A ball starts from rest and accelerates at 0.475 m/s 2 while moving downan incli

Question

A ball starts from rest and accelerates at 0.475 m/s2 while moving downan inclined plane 9.75 mlong. When it reaches the bottom, the ball rolls up another plane,where, after moving 15.85 m,it comes to rest. (Assume positive direction points down the firstplane and up the second plane.) (a) What is the speed of the ball at the bottom of the firstplane?

(b) How long does it take to roll down the first plane?

(c) What is the acceleration along the second plane? (Careful withsign!)

(d) What is the ball's speed 7.70 m along the second plane?
(a) What is the speed of the ball at the bottom of the firstplane?

(b) How long does it take to roll down the first plane?

(c) What is the acceleration along the second plane? (Careful withsign!)

(d) What is the ball's speed 7.70 m along the second plane?

Explanation / Answer

(a)The speed of the ball at the bottom of the first planeis v2 - u2 = 2aS u = 0,a = 0.475 m/s2 and S = 9.75 m or v2 - 02 = 2 * 0.475 * 9.75 =9.2625 or v = (9.2625)1/2 = 3.04 m/s (b)Let the time taken to roll down the first plane be ttherefore we get v = u + at or t = (v - u/a) = (3.04 - 0/0.475) = 6.4 s (c)Let the acceleration along the second plane bea1 therefore we get v12 - u12 =2a1S1 v1 = 0,u1 = v = 3.04 m/s andS1 = 15.85 m or a1 = (v12 -u12/2S1) = (02 -(3.04)2/2 * 15.85) = -0.291 m/s2 The negative value of acceleration indicates that the ball hasretarding motion. (d)Let the ball's speed 7.70 m along the second plane bev2 therefore we get v22 - u12 =2a1 * S2 or v2 = [u12 + 2a1* S2]1/2 S2 = 7.70 m/s S2 = 7.70 m/s

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