Dynamics
Section – 1
Newton Laws
1st Law
2nd Law
3rd law
‘F’ vs ‘a’
‘F’ vs ‘p’
F = ma
F=
mv−mu
t
=
∆p
∆t
Impulse
F x t = mv - mu
Section – 2
Momentum & Collisions
Momentum
Collision
P = mv
Elastic
Inelastic
Section – 1
Newton laws Of Motion
1. Newton’s First Law:
Statement:
“Every object tries to maintain the state of rest or uniform motion along the straight
line unless an unbalance/external force acts upon it”
OR conversely
“An unbalance force disturb the state of rest or uniform motion along the straight
line of an object”
Note
According to this law there is no state of rest or uniform motion along straight line in
the universe. Since there is no place where an external force is not present.
This law has the following information’s
Every object possesses a property called INERTIA by which it opposes the change in its
state of rest or uniform motion along straight line. That’s why it is also called the law of
inertia.
Note: Inertia depends upon mass, more the mass greater is the inertia so the mass can
be defined as,
“Mass is the measure of Inertia”.
The law also provides us the definition of force I,e “Force is an agent which brings the
change in the state if rest or uniform motion along straight line”.
2. Newton’s Secind Law:
The second Law can be expressed in two ways
i)
Force vs acceleration:
Ststement:
“The unbalance force always results in an acceleration in its own direction”.
Mathematically
F = ma
Note imp:
Since in A-Level students know about the resolution of force into components, so
diverse problems can be designed from 2nd law.
Note: imp:
How to approach a problem
Steps
1. Choose your airs along & perpendicular to the motion.
2. Identify all the forces acting on the object in problem. This is also called free body
diagram.
3. Resolve those forces into components which make certain angle with the axis.
4. Identify the unbalance force along the direction of motion.
5. Make the eqution of motion i,e F = ma
Type of problems:
1. Motion on an inclined plane:
Smooth Plane
In such case the only two forces are the weight of object ‘mg’ and the normal reaction
force ‘N’. Since ‘mg’ make an angle ‘θ’ with the axis so we resolve it into components i,e
along the motion & perpendicular to it.
From figure, the unbalance force along the motion Fun = mg sinθ
So equation of motion;
mg sinθ = ma
Note:
If
θ = 0 then a = 0 as
If θ = 90o
sin0 = 0
Then a = g free-fall
as
sin90 = 0
If the plane is Rough plane than Perpendicular to motion
N = mg cosθ
And along the motion
mg sinθ – f = ma
2. Motion in a Lift
As shown the only forces acting are the weight ‘mg’ & normal reaction ‘N’.
Case – 1: Lift moving with constant velocity or at rest
In such case
R = mg
or a = 0
Case – 2: Lift moving up with an acceleration ‘a’:
In such case, the Fun = R – mg
So
R – mg = ma
Or
R = ma + mg
Than
R = 2mg
if
a=g
The person will fell twice heavier.
Case – 3: Lift moving down with an acceleration ‘a’:
In such case, the Fun = mg – R
So
mg – R = ma
Or
R = mg – ma
Than
R=0
if
a=g
The person will fell weightless.
Connected bodies Problems:
Note: v.imp
 In such case the force acting on one body has nothing to do with the other body.
 The bodies/system will have same acceleration.
 The tension ‘T’ is always same & acts away from the point of contact
 We can make two equations of motion separately for bodies and solve them
simultaneously for acceleration.
 Or a single equation of motion for both bodies/system.
If we make a single equation of motion for both bodies than remember tension will not
be included as its same and acting opposite so will cancel out, as shown below.
Fun = m2g – m1gsinθ
So
m2g – m1gsinθ = ( m1 + m2 ) a
Or
a=
m2g – m1g sinθ
m1 + m2
It’s the same equation for ‘a’, as shown in the figure.
Another example:
Or in case of single equation,
Fun = m2g
So
m2g = ( m1 + m2 ) a
or
a=
m2g
You can see it’s the same equation as in figure.
m1 + m2
Suspended body problems:
In such case, Fun = mgsinθ
So the equation of motion is,
mg sinθ = ma
&
T = mg cosθ
Newton’s Second Law Force vs Momentum:
Statement:
“Unbalance is equal to the rate of change of momentum in its own direction”
Mathematically
F=
mv – mu
t
Note:
Both statements of second law are actually same in essence, as shown below
F = ma
Or
F=m(
Or
F=
Note: v.imp
v−u
t
mv − nu
t
)
=
∆p
∆t
The following idea if frequently used in exams in which mass change with time is
considered & the velocity remains constant.
