jennifer asked in Science & MathematicsPhysics · 8 months ago

# How would you solve this ?

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Lv 7
8 months ago

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

second 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!