Post a New Question


posted by on .

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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question