Newton’s Laws 2nd Law The acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. The direction of the acceleration is in the direction of the applied net force. (F = ma)
Newton’s Laws 3rd Law Whenever one object exerts a force on a second object, the second exerts an equal and opposite on the first.
Newton’s Laws Forces occur in pairs in every interaction.Use your foot to push against the floor to move, The floor pushes back.You push backwards against the water in a pool, The water pushes back and moves you forward.
Newton’s Laws What happens if the force pair is not obvious? E.g., a rock falling down a cliff? Earth pulls rock Rock pulls earth.
Newton’s Laws E.g., Car accelerating from a stop light? Tyre pushes on road Road pushes on tyre
Newton’s Laws Rifle and bullet Rifle pushes bullet out of barrel. Bullet pushes rifle back into the shoulder, Kickback.
Newton’s Laws Be careful choosing the action/reaction pairs. FN’ is the reaction force to FN FN’ = Force exerted on the table by the statue The reaction force to Fg is not shown.
Newton’s Laws Newton’s Second Law also applies Mass of bullet is small Acceleration is large Mass of rifle is relatively large Acceleration is large
Newton’s Laws 747 Does a 747 move by exhaust gases pushing on the atmosphere? If so, what would it push against in space? Exhaust gases push against the rocket Rocket pushes against the exhaust gas and moves forwards.
Newton’s Laws Waging Tail of Dog Tail wags dog Most noticeable when tail has a large relative mass
Newton’s Laws Why don’t forces cancel? Kick football: it accelerates away from you. There is only one force: By NII, if force applied, object accelerates. What about equal and opposite force? It doesn’t act on the ball: It acts on your foot.
Newton’s Laws Two people kicking the same soccer ball with equal but opposite force? Two forces on the ball. Cancel each other out.
Newton’s Laws TOK Newton created many laws. What makes a law? How does the meaning of the word law change from subject to subject? Laws simplifies concepts and makes things easier to understand. What are we losing by simplification?
Inertial Mass Kick an empty tin can and: it moves. Kick a can full of sand and it: moves a bit Kick a can full of lead and: it hurts, but doesn’t move.
Inertial Mass The lead filled can has more inertia than the sand filled can. The can with the most matter has the greatest inertia. The amount of inertia an object has depends on its mass. The amount of matter in the object.
Inertial Mass The more mass an object has: The more for it takes to change its state of motion. As F = ma: m = F/a Inertial mass is: Ratio of resultant force to acceleration
Inertial Mass An object with 3 times the mass needs: 3 times the force. If your mass is 60 kg, you have: 60 times the inertia than a 1 kg bag of sugar. If you have a mass of 90 kg and: an opponent on the football field has, a mass of 60 kg
Inertial Mass You have 1.5 times the inertia of the opponent. When you collide: Travelling at the same speed, He will get knocked over.
Gravitational Mass The amount of gravitational mass an object has can be determined by experiment. Accelerate objects with: different masses in, a gravitational field.
Gravitational Mass If we conduct the experiment in a vacuum: No frictional forces. Drop a light and heavy object at the same time: They reach the ground at the same time. They must have the same acceleration.
Gravitational Mass The force acting on both objects is: The weight of the object F=W Mass is a measure of gravitational force of attraction of one body on another. One body will attract another body with the same force as the second body attracts the first.
Gravitational Mass No experiments have shown that inertial mass and gravitational mass are not equal. This is called the equivalence principal: Gravitational mass is equivalent to inertial mass.
Gravitational Mass Physicists have thought for a long time that: Mass is the source of gravitation, Has the inertial property of resistance. This overlooks how different the two properties are. It took Einstein to conclude that Inertia and mass are equivalent.
Gravitational Mass The values of the two are numerically equal: ag = 9.81 m s-2 Gravitational Field Strength = 9.81 N kg-1
Weight Commonly, weight is the measure of: The force of gravity on an object. You can use a spring balance to: measure your weight in Panama. Repeat the procedure in Fairbanks Alaska and: the scale will read ½ % more.
