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Properties of Matter question

2021 · 27 Jul · Shift 1 · Q71
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Properties of Matter question

2021 · 27 Jul · Shift 1 · Q71

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A stone of mass 20 g is projected from a rubber catapult of length 0.1 m and area of cross section 10 −-− 6 m2 stretched by an amount 0.04 m. The velocity of the projected stone is ‾\underline{\hspace{2cm}}​ m/s. (Young's modulus of rubber = 0.5 ×\times× 109 N/m2)
Numerical answer
View written solutionFree

Correct answer: 20

  1. Given data
  • Mass of stone: m=20 g=0.02 kgm = 20\text{ g} = 0.02\text{ kg}m=20 g=0.02 kg
  • Length of rubber: L=0.1 mL = 0.1\text{ m}L=0.1 m
  • Cross-sectional area: A=10−6 m2A = 10^{-6}\text{ m}^2A=10−6 m2
  • Extension: ΔL=0.04 m\Delta L = 0.04\text{ m}ΔL=0.04 m
  • Young's modulus: Y=0.5×109 N/m2Y = 0.5 \times 10^9\text{ N/m}^2Y=0.5×109 N/m2

We need the speed of the stone when released.

  1. Find the elastic force constant of the rubber

For a wire/string-like elastic body,

Y=stressstrain=F/AΔL/LY = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}Y=strainstress​=ΔL/LF/A​

So,

F=YAΔLLF = Y A \frac{\Delta L}{L}F=YALΔL​

This is of the form F=kΔLF = k\Delta LF=kΔL, hence

k=YALk = \frac{YA}{L}k=LYA​

Substitute values:

k=(0.5×109)(10−6)0.1k = \frac{(0.5\times 10^9)(10^{-6})}{0.1}k=0.1(0.5×109)(10−6)​ k=0.5×1030.1=5×103 N/mk = \frac{0.5\times 10^3}{0.1} = 5\times 10^3\text{ N/m}k=0.10.5×103​=5×103 N/m

So,

k=5000 N/mk = 5000\text{ N/m}k=5000 N/m
  1. Elastic potential energy stored

Energy stored in the stretched rubber:

U=12k(ΔL)2U = \frac{1}{2}k(\Delta L)^2U=21​k(ΔL)2 U=12(5000)(0.04)2U = \frac{1}{2}(5000)(0.04)^2U=21​(5000)(0.04)2

Now,

(0.04)2=0.0016(0.04)^2 = 0.0016(0.04)2=0.0016

Thus,

U=12×5000×0.0016=4 JU = \frac{1}{2}\times 5000 \times 0.0016 = 4\text{ J}U=21​×5000×0.0016=4 J
  1. Convert elastic energy into kinetic energy

Assuming all elastic potential energy goes into the kinetic energy of the stone,

12mv2=4\frac{1}{2}mv^2 = 421​mv2=4

Substitute m=0.02m=0.02m=0.02 kg:

12(0.02)v2=4\frac{1}{2}(0.02)v^2 = 421​(0.02)v2=4 0.01v2=40.01v^2 = 40.01v2=4 v2=400v^2 = 400v2=400 v=20 m/sv = 20\text{ m/s}v=20 m/s
  1. Final answer

The velocity of the projected stone is

20 m/s\boxed{20\text{ m/s}}20 m/s​

This matches the stored correct answer.

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