Weekly Challenge 33: Crazy Cannons

7057 /www/nrich/html/content/id/7057/ 1 2011-05-09T09:24:11 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/5987&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/5952&part=">Try this next</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html">Ask NRICH</a> </li> <li> <a href="http://iopscience.iop.org/0305-4624/19/5-6/301/pdf/ptv19i5-6p172.pdf"> Read all about it</a> </li> <li> <a href="http://nrich.maths.org/7315&part=solution">Last week's solution</a> </li> </ul> <div> Two cannons both fire balls at speed $100$ ms$^{-1}$. One of the cannons is fixed at an angle of $45^\circ$ to the horizontal and othe 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. Show that that $T$ must lie within a range of values and that both $T$ and $D$ must be greater than zero. </div> <div class="framework"> 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. </div>   </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. 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 <mdo:image height="361" width="491" src="DvsT.PNG" alt=""></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. 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 <mdo:image height="361" width="491" src="DvsT.PNG" alt=""></mdo:image> </mdoxml> 2 4 0 0 0 0 1 Weekly Challenge 33: Crazy Cannons A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge Stage 5 Mechanics mapping document Kinematics M2