Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A object of mass $5.0\;g$ is projected into air using a rubber string of cross sectional area of $1;mm^2$. The length of the chord when it was unstretched is 10.0 cm. It is then stretched to 12 cm to project the object . The young's modulus of rubber is $5.0 \times 10^8 n/m^2$. The velocity of projectile is (assuming all the energy of string is transferred to projectile)

\[\begin {array} {1 1} (a)\;10\;m/s \\ (b)\;20\;m/s \\ (c)\;30\;m/s \\ (d)\;40\;m/s \end {array}\]
Can you answer this question?

1 Answer

0 votes
The rubber chord has a force constant k given by $k=\large\frac{yA}{L}$
$\qquad= \large\frac{5.0 \times 10^8 \times 1.0 \times 10^{-6}}{0.1}$
$\qquad= 5.0 \times 10^3 N/m$
Now from conservation of energy theorem.
The elastic potential energy of string is equal to the kinetic energy of the projectile
$\large\frac{1}{2} $$k ( \Delta l)^2=\large\frac{1}{2}$$mv^2$
Where $\Delta L$ is the extension in the string
$\therefore v=\sqrt {\large\frac{k}{m}}$$\Delta l$
$\qquad= \sqrt {\frac{5.0 \times 10^3}{5.0 \times 10^{-3}}}$$ 2 \times 10^{-2}$
$\qquad=20 m/s$
Hence b is the correct answer. 


answered Oct 7, 2013 by meena.p
edited Feb 18, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App