Slingshot

7664 /www/nrich/html/content/id/7664/ 1 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> A slingshot is a small projectile weapon which consists of a Y-shaped frame held in one hand, with two rubber strips attached to the uprights. <mdo:image alt="" src="slingshott1.bmp" style="width: 170px; height: 240px; float: right;"></mdo:image> Rubber strips obey Hooke's law which says that when a rubber strip is extended by a small amount $\delta x$, the force it exerts is found to be $F = k\delta x$ where $k$ is a force constant. Suppose that you want to shoot a stone of mass $m = 50$g such that it will go to the river which is at a distance of 50 meters. Moreover, it is given that $L = 12$cm, $H = 10$cm, $k = 200$N/m and $g = 9.81 m/s^2$. How much do you need to extend the strips (i.e. find $x$) in order to make a shot into the river if the stone is fired at an angle of 45° with the horizontal and the length of the unstreched strip is less than $\sqrt{L^2 - H^2}$? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We use conservation of energy to find the speed of the stone. All the energy stored in the elastic bands is transferred to the stone as kinetic energy. Energy stored in the elastic bands is $$E =2 \frac{k(\sqrt{x^2 + (\sqrt{L^2 - H^2})^2} - \sqrt{L^2 - H^2})^2}{2}$$ because we have two of them and the amount by which one is extended is $$\sqrt{x^2 + (\sqrt{L^2 - H^2})^2} - \sqrt{L^2 - H^2}$$ Suppose that the initial speed of the stone is $v$ then it has a kinetic energy $E = \frac{mv^2}{2}$. The stone is fired at an angle of 45°. The initial speed of a projectile and the distance it travels is related by $d = \frac{v^2}{g}$. It is given that $d = 50$m. Thus, the kinetic energy is $E = \frac{mgd}{2}$. By the conservation of energy $${k(\sqrt{x^2 + L^2 - H^2} - \sqrt{L^2 - H^2})^2} = \frac{mgd}{2}$$ Now, the numbers $L = 0.12$m , $H = 0.10$ m , $m = 0.05$kg , $k = 200$N/m , $d = 50$ m, $g = 9.81$ m/s^2 could be plugged in to find $x$ or we can simplify to get $$x = \sqrt{\frac{mgd}{2k} + \sqrt{(L^2 - H^2) \frac{2mgd}{k}}}$$ Thus, we have that $x = 31$ cm. <div> </div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> If a projectile is fired at an angle of 45° to the horizontal then the initial speed of the projectile and the distance it travels is related by $d = \frac{v^2}{g}$. Use the conservation of energy in order to find the speed of the stone. The energy stored in a strip which is extended by $\delta x$ is equal $E = \frac{k \delta x^2}{2}$. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We use conservation of energy to find the speed of the stone. All the energy stored in the elastic bands is transferred to the stone as kinetic energy. Energy stored in the elastic bands is $$E =2 \frac{k(\sqrt{x^2 + (\sqrt{L^2 - H^2})^2} - \sqrt{L^2 - H^2})^2}{2}$$ because we have two of them and the amount by which one is extended is $$\sqrt{x^2 + (\sqrt{L^2 - H^2})^2} - \sqrt{L^2 - H^2}$$ Suppose that the initial speed of the stone is $v$ then it has a kinetic energy $E = \frac{mv^2}{2}$. The stone is fired at an angle of 45°. The initial speed of a projectile and the distance it travels is related by $d = \frac{v^2}{g}$. It is given that $d = 50$m. Thus, the kinetic energy is $E = \frac{mgd}{2}$. By the conservation of energy $${k(\sqrt{x^2 + L^2 - H^2} - \sqrt{L^2 - H^2})^2} = \frac{mgd}{2}$$ Now, the numbers $L = 0.12$m , $H = 0.10$ m , $m = 0.05$kg , $k = 200$N/m , $d = 50$ m, $g = 9.81$ m/s^2 could be plugged in to find $x$ or we can simplify to get $$x = \sqrt{\frac{mgd}{2k} + \sqrt{(L^2 - H^2) \frac{2mgd}{k}}}$$ Thus, we have that $x = 31$ cm. <div> </div> </mdoxml> 2 3 0 0 0 0 0 Slingshot