# positive work, (b) negative work, (c)

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 CSULA FINAL EXAM (Fall 2007) SOLUTION PHYSICS 211 CONCEPTUAL QUESTIONS The Gravitational Force exerted by the Earth on the Moon holds the Moon in a circular orbit. Does Gravity do: (a) positive work, (b) negative work, (c) no work on the Moon, (d) impossible to determine? (c). Since Fg on the Moon acts toward the earth to provide ac inward. The Moon’s displacement at any moment is tangent to the circle, parallel to velocity and perpendicular to FgFg and the displacement is 90° (cos90° = 0), and the work done by Fg on the moon is therefore W = Fdcos90° = ZERO! Two water slides are shaped differently, but they start at the same height, h. Two riders, Paul and Kathleen (different masses), start from rest at the same time and at the same height h but on different slides. (The figure shows them at different heights because it shows them after they have started down) The slides are frictionless. Which rider is traveling faster at the bottom? Using the conservation of mechanical energy:  Both will have the same speed at the bottom. Since speed is independent of the mass. A particle moves in a circular path of radius r with speed v. If its speed increases to 2v while traveling the same circular path, by what factor will the centripetal acceleration of the particle change? By a factor of 4. Because: ac = v2/R. Then increasing the speed by 2v makes the final value for centripetal acceleration equal to: ac = (2v)2/R = 4 (v2/R) After a ball is thrown upward and is in the air, its speed (a)(b) decreases (c) increases and then decreases (d) decreases and then increases (e) remains the same. (d) Decreases and then increases. While the ball is rising, it is slowing down. After reaching the highest point, the ball begins to fall and its speed increases. Solve A glider of mass m is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Think of the launcher as a light spring of force constant kx. The glider is released from rest. Show that the glider attains a final speed of: . Show that the impulse imparted to the glider is given by the expression Assuming that the road of the figure has a uniform linear mass density  = M/L, show that its center of mass XCM = L/2 But  = M/L Þ Show that the energy required to move an object of mass m from the Earth’s surface to an altitude twice the Earth’s radius (RE) is given by , Since: , Then: The bead is released from a height h = 4R. Using conservation of the mechanical energy, show that the speed at point A is given by: The force acting on a particle varies as in the figure. Find the total work done by the force on the particle along its motion. A hockey puck on a frozen pond is given initial speed vx (to the right) with xi = 0. The puck always remains on the ice and finally stops. FBD of the hockey puck. Equations of motion for x and y direction Using equations from part B show that: μk = − ax/g The Ballistic Pendulum is an apparatus used to measure the speed of a fast moving projectile, such as a bullet. A bullet of a mass m is fired into a large block of wood with mass M suspended from some light wires. The bullet embeds in the block, and the entire system swings through a height h. Write an equation for conservation of the linear momentum between points A and B. Before: m1v1 + m2v2 ≡ m1v’1 + m2v’2 Where: v1 = v1A , v2 = 0 , v’1 = v’2 = vB Þ m1v1A = (m1 + m2 )vB Write an equation for conservation of mechanical energy of the system Bullet-Block between points B and C. After: ½(m1 + m2 )v12 + (m1 + m2 )gyi ≡ ½(m1 + m2 )v’12 + (m1 + m2 )gyf Where: v1 = vB , yi = 0 , v’1 = 0 = vB , yf = h Þ ½(m1 + m2 )vB2 + 0 = (m1 + m2 )gh + 0 Show that the system Bullet-Block attains a final speed. From part B, solving for vB: ½vB2 = gh  Show that the bullet has an initial speed given by the expression: . Replacing vB into equation from part A and solving for v1A: m1v1A = (m1 + m2 )( ) When two objects of unequal mass (m2 > m1) are hung vertically over a frictionless pulley of negligible mass, as the figure, the arrangement is called Atwood Machine. FBD for each mass. Equation of motion for y-direction for each mass. ΣFy = T – m1g = m2a (1) ΣFy = T – m2g = – m2a (2) Determine the magnitude of the acceleration (a). From (1): T = m1g + m1a (3) From (2): T = m2 g – m2 a (4) Equating: (3) = (4) and Solving for a m1g + m1 a = m2g – m2a Þ m1a + m2a = m2g – m1g Þ a (m1 + m2) = (m2 – m1)g Þ (5) Determine the tension (T) in the light cord. Substituting (5) into (3): T = m1g + m1a (3) Þ

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