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hw16 - T-mgH =-1 2 mv 2 3 gH = 1 2 me 4(2 gH H = e 4 H...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 16 (2.145, 2.155) 2.145 Lets first find the speed at which the ball hits the ground first time. Work done = Change in kinetic energy W = Δ T mgH = 1 2 mv 2 bottom , 1 v bottom , 1 = p 2 gH The ball now impacts and the resultant bounce back velocity is v 2 = e p 2 gH . It rises and falls and hits the ground again with the same velocity v 2 . It then rebounds with a velocity v 3 = e 2 p 2 gH . Now to find the height to which it rises we again make use of conservation of energy
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Unformatted text preview: T-mgH =-1 2 mv 2 3 gH = 1 2 me 4 (2 gH ) H = e 4 H 2.155 Considering the conservation oF momentum along the x direction m B v B + m C v C = m BC v BC 2(100cos30 ◦ ) + 3(0) = 5 v BC v BC = 20 √ 3 f t / sec The cart (with the ball in it) hits the spring with this speed. Using conservation oF energy W = Δ T-1 2 k δ 2 = 1 2 m f v 2 f-1 2 m i v 2 i-1 2 1500 δ 2 =-1 2 5(20 √ 3) 2 δ = r 6000 1500 δ = 2 f t 1...
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