(a) [tex]-39.4^{\circ}[/tex]
Let's take the initial direction (before the collision) of the cue ball has positive x-direction.
Along the y-direction, the total initial momentum is zero:
[tex]p_y =0[/tex]
Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:
[tex]0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2[/tex]
where
m is the mass of each ball
[tex]v_1= 4.60 m/s[/tex] is the velocity of the cue ball after the collision
[tex]v_2 = 3.40 m/s[/tex] is the velocity of the second ball after the collision
[tex]\phi_1=28.0^{\circ}[/tex] is the angle of the cue ball with the x-axis
[tex]\phi_2[/tex] is the angle of the second ball
Solving for [tex]\phi_2[/tex], we find the angle between the direction of motion of the second ball and the original direction of motion:
[tex]sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}[/tex]
(b) 6.69 m/s
To find the original speed of the cue ball, we analyze the situation along the horizontal direction.
First, we calculate the total momentum along the x-direction after the collision, which is:
[tex]p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m[/tex]
The initial total momentum along the x-direction as
[tex]p_x = m u [/tex]
where
m is the mass of the cue ball
[tex]u[/tex] is the initial velocity of the cue ball
The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:
[tex]mu = 6.69 m\\u = 6.69 m/s[/tex]
