Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
0 votes

A block of mass '3 Kg' is dropped from a height of $\;\large\frac{10}{3}\;$mts onto a mass '7 Kg' which is attached to spring of spring constant 3 N/m initially at rest . The 3 Kg mass sticks to the 7 Kg mass and then the spring starts to stretch . Find the maximum extension of the spring . (Consider inelastic collision (e=0))


Can you answer this question?

1 Answer

0 votes
Answer : (a) $\;0.446\;m$ \[\] Explanation :As the moment before the 3 Kg block touches ,the 7 Kg block , its velocity is\[\]$gh=\large\frac{mv^2}{2}$$v^2=2 \times g \times h$$v^2=2 \times 10 \times 5$$v=10 m/s$After collssion the velocity is \[\] $3 \times 10 =(7+3) v$$v=\large\frac{3\times10}{10}=3 m/s\;.$
By conservation of energy
$10\times 10 \times x + \large\frac{ 3 \times x^2}{2}=\large\frac{10 \times 3 \times 3 }{2}$
$x=\large\frac{-200 \pm \sqrt{41080}}{6}$
$=0.446 \;m$
Therefore , The spring extends by 0.446 m .
answered Mar 10, 2014 by yamini.v
edited Mar 10, 2014 by yamini.v

Related questions

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