A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section $10^{-6}\mathrm{~m}^{2}$ stretched by an amount 0.04 m. The velocity of the projected stone is ______ $\mathrm{m}/\mathrm{s}$. (Young's modulus of rubber $=0.5\times 10^{9}\mathrm{~N}/\mathrm{m}^{2}$)

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Elasticity & SHM

A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section

1 0^{- 6} m^{2}

stretched by an amount 0.04 m. The velocity of the projected stone is ______

m / s

.
(Young's modulus of rubber

= 0.5 \times 1 0^{9} N / m^{2}

)

Answer: 20

The pull in the stretched rubber follows from Young's modulus,

F = \frac{Y A Δ L}{L} = \frac{0.5 \times 1 0 ^{9} \times 1 0 ^{- 6} \times 0.04}{0.1} = 200 N

The rubber is Hookean over the stretch, so the force grows linearly from zero and the stored elastic energy is

U = \frac{1}{2} F Δ L = \frac{1}{2} \times 200 \times 0.04 = 4 J

This energy is handed over as kinetic energy of the stone of mass

0.02

kg, so

\frac{1}{2} m v^{2} = 4

gives

v = \frac{2 \times 4}{0.02} = 400

The stone is projected with a speed of

20

m/s.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer