Tmmkpb=+=20π.44 .s(b) The speed before the collision (since it is at its maximum, passing through equilibrium) is v0= xmω0where0= 2π/T0; thus, v0= 3.14 m/s. Using momentum conservation (along the horizontal direction) we find the speed after the collision. =+=2.61.0Vvmbm/sThe equilibrium position has not changed, so (for the new system of greater mass) this
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.