A billiard ball of mass mA=0.400kg moving with speed vA=180m/s strikes a second ball, initially at rest, of mss mB=0.400kg. As a result of the collision, the first ball is deflected off at an angle of 30.0° with a speed v'A=1.10m/s.
(a) Taking the x axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions separately.
(b) Solve these equations for the speed v'B, and angle θ, of ball B. Do not assume the collision is elastic.
2007-02-25 10:52:07 · 3 answers · asked by Anonymous in Education & Reference ➔ Homework Help
Ball A has initial momentum in the x axis of .4kgx1.8m/s or .72kgm/s. It's y axis momentum is 0.
Ball B has no initial momentum.
The final momentum of the 2 balls combined must be equal to the initial momentum of them. Because neither ball starts with any y axis momentum, the combined momentum must stay at zero, meaning that the momentum of ball A (in the y axis) must be the same as the momentum of ball B, (in the y axis) but in the opposite direction. (again, in the y axis) Since ball B gains momentum, ball A must lose the same amount. Whatever momentum is left for ball A, after balancing the y axis, stays as x axis momentum.
That's how you find the answer. You still need to do the actual math yourself, unless someone else out there is nicer than I am.
2007-02-25 11:24:33 · answer #1 · answered by Thisisnotmyrealname 2 · 0⤊ 0⤋
well.... lets just say you gotta 50/50 chance of being right
2007-02-25 10:55:00 · answer #2 · answered by Susanna 3 · 0⤊ 1⤋
Hmmmmmmm ummmmmm well maybe it involves an equation? Look it up on google idk I suck at math
2007-02-25 10:56:26 · answer #3 · answered by Sephiroth~The One Winged Angel~ 5 · 0⤊ 2⤋