Chapter 3: Force, Impulse and Momentum

Chapter 3: Force, Impulse and Momentum

3.1 Force

What is a FORCE?

When we think of force

We usually imagine a push or a pull exerted on some object.

Exert a force on a ball when you throw it or kick it.

Exert a force on a chair when you sit down on it.

What happens to an object when it is acted on by a force?

It depends on the magnitude and direction of the force.

Force is a vector quantity; thus , we denote it with a directed arrow, just as we do for velocity and acceleration.

Early concept of Force

The early scientist introduced; force as the concept of a field .

The corresponding forces are called field force.

Example; when a mass, m is placed at some point, P near a second mass, M ,we say that m interacts with M by virtue of the gravitational field that exists at P.



Fundamental force

The known fundamental force in nature are all field force.

The fundamental force are :

•strong nuclear force between subatomic particles.

•electromagnetic force between electric charges at a rest or in motion.

•weak nuclear force, which arise in certain radioactive decay processes

•and gravitational force. The force of gravitational attraction between two objects.

ISAAC NEWTON
1642 – 1727

I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the seashore and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me.

The whole burden of philosophy seems to consist in this… from the phenomena of motions to investigate the forces of nature and then from these forces to explain their nature.

3.2 Newton’s Laws of Motion

• Newton’s First Law of Motion

 Inertia

• Newton’s Second Law of Motion

 Mass and Weight

 Center of Mass and Center of Gravity

• Newton’s Third Law of Motion

 Normal Force

 Friction – Static friction and Kinetic frictiont velocity if there is no net external force between the object and environment.

Newton’s First Law of Motion

States:

* An object at rest stays at rest and an object in motion continues in motion with constan

In equation:

Garfield testing
Newton’s First Law of Motion

Inertia

Definition :

Inertia is the tendency of an object to resist any changes in its state of rest or motion.

In other words : ‘Objects’ tend to keep on doing what they’re doing.

Inertia is solely dependent upon mass of the object.

–Object with more mass has more inertia meaning that it has more tendency to resist changes in its state of motion.

Snoopy experiencing Inertia

Experiments about Inertia

Real Experiences of Inertia

Pull Force

Push Force

Newton’s second Law of Motion

States:

• The accelation of an object is directly proportional to the net force acting on it an inversely proportional to its mass.

Restate:

• The rate of change of momentum with time is proportional to the net applied force and is in the same direction.

Note :

The direction of the acceleration is the same as the direction of the applied net force.

S.I. unit of force :

1 Newton is the force that produces an acceleration of 1 m s-2 when acting on a 1 kg mass.

Therefore, 1 N = 1kg m s-2

Mass

Mass is an inherent property of a body.

Mass is a quantitative measure of the inertia of a body.

Mass is the force required per unit of acceleration produced.

The value of mass is independent of location.

Mass is a scalar.

S.I. unit of mass is kilogram.

Weight

Weight is the force exerted on an object by a gravitational field.

W = mg = Fg

Weight varies slightly with altitude because weight depends on the strength of the gravitational field.

gearth gmoon

Weight is a vector.

S.I. unit of weight is Newton or kg m s-2.

Center of Mass

Defination:

• The point at which all the mass can be considered to be ‘concentrated’.

Note:

• A force that exerts on the center of mass will cause the object to make a translational motion.

Center of Gravity

Defination:

• The point at which all the weight exerts on the object.

Note:

• Center of mass and center of gravity for a uniform and simetrical object is at the center point of the object.

Example :

An object with two masses m1 and m2 at the each end of a light rod, d in length. Find the center of mass of the object using m1 and m2 and d.

Solution:

Defination:

• The point at which all the mass can be considered to be ‘concentrated’.

Note:

• A force that exerts on the center of mass will cause the object to make a translational motion.

d

m1 m2

Axcm

Center of mass from A,

=

Example:

An object with two masses m1 and m2 at the each end of a light rod , d in length. Find the center of mass of the object using m1 and m2 and d.

Newton’s Third Law of Motion

States :

For every action (force) there is a reaction (opposing force) of equal magnitud and straight but opposite in direction.

Explaination :

Whenever one object exerts a force FAB on a second object. The second exerts an equal and opposite force FBA on the first.

FAB = -FBA

Examples :

When you push on the wall it will push back with the same force.

