logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...