Weight Fly to Mars and the reading would be: 1/3 of the original reading. The problem occurs because: we are measuring the force, trying to stretch the spring.
Weight The force resisting the stretching of the spring is: The same while, The force of gravity is not. This is a non-equilibrium situation. Use a Beam Balance and the reading will be different.
Weight If two objects have the same weight in Panama: The scale will come to a balance position. Moving to Fairbanks: Both objects will weigh less and, Still come to a balance position. It also works on Mars.
Weight If we use F = ma to determine the weight: We get a particular value.If we use a scale on a balance: We get a different value.The term weight: Can be ambiguousMake sure you read the ‘weight’ in an equilibrium situation
Difference Between Mass & Weight Weight is a measure of the gravitational attraction between bodies. Mass is the amount of matter in that same body.
Difference Between Mass & Weight Mass does not vary: It has the same number of atoms in it. Weight does vary: It depends on how strong the gravitational force of attraction is. Weight can be related to mass by the formula: W = ma
Momentum We have previously considered single particles. We now need to consider the interaction of two particles.
Momentum Newton’s 3rd law describes how: there is an equal and opposite force, in every interaction. The force which the earth pulls the moon around in a circular orbit: is matched by an equal and opposite gravitational force, exerted by the Moon on the Earth.
Momentum To investigate the force acting in an interaction between particles: we must make sure the interaction is, isolated from any external influences. Examples of isolated systems are:
Momentum The contents of a thermos are: thermally isolated from the atmosphere. A sound proofed room is: acoustically isolated.
Momentum on an air track. Consider two gliders colliding If the force applied on glider 2 by glider 1 is F1, then: the force on glider 1 by glider 2 is, according to Newton’s 3rd law; -F2.
Momentum law;From Newton’s 2 ndm1a1 = -m2a2and as acceleration is defined as : v v1 u1 a1 t tand acceleration by glider 2 is; v2 u 2 a2 t
Momentum that:We have assumed the time taken for the collision is, the same for both gliders.The equation now becomes: v1 u1 v2 u2 m1 m2 t tMultiplying both sides by t:
Momentum m1 (v1 u1 ) m2 (v2 u2 ) m1v1 m1u1 m2v2 m2u2Re-arranging so all quantities before the collision are on one side of the equation: m1u1 m2u2 m1v1 m2v2
Momentum This tells us that if we add the vectors before the collision: it will equal the vectors after the collision.The quantity mv is conserved: When no external forces act.
Momentum To simplify matters, we call the quantity mv: Momentum. The momentum p of an object is defined as: the product of its mass and velocity, p = mv. The units of momentum are: kg m s-1.
Momentum We stated previously, the quantity mv, or momentum, is conserved. This law can be stated: The total momentum of an isolated system is constant and: is unaffected by the interaction of it’s parts.
Cons of Momentum Law Provided no external forces act, the total momentum of a system of interacting particles remains constant, despite the interaction of its parts.
Momentum Watch as a 2 kg brick is dropped on a 3 kg loaded cart.
Momentum What changes when the cart is only 1 kg and the brick is 2 kg?
Example A neutron of mass 1.67 x 10-27 kg collides head on with a stationary nitrogen nucleus of mass 23.1 x 10- 27 kg. The initial velocity of the neutron is 1.50 x 107 ms-1 and it rebounds after the collision with a velocity of 1.30 x 107 ms-1 in the opposite direction. What is the final velocity of the nitrogen nucleus?
