A rubber band catapult has initial length 2 ~cm and cross-sectional area 5 ~mm ^2 . It is stretched to 2 ~cm and then released to project a stone of mass of 20 ~g . The velocity of projected stone is (Young's modulus of rubber =5 10^8 Nm ⁻² )

Step-by-step solution

According to work-energy theorem, K E of stone = Elastic potential energy of rubber band. Now, for rubber band; Young's modulus =5 10^8 ~N / m ^2 Length, L=2 10⁻² ~m Change of length, L=2 10⁻² ~m Area of cross-section, A=5 10⁻⁶ ~m ^2 Elastic potential energy of rubber band aligned & = 1 2 Y ( strain )^2 volume & = 1 2 Y ( L L )^2 A L & = 1 2 5 10^8 ( 2 10⁻² 2 10⁻² )^2 5 10⁻⁶ 2 10⁻² & =25 ~J aligned This energy is given to stone (mass =20 ~g ); KE (stone) = E.P.E (Rubber band) 1 2 m v^2=25 1 2 20 10⁻³ v^2=25 array l