Calculating Ball Height from Horizontal Force: Newton's Second Law Explained

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The discussion revolves around calculating the height (h) at which a horizontal force (F) is applied to a billiard ball, leading to a change in its velocity. The problem states that the initial velocity is v0 and the final velocity is (9/7)v0, with the goal of proving that h equals (4/5)R, where R is the ball's radius. Participants express confusion about integrating torque and the choice of the point of momentum, debating whether to use the center or the contact point for calculations. The importance of angular momentum conservation is emphasized, with suggestions to consider the contact point to simplify the analysis. Overall, the conversation seeks clarity on applying Newton's second law and momentum principles to solve the problem effectively.
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Exams tomorow! please help

Homework Statement



A billiard ball is hit with a horizontal force F. The force hits the ball with the height h over the ball's center. the initial velocity is v0 and the the velocity afterward is (9/7)v0. Show that the height h is (4/5)R, where R is the radii of the ball.

Homework Equations



Newtons second law: Fdt=mdv

momentum: t=r x F=Iw

The Attempt at a Solution


I don't really understand this. please help. Thank you very much
 
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hi ippo90! :smile:

initially, the ball will slide (at velocity v0) while spinning, but after a short time, friction will reduce the spin, and the ball will start rolling (at velocity (9/7)v0) :wink:
 


Thank you for the reply. The problem is I don't know where to start. Can you help me?
 


just ignore my last post. I thought about what you said in you post, and it came out=) Thank you for your kind help.

one thing I not sure of is that when I integrated the torque i got something like

hFt=Iw

where h is height, t is time and w is the angul. velocity. here I chose the center as the point of momentum. but afterward I have to use that the angular momentum (L) is conserved. but this means that I have to choose the contact point as the poiint of momentum. why can I do this (why not)?
 


ippo90 said:
just ignore my last post. I thought about what you said in you post, and it came out=) Thank you for your kind help.

one thing I not sure of is that when I integrated the torque i got something like

hFt=Iw

where h is height, t is time and w is the angul. velocity. here I chose the center as the point of momentum. but afterward I have to use that the angular momentum (L) is conserved. but this means that I have to choose the contact point as the poiint of momentum. why can I do this (why not)?

It would be nice to see the problem worked out in detail.
 
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