lecturenotes-Feb22

lecturenotes-Feb22 - Center of Mass Problem A 60 kg man...

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A 60 kg man stands at the edge of a raft of mass 15 kg that is 10 meters long (the raft has a uniform density). The edge of the raft is against the shore of the lake. The man walks toward the shore, the entire length of the raft. How far from the shore does the raft move? Center of Mass Problem
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Quick Review: Velocity of COM In a closed, isolated system the COM velocity ( ) of the system is CONSTANT. Why? v com P tot = M v com = ( m 1 + m 2 ) v com P tot = p 1 i + p 2 i v com = P tot ( m 1 + m 2 ) = p 1 i + p 2 i ( m 1 + m 2 ) = p 1 f + p 2 f ( m 1 + m 2 ) Constant !! F net = d P tot dt = 0 Two bodies that form a closed, isolated system undergo an elastic collision in 1D. Which of the three choices best represents the position-versus-time ( x-t plot) of those bodies and their center of mass velocity ( v com )?
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m v p Linear momentum of a particle of mass and velocity The SI unit for linear momentum is the kg.m is defined a /s s . . p mv p m v = p mv = The time rate of change of the linear momentum of a particle is equal to the magnitude of net force acting on t Below we will prove the fol he particle and has the dir lowing statem ection of the ent: for ( ) net net In equation form: . We will prove this equation using Newton's second law: This equation is stating that the linear momentum of a particle can be c c a . h e n dp F dt dp d dv p mv mv m ma F dt dt dt = = = = = = ged only by an external force. If the net external force is zero, the linear momentum cannot change: net dp F dt = 9.4 Linear Momentum of a Particle
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Skaters Push Off In this (and all) “collisions” momentum is conserved: Is this “collision” elastic or inelastic? It has to be inelastic !! P initial = P final Two skaters of m 1 =54 kg and m 2 =88 kg push off and the woman moves off with v f1 =2.5 m/s. What is the velocity of the man? 0 = m 1 v 1 f + m 2 v 2 f v 2 f = m 1 v 1 f m 2 v 2 f = (54 kg )(2.5 m ) 88 kg = 1.5 m KE initial = 0 KE final = 1 2 m 1 v 1 f 2 + 1 2 m 2 v 2 f 2 = 1 2 (54)(2.5) 2 + 1 2 (88)( 1.5) 2 = 268 J Energy is not conserved here!
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Quick Review: Impulse & Collision Impulse = change in linear momentum Impulse-momentum theorem J F net ( t ) dt t i t f p f p i = Δ p = J Vector! Must satisfy for each direction! Collisions: Δ P = P f P i = 0 Δ KE = K f K i = 0 Δ P = P f P i = 0 elastic inelastic
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It is well known that bullets and other missiles fred at Superman simply bounce oFF his chest. Suppose that a gangster sprays Superman's chest with 10 g bullets at the rate oF 100 bullets/ min, the speed oF each bullet being 700 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude oF the average Force on Superman's chest From the stream oF bullets?
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lecturenotes-Feb22 - Center of Mass Problem A 60 kg man...

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