Question

A cue ball traveling at 5.93 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.

(a) Find the angle between the velocity vectors of the two balls after the collision.

(b) Find the speed of each ball after the collision.

cue ball
target ball

Answer 1

The velocities of the cue ball and target ball are V_c and V_t respectively and mass of the each ball be m and angle between these two vectors be , and momentum of conservation given by

(1)

(2)

Conservation of energy gives

(3)  

By squaring and adding equation (1) and (2) , we get

By using above equation (3) in above gives

By substituting in equation (2)

substituting above value in equation (3)

V_t = 3.587 m/s

V_c = 4.721 m/s

(a) the angle between velocity vectors is 90^o

(b) The speed of each ball after the collision are V_t = 3.587 m/s and V_c = 4.721 m/s