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# 2 momentum notes (all)

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Algebra Based Basic Momentum & Impulse Notes

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• mv + mv = mv + mv
• mv + mv = mv + mv
• mv + mv = mv + mv
• mv + mv = mv + mv
• mv + mv = mv + mv
• ### 2 momentum notes (all)

1. 1. MomentumMomentum Chapter 6Chapter 6
2. 2. Momentum (Symbol = p) [Unit = kg m/s]  Inertia in motion  The Quantity of Motion Momentum = Mass • Velocity p = m • v Newton’s 2nd Law Impulse = change in momentum ( ) f i f i f i v vvF ma a t t v v t F m t t Ft mv mv −∆= = = − • = • = − Impulse (Symbol = I)[Unit = Nsec]  A force in a certain time Concept: In order to change momentum, we need a force in a certain amount of time! f ipI Ft mv mv=∆= = −
3. 3. f ipI Ft mv mv=∆= = − Changing Momentum Difference between hitting concrete bridge vs. yellow barrels full of pea gravel Seat belt vs. Air bag Δp = Ft Δp = Ft 0 - m v = FT 0 - m v = Ft http://www.regentsprep.org/Regents/physics/phys01/impulse/default.htm Momentum is a vector, so direction is important!
4. 4. A 4.0 x 104 N rocket is at rest when it fires its thrusters. If the thrusters put out an average force of 5.0 x 103 N over 20 sec, then (a) What is the impulse the rocket experiences? (b) What is the rocket’s final velocity? (a) Impulse I = F t I = 5000 (20) I = 100,000 Nsec (b) v I = ∆p = mvf – mvi 100,000 = 4080 vf – 4080(0) vf = 24.5 m/s
5. 5. Conservation of MomentumConservation of Momentum Law of Conservation of MomentumLaw of Conservation of Momentum  Momentum cannot be created or destroyedMomentum cannot be created or destroyed Momentum In = Momentum outMomentum In = Momentum out ppinin = p= poutout 2 Types of Collisions Elastic  Energy is conserved Inelastic  Energy is lost to heat, sound, etc. Since we work in a happy, ideal world, we will deal with all elastic collisions. This is for Collisions!
6. 6. 1.1. They hit and all the momentum is transferredThey hit and all the momentum is transferred mA = 4 kg vA = 3 m/s mB = 2 kg vB = 0 m/s Before Collision mA = 4 kg vA = 0 m/s mB = 2 kg vB = ? m/s After Collision pin = pout mAvA + mBvB = mAvA + mBvB (4)(3) + (2)(0) = 4(0) + 2(vB) vB = 6.00 m/s
7. 7. 2.2. They hit stick togetherThey hit stick together 4 kg 3 m/s 2 kg 0 m/s Before Collision 4 kg ? m/s 2 kg ? m/s After Collision pin = pout mAvA + mBvB = (mA+mB)vAB (4)(3) + (2)(0) = (4+2)vB vAB = 2.00 m/s
8. 8. 3.3. They hit and bounce awayThey hit and bounce away 4 kg 2 m/s 2 kg -3 m/s Before Collision 4 kg -1 m/s 2 kg ? m/s After Collision pin = pout mAvA + mBvB = mAvA + mBvB (4)(2) + (2)(-3) = (4)(-1) + (2)vB vAB = 3.00 m/s ? ? ?
9. 9. 3.3. They hit and bounce awayThey hit and bounce away Before Collision After Collision pin = pout mAvA + mBvB = mAvA + mBvB (4)(2) + (2)(-3) = (4)(-1) + (2)vB vAB = 3.00 m/s 4 kg 2 m/s 2 kg -3 m/s 4 kg -1 m/s 2 kg ? m/s ? ?
10. 10. 4.4. Both start with zero velocityBoth start with zero velocity 4 kg 0 m/s 2 kg 0 m/s Before Collision 4 kg ? m/s 2 kg 20 m/s After Collision pin = pout mAvA + mBvB = mAvA + mBvB (4)(0) + (2)(0) = (4)(vA) + (2)(20) vA = -10.0m/s