Question

A cue ball traveling at 4.50 m/s along the +x axis collides elastically with another billiard ball initially at rest. After the collision, the cue ball’s velocity is 25o below the x-axis. Both balls have the same mass. Derive 3 equations where the only unknowns are the final speed of each ball and the direction of the billiard ball that was initially at rest (v1f, v2f and 2f). For a few extra credit points on your homework, solve for v1f, v2f if 2f=35o .

Find the final speed and direction of each ball. (You can stop once you have three equations with three unknowns –> v1f, v2f and 2.

Answer 1

Conservation of momentum in X direction

m1u1 + m2u2 = m1v1f cos(25) + m2v2f cos(theta2f)

u2 = 0

m1 = m2

4.5 = v1f cos(25) + v2f cos(theta2f) ---- (1)

Conservation of momentum in y direction

m1v1f sin (25) = m2v2f sin(theta2f)

v1f sin(25) = v2f sin(theta2f) ----- (2)

Conservation of energy

1/2 m1u1^2 = 1/2 m1v1f^2 + 1/2 m2v2f^2

u1^2 = v1f^2 + v2f^2

v1f^2+v2f^2 = 4.5*4.5 ------ (3)

These are the three equations

Second part

v2f = 2.67 m/s

v1f = 3.623 m/s

If theta2f = 35 degrees

Actually speaking the angle cannot be 35 degrees to satisfy all three equations