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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.

enter image description here

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?

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    $\begingroup$ The strategy seems correct to me! But I didn't check the computation. :) In fact, we get two equations from $m{\bf{v}}=m_1{\bf{v}}_1+m_2{\bf{v}}_2$. Then we solve for the two unknowns in the vector ${\bf{v}}_2$. The unknowns can be the magnitude and angel or the components in $x$ and $y$ direction. :) $\endgroup$ – H. R. Mar 3 '16 at 19:20
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    $\begingroup$ In (2), you know $\theta$ so you can get $v_2$ immediately. This simplifies things alot. I didn't look in detail at your last equation, but you're definitely making a math error. $\endgroup$ – bob.sacamento Mar 3 '16 at 19:45
  • $\begingroup$ How do we know $\theta$ in (2) ? @bob.sacamento $\endgroup$ – Mary Star Mar 3 '16 at 19:46
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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

$$\begin{align} \|{\bf{v}}_2\| &= \frac{\sqrt{13-6\sqrt{2}}}{2} \\ \theta &= - \left[\arctan\left(\frac{3\sqrt{2}}{4-3\sqrt{2}}\right)+\pi\right] \end{align}$$

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  • $\begingroup$ The magnitude of $v_2$ is $$|v_2|=\sqrt{\left (\frac{4-3\sqrt{2}}{4}\right )^2+\left (\frac{3\sqrt{2}}{4}\right )^2}=\sqrt{\frac{16-24\sqrt{2}+18+18}{16}}=\frac{\sqrt{52-24\sqrt{2}}}{4}$$ right? To find the angle that the vector $v_2$ makes with the $x$-axis, do we use the dot product of $v_2$ and the unit vector along the $x$-axis, $(1,0)$ ? We would have the following: $$v_2\cdot (1,0)=|v_2|\cos \theta \Rightarrow \cos\theta=\frac{\frac{4-3\sqrt{2}}{4}}{\frac{\sqrt{52-24\sqrt{2}}}{4}}=\frac{4-3\sqrt{2}}{\sqrt{52-24\sqrt{2}}}$$ Is this correct? $\endgroup$ – Mary Star Mar 3 '16 at 21:09
  • $\begingroup$ I added the final answer computed by Maple. :) Your answer seems to be OK. for the magnitude you can factor a $4$ under the radical in numerator and then simplify to get my result in the answer. :) $\endgroup$ – H. R. Mar 3 '16 at 21:13
  • $\begingroup$ Ah ok... I read again the formulation of the question and I go stuck... $$$$ "The one ball get soared with velocity $m/s$ and with angle $45^{\circ}$ to the direction that has the biggest ball at the moment of the crash" $$$$ Why does it soared to the direction that has the biggest ball at the moment of the crash? Isn't the direction of the biggest ball at the $x$-axis? But none of the balls after the crash is on the $x$-axis... Could you clarify it to me? $\endgroup$ – Mary Star Mar 3 '16 at 21:58
  • $\begingroup$ @MaryStar: The direction of the biggest ball exactly before the impact (which is the $x$ axis) is used as a reference to clarify the direction of one of the other two small balls after the impact. In fact, to solve impact problems we need some assumptions that relates some quantities before and after the impact. That's all. :) $\endgroup$ – H. R. Mar 3 '16 at 22:06
  • $\begingroup$ But the two smaller balls aren't on the $x$-axis after the impact, are they? $\endgroup$ – Mary Star Mar 3 '16 at 22:10

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