Two identical steel cubes (masses $50\,g$,side $1\,cm$) collide head-on face to face with a speed of $10\,cm/s$ each. Find the maximum compression of each. Young's modulus for steel $Y = 2 \times 10^{11}\,N/m^2$.
(D) Given: $m = 50\,g = 0.05\,kg$,$L = 1\,cm = 0.01\,m$,$v = 10\,cm/s = 0.1\,m/s$,$Y = 2 \times 10^{11}\,N/m^2$.
At maximum compression,the kinetic energy of the cubes is converted into elastic potential energy.
The effective spring constant $k$ for a cube is given by $F = k \Delta L = Y A \frac{\Delta L}{L}$.
Thus,$k = \frac{YA}{L} = \frac{Y L^2}{L} = YL$.
Total initial kinetic energy $KE = 2 \times (\frac{1}{2} m v^2) = m v^2 = 0.05 \times (0.1)^2 = 5 \times 10^{-4}\,J$.
Total elastic potential energy at maximum compression $\Delta L_{max}$ for two cubes is $PE = 2 \times (\frac{1}{2} k (\Delta L_{max})^2) = k (\Delta L_{max})^2$.
Equating $KE = PE$: $m v^2 = (YL) (\Delta L_{max})^2$.
$\Delta L_{max} = \sqrt{\frac{m v^2}{YL}} = \sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.01}} = \sqrt{2.5 \times 10^{-13}} \approx 5 \times 10^{-7}\,m$.
At maximum compression,the kinetic energy of the cubes is converted into elastic potential energy.
The effective spring constant $k$ for a cube is given by $F = k \Delta L = Y A \frac{\Delta L}{L}$.
Thus,$k = \frac{YA}{L} = \frac{Y L^2}{L} = YL$.
Total initial kinetic energy $KE = 2 \times (\frac{1}{2} m v^2) = m v^2 = 0.05 \times (0.1)^2 = 5 \times 10^{-4}\,J$.
Total elastic potential energy at maximum compression $\Delta L_{max}$ for two cubes is $PE = 2 \times (\frac{1}{2} k (\Delta L_{max})^2) = k (\Delta L_{max})^2$.
Equating $KE = PE$: $m v^2 = (YL) (\Delta L_{max})^2$.
$\Delta L_{max} = \sqrt{\frac{m v^2}{YL}} = \sqrt{\frac{5 \times 10^{-4}}{2 \times 10^{11} \times 0.01}} = \sqrt{2.5 \times 10^{-13}} \approx 5 \times 10^{-7}\,m$.
Explore More
Similar Questions
An elevator cable can have a maximum stress of $7 \times 10^7\,N/m^2$ for appropriate safety factors. Its maximum upward acceleration is $1.5\,m/s^2$. If the cable has to support the total weight of $2000\,kg$ of a loaded elevator,the minimum area of cross-section of the cable should be ....... $cm^2$ $(g = 10\,m/s^2)$.
Difficult
View SolutionRead the following two statements below carefully and state,with reasons,if it is true or false.
$(a)$ The Young's modulus of rubber is greater than that of steel;
$(b)$ The stretching of a coil is determined by its shear modulus.
$(a)$ The Young's modulus of rubber is greater than that of steel;
$(b)$ The stretching of a coil is determined by its shear modulus.
Medium
View Solution$A$ sphere of mass $4 \,kg$ is attached to one end of a steel wire of length $1 \,m$ and radius $1 \,mm$. It is whirled in a vertical circle with an angular velocity $10 \,rad \,s^{-1}$. If the sphere is at the lowest point of its path,the elongation in the wire is . . . . . . $(g=10 \,ms^{-2}, Y_{\text{steel}}=20 \times 10^{10} \,Nm^{-2})$ (in $\,mm$)
$A$ string of cross-sectional area $4 \times 10^{-6} \, m^{2}$ and length $0.5 \, m$ is connected to a rigid body of mass $2 \, kg$. The body is rotated in a vertical circular path of radius $0.5 \, m$. The body acquires a speed of $5 \, m/s$ at the bottom of the circular path. The strain produced in the string when the body is at the bottom of the circle is $\dots \times 10^{-5}$. (Use Young's modulus $Y = 10^{11} \, N/m^{2}$ and $g = 10 \, m/s^{2}$)
DifficultJEE MAIN 2022
View Solution$A$ wire of length $5 \,m$ is twisted through $30^{\circ}$ at its free end. If the radius of the wire is $1 \,mm$,the shearing strain in the wire is: (in $^{\circ}$)
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo