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Momentum

Momentum. Momentum can be defined as "mass in motion.". Momentum = mass * velocity. p = m * v. Units are. kg*m/s. Momentum is a vector quantity. It has both magnitude and direction. A 130 kg defender moving forwards at 9 m/s will have. (130 kg) (9m/s).

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Momentum

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  1. Momentum Momentum can be defined as "mass in motion." Momentum = mass * velocity p = m * v Units are kg*m/s Momentum is a vector quantity. It has both magnitude and direction A 130 kg defender moving forwards at 9 m/s will have (130 kg) (9m/s) = 1170 kgm/s of forward momentum pD = 1170 kgm/s A 84 kg running back moving sideways at 6 m/s will have (84 kg) (6m/s) = 504 kgm/s of sideways momentum pB 504 kgm/s What happens when the defender tackles the running back?

  2. Conservation of Momentum If there are no outside forces acting on an object its momentum will stay the same. Friction is an outside force that will change the momentum of an object In a collision between two objects the total momentum just before the collision will be the same just after the collision Total Momentum just before collision = Total Momentum just after collision pDi + pBi= pDf + pBf mDvDi + mBvBi = mDvDf + mBvBf If the two objects stick together after they collide there is only one final velocity mDvDi + mBvBi = mDvf + mBvf= (mD + mB)vf p + If the collision occurs in a straight line you must assign velocity positive or negative p -

  3. Conservation of Momentum A 130 kg defender moving forwards at 9m/s tackles a 84 kg running back moving towards him at 6 m/s. If a sticky (perfectly inelastic) collision takes place what velocity do the players move off after the collision? mDvDi + mBvBi = (mD + mB)vf p + (130kg)(+9 m/s) + (84kg)(-6m/s) = (130kg + 84kg) vf p - +666 kgm/s = (214kg) vf +3.12 m/s = vf If the collision is at an angle you must add the initial momentum vectors tip to tail and use the parallelogram rule to find the total momentum pBi pDi pTotal I = pTotal f

  4. Conservation of Momentum A 130 kg defender moving forwards at 9m/s tackles a 84 kg running back moving sideways at 6 m/s. If a sticky (perfectly inelastic) collision takes place what velocity (speed and direction) do the players move off after the collision? pBi = 504 kgm/s pDi = 1170 kgm/s pTotal I = pTotal f pTotal f =  (pBi2 + pDi2)  pTotal f =  ((504 kgm/s)2 + (1170kgm/s)2) Using scale diagram: 1cm = 100 kgm/s and measure the angle with a protractor pTotal f = 1273.9 kgm/s (mD + mB) vf = 1273.9 kgm/s vf = 1273.9 kgm/s / (130kg + 84kg)  = 23.30 vf = 5.95 m/s = 6 m/s

  5. Explosions In an explosion internal forces are responsible for the object breaking apart. Because the pieces impart equal and opposite forces on each other (Newton’s third law) these internal forcescannot provide a net change in momentum so momentum must be conserved in explosions Consider a falling firecracker that explodes into two pieces. The momenta of the fragments combine by vector rules to equal the original momentum of the falling firecracker. Note: If an object is at rest and then explodes the total momentum of all the pieces just after the explosion must all add up to zero!

  6. Impulse An impulse is a force exerted over a finite period of time A net impulse will change the momentum of an object in accordance with Newton’s second law of motion FNET = ma = m (v / t) therefore FNET t = m v =  (mv) = p A 0.2 kg baseball traveling at 40 m/s is hit and returned at 50 m/s in the opposite direction. If the ball and bat are in contact for 0.002s determine the average force on the ball p + mball = 0.2 kg vbi = 40 m/s vBf = -50 m/s t = 0.002s Fav = ? p - Fav t = m v = m (vBf - vBi) therefore Fav = m (vBf - vBi) / t Fav = (0.2 kg) (-50 m/s - 40 m/s)/ (0.002s) = - 9000N the area under the f-t curve represents the change in momentum or impulse applied to the object INET= Fav t = Area

  7. Impulse There are two ways of changing the momentum of an object. It can be changed with a large force exerted over a short period of time or with a smaller force exerted over a longer period of time Extending the time for which a force acts is especially seen in sports - and vehicle safety - shocks, padded dash board, air-bags, seat belts, crush zones, safety barriers around bridge columns pads, mats, gloves, following through The strings on a tennis racket bend and the ball compresses. This gives a longer time for the rebounding forces to act, thus increasing the velocity of the ball. The front end of a modern car is designed to cave in in order to extend the time of the collision. This reduces the collision force experienced by the occupants of the car. Note: the force does not increase by having strings. This depends on the swing.

  8. Impulse and Conservation of Momentum According to Newton’s third law when two object interact FA= - FB Because the two objects interact for the same period of time FAt= - FB t This can be written in terms of impulse as…. IA= - IB Written in terms of changes in momentum…. pA= - pB acar >> aTruck ICar= - ITruck A massive truck and a small car are involved in a collision. Which object has the largest I) impulse ii) change in momentum iii) acceleration during the collision? i) Both vehicles have the same impulse acting on them but in opposite directions ICar= - ITruck mCarvCar = MTruckvTruck ii) Both vehicles have the same change in momentum but in opposite directions iii) The change in velocity of the car is greater than the change in velocity of the truck because of its smaller inertia (mass). The car’s acceleration is therefore greater than the truck’s acceleration. The car’s passengers would therefore feel the acceleration more than the truck driver.

  9. Deriving Conservation of Momentum According to Newton’s third law when two object interact FA= - FB Because the two objects interact for the same period of time FAt= - FB t This can be written in terms of impulse as…. IA= - IB Written in terms of changes in momentum…. pA= - pB mA (vAf - vAi) = - mB (vBf- vBi) Substitute for p…. mAvAf - mAvAi = - mBvBf+ mBvBi rearrange…. mAvAf+mBvBf= mAvAi + mBvBi ptotal f= ptotal i The total momentum after the collision = The total momentum before the collision If a shotgun of mass 3.0 kg fires shot having a total mass of 0.05kg with a muzzle velocity of 525 m/s, what is its recoil velocity? mg = 3.0 kg ms = 0.05 kg ptotal I = 0 kgm/s vsf = 525 m/s vgf = ? ptotal I = ptotal f = mgvgf + msvsf = 0 kgm/s vgf = - (0.05 kg) (+525 m/s)/ (3.0 kg) (3.0 kg) vgf + (0.05 kg) (+525 m/s)= 0 kgm/s vgf = - 8.75 m/s = - 9 m/s

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