Particle Collision Vector Question

[Pablo1204]

When I attemped this question at first I didn't realise that you could just let And then solve the two equations simultaneously. Instead I tried solving for the value of v such that the magnitude of AB was equal to zero which isn't as elegant but I still want to see it work. I got to this equation I had to put this into Wolframalpha but didn't get a numerical value for v. However, plotting this function on Desmos it can be seen that v = 4 is the only value of v such that the function has an x-int. So how can we find this value of v?

[RuiAce]

Note: For this question, providing that \( \overrightarrow{AB} = [-56 + t(v+3)]\mathbf i + [8 + t(7v-29)]\mathbf j \) would've been helpful background information, to save a bit of time having to do part a).

The reason why your method doesn't work (at least, at first glance!) is because you've actually discarded information. The magnitude helps summarise information about both the \(\mathbf{i}\) and \(\mathbf{j}\) components here; in particular it gives you the vector's length. But the length is now just only one piece of information, whereas earlier you had both components, and hence two pieces of information.

Therefore, in principle, you will no longer be able to complete the question. One way to navigate around this is to bring back any of the variables that you discarded. If that were the \(\mathbf i\) component, it would mean putting \(-56 + t(v+3)=0\) into the equation. Ordinarily, this would be the only way to carry on (and it may suck - I'm not sure, but you might get a polynomial of degree 3 or higher).

But, and this is in my opinion a huge but: in this case, your working is salvageable. I can't fully recall what's taught in spesh anymore, so bear with me if I stray a bit outside the syllabus, but the following approach will still allow you to recover \(v=4\). This is despite that you took a bad turn computing the magnitude. Note that the subsequent methods are also extra, and the original method of setting \( \overrightarrow{AB} = \mathbf 0\) is superior; this merely helps salvage what you did.

\(\overrightarrow{AB}\) is of course nothing more than a vector. And it changes with respect to time. But \(\overrightarrow{AB}\) does not have \(t^2\), \(t^3\), \(\sqrt{t}\), \(e^t\), \(\sin t\), or any super ugly functions within it. Rather, it only has \(t\) itself, and constants. Therefore, \(\overrightarrow{AB}\) changes linearly with respect to time. Equivalently, it is a linear function of \(t\).

Because it is linear, \(\overrightarrow{AB}\) will never be the same vector twice. Therefore, if \(\overrightarrow{AB}\) ever happens to be the zero vector, then it will happen for one and only one value of \(t\). (This is not obvious, and requires further thought. I've attached a GeoGebra simulation here, where I arbitrarily chose A to start from the origin, and B to start from (-56, 8). You can move the slider for \(t\).) This motivates the following: Any equation in terms of \(t\) should only have one unique solution for \(t\).

Now revisit your equation above, which is a quadratic. We can then deduce that this quadratic also has only one solution for \(t\). Therefore, we may be tempted to treat \(v\) as a constant for now, and write it explicitly as a quadratic in terms of \(t\). This gives
\[ 0 = (50v^2 - 400 v + 850)t^2 - 800t + 3200. \]
Recall that the quadratic formula \( x=\frac{-b \pm \sqrt{\Delta}}{2a} \), where \(\Delta = b^2-4ac\), is used to explicitly compute the roots of a quadratic equation. But furthermore, the discriminant \(\Delta = b^2-4ac\) tells you how many solutions there are. If \(\Delta = 0\), then we have one unique solution.

We use that here. Taking the above as a quadratic equation in terms of \(t\), setting its discriminant to zero gives
\[ 640000 - 12800(50v^2 - 400v+850) = 0. \]
You can play around with this quadratic to see that it reduces to the super elegant \( -640000(v-4)^2 = 0 \), which clearly has the unique solution \(v=4\). Which is what you were looking for.