In the shown figure, a particle of mass $(M / 10)$ strikes the block of mass M (which was hanging in rest as shown in figure) with velocity $v_{0}$ and gets attached to it. For what velocity $v_{0} ({in ms}^{- 1})$ , the block B just able to leave the ground?
(Given: $M = 100 gm, K = 880 N/m$ , $g = 10 m / s^{2}$ )

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

From C.L.M.
$\frac{M}{10} v_{0} = \frac{11}{10} Mv \Rightarrow v = \frac{v_{0}}{11}$ … (i)
For block B to leave ground
$K (x_{0} + x) = 2 Mg (where x_{0} = \frac{Mg}{K})$ $∴ x = \frac{Mg}{K}$ … (ii)
Question Image
From COE :
$\frac{11}{10} Mgx - [\frac{1}{2} K {(x_{0} + x)}^{2} - \frac{1}{2} {Kx}_{0}^{2}] = 0 - \frac{1}{2} \times \frac{11}{10} {Mv}^{2} {solving v}_{0} = g \sqrt{\frac{88 M}{K}}$

putting the values we get $v_{0} = 1 {ms}^{- 1}$