When little Jim pushes little Tim, little Tim pushes back with the same force. The boy with the better grip on the ground will keep from falling.

In the game of tug of war, when one side pull on the other side, the other side pulls back with the same force.

Normal Force

Definition :

The contact force exerted by a surface onto a body resting or sliding on the surface and acts perpendicularly to the surface.

N N



Friction

Friction originates from forces between atoms and molecules when surfaces are in contact.

Example :

–Friction occurs when a body moves on a rough surface or through a fluid medium (water, air, etc).

The direction of the friction is parallel to the surface in contact and opposite to the direction of in which an object wants to move.

Friction is a retarding force that resists motion on a surface.

N

direction of f motion

f N

f = N

is the coefficient of friction.

is depends on the objects involve and on the condition of the surface.

Two Types of Friction

• Static Friction

• Kinetic Friction

Static Friction

Static friction, fs is the force of friction between two objects when there is no motion.

fs changes with the external force, Fext.

fssN where

–s = the coefficient of static friction

–N = the magnitude of the normal force

Equality holds when the object is at the point of slipping.  fs(max) = sN

Example :

Consider a block on a rough surface. Apply an external force to the block.

–if Fextfs (max) the object won’t move.

–as Fext increases, fs will increase until it reaches its maximum value.

–When Fext = fs (max) the block will start to move which is called the point of slipping.

–Once the force starts to move the force of friction is given by kinetic friction, fk .

–

Kinetic Friction

Kinetic Friction, fk is the force of friction between two objects when there is motion.

fk = kN where

–k = the coefficient of kinetic friction

–N = the magnitude of the normal force

k is nearly independent of the velocity of the object under consideration.

fk is approximately constant for any given pair of materials.

Note :

Values of s and k depend on the nature of the surfaces that are in contact.

Usually ks fk < fk .

–rubber on concrete s = 1.0, k = 0.8

–waxed wood on wet snow s = 0.14, k = 0.10

s and k are nearly independent of the area of contact between the two surfaces.

3.3 Linear momentum
and its
Principle of Conservation

• Application of Newton’s first and second laws of motion in linear momentum
• Linear momentum
• Principle of conservation of linear momentum

Application of Newton’s First and Second Law’s of Motion

• From Newton’s Second Law:

Case1

• Object at rest or in motion with constant velocity but with changing mass.
• Example: Rocket

Case2

• Object with constant mass but changing velocity.
• Example: Rocket

Case 3

• Object at rest or in motion with constant velocity and mass.

• Newton’s First Law of Motion

Linear Momentum

Definition :

Momentum = Mass x Velocity

If F = 0  p = mv = constant

Momentum is a vector.

The direction of the momentum is the same as the direction of the velocity.

S.I. unit of momentum is kg m s-1 or N s.

Example :

Find the magnitude of the momentum of a cricket ball of mass 420 g thrown at 20 m s-1.

Solution :

Given m= 0.42 kg

v= 20 m s-1

Momentum= mass x velocity

p= mv

= 0.42 x 20 N s

= 8.4 N s

Example :

A 1.5 kg ball was kicked with initial velocity of 40 m s-1 at the angle of 30 with the horizontal line. Calculate the initial momentum of the ball and also the horizontal and vertical components of the initial momentum.

Solution:

vyv

Given m= 1.5 kg

v = 40 ms-1300

vx

Momentum, p = mv

= 1.5 x 40 kg ms-1

= 60 kgms-1

Horizontal component of the momentum :

px= mvx

= mv kos 

= 1.5 (40) kos 30

= 51.96 kg m s-1

or px= p kos 

= 60 kos 30

= 51.96 N s

Vertical component of the momentum :

py= mvy

= mv sin 

= 1.5 (40) sin 30

= 30 kg m s-1

or py= p sin 

= 60 sin 30

= 30 N s

Principle of Conservation of
Linear Momentum

States :

When the net external force on a system is zero, the total momentum of that system is constant.

Or

Provided there are no external forces acting on a system, the total momentum before collisions equals the total momentum after collisions.

Newton’s First Law of Motion :

F = 0  p = 0  p = constant

Expressed symbolically :

 initial momentum =  final momentum

pi= pf

For a collision involving two bodies :

m1u1+ m1u2 = m1v1 +m2v2

3.4 ELASTIC & INELASTIC COLLISION

3.4.1 Collision

* Collision is a process in which the colliding parties interact with each other very strongly and briefly such that all the other forces can be ignored in this process.

