A cue ball initially moving at 3.4 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball's final speed is 0.94 m/s at an angle of θ with respect to its original line of motion? Find the eight ball's speed after the col - lision. Assume an elastic collision (ignoring friction and rotational motion).

Answer in units of m/s.

speed of eight ball speed after the collision is 3.27 m/s

Explanation:

given data

initially moving v1i = 3.4 m/s

final speed is v1f = 0.94 m/s

angle = θ w. r. t. original line of motion

solution

we assume elastic collision

so here using conservation of energy

initial kinetic energy = final kinetic energy ... 1

before collision kinetic energy = 0.5 * m * (v1i) ²

and

after collision kinetic energy = 0.5 * m * (v1f) ² + 0.5 * m * (v2f) ²

put in equation 1

0.5 * m * (v1i) ² = 0.5 * m * (v1f) ² + 0.5 * m * (v2f) ²

(v2f) ² = (v1i) ² - (v1f) ²

(v2f) ² = 3.4² - 0.94²

(v2f) ² = 10.68

taking the square root both

v2f = 3.27 m/s

speed of eight ball speed after the collision is 3.27 m/s