plugin-solution_phy_7_2007

# plugin-solution_phy_7_2007 - Solutions for the Physics...

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Unformatted text preview: Solutions for the Physics 7A-AB Final, Fall 2007 Damien Martin December 15, 2007 Question 1 Two masses are connected by a rope, which is pulled over two pulleys as shown below. The masses are placed as in the diagram below, but start to move when both masses are “released” (i.e. the person holding the masses in place lets go of both masses). The rope does not stretch as the masses move, and we may neglect any change of energy of the pulleys. 2 kg 1 kg a) Indicate if the 1 kg mass goes up or down once released. [1%] Solution: The 1 kg mass must move up. Moving down would mean the 2 kg mass would move up because they are connected by a string, and so have to move equal amounts in opposite directions. Moving the 1 kg mass down and the 2 kg mass up would imply Δ PE > ⇒ Δ KE < 0. As KE i = 0 this would force KE to be negative! Therefore the 1 kg mass must move down (or stay in the same place). (An alternative explanation involves looking at the magnitudes of forces on the two blocks). b) Indicate if the 2 kg mass goes up or down once released. Solution: Down (see reasoning for part 1). c) When the 1 kg mass is released and has moved 10 cm, how far has the 2 kg mass travelled? We will refer to this state the final state in subsequent problems. [2%] Solution: The rope does not stretch, so both masses must move the same amount in opposite directions. Therefore the 2 kg mass must have moved 10 cm (down). d) What is the speed of the 1 kg mass moving in the final state? [10%] Solution: We start by noting that it is a closed system, and that because the rope does not stretch the velocities of the two objects must be the same. Let us summarise our findings on an energy- interaction diagram (not necessary for credit, but the energy conservation piece must be there!) KE (1kg) KE (2 kg) v i = 0 v i = 0 v f =?? v f =?? PE (1 kg) h f = h i + 0 . 1 m h PE (2 kg) h f = h i- . 1 m h v v Δ KE 1 + Δ P E 1 + Δ KE 2 + Δ P E 2 = 0 [+] [+] [+] [-] Putting in values: 1 2 m 1 v 2 + m 1 g (0 . 1 m) + 1 2 m 2 v 2 + m 2 g (- . 1 m) = 0 1 2 ( m 1 + m 2 ) v 2 = ( m 2- m 1 ) g (0 . 1 m) v = s 2( m 2- m 1 ) g (0 . 1 m) m 1 + m 2 = s 2(1kg)(10m/s 2 )(0 . 1 m) 3 kg = 0 . 816 m/s e) What is the speed of the 2 kg mass moving in the final state? [4%] Solution: As the string does not stretch. both the blocks must be going the same speed. Therefore the speed is also 0.816 m/s. Question 2 One mole of a diatomic (ideal) gas undergoes a cyclic process A → B → C → A , where A , B and C are the points shown in the PV diagram. Volume (m 3 ) C A B 1 1.5 2 1 1.2 a) Does the internal energy increase, decrease or stay the same over one cycle? (i.e. pick one) [2%] Solution: Internal energy is a state function; over a cycle the initial and final points are the same....
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plugin-solution_phy_7_2007 - Solutions for the Physics...

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