* The interaction force, even though brief, depends strongly on time.

* The average force of a collision process can be calculated by dividing the impulse, J by the time interval of the collision, t ;

* In a collision process between two bodies m1 and m2, the external forces can be ignored, and as a result the linear momentum before and after the collision is conserved.

m1u1 + m2u2 = m1v1 + m2v2

where ; u1 and u2 are velocities of m1 and m2 before collision

v1 and v2 are velocities of m1 and m2 after collision

* On the other hand, the kinetic energy may or may not be conserved in a collision.

* Two types of collisions are elastic collision and inelastic

collision.

Example

An 1800-kg truck stopped at a traffic light is struck from the rear by a 900-kg car and the two become entangled. If the smaller car was moving at 20 m/s before the collision, what is the speed of the entangled mass after the collision ?

Solution

Conservation of momentum

3.4.2 Elastic collision
* An elastic collision is that in which the momentum of the system as well as kinetic energy of the system before and after collision is conserved.
Conservation of momentum :
m1u1 + m2u2 = m1v1 + m2v2
Conservation of kinetic energy :
½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v2 2

* Elastic collision in one dimension

Consider ; m1 and m2 = massesof two non-rotating spheres u1 and u2 = velocities of m1 and m2 before collision

v1 and v2 = velocities of m1 and m2 after collision

Let u1 is greater than u2 and are in the same direction:

Momentum of the system before collision = m1u1 + m2u2

Momentum of the system after collision = m1v1 + m2v2

According to the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

m1v1 – m1u1 = m2u2 – m2v2

m1(v1 – u1) = m2(u2 – v2) ------(1)

Similarly

K.E of the system before collision = ½ m1u12 + ½ m2u22

K.E of the system after collision = ½ m1v12 + ½ m2v22

Since the collision is elastic, so the K.E of the system before and after collision is conserved.

Thus

½ m1v12 + ½ m2v22 = ½ m1u12 + ½ m2u22

½ (m1v12 + m2v22) = ½ (m1u12 + m2u22)

m1v12-m1u12 = m2u22-m2v22

m1(v12-u12) = m2(u22-v22)

m1(v1+u1) (v1-u1) = m2(u2+v2) (u2-v2) ------(2)

Dividing equation (2) by equation (1);

v1+ u1 = u2 + v2

From the above equation

v1 = u2 + v2 -u1 ______(3)

v2 = v1 + u1 -u2 ______(4)

Putting the value of v2 in equation (1)

m1 (v1-u1) = m2 (u2-v2)

m1 (v1-u1) = m2{u2-(v1+u1-u2)}

m1(v1-u1) = m2{u2-v1-u1+u2}

m1(v1-u1) = m2 { 2u2-v1-u1}

m1v1-m1u1= 2m2u2-m2v1-m2u1
m1v1+m2v1= m1u1-m2u1+2m2u2

v1(m1+m2) = (m1-m2)u1+2m2u2

* Elastic collision in two dimensions

Elastic collision in two dimensions can be analyzed by using the fact that momentum is a vector quantity.

Consider a glancing collision between two spheres of mass m1 and m2 ;

v1 , v2 = velocities before collision ,

v’1 , v’2 = velocities after collision ,

m2 is initially at rest,  v2 = 0

the initial velocity is along the x axis,

 the initial momentum along the y axis = 0

Conservation of momentum gives ;

momentum along x : m1v1x = m1v’1x + m2v’2x

m1v1 = m1v’1 cos 1 + m2v’2 cos 2

momentum along y : 0 = m1v’1y + m2v’2y

m1v’1 sin 1 = m2v’2 sin 2

Conservation of energy gives ;

kinetic energy : ½ m1v12 = ½ m1v’12 + ½ m2v’22

m1v12 = m1v’12 + m2v’22

Example 1

A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck after the collision.

Solution

Example 2

A 200 g tennis ball moving with a speed of 15 m/s collides with a stationary ball of 800 g in an elastic collision. The tennis ball is scattered at an angle of 45o from its original direction with the speed of

5 m/s. Find the final speed (magnitude and direction) of the struck ball.

