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lab3a10 - Expt 3 Conservation of Energy and Projectile...

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1 Physics E-1a Expt 3: Conservation of Energy Fall 2010 and Projectile Motion Introduction Preparation: Before coming to lab, read this lab handout and the suggested reading in Giambattista (Chapter 6, sections 6.1 - 6.5). Then complete your Pre-lab Questionnaire. Be sure to bring your completed Pre-Lab Questionnaire to lab along with this handout, writing paper, a calculator, and your copy of the Lab Companion. Post Lab questionnaire: At the beginning of the lab section, you will be given an additional handout with a series of questions to be answered and handed in at the end of the experiment. Try to answer these questions with one or two concise sentences. For this experiment you will also hand in your target. Experimental Setup : The apparatus shown in the figure consists of a steel ball, suspended to swing like a pendulum. A mechanism mounted at the edge of the bench releases the ball. The ball then becomes a free projectile, and lands on the floor a distance R from the launch point. Details are given later. V o release start h 1 h 1f h 2 H R ! Objective: You will attempt to predict the range of the steel ball by invoking the law of conservation of energy to determine v o and kinematics of trajectory motion to predict R.
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2 Theory Let us begin by assuming that the energy loss due to friction is negligible. Since the ball is at rest at h 1 , by conservation of mechanical energy, [potential energy at h 1 ] = [potential energy at h 2 ] + [kinetic energy at h 2 ], or mgh 1 = mgh 2 + (1/2)mV o 2 [1] Therefore V o = 2 g (h 1 ! h 2 ) [1a] The components of velocity at the moment of release are therefore V horiz = cos ! 2 g (h 1 " h 2 ) and V vert = sin ! 2 g (h 1 " h 2 ) . [2] If the ball is in flight for a time t, then the range R is R = (V horiz )t [3] and, for the height at the launch point H, we can write (assuming down is positive) H = 1 2 gt 2 ! V vert ( ) t [4] Solve equation (4) for t: t = V vert ± (V vert ) 2 +2gH g [4a] Substitute [4a] into equation [3] and, making use of equation [2], we have the result R = 2 cos ! [sin ! (h 1 " h 2 ) + sin 2 ! (h 1 " h 2 ) 2 +H(h 1 " h 2 ) ] [5]
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3 We began with Eq. [1] by assuming frictional energy losses to be negligible. However, if you let the steel ball make a complete swing (forth and back), you will notice that it does not return to the same height. Rather, it returns to some smaller height, h 1f . Air resistance and friction in the pendulum suspension contribute a small but noticeable loss in mechanical energy — designate this energy loss as E f . The fact that the steel ball does not return to the same height suggests a way in which we may determine E f experimentally: The difference in height (h 1 - h 1f ) will give us the loss in potential energy in one complete swing, which we can attribute to the frictional loss: E f = mg(h 1 - h 1f ) In the actual experiment, the steel ball does not swing forth and back — it is released on the forward swing. Therefore, we shall assume that in the forward swing only one-half of this energy is lost. 1 We now rewrite Eq. [1] to take into account this frictional energy loss:
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