2LINEAR MOMENTUM p=mv Momentum = Mass x Velocity The SI unit for momentum is kg·m/sMomentum and velocity are in the same directionIs a vector
3Using the equation p=mv At the same velocity, as mass increases – momentum increasesAt the same mass, as velocity increases – momentum increases
4ExampleYou are driving north, a deer with mass of 146 kg is running head-on toward you with a speed of 17 m/s. Find the momentum of the deer.p=mvp=146 kg * 17 m/sp= 2500 kg·m/s to the south
5CHANGING MOMENTUM A change in momentum takes time and force For example soccer – when receiving a pass it takes force to stop the ballIt will take more force to stop a fast moving ball than to stop a slow moving ballA toy truck and a real truck moving at the same velocity, it will take more force to stop the real truck than to stop the toy truck
6IMPULSE Impulse is the applied force times the time interval FΔt Impulse = FΔtForce is reduced when the time interval increasesExample – giving a little when you catch a ballTrampoline
7IMPULSE-MOMENTUM THEOREM From Newton’s second law (it will never go away)And the equation for acceleration: a=v/tWe can find the equation for force in terms of momentumF=Δp/ΔtForce = change in momentum / time intervalWe can also rearrange this equation to find the change in momentum in terms of external net force and timeΔp=FΔt, Δp=FΔt=mvf-mvi
8ExampleA 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in s. What is the force exerted on the ball by the receiver?
9p=mvΔp=FΔt=mvf-mvivi= 15 m/svf=0.0 m/st=0.020 sF*0.020 s = 0 – 0.50 kg*15 m/sF= 380 N
10IMPULSE-MOMENTUM THEOREM AND STOPPING DISTANCE The impulse-momentum theorem can be used to determine the stopping distance of a car or any moving objectUse Δp=FΔtSo Δt=Δp/F=mvf-mvi/FThen when you find time you can usedistance = average velocity * timeΔx=1/2(vf+vi) Δt
11ExampleA 2240 kg car traveling to the west slows down uniformly to rest from 20.0 m/s. The decelerating force on the car is 8410 N to the east. How far would the car move before stopping?Δt=Δp/F=mvf-mvi/FΔt=0-2240kg*20.0m/s / 8410 NΔt = 5.33 sΔx=1/2(vf+vi) ΔtΔx=1/2(0+20.0m/s)*5.33x=53.3 m to the west
13What happens to momentum when two or more objects interact? First you have to consider total momentum of all objects involved. This is the sum of all momentumsLike energy momentum is conservedConservation of Momentum:Total initial momentum = total final momentumm1v1i + m2v2i = m1v1f + m2v2f
14ExampleA boy on a 2.0 kg skateboard initially at rest tosses an 8.0 kg jug of water in the forward direction. If the jug has a speed of 3.0 m/s relative to the ground and the boy and the skateboard move in the opposite direction at 0.60 m/s, find the mass of the boy.
15m1v1i + m2v2i = m1v1f + m2v2fm1=mass of jug = 8.0 kgm2= mass of boy and skateboard= 2.0 kg + xV1i = 0, v2i = 0V1f = 3.0 m/s, v2f=0.60 m/s0=8.0kg*3.0m/s forward + (2.0kg +x)*.60m/s back24 kg·m/s = (2.0 kg + x)*.60 m/s40 kg = 2.0 kg + xx= 38 kg
16Newton’s Third law and Collisions If the forces exerted in a collision are equal and oppositeAnd the time each force is exerted would be the sameThan the impulse on each object in a collision would be equal and oppositeSince impulse is equal to the change in momentum than the change in momentum would be equal and oppositeSo if one object gained momentum after a collision than the other object must lose the same amount of momentum
17CollisionsWhen two objects collide and then move together as one mass, the collision is called aPerfectly inelastic collisionThese are easy situations because the two objects become pretty much one object after the collisionSo the conservation of momentum equation becomesm1v1i + m2v2i = (m1 + m2)vfThe two objects will have the same final velocity
18ExampleA grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg grocery cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and the bag?
19m1v1i + m2v2i = (m1 + m2)vfkg * 5.5 m/s = (18kg + 9kg)vf50. kg m/s = 27*vf1.9 m/s
20Kinetic Energy and Inelastic Collisions Total kinetic energy is not conserved in inelastic collisions, it does not remain constantSome of the energy is converted into sound energy and internal energy as the objects deforms during the collisionThis is why it is called inelastic, elastic usually means something that can keep its shape or return to its original shapeIn physics elastic means that the work done to deform an object is equal to the work done to return to its original shapeIn inelastic collisions some of the work done on the inelastic material is converted to other forms of energy such as heat or sound
21ExampleA 0.25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target.What is the final velocity of the combined mass?What is the decrease in kinetic energy during the collision?
22m1v1i + m2v2i = (m1 + m2)vfkg *12 m/s = (0.25 kg+ 6.8 kg)vvf=0.43 m/s to the westKEi = 0+ ½ 0.25*(12m/s)2= 18 JKEf = ½ (7.1 kg)*(0.43 m/s)2= .66 JChange in kinetic = 17 J
23Elastic CollisionsIn an elastic collision, two objects collide and return to their original shapes with no loss of total kinetic energy.The two objects move separatelyBoth total momentum and total kinetic energy are conserved
24Since both are conserved than these equations apply m1v1i + m2v2i =m1v1f +m2v2f½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2
25ExampleA 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head-on collision with a 14.0 kg raft moving to the right at 16.0 m/s. After the collision, the raft moves to the left at 14.4 m/s. Disregard any effects of the water.Find the velocity of the canoe after the collision.Verify your answer by calculating the total kinetic energy before and after the collision.
26Collisions in Two or more Dimensions Conservation of momentum still appliesUse your vectors.Momentum is conserved in all directionspxi=pxfpyi=pyf
28Center of Mass (CM)The point at which all mass of an object can be considered to be locatedAn object can be considered a point or small particle no matter what the size or shape of the object
29Center of mass moves just as a particle moves no matter what the object does
30Finding the center of mass of an object Define a coordinate system(preferably one that would make the math easy)Center of mass can be found by finding the sum of all the masses times their respective distance from the defined origin and dividing by the sum of all the masses
32Center of gravity (CG)The point at which the force of gravity can be considered to actGravity acts on all parts of the objectBut for determining translational motion we can assume that gravity acts in one particular spotSame spot as the center of mass