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November 23, 2014

November 23, 2014

Posted by **Nat** on Thursday, March 8, 2007 at 10:04pm.

b) What is the original speed of the cue ball?

a) would I use

m_ax*v_ax= (m_a+M_b)Vcos theda and do the same for the y axis But what about the masses? and what would the V be equal to?. I'm not completely sure what to do though.

b) What is the equation I would use? Is it(m_a)(v_a)+(m_b)(v_b)=1/2(m_a+m_b)V^2?

Can you please clearify each step of this problem? The part that really confuses me is that the masses are of the same mass. I know they would cancel but i'm more confortable when they have values. I should be getting 41 degrees for a) and for b) 4.76 m/s

The total momentum perpendicular the original cue ball direction remains zero.

3.5 sin 22 = 2.0 sin A

sin A = 1.75 sin 22 = 0.65556

A = 40.96 degrees

Choose original cue ball speed to require conservation of forward momentum

V = 2 cos 40.96 + 3.5 cos 22 = 4.755

The collision turns out to be inelastic since there is less final kinetic energy than initially. In reality, this type of collision would be unlikely for billiard balls.

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