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Unformatted text preview: practice 08 – ALIBHAI, ZAHID – Due: Mar 19 2007, 4:00 am 1 Question 1 part 1 of 3 10 points Two small spheres of mass m 1 and m 2 are suspended from the ceiling at the same point by massless strings of equal length ℓ . The lighter sphere is pulled aside through an angle of θ i from the vertical and let go. The acceleration of gravity is 9 . 8 m / s 2 . g Before m 1 ℓ θ i m 2 g After θ f At what speed will the lighter mass m 1 hit the heavier mass m 2 ? 1. v 1 i = { g ℓ [cos θ i ] } 2. v 1 i = { 2 g ℓ [cos θ i ] } 1 2 3. v 1 i = { 2 g ℓ [1 cos θ i ] } 4. v 1 i = { 2 g ℓ [1 cos θ i ] } 1 2 correct 5. v 1 i = { g ℓ [1 cos θ i ] } 6. v 1 i = { g ℓ [1 cos θ i ] } 1 2 7. v 1 i = { 2 g ℓ [cos θ i ] } 8. v 1 i = { g ℓ [cos θ i ] } 1 2 Explanation: The velocity just before the collision v i can be determined by energy conservation. When particle 1 is at its initial condition, it is at rest and displaced by an angle θ i from the vertical. The total energy is all potential and is given by U i = m 1 g ℓ (1 cos θ i ) where ℓ (1 cos θ i ) is the distance above the lowest point. Just before the collision, the en ergy of sphere 1 is all kinetic energy, 1 2 m 1 v 2 1 i . Equating the two energies gives 1 2 m 1 v 2 1 i = m 1 g ℓ (1 cos θ i ) . Solving for v 1 i gives v 1 i = { 2 g ℓ [1 cos θ i ] } 1 / 2 . Question 2 part 2 of 3 10 points After lighter sphere is let go and collides with the heavier sphere at the bottom of its swing, two spheres immediately bind to gether. What is conserved in this collision process? Let E = mechanical energy; P = momentum. 1. E 2. Both E and P 3. P correct 4. Neither E nor P Explanation: This is a perfectly inelastic collision. The speed of the two spheres after collision is de termined by momentum conservation. Question 3 part 3 of 3 10 points After the lighter sphere is let go and col lides with the heavier sphere at the bottom of its swing, the two spheres immediately bind together. practice 08 – ALIBHAI, ZAHID – Due: Mar 19 2007, 4:00 am 2 What is the speed V f of the combined sys tem just after the collision? 1. V f = parenleftbigg m 1 + m 2 2 m 2 parenrightbigg v 1 i 2. V f = parenleftbigg m 1 + m 2 2 m 1 parenrightbigg v 1 i 3. V f = parenleftbigg m 2 m 1 + m 2 parenrightbigg v 1 i 4. V f = parenleftbigg m 1 + m 2 m 2 parenrightbigg v 1 i 5. V f = parenleftbigg 2 m 2 m 1 + m 2 parenrightbigg v 1 i 6. V f = parenleftbigg 2 m 1 m 1 + m 2 parenrightbigg v 1 i 7. V f = parenleftbigg m 1 m 1 + m 2 parenrightbigg v 1 i correct 8. V f = parenleftbigg m 1 + m 2 m 1 parenrightbigg v 1 i Explanation: This is a completely inelastic collision. The speed of the two spheres after collision is de termined by momentum conservation m 1 v 1 i = ( m 1 + m 2 ) V f (1) where m 1 is the mass and v 1 i is the initial velocity of sphere 1 just before the collision, m 2 is the mass of sphere 2, and V f is the velocity of the combined spheres just after the collision. Note: v 2 i = 0....
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 Spring '08
 Turner
 Kinetic Energy, Mass, Momentum, Work, Light, m/s

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