CU PHYS221 Practice Test 3 Solution - vC

Bob is roller blading down leslie campbell avenue

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9. Bob is roller-blading down Leslie Campbell Avenue with a velocity of 4.00 m/s East when he collides with his friend Stanley roller-blading in the opposite direction with a speed of 2.00 m/s. If 45.0 kg Bob and 60.0 kg Stanley remain entangled together after their collision, what is their final velocity? Find : v f Know : m B = 45.0 kg v B-i = +4.00 m/s x m S = 60.0 kg v S -i = −2.00 m/s x Perfectly Inelastic collision Sketch : Equation(s) : ∑ p initial = ∑ p final p = m v 8 Version C v Bob-i v Stanley-i Final Initial
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PHYS221 Practice Test 3 Solve : This is a conservation of momentum problem for a one dimension perfectly inelastic collision If we define East as the direction of + x , and use the subscripts “B” for Bob and “S” for Stanley, the initial velocities are: “ v B-i = +4.00 m/s x ” and “ v S-i = −2.00 m/s x Write conservation of momentum equation in the x direction for the perfectly inelastic collision m B v B-i + m S v S-i = (m B + m S ) v f Solve for v f : v f = m B v B-i + m S v S-i (m B + m S ) Substitute known values “m B = 45.0 kg”, “ v B-i = +4.00 m/s x ”, “m S = 60.0 kg”, “ v S-i = −2.00 m/s x ” to solve for “ v f ”: v f = (45.0 kg)(+4.00 m/s) + (60.0 kg)(−2.00 m/s) = +0.57 m/s x {(45.0 kg) + (60.0 kg)} Since we defined East as the direction of + x , v f = 0.57 m/s East
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