**Extension of a string**

One end of a light elastic string of natural length `l` , passing through a small smooth ring of mass `m`, is attached to a point `O` on the ceiling of the room. A particle `P` of mass `M` attached to the other end of the string hangs in equilibrium, with the ring being held at rest at the point `O` . If `2Mg` is the modulus of elasticity of the string, the extension of the string in the equilibrium position is `l/2` .

Q: The ring, now released from rest at `O`, slides vertically downward along the string, under gravity, collides and coalesces with `P`. Show that the composite body consisting of the ring and the particle will begin to move vertically downward with velocity `(m / (M + m))* sqrt (3gl)`.

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When the 'system' is in equilibrium, the string is extended by l/2 (answer to part A). Therefore, its present length is l + l/2 = (3/2)l.

The ring, then, has (3/2)l to fall before it reaches the particle P, so

`d = (3l)/2`

Now, `d = ut + (at^2)/2` where `u` is the initial velocity and `a` is acceleration.

Here, `u=0` and `a = g` (acceleration due to gravity) so we have that

`at^2 = 3l` `implies t = sqrt((3l)/a) = sqrt((3l)/g) `

Now, velocity is the derivative of distance with respect to time t, so

`v = u + at`

Here, `v = g t`

When `t = sqrt((3l)/g)` the ring reaches the particle P and their conjoined mass becomes `m+M`. The resultant downward force is their conjoined mass x g minus the current tension in the string `Mg`, ie `mg`.

Using `F = mu a` where `mu` represents mass and `a` is the acceleration we are interested in, we have

`mg = (m+M)a` so that

`a = (m/(m+M))g`

Just before the ring hits P the velocity of the ring is `g t = sqrt(3gl)` . However, the instant the ring hits P, the acceleration of the conjoined body suddenly increases to `a = (m/(m+M))g` so that the velocity at that instant is

`(m/(m+M))sqrt(3gl)` **answer as required**

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