NEWTONS 2st LAW Momentum P = Mass Velocity = mv P does not change if there is no external force.This is called law of conservation of momentumInertial propertyDefinition Rate of change of momentum is proportional to the applied external imbalance force & takes place in direction of force

Linear Momentum Momentum is a vector; its direction is the same as the direction of the velocity.

NEWTONS 2st LAWIf the net force acting on a system is zero, then the total momentum of the system is always conserved ( does not change irrespective of collision & movement as of inner objects)The Definition will also meanIf there is no external force total linear momentum of system is always conserved

Change in MomentumChange in momentum: Dp = pafter - pbeforeTeddy Bear: Dp = 0 - (- mv ) = mv Bouncing Ball: Dp = mv - ( -mv ) = 2mv

Question 1A 10 kg cart collides with a wall and changes its direction. What is its change in x-momentum Dpx? -30 kg m/s -10 kg m/s 10 kg m/s 20 kg m/s 30 kg m/s

Law of Conservation of Linear MomentumIf the net force acting on a system is zero, then the total momentum of the system is always conserved ( does not change irrespective of collision & movement as of inner objects)

In sciences isolated system meansA physical system that does not interact with its surroundings. No external imbalance force can actIts total energy and mass stay constant. They cannot enter or exit, but can only move around inside.Example of isolated system

Impulse of force ( I )

Impulse of force : In physics, an impulse is defined as product of force & time for which the force actsWhen a force is applied to a rigid body it changes the momentum of that body.A small force applied for a long time can produce the same momentum change as a large force applied brieflyHence it is the product of the force and the time for which it is applied that is important. The impulse is equal to the change of momentum. W

Understanding Impulse : Let force F act on object for t secsThe quantity Force x time is known as impulse.But the quantity mv is change in momentum, Hence Impulse = Change in momentum= change in velocity due to F in time t in time

ImpulseImpulse is a vector, in the same direction as the average force.

I = F t = Area under F & .t curve = movement of forceConstant force

Leaner variable force forceI = Fmax t = Area of ABC under F & t curve = movement of force

Non linear variable force force

Microscopic view of a bounce.Momentum and Impulse in collisionBatBall

Microscopic view

Momentum and ImpulseMicroscopic view of a bounce.

Profile of the force during a collision.

Impulse and Average Force

Rigid ball & soft ball actual high speed video

Example: Hitting a Baseball (1) A 150 g baseball is thrown at a speed of 20 m/s. It is hit straight back to the boller at a speed of 40 m/s. The interaction force is as shown here. What is the maximum force Fmax that the bat exerts on the ball? What is the average force Fav that the bat exerts on the ball?0.006 sec

Change in movement

0.006 secImpulse = Change in movement= Area under force time curve = Fmax 0.006 = F max 0.003timeForceFmax

Example: A Karate Collision With an expert karate blow, you shatter a concrete block. Consider that you hand has a mass of 0.70 kg, is initially moving downward at 5.0 m/s, and stops 6.0 mm beyond the point of contact.What impulse does the block exert on your hand?What is the approximate collision time and average force that the block exerts on your hand?

Example: A Karate Collision0.7 kg

U = 5m/secV = 0S = 0.006 m.t = ?For handAverage velocity of hand = 5/2 = 2.5m/secHand movement = Average velocity of hand t0.006m = 2.5m/sec t sec.t = 0.006 m / 2.5 m/sec = 0.0024 secS=0.006mConsider hand movement.t = impact time in sec

Solution: A Karate CollisionFmaxFAvgForcetimeImpulse due to hand = I p = Change hand movement = mass change in velocity = 0.7 kg 5m/sec = 3.5 kg m /sec

Average force = I / t = 3.5 /0.0024 = 1458 N

Types of internal interactionsLeaner CollisionMoving objectsGun bulletExplosion etc

Collision of objectsActionReactionAction & reaction are contact forcesTime for which they are active is same Impulse produced by them is equal & apposite Change in movement of both balls is equal but appositeThus net change in movement is zeroHence total momentum before collision will be same as movement after collisionF- F

Same massHeavier to lighter lighter to HeavierThreeballsConservation laws are valid in all types of collisions

Leaner collisionsV1

Leaner collisions

There are two main types of collisions

Elastic collision: Both momentum and kinetic energy are conserved.

Inelastic collision: momentum is conserved but kinetic energy is not conserved.

Inelastic collision

Inelastic CollisionsCollision: two objects striking one another.Inelastic collision: momentum is conserved but kinetic energy is not: pf = pi but Kf Ki.Completely inelastic collision: objects stick together afterwards: pf = pi1 + pi2As kinetic energy is not conserved there is energy loss in collision by many waysSound producedPermanent Deformation of objectsFriction etc

A completely inelastic collision:

Inelastic Collisions Solving for the final momentum in terms of the initial momenta and masses:u1u2

Example 1

Example: Velocity of a BeeA insect with a weight of 0.150 g is seating on one end of a floating wooden stick.Weight of stick is 5.0 grams .The insect runs to the other end of the stick with a velocity Vi relative to still water.The stick moves in the opposite direction with a velocity of 0.120 cm/s. Find the velocity Vi of the bee. Neglect friction (g =10m/sec2)Press Esc for return

Solution to example 1

There is no external force on the system. Hence its total movement is always conservedConsider two events .Insect & wooden piece are stationaryThen both move at constant velocity Initially insect applies force (internal action) on stick for t sec. Reaction acts on insect for t secs,Insect get velocity Vi cm/sec in t secWooden piece get velocity 0.12cm / sec in t secAfter tsecs force = 0 & velocity do not changeConsider insect & wood stick as isolated systemPress Esc for returnPress Esc for return

