Explanation:
We need to apply the conservation law of linear momentum to two dimensions:
Let [tex]p_{1}[/tex] = momentum of the 1st ball
[tex]p_{2}[/tex] = momentum of the 2nd ball
In the x-axis, the conservation law can be written as
[tex](p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f} [/tex]
or
[tex](m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f} [/tex]
Since we are dealing with identical balls, all the m terms cancel out so we are left with
[tex](v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}[/tex]
Putting in the numbers, we get
[tex]1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}[/tex]
[tex] = > (v_{2})_{f} \cos \theta_{2} = 0.703[/tex]
In the y-axis, there is no initial y-component of the momentum before the collision so we can write
[tex]0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f} [/tex]
or
[tex] = > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412[/tex]
Taking the ratio of the sine equation to the cosine equation, we get
[tex] \frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586[/tex]
or
[tex] \theta_{2} = { \tan}^{ - 1} (0.586) = 30.4[/tex]
Solving now for [tex](v_{2})_{f}[/tex],
[tex](v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s} [/tex]
