Question

Conservation of Energy 1 Slingshot /Catapult An ordinary rubber band, similar to the ones we used in lab, has a spring constant of 90 N/m The rubber band is used to launch a small stone. Suppose the rubber band is extended 10 cm beyond its rest position when the stone is launched, and assume that the mass of the stone is 5 g (a) Find the force needed to stretch the rubber band to 10 cm (b) Assuming that all of the energy of the stretched rubber band is transferred to the stone, find the speed of the stone when it is released. (c) Find the kinetic energy and momentum of the stone when it is released. (d) Now suppose that the mass of the stone is doubled, and all other parameters are the same as above. Find the kinetic energy and momentum of the stone in this case.

Answer 1

solution of Rubber band is extend to se'm force applied on the Rubber Band then 1 f = kx where = applied force k= spring constant 1. = gon/m And stored in Rubber Band Potential Energy U={kx² - Force needed to stretch the rubber band f = gox (ool) TF = 9N) . 6 in this case Toral potential Energy will be convert into Kinetic Energy, cie. £ K x ² = 1 m va where m= mass of stone

o= speed of stone kx2 = mu? gox (01)2 = (5x103) * 02 gox 10² x 10-2 --02 5 12 = 180 U 13.416 mis ► Kinetic Energy (K) = 1 mv2 K= * (5X10-3) x (180) K-(0.45) Joule momentum Cb) = mv þ = (5x10-3) X 121416 [p = 6.70 8x10-2 kg m/s]

€ since Energy on this case, spring potential converts into Kinetic Energy and USIK m2 = constances = does not change on charging mas That's why Kinetic Energy will not change and will be (0.45 Joule). since we know that k = ķ 2m where K= Kinetic Energy p= momentum m = mass since Initial Kinetic Energy = Final Kinetic Energy ami ame pia Pf a initial momentum = 6.708 X10-2 final momentum

L . , mi-inisial mass= 54103 Kg me = Final mas = 10x103 kg . Then putting values (6708x10-2) = b 24 (5x1043) 24 (1041043) V = %)*(6.409 4103) This = 9.406 xlo2 kg. m/s Here momentum changes and becomes g. 486 4102 kg m/s