Consider insect & wood stick as isolated systemAs both insect & wooden piece at at rest net momentum of system is = 0In second case movement of system = ( wwood /g x Vwood ) + ( winsect /g x Vinsect )As momentum is conserved, total movement = 0( wwood x Vwood ) + ( winsect x Vinsect ) = 0wwood x Vwood ) = ( winsect x Vinsect )--(g cancelled) 5 x 0.12 = 0.15 x Vi , Vi = 4cm/secVi0.12 cm/sec+ve

A car crashes into a wall at velocity 25 m/s and is brought to rest in 0.1 s. Calculate the average force exerted on a 75 kg mass test dummy by the seat belt. Also calculate dummies average accelerationExample 2

Visualizing the problem

We focus our attention on the dummy and forget the car.initial velocity of dummy v0 = 25 m/s)Mass of dummy (m0 = 75 kg)Dummys original momentum = P0P0 = m0 v0P0 = (75 kg) (25 m/s)P0 = 1875 (kg m)/sDummys final momentum = PfPf = mf vfPf = (75 kg) (0 m/s) = 0

P0 = 1875 (kg m)/s 2Pf = 0Change in dummys momentum = 1875 kg m/s.Now when an object interacts with other object it experiences an impulse = Average force x time for which force actsThe same is = Change in objects momentumLet F = seat belt force on dummyF x 0.1 = - 1875F = - 18750( kg m) / s2

Dummys velocity changed by 25 m/s (v), in a time of 0.1 s.v = acceleration timeacceleration = V / timeacceleration =25 m/s/0.1 sacceleration = 250 m/s2 = 25 g

Judy (mass 40.0 kg), standing on slippery ice, catches her leaping dog, Atti (mass 15 kg), moving horizontally at 3.0 m/s. what is the speed of Judy and her dog after the catchExample 3

(a) Before Judy catches her dog(b) After Judy catches her dog

(a) Before Judy catches her dog(b) After Judy catches her dog40 kg Po =Initial momentum = 40 x 0 + 15 x 3 = 45 kg m /sec Pf = (15+ 40 )x Vf = 55 Vf kg m /sec VfVf = 45/55 = 0.818m/sec

A cannon of mass M propels a cannonball of mass m horizontally with velocity Vb. What is the recoil velocity of the cannon? ( Mass of cannon>> mass of cannon ball)

Forces acting on the system are:Force exerted by the cannon on the cannon ball, as the cannon is fired,Equal and opposite reaction force exerted by the cannonball on the cannonThese forces are extremely large, but act only for a short instance in time, we call these impulsive forces.There are no external horizontal forces acting in horizontal direction under consideration.

Hence total horizontal momentum of the system is a conserved Prior to the firing of the cannon, the total momentum is zero ( nothing is initially moving)After the cannon is fired, the total momentum of the system takes the formmM

Hence total horizontal momentum of the system is a conserved Prior to the firing of the cannon, the total momentum is zero (since momentum is mass times velocity, and nothing is initially moving).After the cannon is fired, the total momentum of the system =

Elastic Collisions

In elastic collisions, both kinetic energy and momentum are conserved.Final momentum = initial momentumFinal Kinetic energy = initial kinetic energy

( Subscript 1 = Initial parameters before collision)( Subscript 2 = Final parameters before collision)ConventionBeforecollision

The kinetic energy of a point object (an object so small that its mass can be assumed to exist at one point), or a non-rotating rigid body, is given by the equation For example, one would calculate the kinetic energy of an 80kg mass traveling at 18 meters per second (40mph) as

Final momentumInitial momentumMomentum Conservation in Elastic collision U= velocity before collision V= velocity after collision

Final KEInitial KEKE is also conserved in Elastic collision

Solving these two equations we get

Example: Elastic Collision of Two BlocksA 4.0 kg block moving to the right at 6.0 m/s undergoes an elastic head-on collision with a 2.0 kg block moving to the right at 3.0 m/s. Find their final velocities.m1 = 4 kgm2 = 2 kg

Solution :Elastic Collision of Two Blocksm1 = 4 kgm2 = 2 kg

Solution :Elastic Collision of Two Blocksm1 = 4 kgm2 = 2 kg

Collision of a moving ball with a stationary ball with different weightsm1 = m2m1m2m1 > m2m1 < m2Solid wall

Particular cases 1 : m1 = m2, u2 = 0V1 = 0V2 = U1Before collision: u2= 0 (Stationary)u1As m1=m2

Particular cases 1 : m1 = m2, u2 = 0V1 = 0V2 = U1

Particular cases 2 : m1 > m2, u2 = 0V1 is + ve= 0V2 is + ve

Particular cases 3 : m1 < m2, u2 = 0V1 is - ve= 0V2 is + ve

Particular cases 3 : m1

Particular cases 4 : m1 , m2 tending to , u2 = 0V1 = - u1V2 = 0Elastic collision against wall

End

10 g bullet is fired from a 3.0 kg rifle with a speed of 500 m/s. Find recoil speed of rifle?

An astronaut of mass 60 kg is doing spacewalk to repair a satellite. He asked for a repair manual. You throw it to him at a speed of 4.0 m/s relative to the spacecraft. He is at rest when he catches the 3.0 kg book.CalculateHis velocity just after he catches the book.Initial and final KE of book-astronaut system.Impulse exerted by the book on the astronaut.

(a) velocity just after he catches the book. (b) initial and final kinetic energies of the book-astronaut system. (c) impulse exerted by the book on the astronaut.

important point :isolated systems are not equivalent to closed systems.Closed systems cannot exchange matter with the surroundings, but can exchange energyIsolated systems can exchange neither matter nor energy with their surroundings,

Inelastic Collisions in 1DTherefore, Kf < Ki and there is a net loss of energy in an inelastic collision.