Solution mneutron= 1.67 x 10-27 kg mnitrogen = 2.31 x 10-26 kg uneutron = 1.50 x 107 m s-1 vneutron = -1.30 x 107 m s-1 unitrogen = 0 ms-1 vnitrogen = ? ms-1
Solution v neutron v neutron v nitrogenm neutron m neutron m nitrogen Before After
Solution By the law of conservation of momentum pi = pf pi = mneutron uneutron + mnitrogen unitrogen pi = (1.67 x 10-27) x (1.50 x 107) + 0 pi = 2.505 x 10-20 kg m s-1
Solution pf = mneutron vneutron + mnitrogen vnitrogen pf = (1.67 x 10-27) x (-1.30 x 107) +(2.31 x 10-26) x vnitrogen pf = (2.31 x 10-26 vnitrogen) - 2.171 x 10-20 As pi = pf
Solution 2.505 x 10-20 = (2.31 x 10-26 vnitrogen) - 2.171 x 10-20 2.505 x 10-20+2.171 x 10-20 = 2.31 x 10-26 vnitrogen 4.676 x 10-20 = vnitrogen 2.31 x 10-26 vnitrogen = 2.02 x 106 m s-1 in the initial direction of the neutron
Example A car of mass 750 kg was travelling due north when it collided with a truck of mass 2000 kg which was travelling due east. The police accident investigation squad determined from witnesses that the truck was travelling at a speed of 36 km h-1 and that, after the collision, the car and truck stuck together and moved on the direction N53.1oE.
Example Neglect friction in answering the following questions. a) Was the car exceeding the speed limit of 60 km h- 1? b) What was the common speed of the car and the truck after the collision?
Solution 20000 tan 53.1 o 750 uc 20000 uc 750 x tan 53.1o uc = 20.02 m s-1
Solution – Part (a) Speed of car = 20.0 ms-1 20 x 3600 1000 = 72 kmh-1 The car was exceeding the speed limit.
Solution – Part (b) o 20000 sin 53 .1 2750 x vc T 20000 vc T 2750 x sin 53.1o = 9.09 m s-1 32.7 km h-1
Impulse Application of Newton II During collisions, objects are deformed. F t
Impulse F = ma For constant acceleration: v F m t (v u) F m t
Impulse Ft = mv – mu Ft = p Ft is called the: impulse of the force. This impulse causes the: momentum to change. An impulse is a short duration force: usually of non-constant magnitude.
Impulse Units are the same as those for momentum: kg m s-1 or s N Defined as: the product of the force and the time, over which the force acts. During collisions: t is often very small so, the Fav is often very large.
Newton’s Law 2nd Revisited We can now redefine the term force. Force is the rate of change of: the linear momentum of a body. From this we can derive Newton’s 2nd Law.
Newton’s Law 2nd Revisited p p p f iFnet t t mv muFnet t m(v u)Fnet t
Newton’s Law 2nd Revisited m v Fnet t Fnet = ma Impulse Question
Momentum and Energy The total energy in an isolated system is conserved but: energy can be transferred from one object to another and, can be converted from one form to another.
Momentum and Energy The units are Joules (J) and: it is a scalar quantity and, does not have a direction. In collisions: total energy is always conserved.
Momentum and Energy The kinetic energy will not always remain constant but: may be converted to other forms. This could be: rotational kinetic energy, sound or, heat. There are three types of collisions:
Momentum and Energy 1. Elastic: Momentum is conserved and: no kinetic energy is lost. This occurs on the microscopic scale: such as between nuclei.
Momentum and Energy What happens when a light car rear-ends a heavy truck and collides elastically? Calculate the kinetic energy.
Momentum and Energy Before Ek = ½ mv2 Ek = ½ x 1000 x 202 Ek = 2.0 x 105 J
Momentum and Energy After Ek = ½mv2 + ½mv2 Ek = (½x 1000 x 102) + (½x 3000 x 102) Ek = (50000) + (150000) Ek = 200000 Ek = 2.0 x 105 J Ek is conserved
Momentum and Energy What happens when the heavy truck rear-ends the light car elastically?Calculate the kinetic energy.
Momentum and Energy Before Ek = ½ mv2 Ek = ½ x 3000 x 202 Ek = 600000 J Ek = 6.0 x 105 J
Momentum and Energy After Ek = ½mv2 + ½mv2 Ek = (½x 3000 x 102) + (½x 1000 x 302) Ek = 150000 + 450000 Ek = 600000 J Ek = 6.0 x 105 J Ek is conserved
Momentum and Energy 2. Inelastic: Momentum is conserved but: kinetic energy is lost. All macroscopic collisions are inelastic. Some collisions are almost elastic Billiard balls and, air track/table gliders.
