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

In a game of pool the cue ball strikes another ball of the same mass and initially at rest. After the collision the cue ball moves at 3.50 m/s along a line making an angle of 22 degrees with its original direction of motion. and the second ball had a speed of 2 m/s. a)What is the angle between the direction of motion of the second ball and the original direction of motion of the cue ball?

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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