F=
Or
1
∆𝑡
F=(
( mv )
∆𝑚
∆𝑡
)xv
3. Newton’s Third Law:
Statement:
“Every action force has an equal and opposite reaction force”.
Note:
 The two forces must be of the same nature like gravitation – gravitation or
Electrical – Electrical.
 Also the two forces must act upon separate bodies.
Section – 2
Momentum & Collissions
1. Momentum: ( p = mv )
It tells us about the quantity of motion in a body and is defined as;
Definition:
“It’s the product of mass and velocity”.
Mathematically
Momentum = Mass x velocity
p = mv
It’s a vector quantity
It’s units are N.s or kg ms-1
Law of Conservation Of Momentum:
This is one of the most powerful law in physics.
Statement:
“For an isolated system the total momentum of the two objects before collision is equal
to the total momentum of the two objects after collusion”
OR
“Total momentum of the two objects before collision is equal to the total momentum
of the two objects after collusion provided no external forces are involved”
Note:
Isolated system is the one in which no external forces are involved.
Mathematically
Momentum before collision = momentum after collision
m1u1 + m2 u2 = m1v1 + m2 V2
Proof:
By Newton’s 3rd law during collision
F21 = - F12
m1 v 1 − m1 u 1
Now apply 2nd law
t
=-(
m 2 v 2 − m2 u 2
t
)
Or
m1 v1 − m1 u1 = m2 u2 − m2 v2
By rearranging
m1 u1 + m1 v1 = m2 u2 + m2 v2
Hence
momentum before = momentum after
Momentum In 2-D:
In 2D, the momentum is split into components. We apply the conservation of
momentum along and perpendicular to the motion.
1. Along the motion:
Momentum before collision = Momentum after collision
m1u1 + 0
= m1 v1 cosθ1 + m2 v2 cosθ2
2. Perpendicular to the motion:
Momentum before collision = Momentum after collision
Or
0
= m1 v1 sinθ1 + ( - m2 v2 sinθ2 )
m1 v1 sinθ
= m2 v2 sinθ2
Section – 2
Collision:
In physics a phenomena is said to ba a collision if it has the following features.
1. The colliding bodies must make a close system.
2. It must happen for a very short time.
3. What is before collision must be different from what is after collision.
4. The momentum and total energy are always conserved.
Type Of Collision:
There are two types
1. Elastic collision:
In such collision the kinetic energies, momentum and total energy are always
conserved.
2. Inelastic collision:
In such collision the momentum and total energy are conserved but not the kinetic
energies.
Note: imp
If momentum is not conserved in any situation then that must not be a closed system.
Kinetic Energy In Explosive Situation:
Explosive situation is the one in which a single mass splits into two masses.
For example firing a cannon or gun, or radioactive decay in which a nucleus splits into
daughter nucleus & alpha or beta particle.
Objective:
Our objective is to find out how the kinetic energies are shared.
Note:
1
As
K = m v2
Or
K=
Or
k=
2
1 m2 v 2
2
m
1 p2
as
2 m
p = mv
Now
k1
k2
k1
Or
k2
=
=
1 p21
2 m1
/
1 p21
2 m1
m2
m1
Result:
The kinetic energies are shared in the inverse ratio of masses when a single mass splits
onto two masses.
Sharing Of Total Energy Among Two Masses:
As,
E = k1 + k2
Now we know,
k1
k2
Or
=
k2 =
m2
m1
m1
m2
x k1
Put this in above equation,
E = k1 +
m1
m2
x k1
Or
E = k1 (1 +
Or
E = K1 (
Or
k1 = (
m1
m2
)
m1 + m2
m2
m2
m1 + m2
)
) =
opposite mass
Sum of masses
Similarly.
K2 = (
m1
m1 + m2
)
Newton Law Of Relative velocities (Restitution):
Statement:
v 2−v 1=−e(u 2−u 1).
This formula is Newton's law of restitution.
The coefficient of restitution always satisfies
0 ≤ e ≤ 1.
When e=0, the balls remain in contact after the collision.