If a particle of mass $m$ has velocity $v$, its momentum is $p=mv$.
In a game with balls, one ball of mass $2g$ springs with velocity $2m/s$, it hits two balls, both of which have mass $1g$, and stops.
The one ball get soared with velocity $3m/s$ and with angle $45^{\circ}$ to the direction that has the biggest ball at the moment of the crash, as is shown below.
Supposing that the total momentum is the same before and after the crash, I have to find with what angle and velocity the second ball will move.
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I have done the following:
Let $(u,v)$ be the unit vector in the direction of the desired velocity.
Let $\theta$ is the desired angle.
$(1,0)$ is the unit vector of the $x$-axis.
Then we have that $(u,v)\cdot (1,0)=\cos\theta \Rightarrow u=\cos\theta$.
Since the vector is unit, we have that $u^2+v^2=1 \Rightarrow v^2=1-u^2 \Rightarrow v^2=1-\cos^2\theta \Rightarrow v^2=\sin^2\theta \Rightarrow v=\pm \sin\theta$.
Since the desired vector shows to the negative $y$, we reject $v\sin \theta$, or not?
So, the unit vector is $(u,v)=(\cos \theta , -\sin \theta)$.
Therefore, the velocity vector that we are looking for is $v_2(\cos \theta , -\sin \theta)$, where $v_2$ is the magnitude of the velocity.
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From the Momentum Conservation Principle at the $x$-axis we have the following:
$$4=3\cos \frac{\pi}{4}+v_2\cos\theta \Rightarrow v_2\cos\theta=4-3\frac{\sqrt{2}}{2} \tag 1$$
From the Momentum Conservation Principle at the $y$-axis we have the following:
$$0=3\sin \frac{\pi}{4}-v_2\sin\theta \Rightarrow v_2\sin\theta=3\frac{\sqrt{2}}{2} \tag 2$$
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Therefore, $$v_2^2\cos^2\theta+v_2^2\sin^2\theta=16-12\sqrt{2}+\frac{9}{2}+\frac{9}{2}=25-12\sqrt{2} \Rightarrow v_2^2=25-12\sqrt{2} \\ \Rightarrow v_2=\pm \sqrt{25-12\sqrt{2}}$$
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Is this correct so far?
Here is the method that I prefer to proceed with. First, we employ the conversation of momentum and hence we will get
$$m{\bf{v}}=m_1{\bf{v}}_1+m_2{\bf{v}}_2$$
This is a vector equation. The unknown is the vector ${\bf{v}}_2$ and all the other things including $m, m_1, m_2, {\bf{v}}_1$ are given. So we just solve for ${\bf{v}}_2$ to get
$${\bf{v}}_2=\frac{m}{m_2}{\bf{v}}-\frac{m_1}{m_2}{\bf{v}}_1$$
I can say that we are done now. But to go further, we can employ the numerical values in your example. So we can get
$$\begin{align} {\bf{v}}_2&=\frac{1}{2}({\bf{v}}-{\bf{v}}_1) \\ &=\frac{1}{2}\left[(2{\bf{i}})-3\left(\cos\frac{\pi}{4}{\bf{i}}+\sin\frac{\pi}{4}{\bf{j}}\right)\right] \\ &=\frac{4-3\sqrt{2}}{4}{\bf{i}}-\frac{3\sqrt{2}}{4}{\bf{j}} \\ &=v_{21}{\bf{i}}+v_{22}{\bf{j}} \end{align}$$
Now, I think that one can easily say that what are the magnitude and the angel that the vector ${\bf{v}}_2$ makes with the $x$ axis.
$$\begin{align} \|{\bf{v}}_2\| &= \sqrt{v_{21}^2+v_{22}^2} \\ \cos \theta &= \frac{v_{21}}{\sqrt{v_{21}^2+v_{22}^2}} \\ \sin \theta &= \frac{v_{22}}{\sqrt{v_{21}^2+v_{22}^2}} \end{align}$$
and the final answers will be
$$\begin{align} \|{\bf{v}}_2\| &= \frac{\sqrt{13-6\sqrt{2}}}{2} \\ \theta &= - \left[\arctan\left(\frac{3\sqrt{2}}{4-3\sqrt{2}}\right)+\pi\right] \end{align}$$