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Unformatted text preview: practice 08 – GADHIA, TEJAS – Due: Mar 8 2007, 3:00 pm 1 Question 1 part 1 of 3 10 points A massless, vertical spring of force constant k is attached at the bottom to a platform of mass m p , and at the top to a cup which has a mass m c and the ball collides with it inelastically. The platform rests on a scale. A ball of mass m b is dropped into the cup from a negligible height. Scale m b m c m s = 0 m p h Figure: h is the height of the ball above the bottom of the cup; i.e. , the distance the ball falls. In the first question h is negligible and in the second question h is appreciable. What is the reading W on the scale when the spring is compressed an amount d = m b g k ? 1. W = m p g 2. W = m b g 3. W = m c g 4. W = ( m p + m c ) g 5. W = ( m p + m b ) g correct Explanation: The freebody diagram shows the forces acting on the platform when the spring is partially comparessed. The scale reading W is the force the scale exerts on the platform and is represented on the freebody diagram by vector F n . F n m p vectorg vector F on spring platform We can use Newton’s second law to deter mine the scale reading W in Part 1. We’ll use both conservation of energy and momen tum to obtain the scale reading when the ball collides inelastically with the cup. Applying summationdisplay F y = ma y to the spring when it is compressed a distance d , we have F n m p g F ball on spring = 0 . Solving for F n , we have F n = m p g + F ball on spring = m p g + k d = m p g + k parenleftBig m b g k parenrightBig = m p g + m b g = ( m p + m b ) g . Question 2 part 2 of 3 10 points The ball is dropped from a height h , collides inelastically with the cup, and comes to rest momentarily with the spring compressed. What is the reading W on the scale? 1. W = g m p 2. W = g bracketleftBigg m p + m b radicalBigg 2 k h g ( m b + m c ) bracketrightBigg cor rect 3. W = g m b radicalBigg 2 k h g ( m b + m c ) practice 08 – GADHIA, TEJAS – Due: Mar 8 2007, 3:00 pm 2 4. W = g bracketleftBigg m p m b radicalBigg 2 k h g ( m b + m c ) bracketrightBigg Explanation: Letting the zero of gravitational energy be at the initial elevation of the cup and v b i represent the velocity of the ball just before it hits the cup, use conservation of energy to find this velocity, we have Δ K + Δ U g = 0 , where K i = U g f = 0 . So 1 2 m b v 2 b i mg h = 0 and v b i = radicalbig 2 g h. Using conservation of momentum to find the velocity of the center of mass, we have vectorp i = vectorp f v cm = m b m b + m c v b i = parenleftbigg m b m b + m c parenrightbigg radicalbig 2 g h. Applying conservation of energy to the colli sion to obtain, we have Δ K cm + Δ U s = 0 or , with K f = U si = 0, 1 2 ( m b + m c ) v 2 cm + 1 2 k x 2 = 0 ....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Force, Mass

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