10_25_Momentum_2_Again

10_25_Momentum_2_Again - 10/25 Momentum 2 Again 0 Text:...

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Unformatted text preview: 10/25 Momentum 2 Again 0 Text: Chapter7 (and 6 for Energy) 0 Presentation “Colliding Rockets” delayed a day 0 Upcoming presenters please see me 0 HW “Energy and Momentum” Due Friday 10/26 (do some today) 0 HW “Impulse” Due Monday 10/29 Momentum and the 2nd Law ∆v = a ∆t vf = vi + a mvf t mvi + ma ∆= ∆v mt f = mvi + Fnet ∆ t Impulse Momentum, p We used definition of acceleration and the 2nd law to define momentum and impulse. 2nd Law (No letter for this) Momentum and the 2nd Law mvf = mvi + Fnet ∆ t mvf - mvi = Fnet ∆ t m(vf - vi) = Fnet ∆ t m(∆ v) = Fnet ∆ t (divide by t and get 2nd law) ∆ p = Fnet ∆ t Changes in momentum equal impulse Example: Before 3 kg vi = 4 m/s 3 kg vi = 6 m/s 3 kg vi = 6 m/s ∆ p = m(∆ v) = F∆ t After 3 kg vf = 6 m/s 3 kg vf = 4 m/s 3 kg vf = 4 m/s ∆ p = 30 kg m/s left ? ∆ p = 6 kg m/s left ? ∆ p = 6 kg m/s right ? Momentum A A Mass Speed Before Speed After A 1 kg 8 m/s 1 m/s B 3 kg 2 m/s 5 m/s B B What is ∆ p for block A? A pA,i = 8 kg m/s pA,f = 1 kg m/s ∆ p = 9 kg m/s left What is ∆ p for block B? B pB,i = 6 kg m/s pB,f = 15 kg m/s ∆ p = 9 kg m/s The changes in p are equal and opposite! FA,B = -FB,A 3rd law and Ft = ∆ p. Bullet through block A vB,i = 100 m/s mB = 0.05 kg What is ∆ p for the B bullet? What is ∆ p f= 2block A? or kg m/s p = mv ∆ p = m(∆ v) = Ft mA = 2 kg (vB,i = 0) A vA,f = ? v = 60 m/s B,f A pB,i = 5 kg m/s A pB,f = 3 kg m/s What is pA,f for block A? pA,f ==22kkgm/s (equal and opposite) g m/s (vB,i = 0) ∆p vA,f = 1 m/s Bullet sticks in block A vB,i = 100 m/s mB = 0.05 kg What is ∆ p for the bullet? mA = 2 kg A ∆ p = m(∆ v) = Ft vA+B,f = ? A pB,i = 5 kg m/s ∆ pB = ? pB,f = ? Consider the system of A and B together and ∆ pA+B = 0 Why? There are no external forces on the system of A + B. Then pSYS,i = pSYS,f or ∆ pSYS = 0 Bullet sticks in block A vB,i = 100 m/s mB = 0.05 kg Then pA+B,i = pA+B,f mA = 2 kg A ∆ p = m(∆ v) = Ft vA+B,f = ? A mBvB,i +mAvA,i = (mB + mA)vA+B,f 5 + 0 = (2.05)vA+B,f (All directions to the right) vA+B,f = 2.4 m/s Is energy conserved? No. Mechanical energy was lost in friction between the bullet and block. (Inelastic collision) ...
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This note was uploaded on 12/21/2010 for the course PHYS Physics 21 taught by Professor Larrywatson during the Fall '08 term at North Dakota.

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10_25_Momentum_2_Again - 10/25 Momentum 2 Again 0 Text:...

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