Post a New Question


posted by .

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

A trapeze artist weighs 8ooN. The artist is momentarily held to one side of a wing by apartnr so that both of the swing ropes are at an angle of 30 degrees with the vertical. in such a condition of static equilibrium, what is the horizontal force being applied by the partner?

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question