ps02sol - 8.01T Problem Set 2 Solutions Fall 2004 Problem...

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8.01T Problem Set 2 Solutions Fall 2004 Problem 1. Measurement of g . a) The ball moves in the vertical direction under the influence of the con- stant force of gravity 1 . Hence in our approximation the ball undergoes one- dimensional motion with constant acceleration g . Let us introduce vertical axis x and choose the positive direction to point up. We then place the origin of the coordinate system at the bottom of the lower pane of glass and choose to measure the time from the moment when the ball rises past the origin, i.e. we choose x 0 = 0 and t 0 = 0 (See Fig. 1). With our choice of initial position Figure 1: and time, the equation for x ( t ) is gt 2 x ( t ) = v 0 t (1) 2 and the graph of x ( t ) should look like the one on Fig. 1 The velocity of the ball is given by v ( t ) = v 0 gt (2) b) Using Fig. 1, it is easy to understand that the time the ball travels across the window, i.e. the interval of time between the moment it rises past the bottom of the lower plane of glass and disappears past the top of the upper pane, is = ( T A T B ) / 2. Since the height of the glass is h , we obtain, from Eq. (1) g 2 h = v 0 (3) 2 1 We neglect the air resistance and the variation of the gravitational force over the height of the ball’s trajectory 1
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8.01T Problem Set 2 Solutions Fall 2004 To find the second equation, we notice that the highest point of the trajectory corresponds to v = 0 and t = T A / 2. Thus we can find v 0 as v 0 = gT A / 2 (4) We now substitute this result back into Eq. (3) and obtain 2 h 8 h g = ( T A ) = T 2 A T 2 B (5) Problem 2. Track event.
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This note was uploaded on 04/22/2011 for the course PHYS 1441 taught by Professor White during the Spring '08 term at UT Arlington.

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ps02sol - 8.01T Problem Set 2 Solutions Fall 2004 Problem...

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