar or vector.
Fg
Formula: g  m
Gravitational Field Strength
What is the mass of an object if it weighs
50.0 N near the earth’s surface?
g
= 9.8 m/s²
Fg
= 50.0 N
m
=?
g 
m 


Fg
m
Fg
g
50.0 N
2
9.8m / s
5.1kg
Normal Force
On a flat surface, the normal force is equal
to the force due to gravity (Fg).
Fn = Fg = ma
Fn
Fg
Frictional Force
Forces that oppose motion.
Dependent on the normal force (Fn) and
the nature of the two surfaces.
Is NOT dependent on surface area.
Frictional Force
Ff = μFn
Ff = Frictional force
Fn= Normal force
μ = Coefficient of friction
Ff
Fn
Frictional Force
A 25 kg object is pulled along a horizontal
surface at a constant acceleration of 2.5
m/s². The coefficient between the two
surfaces is 0.14. What is the frictional
force?
Ff
Fn
μ
=?
= ma
= 0.14
Ff
= μFn
= μma
= (0.14)(25 kg)(2.5 m/s²)
= 8.8 N
Elastic Force – Hooke’s Law
States that: the elastic force is directly
proportional to the distortion of the
material.
Elastic force is the force an object has to
restore itself.
The equilibrium position is the position
where the material will restore itself to.
The distortion is the distance from the
equilibrium position to the stretched
material.
Elastic Force – Hooke’s Law
Equilibrium Position
Distortion
Clamp
Elastic Material
Elastic Force – Hooke’s Law
Formula: Fs = -kx
Fs = elastic force
k = spring constant – unit : N/m
x = distortion
- represents the direction of the elastic
force.
Unit is in newton(N)
Elastic Force – Hooke’s Law
A 2.50 kg mass on a spring is 0.300m from
its equilibrium position. If the spring
constant is 32.0 N/m, what is the elastic
force acting on the mass?
Fs
k
x
=?
= 32.0 N/m
= 0.300 m
Fs
= -kx
= -(32.0 N/m)(0.300 m)
= -9.60 N
Newton’s Third Law of Motion
States that: for every action (force) there is
an equal but opposite reaction (force).
For an example - a book is placed on a
table. The book exerts a force (due to
gravity) on the table. An equal force is also
exerted from the table to the book.
Both forces cancel out which explains why
the book does not accelerate.
Momentum
Law of Conservation of Momentum:
momentum will remain constant unless an
unbalanced force acts upon it.
Momentum is the product of mass and
velocity.
P = mv
Unit is in kg · m/s.
Momentum
Calculate the moment of a 15.3 kg object
traveling and a velocity of 30.0 m/s east.
P=?
P
= mv
m = 15.3 kg
= (15.3 kg)(30.0 m/s)
v = 30.0 m/s
= 459 kg · m/s
Impulse
The Change in momentum.
Δp = mΔv
Unit is in kg · m/s
Impulse
A net force of 31.6 N south is used to
accelerate a 15.0 kg object uniformly from
rest to 10.0 m/s. What is the change in
momentum?
Δp = ?
Δp
= mΔv
m = 20.0 kg
= (20.0 kg)(10.0 m/s)
Δv = 10.0 m/s
= 300 kg · m/s
Elastic Collision
Collision of two objects.
Do not stick.
A 0.50 kg ball is travelling east at a velocity
of 9.5 m/s when it collides with a 0.35 kg
object travelling west at a velocity of 7.0
m/s. After the collision the 0.50 kg ball is
travelling west at a velocity of 4.0 m/s.
What is the velocity of the 0.35 kg ball after
collision?
Elastic Collision
Before collision
1
2
m = 0.50 kg
m = 0.35 kg
v = 9.5 m/s
v = -7.0 m/s
p = 4.75 kg·m/s
p = -2.45 kg·m/s
Pbefore = 2.30 kg·m/s
Elastic Collision
After collision
1
2
m = 0.50 kg
m = 0.35 kg
v = -4.0 m/s
v=?
p = -2.0 kg·m/s
p = 4.3 kg·m/s
Pafter = 2.30 kg·m/s
Elastic Collision
After collision
1
2
m = 0.35 kg
v=?
p = 4.3 kg·m/s
Pafter = 2.30 kg·m/s
p  mv
p
v
m
4.3kg  m / s

0.35kg
 12m / s
Inelastic Collision
Two objects collide.
Both objects stick to each other (masses are
added).
A 2.1 x 10² kg car is travelling east at a
velocity of 52 km/h collides with a 3.1 x
10² kg car travelling west at a velocity of 20
km/h. During collision, both cars lock
together. What is the velocity of the locked
cars after the collision?
Inelastic Collision
Before collision
1
2
m = 2.1 x 10² kg
m = 3.1 x 10² kg
v = 52 km/h
v = -20 km/h
p = 10920 kg · km/h
p = -6200 kg · km/h
Pbefore = 4720 kg · km/h
Inelastic Collision
After collision
1+2
m = 5.2 x 10² kg (add both masses)
v = ? km/h
p = 4720 kg · km/h
Pafter = 4720 kg · km/h
p  mv
p
v
m
4720kg  km / h

5.2  10 2 kg
 9.1kg  km / h
Explosions
Objects are combined before explosion
and are separated after explosion.
A 0.125 kg bullet is fired from a 5.5 kg gun.
If the velocity of the bullet is 325 m/s,
what is the recoil velocity of the gun?
Explosions
Before collision
1+2
m = 5.625 kg
v = 0 m/s
p = 0 kg · m/s
Pbefore = 0 kg · m/s
Explosions
After explosion
1
2
m = 5.5 kg
m = 0.125 kg
v = ? m/s
v = 325 m/s
p = -40.625 kg · km/h
p = 40.625 kg · m/s
Pafter = 40.625 kg · m/s
Explosions
After explosion
1
2
p  mv
m = 5.5 kg
p
m
 40.625kg  km / h

5.5kg
 7.4km / h
v 
v = ? m/s
p = -40.625 kg · km/h
Pafter = 40.625 kg · m/s
Conclusion
1st law of motion: if the body is at rest, it
will remain at rest. If the body is moving
with a constant velocity, it will continue to
do so.
2nd law of motion: ΣF = ma
3rd law of motion: For every reaction there
is an equal and opposite reaction.
Hooke’s Law: Fs = -kx
Conclusion
Normal force: Fn = Fg
Frictional force: Ff = μFn
Law of universal gravitation:
Gravitational field strength:
Fg 
g
Gm1m2
r2
Fg
m
Conclusion
Momentum: P = mv, momentum is
conserved.
Impulse: Δp = mΔv = Fnett, change in
momentum
Elastic: Collisions that do not stick.
Inelastic: Collisions that stick.
Explosions: Separate after explosion.
References
Physics 11 Student Notes and Problems
(SNAP)
Fundamentals of Physics, Halliday Resnick,
Third Edition