Question

A cue ball traveling at 8.0 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° 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	____m/s
target ball	_____m/s

Answer 1

here,

the initial velocity of cue ball , u = 8 m/s

let the final velocity of cue ball and target ball be v1 and v2

using conservation of momentum along original line of motion

m * u = m * v1 * cos(30) + m * v2 * cos(theta)

8 = v1 * cos(30) + v2 * cos(theta) .....(1)

and

using conservation of momentum along a line perpendicular to original line of motion

0 = m * v1 * sin(30) - m * v2 * sin(theta)

0 = v1 * sin(30) - v2 * sin(theta) .....(2)

and

using conservation of kinetic energy

0.5 * m * u^2 = 0.5 * m * v1^2 + 0.5 * m * v2^2

8^2 = v1^2 + v2^2 ....(3)

from (1),(2) and (3)

v1 = 1.23 m/s

v2 = 7.9 m/s

theta = 4.46 degree

a)

the angle between the velocity vectors of the two balls after the collision= theta + 30 degree = 34.46 degree

b)

final speed of cue ball is 1.23 m/s

the final speed of target ball is 7.9 m/s