A mass `M` is in static equilibrium on a massless vertical spring as shown in the figure. A ball of mass `m` dropped from certain height sticks to the mass `m` after colliding with it. The oscillations they perform reach to height `'a'` above the original level of spring `&` depth `'b'` below it .

`(a)` Find the force constant of the spring.
`(b)` Find the oscillation frequency.
`(c)` What is the height above the initial level from which the mass `m` was dropped ?

Answer

A, B

Explanation

(i) Let `x_(0) = (Mg)/(k)` = initial compression in spring
COME : `1/2k(x_(0) + b)^(2) = 1/2 k(a-x_(0))(2) + (m+M)g(b+a)`
`rArr k = (2mg)/(b-a)`
(ii) `k = (2mg)/((b-a)) rArr (m+M)omega^(2) = (2mg)/(b-a)`
`rArr f = 1/(2pi)sqrt((2mg)/((b-a)(m+M)))`
(iii) Let `h =` initial height of m over the pan
`v =` common velocity of `(m+M)` after collision
COLM `: msqrt((2gh)) = (m+M)v"......"(i)`
COME : `1/2kx_(0^(2)) + 1/2(m+M)v^(2) + (m+M)gb = (1)/(2)k(x_(0) + b)^(2)`
`rArr = h = ((M+m)/(m))((ab)/(b-a))`