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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
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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))

$(a)\;0.446\;m\qquad(b)\;0.518\;m\qquad(c)\;0.357\;m\qquad(d)\;None\;of\;these$

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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
$mgx+\large\frac{Kx^2}{2}=\large\frac{mv^2}{2}$
$10\times 10 \times x + \large\frac{ 3 \times x^2}{2}=\large\frac{10 \times 3 \times 3 }{2}$
$100x+\large\frac{3x^2}{2}=45$
$200x+3x^2=90$
$3x^2+200x-90=0$
$x=\large\frac{-200 \pm \sqrt{41080}}{6}$
$=\large\frac{202.68-200}{6}$
$=0.446 \;m$
Therefore , The spring extends by 0.446 m .
answered Mar 10, 2014 by yamini.v
edited Mar 10, 2014 by yamini.v
 

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