Newton’s Laws Of Motion• The First LawEverybody continues in its state of rest or

uniform motion in straight line unless acted upon by an imbalanced external force to make a change

The law is all about reluctance.1.One effect of a force is to alter the dimension

or shape of a body on which it acts.2. Other is alter the motion state of the body

Effects of Newton's first law of motion

1. An object at rest begins to move only when it is pushed or pulled

2. Electrons in a conductor begin to move or drift when a potential difference or electromotive force is applied in a closed circuit

3. A body changes its direction in straight line when a force is applied for example in circular motion

Effects of Newton's first law of motion cont….…

4.Passengers in a bus or car are jerked forward when the bus stops suddenly because they continue in their state of motion until brought to rest by some external forces. The opposite is true when the bus starts moving suddenly

So an external force applied on a moving object changes the velocity in magnitude and direction . Question :Why does a cat fail to catch a rate just negotiating a corner?

Newton’s second law of motion

• The rate of change of momentum p is proportional to the force applied to make the change and takes place in a straight line

• Consider a body having initial velocity u at time t= 0 and reaches a velocity v after a time t. Change in momentum is

The rate of change of momentum is

mumvp

maFkkmadtdvmmumvF

tp

1,

⟹∑FTo apply equation i above, you need to: 1) Choose a system2) Draw all the forces on the system3) Choose a coordinate system, one of the

system in the direction of the acceleration 4) Set up Newton's 2nd laws in both direction i.e.

Application of Newton's second law of motion

a) A block on a horizontal plane

Reaction (R)

Frictional force F

Weight(mg) of a body

Tension (T)

A Block On A Horizontal Plane Cont…

i. If the surface is frictionless , the tension T is the only force which causes a block to move and the forces acting on the system are as follows . In X –direction : This is from Newton’s second law of motion and in Y – direction : ,but acceleration along y is zero since the block is accelerating along x . Hence we have

)1.....(maT

mamgRmaWR

)2.....(,0 mgRmgRmgR

A block on a horizontal plane cont…

ii.If the surface is rough , the net force which causes a block to move is

and the forces acting on the system are as follows . In X –direction : This is from Newton’s second law of motion and in Y – direction : ,but acceleration along y is zero since the block is accelerating along x . Hence we have

)1.....(maFT f

mamgRmaWR

)2.....(,0 mgRmgRmgR

fFT

(b) A Block On An Inclined Plane

• Consider the figure below

Frictional fo

rce F

sinmg

mg

sinmg

cosmg

R

A Block On An Inclined Plane Cont..

i. If no friction the forces acting on the system are in y –direction and

in x direction.

ii. If the plane is rough or has friction then vertically while horizontally

if there is a free fall since

cosmgR

sinsin gamamg

cosmgR )cos(sincos,sin gamgFmaFmg

090 ga

Tension And Acceleration Of Connected Bodies

• Consider the diagram below

T

T

𝑚1

𝑚2

a

𝑚1𝑔

R

𝑚2𝑔

Tension And Acceleration Of Connected Bodies Cont…

• Consider the forces acting in each body separately

• For body m1 along x direction

• For body m2 we have vertical component

• , solve for a in 1&2

1....1amT

2...22 amTgm

gmm

ma )(21

2

Tension And Acceleration Of Connected Bodies Cont…

• From equation 1 we find that the expression for tension is

PROBLEM Show that the coefficient of friction or constant

of proportionality of a body moving along an inclined plane at an angle θ is

gmmmmT )(

21

21

tan

Examples

1.Find the tension T and the acceleration if there is no friction force on the pulley

𝑚1

𝑚2

T

8kg

2kg

T

SOLUTION

μ=0,Let The motion is downward towards

⟹a⟹

⟹ a=Solving for tension T Consider the mass From ∑F But ∑F=T ⟹T⟹T=2kg×8m/

2.Find the tension and the acceleration

Given the figure below

T

6 kg

6 kg

12 N

solution

Let ∑F⟹T-12N=⟹ T-12N=6a……………….1g-T=a 6×10-T=6a…………………..2 adding the two above,60N-12N=12a⟹a=

∴𝑎=4𝑚/𝑠2Solving for tension by using either equation 1 or 2 From equation 1,T-12N=6a,but ⟹T-12N=6kg× ⟹T-12N=24N

Frictional ForceIs the force that prevents a body from sliding or

motion.• Types of frictional forcea. Static friction: Is the force which opposes a

body at rest from starting motionb. Dynamic friction: Is the force that opposes a

body in motion from continuing to moveExplain conspicuously the importance of friction

in our daily life especially in engineering life

Art wood machineIs a system of pulleys and masses m1 and m2 on

sides 1 and 2 respectively connected by a negligible mass string. Consider the figure below

m2m1

TT

m2gm1g

Art Wood Machine• From the figure above And Solving 1 and 2 we obtainAcceleration And tension is

1....11 amgmT 2......22 amTgm

gmmmma )(12

12

gmmmmT )2(

21

21

Examples

1.

Find the tension and the acceleration

3kg7kg

T

solution

Let masses Applying Newtonʼs 2nd law ∑F=maT-⟹ T-30

⟹70

Adding 1&4 70N⟹a= 4m/

Using equation 4 to find tension 70 but a= 4m/ ⟹ 70 ⟹70N-28N=T

Linear Momentum P• Is the product of mass and velocity of a moving

body . i.e. • From Newton's 2nd law of motion ,F=ma where

a=

Is called the impulse of force defined as the product of force and time interval or as the change in momentum

mvp

mdvFdtdtdvmmaF ,

Note: In collisions the total momentum of colliding bodies is always conserved.

