Starting with an initial speed of 5.00 m/s at a height of h = 0.330 m, the m1 = 1.45 kg ball swings downward and strikes the m2 = 4.40 kg ball that is at rest, as the drawing shows.
(a) Using the principle of conservation of mechanical energy, find the speed of the 1.45-kg ball just before impact.
i know this = 5.61 m/s
(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
(1.45-kg ball)= i know this is -2.829 m/s
*(4.40-kg ball)=?
(c) How high does each ball swing after the collision, ignoring air resistance?
*(1.45-kg ball)=?
* (4.40-kg ball) =?
i can't figure out the starred ones
2007-10-19 11:22:07 · 1 answers · asked by samiam389 1 in Science & Mathematics ➔ Physics
b. From conservation of momentum,
P = m1v1+m2v2 = m1v1'+m2v2'
From conservation of energy,
v1-v2 = v2'-v1'
solving, v1' = -2.8290 m/s, v2' = +2.7810 m/s
c. mv^2/2 = mgh so h = v^2/(2g). U-do-it.
2007-10-19 14:45:05 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