Two balls, of masses mA = 36 g and mB = 76 g are suspended. The lighter ball is pulled away to a 60° angle with the vertical and released.
(a) What is the velocity of the lighter ball before impact? (Take the right to be positive.)
I got the answer to this and it is 1.71, what I can't figure out is the next questions.
(b) What is the velocity of each ball after the elastic collision?
(c) What will be the maximum height of each ball (above the collision point) after the elastic collision?
This is due soon, so any help is greatly appreciated.
2007-11-15 08:44:15 · 1 answers · asked by mookle_x 1 in Science & Mathematics ➔ Physics
with elastic collisions, the equations are:
v1 = u1*(m1-m2)/(m1+m2) + u2*2*m2/(m1+m2)
v2 = u1*2*m1/(m1+m2) + u2*(m2-m1)/(m1+m2)
where v1 and v2 are velocities after impact and u1 and u2 are before impact.
2007-11-15 10:13:41 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