Question

Billiard balls of identical mass roll on a table, and we can ignore the small friction between the balls and the table. The cue ball travels east at a speed of 7.21 m/s and hits the 8 ball which is initially stationary. After the collision the cue ball travels at a speed of 3.05 m/s 29.5 degrees south of its initial eastward heading. What is the direction of the 8 ball in degrees measured counterclockwise from East (+X)? Enter only the number to three significant figures. (Let east be +X, north be +Y, south be -Y, west be -X.)

Answer 1

Given data, m. = mg = m, u = 7.21 m/s, Ug = 0 m/s, v. = 3.05m/s, and 0 = 29.5° Draw a free body diagram for the collision between the cue ball and 8-ball is as follows, Vey mg Vox т. uc mg V.cos m v sin e Use the law of conservation of momentum in horizontal direction as follows, m.u. + mg Ug = m. v. cos 0 +mg Vex Substitute the values in above equation. m(7.21 m/s)+m(0 m/s)= m(3.05 m/s) cos 29.5° + mvxx 7.21 m/s = (3.05 m/s) cos 29.5° + Vęx

= (3.05 m/s)(0.8704)+ Vex Vex = (7.21 m/s)-(2.6546 m/s) = 4.55 m/s Vsy From the law of conservation of momentum in vertical direction, m.v. sin 29.5º = mg Vsy m(3.05 m/s)(0.4924)= mvsy (3.05 m/s)(0.4924)= vey = 1.50 m/s The resultant of the magnitude of the velocity of the 8-ball is, V$ = VVS + vi = (4.55 m/s)? +(1.50 m/s)? = 4.80 m/s Therefore, the magnitude of the final velocity of the 8-ball after the collision is 4.80 m/s 2 +vi 8y