Question

A rubber cord has a cross -sectional area 1 mm^(2)and total unstretched length10.0cm. It is streched to 12.0 cm and then released to project a missile of mass 5.0 g.Taking young's modulus Y for rubber as 5.0xx10^(8) N//m^(2).Calculate the velocity of projection .

Answer 1

Hence the velocity of projection is  141.3 m/s

Explanation:

12mv^2 = 1 / 2 × stress × strain × volume

mv^2 = stress × strain × volume

mv^2 = (strain)^2 × Y × volume

mv^2 = (△L / L)^2 × Y × AL

v = √(△L)^2 YA / Lm

Calculation of velocity:

V = √ 0.02 x 5.0 x 10^8 x 1 x 10^-6 m^2 / 0.1 m x 5 x 10^-3 Kg

V = √ 0.1 x 10^2 / 0.5

√  = √ 10 x 10^3 / 0.5

V = √ 10000 / 0.5

V = √ 20000 = 141.3 m/s

Hence the velocity of projection is  141.3 m/s