Test 2 sol

# Test 2 sol - momenta of the two carts must add to zero 0 =...

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Solutions for Test Two 1)Work is defined as force times the distance over which the force operates: W = F x D W = (70 N) x (20 m) W = 1400 joules 2)Using the definition of kinetic energy, E = ½ mv 2 Change in energy = ½ m (v 2 ) 2   –  ½ m (v 1 ) 2 = ½ m [ (v 2 ) 2  – (v 1 ) 2  ] = ½ (10 kg) [ 1600 (m/s) 2  – 900 (m/s) 2  ] = 3500 kg (m/s) 2   =  3500 joules 3)By the law of conservation of momentum, the momentum of all the objects before the  release must equal the momentum of all the objects after the release.  Since the two  carts are not moving beforehand, the initial momentum is zero, so afterwards the

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Unformatted text preview: momenta of the two carts must add to zero: 0 = m 1 v 1 + m 2 v 2 But, m 1 = 6m 2 0 = 6m 2 v 1 + m 2 v 2- 6m 2 v 1 = m 2 v 2- 6 v 1 = v 2 So, the lighter cart rolls 6 times faster (and the negative sign indicates that it rolls in the opposite direction). 4)Use Newton's Second Law, and divide one equation by the other: F = M 1 x A 20F = M 2 x A “F” and “A” are the same in each equation (since the boxes accelerate at the same rate), so they cancel out when you divide: 1/20 = M 1 / M 2 The mass ratio of the two boxes is 1/20....
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## This note was uploaded on 02/13/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Test 2 sol - momenta of the two carts must add to zero 0 =...

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