PHY11T2MOMENTUM - IMPULSE&MOMENTUM Impulse(J):

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IMPULSE & MOMENTUM
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Impulse (J) :   Refers to the product of the force and the time  interval it acts on the body.  J = F(t 2  – t 1 ) = F(Δt) where : F – applied constant force ( in Newtons) Δt – time interval (in seconds) J – Impulse ( in Ns) F m m v 1 v 2 t 1                                        t 2
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Momentum (p) :   Defined as the product of the mass and the  velocity of an object. p = mv where : m – mass (kg) v – velocity (m/s) p – momentum (kg-m/s) F m m v 1 v 2 t 1                                        t 2
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Both Momentum & Impulse are vector quantities, thus they have both horizontal & vertical components. And follow standard sign conventions X - component p x = mv x J x = F x (t 2 – t 1 ) = F x (Δt) Y - component p y = mv y J y = F y (t 2 – t 1 ) = F y (Δt) p = p x 2 + p y 2 J = J x 2 + J y 2
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F m m v 1 v 2 t 1                                        t 2 J = Δp FΔt  = mv – mv 1   “The change in momentum at any time interval equals the  impulse of the force applied during that time interval.” F x Δt  = mv 2x  – mv 1x   F y Δt  = mv 2y  – mv 1y  
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1. A baseball has a mass of 0.2kg.  (a) If the velocity of a pitched ball has a magnitude of 35m/s, and after the ball is batted the velocity is 55m/s in the opposite direction, find the change in momentum of the ball and the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2 ms (2x10 -3  s), find the average force applied by the bat. Given : Required : a) Δp = mv 2  – mv 1 Δp = (0.2kg)(+55m/s) – (0.2kg)(-35m/s) m = 0.2 kg v = 35 m/s v = 55 m/s F Δp = 18 kg-m/s J = Δp = 18 Ns b) F = ? With Δt = 2x10 -3 s J = FΔt 18 Ns = F(2x10 -3 s) 18 Ns/ (2x10 -3 s) = F 9,000 N = F F = 9,000 N or 9 kN SAMPLE PROBLEMS
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2.  Prob.8-12  ]  A  bat  strikes a  0.145  kg baseball.  Just before  impact, the ball is traveling
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PHY11T2MOMENTUM - IMPULSE&MOMENTUM Impulse(J):

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