Question

A boy's catapult is made of rubber cord which is 42 cm long, with a 6 mm diameter of cross-section and negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in the area of the cross-section of the cord while stretched. Young's modulus of rubber is closest to ______.

पर्याय

• 10⁶ Nm^-2

• 10⁴ Nm^-2

• 10⁸ Nm^-2

• 10³ Nm^-2

Answer 1

A boy's catapult is made of rubber cord which is 42 cm long, with a 6 mm diameter of cross-section and negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms^-1. Neglect the change in the area of the cross-section of the cord while stretched. Young's modulus of rubber is closest to 10⁶ Nm^-2.

Explanation:

Given: The length of the rubber cord of the catapult is L = 42 cm, The cord's mass is negligible, and its cross-sectional diameter is D = 6 mm = 0.006 m, The stone's mass is m = 0.02kg,

The boy pulls the catapult by a distance of length = ΔL = 20 cm and the stone's initial velocity when it leaves the catapult is v = 20 m/s.

To find: The rubber cord's Young's modulus.

The potential energy stored by a stretched rubber cord is as follows:

`1/2Y((DeltaL)/L)^2AL = 1/2Y(DeltaL)^2/Lpi(D/2)^2` .......(i)

A is the area of the cross-section of the cord with diameter D in equation (i) The stone receives kinetic energy from the potential energy stored in the stretched cord.

Kinetic energy of stone = `1/2mv^2` ....(ii)

Because energy is always conserved, solve equations (i) and (ii),

`1/2Y(DeltaL)^2/Lpi(D/2)^2 = 1/2mv^2`

Put the values provided in the equation above.

`1/2Y(0.2 xx 0.2 xx pi xx 0.006 xx 0.006)/(0.42 xx 4)`

= `1/2 xx 0.02 xx 20 xx 20`

Y ≈ 10⁶ N/m²