- NCSLv 78 months agoFavourite answer
a) begin by conserving "vertical" momentum, which is zero. Make "to the right" a negative angle:
0 = M*v*sin53º + M*V*sin-37º
where v is the velocity of the second ball.
v = V*sin37º / sin53º = 0.7536V
conserve momentum in horizontal direction, substituting for v:
M*4m/s = M*0.7536V*cos53º + M*V*cos-37º
4 m/s = 1.252V
V = 3.2 ms ◄ originally stationary ball
v = 0.7536*3.2m/s = 2.4 m/s ◄ originally moving ball
b) this is basically just using this same process iteratively.
However, we must make an assumption in order to find a solution -- elastic collisions. We know that for bodies of equal mass, an elastic collision results in a scatter angle of 90º (as in part a). Otherwise we don't have enough data to solve for the velocities and the angles.
conserve "vertical" momentum for the cue-9 collision:
0 = M*v*sin20º + M*V*sin-70º
where v is the velocity of the cue ball
v = V*sin70º / 20º = 2.747V
and so horizontally
M*6m/s = M*(2.747V)*cos20º + M*V*cos-70º
6 m/s = 2.924V
V = 2.05 m/s ≈ 2.1 m/s ◄ 9-ball
v = 2.747V = 5.6 m/s ◄ cue ball after 1st collision
conserve "vertical" momentum
0 = M*v*sin40º + M*V*sin-50º
where v is new post-collision velocity of the cue ball
v = V*sin50º/sin40º = 1.192V
and horizontally, substituting for v:
M*5.6m/s = M*(1.192V)*cos40º + M*V*cos50º
5.6 m/s = 1.556V
V = 3.6 m/s ◄ 8-ball
v = 1.192V = 4.3 m/s ◄ cue ball, final
But check the math!
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- Anonymous8 months ago
Open up photoshop and go to image > rotate canvas
Then save the new image as a pdf or a jpg.