Question

A cue ball traveling at 4.25 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	m/s
target ball	m/s

Answer 1

given

v = 4.25 m/s

mass of both masses = m

theta = 30 deg

a. let angle of the other ball after collision with the initial direction of motion = phi

also, let speeds after collision be u1, u2

then from conservation of momentum

mv = mu1*cos(30) + mu2*cos(phi)

mu1*sin(30) = mu2*sin(phi)

u1 = 2u2*sin(phi)

u1*cos(30) + u2*cos(phi) = 4.25

u1^2/4 + (4.25 - u1*cos(30))^2 = u2^2 = u1^2/4 + 18.0625 + 0.75u1^2 - 7.361u1

u2^2 = u1^2 - 7.361u1 + 18.0625

from conservation of erngy

0.5*m*4.25^2 = 0.5*m*u1^2 + 0.5*m*u2^2

18.0625 = u1^2 + u2^2 = 2u1^2 - 7.361u1 + 18.0625

u1 = 3.6805 m/s

u2 = 2.12518 m/s

phi = 59.98 deg

hence angle bw balls after collision = phi + 30 = 89.9883 deg