Crazy cannons
Two cannons are fired at one another and the cannonballs collide... what can you deduce?
Problem
Two cannons both fire balls at speed $100 \mathrm{ms}^{-1}$. One of the cannons is fixed at an angle of $45^\circ$ to the horizontal and the other is fixed at angle $30^\circ$ to the horizontal. The cannons are set up facing each other at a distance $D$ from each other. One cannon is fired and then, $T$ seconds later, the other cannon fired.
The balls subsequently strike each other in mid air.
Assuming that the cannon at an angle of $45^\circ$ is fired first, show that that $T$ must lie within a range of values and that both $T$ and $D$ must be greater than zero.
What if it was the cannon at an angle of $30^\circ$ which was fired first?
Did you know ... ?
The mathematics used in this question, along with an understanding of how gravity works at large distances, is sufficiently complex to send rockets to the moon and the planets of our solar system.
Getting Started
Put the first cannon at the origin $(0, 0)$ and the second cannon at the point $(D, 0)$. Can you write down equations for the $x$- and $y$-coordinates of each of the cannon balls modelling $g$ as $10 \mathrm{ms}^{-2}$?
Student Solutions
Put the first cannon at the origin $(0, 0)$ and the second cannon at the point $(D, 0)$.
Using a constant acceleration of $-g$ in the $y$-direction and $0$ in the $x$-direction it is a simple matter to write down the positions of each cannon ball at a time $t> T$ if we make use of the formula $s=ut+\frac{1}{2}at^2$.
$$
\begin{eqnarray}
(x_1, y_1)&=&\left(100\cos(45^\circ)t, 100\sin(45^\circ)t-\frac{1}{2}gt^2\right)\cr
(x_2, y_2)&=&\left(D-100\cos(30^\circ)(t-T), 100\sin(30^\circ)(t-T)-\frac{1}{2}g(t-T)^2\right)
\end{eqnarray}
$$
As with all mechanics problems, the first part involves a careful setup of the equations. Once I have checked these carefully (... OK, that's done...) we can proceed with the algebra to resolve the equations.
Since I know that the two cannon balls strike each other the plan of attack is to equate the two $x$ and $y$ coordinates. Modelling $g$ as $10 \mathrm{ms}^{-2}$, I find that
$$
50 \sqrt{2}t = D-50\sqrt{3}(t-T)
$$
and
$$
50\sqrt{2}t-5t^2=50(t-T)-5(t-T)^2\;.
$$
After some rearrangement, the second of these equations gives me
$$
\begin{eqnarray}
\left(10(\sqrt{2}-1)-2T\right)t &=& -T^2-10T\cr
\Rightarrow t = \frac{T^2+10T}{2T-10(\sqrt{2}-1)}\;.
\end{eqnarray}
$$
Since for a collision to occur we must have $t> 0$, which implies that
$$
T> 5(\sqrt{2}-1)\;.
$$
Thus, there is a minimum value of $T$ (which might be greater than $5(\sqrt{2}-1)$; it is not less than this value). Now, for a collision to occur in the air the $y$ coordinate at the point of collision must be positive. The expression for the first cannon ball quickly gives us the inequality
$$ t< 10\sqrt{2}\;.$$
This gives us a more complicated inequality for $T$ as
$$
\frac{T^2+10T}{2T-10(\sqrt{2}-1)}< 10\sqrt{2}\;.
$$
Rearranging we see that
$$T^2+10(1-2\sqrt{2})T+100(2-\sqrt{2})< 0\;.$$
Values of $T$ which satisfy this equation are those lying between the two roots
$$
T_{1, 2} = \frac{10(2\sqrt{2}-1)\pm\sqrt{(10(1-2\sqrt{2})^2-4(100(2-\sqrt{2}))}}{2}\;.
$$
Thus,
$$
10(\sqrt{2}-1) < T< 10\sqrt{2}\;.
$$
I used a spreadsheet to plot the values of $D$ against $T$. The range of permissible values is
Teachers' Resources
Why do this problem?
This problem is an engaging application of projectile motion; although the situation is fairly straightforward to comprehend and visualise, it is rather more challenging to investigate mathematically. Students will need to consider not only motion in horizontal and vertical directions for the two cannon balls but also the time when a particular position is reached. Parametric equations and careful consdieration of the different starting points of the two cannon balls are essential.
Possible approach
Students could be asked to consider the paths of two cannon balls fired toward each other. How do we know if the paths cross at all? Are there any restrictions needed to ensure that their paths cross? Does the intersection of the paths of the two cannon balls indicate that they necessarily collide? Why or why not?
Once students have started to make sense of the situation being presented, they could be asked to write equations for horizontal and vertical components of the position of the first cannon ball and then the second, taking into account both its starting position and the delay before the second cannon is fired. Which quantities need to be equal for the cannon balls to collide? What values can $T$ and $D$ take in these circumstances? Potential solutions can be tested using graph-drawing packages.
Key questions
- How do we know if the paths of two projectiles cross? Are there any restrictions needed to ensure that their paths cross?
- Does the intersection of the paths of two projectiles indicate that they necessarily collide? Why or why not?
- How are the horizontal and vertical positions of a projectile related to its initial speed and angle of projection?
Possible extension
Consider the case where the cannon at an angle of $30^\circ$ is fired first? Can you draw a diagram to show roughly where in the trajectory of each cannon ball the collision occurs? Can you adapt your equations and algebra from the first situation to find new ranges for $T$ adn $D$?
Use graph drawing software to show that not only do the cannon balls' paths cross, but also that the cannon balls reach a particular point at the same time (so that they collide) for given values of $T$ and $D$ in each of the two cases.
Possible support
Provide students with suitable equations of constant acceleration.