Solution

Conservation of momentum along x :

m1v1 = m1v’1 cos 1 + m2v’2 cos 2

0.2(15) = 0.2(5 cos 45o )+ 0.8 (v’2 cos 2 )

v’2 cos 2 = 3 - 0.2(5 cos 45o )

0.8

v’2 cos 2 = 2.866 (i)

Conservation of momentum along y :

m1v’1 sin 1 = m2v’2 sin 2

0.2(5 sin 45o ) = 0.8 v’2 sin 2

v’2 sin 2 = 0.2(5 sin 45o )

0.8

v’2 sin 2 = 0.884 (ii)

3.4.3 Inelastic collision

* An inelastic collision is that in which the momentum of the system before and after collision is conservedbut the kinetic energy before and after collision is not conserved.

Conservation of momentum,

m1u1 + m2u2 = (m1+ m2) v

Kinetic energy is not conserved,

 K.E before collision  K.E after collision

½ m1u12+ ½ m2u22  ½ ( m1 +m2)v2

( some of the kinetic energy is transformed into other forms of energy such

as heat or sound)

* Inelastic collision in one dimension

In a completely inelastic collision the two objects stuck together after collision.

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1 + m2 ) v

If m2 is at rest, u2 = 0 ;

Kinetic energy before collision , Ki = ½ m1u12

Kinetic energy after collision, Kf = ½ (m1+ m2) v2

* Ballistic pendulum

Ballistic pendulum is a device invented by Benjamin Robins in 1742 to measure the speed of a bullet.

The block is at rest, so its velocity is zero.

Conservation of momentum;

mvi = (M + m) vf (i)

Kinetic energy of both bullet and the block after collision;

K.E = ½ (M+ m) vf2

Conservation of energy;

Kinetic energy= Gravitational potential energy

½ (M+ m) vf2= (M + m )g h

vf=  2 g h  (ii)

Example 1

In a ballistic experiment, suppose that ;

h = 5.00 cm,

m = 5.00 g

M = 1.00 kg.

Find the (a) initial speed of the bullet, vi

(b) the loss in energy due to the collision

[ g = 9.81 m/s2 ]

Solution

(b) The loss in energy due to the collision, K

K = Ki – Kf

= ½ mvi2 - ½ (m + M) vf2

= ½ mvi2 - ½ (m + M) (2gh)

= ½ {(5x10-3)(199)2 -(5x10-3 + 1.00) (2)(9.81)(5x10-2)

= 99.0 - 0.5

= 98.5 J

Example 2

Granny (m=80 kg) whizzes around the rink with a velocity of 6 m/s. She suddenly collides with Ahmad (m=40 kg) who is at rest directly in her path. Rather than knock him over, she picks him up and continues in motion without "braking." Determine the velocity of Granny and Ahmad. Assume that no external forces act on the system so that it is an isolated system.

Solution

Example 3

Two cars approaching each other along streets that meet at a right angle collide at the intersection. After the crash, they stick together. If one car has a mass of 1450 kg and an initial speed of 11.5 m/s and the other has a mass of 1750 kg and an initial speed of 15.5 m/s, what will be their speed and direction immediately after impact ?

Solution

The x component of the vector;

m1v1 = (m1+ m2) vx

The y component of the vector,

m1v1 = (m1+ m2) vy

3.4.4 Coefficient of Restitution

* Coefficient of restitution is the ratio of the differences in velocities before and after the collision

* The coefficient of restitution will always be between zero and one,

( 0 < ek < 1 )

* A perfectly elastic collision has a coefficient of restitution of 1,

ek = 1

* A perfectly inelastic collision has a coefficient of restitution of 0,

ek = 0

3.5 Impulse

The impulse of the force F equals the change in the momentum of the particles

When a baseball hits a bat

•or when two billiard balls collide , they exert forces on each other over a very short time interval. Forces of this type , which exist only over short time , are often called impulsive forces.

•According Newton’s second law

F = Δ (mv) = Δp

Δt Δt

where F is the net force applied to an object and Δp is its change

in momentum during a time Δt

F Δt = Δp

The quantity F Δt is called the impulse . It is the product of the force

and the time interval Δt over which the force acts.

Graph of force vs time

•Average and instantaneous force during a typical brief collision between two moving bodies.

The area under the curve of force versus time is equal to the impulse.

•since the area of the rectangle whose height is the average force equals the area under the curve , we can replace the instantaneous force by the average force to obtain the impulse.