Answer
$v_{f1}=0.65m/s$ and $\theta=10^o$
Work Step by Step
1) On the $y$ axis:
- Ball 1 has mass $m_1=0.15kg$, initial vertical velocity $v_{01}=-v_{01}\cos50=-0.9\cos50=-0.58m/s$ and final vertical velocity $v_{f1}=+v_{f1}\sin\theta (m/s)$.
- Ball 2 has mass $m_2=0.26kg$, initial vertical velocity $v_{02}=-v_{02}\cos50=0$ and final vertical velocity $v_{f2}=-0.7\sin35=-0.4m/s$.
Using conservation of momentum, we have
$$m_1v_{01}+m_2v_{02}=m_1v_{f1}+m_2v_{f2}$$ $$-0.087+0=0.15v_{f1}\sin\theta-0.104$$ $$0.15v_{f1}\sin\theta=0.017$$ $$v_{f1}\sin\theta=0.113 (1)$$
2) On the $x$ axis:
- Ball 1 has initial horizontal velocity $+0.9\sin50=+0.69m/s$ and final horizontal velocity $+v_{f1}\cos\theta (m/s)$.
- Ball 2 has initial horizontal velocity $v_{02}=+0.54m/s$ and final horizontal velocity $+0.7\cos35=+0.57m/s$.
Using conservation of momentum, we have
$$0.15(+0.69)+0.26(+0.54)=0.15(+v_{f_1}\cos\theta)+0.26(+0.57)$$ $$0.104+0.140=0.15v_{f_1}\cos\theta+0.148$$ $$0.15v_{f_1}\cos\theta=0.096$$ $$v_{f_1}\cos\theta=0.64 (2)$$
Divide (2) over (1): $$\frac{v_{f_1}\sin\theta}{v_{f_1}\cos\theta}=\tan\theta=\frac{0.113}{0.64}=0.177$$ $$\theta=10.037^o$$
So $$v_{f1}=\frac{0.113}{\sin10.037}=0.65m/s$$
