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Chapter 7 Impulse and Momentum 1) Definition p = mv 2) Impulse For a constant force, r J Ft From the 2nd law, r r (mv) p F , t t so r r J Ft p If the force is not constant, use the average force r r J Ft p 7) Conservation of momentum in 2d r r r p1 p2 p1 p2 v1 v2 r r r r m1v1 m2 v2 m1v1 m2 v2 m1v1x m2 v2 x m1v1x m2 v2 x m1v1y m2 v2 y m1v1y m2 v2 y If initial conditions are known, this gives 2 equations, with 4 unknowns, so more information is needed. v1’ v2’ 7) Conservation of momentum in 2d a) Inelastic collision: p1’ = p2’ = p’ reduces unknowns to px’ and py’ Example: Find v’ if m1 = 1450 kg, m2 = 1750 kg, v1 = 11.5 m/s, v2 = 15.5 m/ss • conservation of x momentum m2 v2 m1 v1 v ' vx 2 vy 2 9.95 m/s tan vx x y vy m1v1x m2 v2 x m1v1x m2 v2 x 58.4º v’ m1v1x 0 (m1 m2 )vx m1 vx v1x 5.21 m/s m1 m2 • conservation of y momentum m1v1y m2 v2 y m1v1y m2 v2 y 0 m2 v2 y (m1 m2 )vy vy m2 v2 y 8.48 m/s m1 m2 7) Conservation of momentum in 2d b) Elastic collision: Energy conservation adds 3rd equation: 1 2 mv12 12 mv2 2 12 mv1 2 12 mv2 2 The last condition is determined by the shape & location of impact: e.g. F For a billiard ball collision, the angle of the object ball is determined by the line through the centres at the point of contact. 7) Conservation of momentum in 2d b) Elastic collision: Energy conservation adds 3rd equation: v2’ Example: Cue ball angle: F v1 Conservation of momentum r r v1 v1 v2 v2’ v1’ v1 1 • identical masses • elastic collision • m2 initially at rest Find 1 2 2? Conservation of energy v12 v1 2 v2 2 v1’ Therefore, by Pythagoras 90º 8) Centre of Mass a) Acceleration and force: The centre-of-mass of a system of particles (or 3d object) reacts to the sum of the forces like a point particle with a mass equal to the total mass of the particles. Ftotal r mtotal aCM If the total force is zero, the centre-of-mass does not accelerate. (If there are no external forces, the internal forces sum to zero by the 3rd law.) r r r r r r r For 2 masses, Ftotal F1 F2 , where F1 m1a1 , and F2 m2 a2 are the forces on the two masses r r m a m2 a2 aCM 1 1 m1 m2 8) Centre of Mass b) Momentum and velocity Since a = v/t, r r m1 v1 m2 v2 r vCM m1 m2 But, since vCM = 0 if v1 = 0 and v2 = 0, this becomess vCM r r m1v1 m2 v2 m1 m2 The numerator represents the total momentum, which is conserved in the absence of external forces, so again, the centre-of-mass velocity is constant. Note that the CM momentum is simply equal to the total momentum: r r r (m1 m2 )vCM m1v1 m2 v2 8) Centre of Mass c) Position of the centre of mass Since v = x/t, the above gives r r m1 x1 m2 x2 r xCM m1 m2 If 2 particles coincide, they also coincide with the CM, so xCM r r m1 x1 m2 x2 m1 m2 In one dimension, For m1 = 5.0 kg, m2 = 12 kg, x1 = 2.0 m, and x2 = 6.0 m, xCM = 4.8 m