A solid, uniform ball of mass 10.7 , and radius 0.43 , rolls without slipping up
ID: 1702800 • Letter: A
Question
A solid, uniform ball of mass 10.7 , and radius 0.43 , rolls without slipping up a hill of height 27.7 , as shown in the figure . At the bottom of the hill the ball has a velocity of 31.6 . At the top of the hill, it is moving horizontally, and then it goes over the vertical cliff.(a) What is the total initial kinetic energy of the ball at the botton of the hill? Don't forget to use the numbers in the introduction not the picture. (J)
(b) How fast is it moving just before it lands?
Pleas help me. I've attempted this problem several times, but still haven't been able to figure it out.
Explanation / Answer
The mass of the ball M = 10.7kg
the radius of the ball R = 0.43m
the heigth of the hill h = 27.7m
the initial velocity of the ball vi = 31.6m/s
(a) The total kinetic energy
Ki = 1/2 I^2 + 1/2mv^2
= (0.5)(2/5MR^2)(v/R)^2 + (0.5)Mv^2
= (1/5)(Mv^2) + (0.5)Mv^2
= 0.7Mv^2
= (0.7)(10.7kg)(31.6m/s)^2
= 7479 J
The final translational velocity of the ball before it hits the ground is the vector sum of the X and Y components of velocity after the ball leaves the cliff.
Then you must find the time it takes for the ball to fall to the ground from 28m:
Since -27.7 m = -4.9m/s2 (t2)
This gives: t = 2.38 s
From law of cosnervation of energy
Ki = Uf + Kf
then Kf = Ki - Uf = 7479J - Mgh
= 7479 - (10.7)(9.8)(27.7)
0.7Mv^2 = 4574.378
then the final speed vf = (4574.378)/(10.7)(0.7)
= 24.7 m/s
Y-component is: 9.8m/s2 * 2.38 s = 23.32m/s
So the final velocity of the ball is:
VTf = [(23.32m/s)2 + (24.7m/s)2]
= 33.9 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.