A13 (2.5) You want to build a very complicated timer that launches a steel ball

Question

A13 (2.5) You want to build a very complicated timer that launches a steel ball bearing straight upward from an initial position. The goal is for the ball bearing to land back at its initial position at exactly 0.780 s after it was first launched. (a) What initial velocity must be used for this to work properly? (Ignore air resistance.) (b) How high will the ba bearing travel? Answers: (a) 3.82 m/s, (b) 0.745 m A14 (2.4) As you found in exercise A13, the initial velocity of the ball bearing must be 3.82 m/s The mechanism you develop to launch the ball is in contact with the ball for a distance of 25.4 mm. Assume that this mechanism produces a constant acceleration on the ball bearing that takes it from rest to the required velocity of 3.82 m/s. (a) Determine the required constant acceleration. (b) How much time was the ball bearing in contact with the mechanism? Answers: (a) 287 m/s2 (b) 0.0133s

Explanation / Answer

13)

As the ball returns with same initial speed downwards, using first k.e

v = u + at
-u = u -gt

u = gt/2 = 9.8 x 0.78 / 2 = 3.822 m/s

using third k.e

v^2 = u^2 + 2as

0 = u^2 -2gh

h = u^2/2g = 3.822^2 / 19.6 = 0.745m

14)

a)

using third k.e

v^2 = u^2 + 2as

3.822^2 = 0 + 2a x 0.0254 ... here s= 0.0254m because that's the distance for which accn acted

a = 287 m/s^2

b)

t = v-u / a ... first k.e

= 3.822 - 0 / 287 = 3.822/287 = 0.0133 sec

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