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MOMENTUM

MOMENTUM. Newton’s 3 rd Law. For every action there’s a reaction. For every action Force, there’s a reaction Force. For every action force ON AN OBJECT , there’s an equal & opposite reaction force BY THAT OBJECT. Momentum. Momentum represents the potential to “ do damage ”.

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MOMENTUM

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  1. MOMENTUM

  2. Newton’s 3rd Law • For every action there’s a reaction • For every action Force, there’s a reaction Force • For every action force ON AN OBJECT, there’s an equal & opposite reaction force BY THAT OBJECT.

  3. Momentum • Momentum represents the potential to “do damage” • Momentum is the product of mass & velocity. • Momentum is conserved in a closed system. • Represented by p in an equation Formula: p = m v Units: = kg m s kg m s

  4. Time for the fun part Practice Problem 1 A 200 kg astronaut is moving @ 4m/s, calculate his momentum. GivenFormulaEquation P= (200 kg) (4 m/s) m= 200 kg P= m v P= 800 kg m/s v= 4 m/s P = ?

  5. Impulse • Impulse is the product of force & time. • Impulse represents the change in momentum. • Represented by J Formula: J = F t Units: = Ns N* s Wow- same as momentum s kg m s2 kg m s

  6. Now, put it all together… And you get  P= P-Po= m v = F t= J

  7. Practice problem 2How long would a 5 N force have to act on the astronaut in problem #1 in order to reverse his velocity & make it the same magnitude?GivenFormulaEquation t= (-800 kg m/s - 800 kg m/s) (-5N) F t=P-Po F= -5 N m= 200 kg t= (P-Po) F Po = 800 kg m/s t= (-1600 kg m/s) (-5N) P= -800 kg m/s v= -4 m/s vo= 4 m/s t=320 s t= ?

  8. Or Try…… Practice problem 2How long would a 5 N force have to act on the astronaut in problem #1 in order to reverse his velocity & make it the same magnitude?GivenFormulaEquation t= (200 kg)(-8m/s) /(-5N) F t= m v  v=-8 m/s t= (-1600 kg m/s)/(-5N) t= (m v)/F m= 200 kg t= 320s F= -5 N t= ?

  9. 5) A 100 Kg man goes from 70 m/s to rest in 0.05 s when he hits thedash in a crash. Calculate the force on his body. • 5A.) GivenFormulaEquation vo = 70 m/s Δv = v - vo Δv = v - vo v = 0 m/s Δv = (0 m/s) – (70 m/s) t = 0.05 s Δv = - 70 m/s m = 100 Kg F = ? Ft=mΔvF = (m Δv) / t F =(100 Kg)(-70 m/s)/(0.05 s)F = (-7000 Kg m/s) / (0.05 s)F = -140,000 N

  10. 6.)  If a seatbelt was used in problem #5, the collision time could maybe be increased to 0.5 s. Recalculate the force on his body. 6A.) GivenFormulaEquation vo = 70 m/s Δv = v – vo Δv = v – vo v = 0 m/s Δv = (0 m/s) – (70 m/s) t = 0.5 s Δv = - 70 m/s m = 100 Kg F = ? F = (m Δv) / t F = (100 Kg)(-70 m/s)/(0.5 s) F = (-7000 Kg m/s) / (0.5 s) F = -14,000 N *NOTICE: Bigger time (10x bigger!) means smaller force (1/10!)

  11.  7) If an airbag was used as well, the collision time would be increased to 1 s. Find the force if an airbag is used. 7A.) GivenFormulaEquation vo = 70 m/s Δv = v - vo Δv = v - vo vf = 0 m/s Δv = (0 m/s) – (70 m/s) t = 1 s Δv = - 70 m/s m = 100 Kg F = ? F = (m Δv) / t F = (100 Kg)(-70 m/s)/(1 s) F = (-7000 Kg m/s) / (1 s) F = -7,000 N

  12. 8)Which of the previous 3 people received the greatest impulse? 8A) All received the same impulse. #5 #6 #7 J=Ft J=Ft J =Ft =-140,000 X 0.05 =-14,000 X 0.5 =-7,000 X 1 =-7,000 kgm/s =-7,000 kgm/s =-7,000 or since F t = Δp = m Δv you can show that all three had a mass of 100 kg and a Δv of –70 m/s they all had the same Δp and therefore the same impulse!! The same job was done-Just in a different way.

  13. The Fun StuffCollisions and Explosions • Pbefore should equal Pafter in all collisions and negative collisions if you look at the whole system. • “system” includes all objects which interact • Is your system closed (no outside forces)? • Yes- then the amount of P you have will be the same

  14. Collisions 1 Car A is going East @ 20 m/s and has a mass of 1000kg. Car B is going West @ 20 m/s and has a mass of 1000kg. Calculate their v after the collision (they stick together). Before: After: v=? Car A Car B m= 2000kg v= -20m/s v= 20m/s m = 1000kg P=0 m = 1000kg v= P/m P=-20,000kgm/s P=20,000kgm/s v= 0/2000kg Total P for the system= 0 kgm/s v= 0 m/s

  15. Collisions 2 Car A has a mass of 1500 kg and is traveling at a velocity of 20 m/s East. Car B has the same mass but is traveling at a velocity of 10 m/s West. Then they collide and stick together. Find their final velocity. Before: A(East (+))B(West) After: A+B v= -10m/s v= 20m/s v=? m= 1500kg m= 1500kg m= 3000kg P= -15,000kgm/s P= 30,000kgm/s P= 15,000kgm/s v= P/m Total P= 15,000kgm/s v=15,000kgm/s / 3000kg v= 5m/s East

  16. Collisions 3 A fly (0.005kg) is flying East @ 10 m/s. It collides head on with a 2,000kg Westbound car going 25m/s (speed limit). Of course they stick. Find the v. Before FlyCar After P= -49,999.95 kgm/s m= .005 kg m= 2,000 kg m= 2,000.005 kg v= 10 m/s v= -25 m/s v=? P= .05 kgm/s P= -50,000 kgm/s v= P/m Total P= 50,000.05 kgm/s? v= -24.9999 m/s No, Total P= -49,999.95 kgm/s Or v= 24.9999 m/s West Bet he doesn’t have the guts to do that again!!

  17. Explosions 1 A 4.3kg rifle fires a 94 gram bullet with a 1.167E-3 second 57,228.7 N force. Find v of the bullet and the gun. The total momentum before is zero. After GunBullet F= 57,228.7N F= -57,228.7N m= 4.3kg m= 94g v= ? v=? t= 1.167E-3 s t= 1.167E-3 Ft= m v v= Ft/m v= Ft/m =57,228.7N(1.167E-3s)/4.3kg =-57,228.7N(1.167E-3)/.094kg = -710m/s =15.5 m/s

  18. Using the information from the last problem, find out what would happen if you attached a 80kg person to the gun. What if…. .79 m/s

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