Momentum and Energy What happens in a head-on where the collision is inelastic? Calculate the kinetic energy.
Momentum and Energy Before Ek = ½mv2 + ½mv2 Ek = (½x 1000 x 202) + (½x 3000 x 202) Ek = 20000o + 600000 Ek = 800000 J Ek = 8.0 x 105 J
Momentum and Energy After Ek = ½mv2 + ½mv2 Ek = (½x 1000 x 102) + (½x 3000 x 102) Ek = 50000 + 150000 Ek = 200000 J Ek = 2.0 x 105 J Ek is not conserved
Momentum and Energy 3. Perfectly inelastic: Momentum is conserved but: kinetic energy is lost and, the bodies stick together, after the collision. Elastic & Inelastic Collisions
Momentum and Energy What happens when a light car rear-ends a heavy truck and collides and sticks together? Determine the kinetic energy.
Momentum and Energy Before Ek = ½ mv2 Ek = ½ x 1000 x 202 Ek = 200000 J Ek = 2.0 x 105 J
Momentum and Energy After Ek = ½mv2 + ½mv2 Ek = (½x 3000 x 52) + (½x 1000 x 52) Ek = 37500 + 12500 Ek = 50000 J Ek = 5.0 x 104 J Ek is not conserved
Momentum and Energy What happens when the truck rear ends the car and they stick together? Calculate the kinetic energy.
Momentum and Energy Before Ek = ½ mv2 Ek = ½ x 3000 x 202 Ek = 600000 J Ek = 6.0 x 105 J
Momentum and Energy After Ek = ½mv2 + ½mv2 Ek = (½x 3000 x 152) + (½x 1000 x 152) Ek = 37500 + 12500 Ek = 337500 J Ek = 3.4 x 105 J Ek is not conserved
Example Show that the collision between the neutron and the nitrogen nucleus in the earlier example is elastic.
Solution mnitrogen = 2.31 x 10-26 kg uneutron = 1.50 x 107 m s-1 unitrogen = 0 ms-1 vneutron = -1.30 x 107 m s-1 vnitrogen = 2.02 x 106 m s-1 Ekinitial = ½mneutronu2neutron + ½mnitrogenu2nitrogen
Solution = ½(1.67 x 10-27) x (1.50 x 107)2 + ½(2.31 x 10-26) x 02=1.88 x 10-13 JEkfinal = ½mneutronv2neutron + ½mnitrogenv2nitrogen= ½(1.67 x 10-27) x (1.30 x 107)2 + ½(2.31 x 10-26) x(2.02 x 106)2=1.88 x 10-13 JSince Ekinitial = Ekfinal , the collision is elastic.
Example An atom of helium gas moving with a speed of 3.0 x 102 ms-1 collides elastically with a stationary atom of the same gas. After the collision the first atom moves off at an angle of 30o to its initial direction.a) Find the speed of the incident atom after the collision. b) Find the velocity of the struck atom after collision.
Solution – Part (a) The masses of the incident atom, and the struck atom, ms, are equal since: they are both helium atoms. mi = ms = m The collision is isolated, and so:
Solution – Part (a)p =p i f miui = mivi + msvs i.e. mui = mvi + mvs Thus ui = vi + vs
Solution – Part (a) The collision is elastic, so Eki = Ekf ½miui2 = ½mivi2 + ½msvs2 i.e. ½mui2 = ½mvi2 + ½mvs2 Thus ui2 = vi2 + vs2
Solution – Part (a) Therefore ui is a hypotenuse and = 90o in the vector- addition triangle, by Pythagoras’s theorem. By trigonometry,
Solution – Part (a) vi = uicos30o = 300 x 3/2 = 2.6 x 102 ms-1 Notice that we will always get = 90o just from the masses being equal and the collision elastic.
Solution – Part (b) By trigonometry in the vector diagram, vs = uisin30o = 300 x ½ =150 m s-1 vs= 1.5 x 102 m s-1 at 90o to vi