But their total kinetic energy is not conserved,why?Because some of it is changed to heat or sound energies.A mass m moving with velocity v has kinetic energy equal to Suppose two masses and moving with different velocities and in the same direction collide head on with each other. After collision their velocities change to and respectively

If their collision is elastic, then their total kinetic energy is conserved i.e. ∑K.

From conservation of momentum which applies to any type of collision, the total momentum before collision equals the total momentum after collision i.e. ⟹

The Law Of Conservation Of Linear Momentum

If no external forces acting on a system of colliding bodies , the total momentum remains constant. i.e. the sum of momentum before collision equals the sum of momentum after collision. From Newton's second law of motion=ma but, a= ⟹F ⟹Fdt

But mdv is the change in momentum

i.e

But no external force,i.e F=0

⟹0 ⟹ i.e. Total momentum before collision equals total momentum after collision

The Law Of Conservation Of Linear Momentum Cont…

Again consider two bodies of masses m1 and m2 moving with initial velocities u1 and u2 respectively colliding head on, after collision the two bodies move with velocities

As in figure 1 below,

………………….These bodies are moving in the same direction

𝑚1 𝑚2𝑢1 𝑢2

Figure 1

Case 1:Before collision

The Law Of Conservation Of Linear Momentum Cont…

From Newton's third law F1= -F2

2211221122

2111 )()( vmvmumum

dtuvm

dtuvm

m1

m2

v2

v1

After collision

The Law Of Conservation Of Linear Momentum Cont…

• If after collision the bodies stick together then they move with the same final velocity v and the expression changes .

Problem .Develop the expression for final and initial

momentum relationship of two bodies which stick together after collision

Answer: where v is the combined velocity

NOTE:

Is an equation for inelastic collision.Examples for inelastic collisions are The bullet hitting wood The collision between two objects such that after

collision they stick together and move with the same velocity

Seat belts tied to a person and the sudden application of the break

Car crash e.g the car crashing against a tree

Examples

1.A ball A of mass 0.4kg moving with velocity of 5m/s collides head on with ball B of mass 0.2kg moving with velocity 2m/s.After collision A moves with velocity of 3m/s and B moves with velocity of 6m/s.i)Show that the collision obeys the law of conservation of momentumii)Is the collision elastic or inelastic?

Solution

Mass Velocity Mass Velocity After collision Velocity Velocity

i) Before collision, total momentum

⟹0.4kg But total momentum after collision

⟹0.4kg ⟹2.4kgm/s So, momentum is conserved and hence the law of conservation of momentum is obeyed.

ii)Before collision total kinetic energy =

⟹ ⟹5.4J After collision total kinetic energy ⟹ ⟹5.4J Since ∑∑ then kinetic energy is conserved therefore the collision is elastic

Types Of Collisions In One Dimension

• Elastic collision: kinetic energy is conserved . Bodies separate after collision

• Inelastic collision: here some of kinetic energy is lost . All real collisions belong to this category

• Perfect inelastic collision: a great deal of kinetic energy is lost. This occurs when two bodies stick together during collision

Problems

1)Which has greater momentum a car at rest or a moving person?2)Would a head on collision between two cars be more damaging to the occupants if the cars stuck together or if the cars rebounded upon impact?Assume the time of collision is the same3)A 10kg block travelling at 5m/s collides with a 5kg block travelling at 3m/s in the same direction and they stick together. What are the velocities of the block immediately after collision?

PROBLEM 1• A block whose mass is 5kg rests on a

horizontal surface, what constant horizontal force is required to give it a velocity of 4m/s stating from rest if the frictional force between the block and the surface in contact is 3N? Assume all forces act on centre of gravity of the block and it moves 2seconds

• Answer F = 13N

PROBLEM 2• An elevator and its load have a total mass of

800kg .Find the tension in the supporting cable when the elevator or lift originally moving downwards at 10m/s is brought to rest with a constant acceleration in a distance 25m

• Answer T= 9600N.Since T> W the system is retarding.

PROBLEM 3• An object of mass 2kg is attached to the hooks

of a spring balance and the later is suspended vertically from the roof of a lift . What is the reading on the spring balance when the lift is

a. Ascending with an acceleration of 0.2m/s2 b. Descending with an acceleration of 0.1m/s2

c. Ascending with uniform velocity of 0.15m/sTake g as 10m/s2

PROBLEM 4a. If the mass of bullet is 5g and the mass of the

gun is 10kg and the velocity of the bullet is 300m/s, find the recoil speed of the gun

Answer V= 0.15m/sb. An aero plane flying at 250m/s takes air into its

engine at a rate of 500kg/s . Fuel is burnt at 20kg/s . The engine expels gases at a speed of 600m/s relative to the jet. Calculate the driving force of the engine (187000N)

PROBLEM 5• A 6kg block rests on a smooth table. It is

connected by a string of negligible mass to a block hanging over the end of the table . Assuming motion is downwards ,Find

i. The acceleration of each block(2.5m/s2 )ii. The tension in the connecting string(14.7N)iii. The position of mass of A after 0.4 secondiv. The velocity of mass A at 0.4 seconds

(0.98m/s)

PROBLEM 6• Two blocks of masses 100kg and 80kg inclined

at 30 degrees and 53 degrees respectively are connected by a cord passing over a small frictionless pulley resting on a frictionless planes .

i. Which way will the system move?ii. What is the acceleration of the blocks?(0.76)iii. What is the tension in the cord?

PROBLEM 6In the first second of its flight , a rocket ejects

1/60 of its mass with relative velocity of 2400m/s. What is the acceleration of the rocket?

• Answer a= 30.2