Question

' $A$ ' block of mass $m$ is held at rest on a smooth horizontal floor. A light frictionless, small pulley is fixed at a height of $6 m$ from the floor. A light inextensible string of length $16 m$ , connected with $A$ passes over the pulley and another identical block $B$ is hung from the string. Initial height of $B$ is $5 m$ from the floor as shown in figure. When the system is released from rest, $B$ starts to move vertically downwards and $A$ slides on the floor towards right. (i) If at an instant string makes an angle $θ$ with horizontal, calculate relation between velocity $u$ of $A$ and $v$ of B (ii) Calculate $v$ when $B$ strikes the floor.

Accepted Answer

Given,
Mass of block $m$ is, $6 m$
Mass of block $B$ is, $5 m$
Height of pulley is, $A$
Total length of string is, $L = 16 m$
Length of string from $θ$ to pulley initially is, $u$
Initial horizontal distance of $B$ from $A$ is, $x = \sqrt{z^{2} - y^{2}} = \sqrt{10^{2} - 6^{2}} = 8 m$
From the triangle $x y z$ , we can conclude that, $\cos θ = \frac{x}{z} = \frac{8}{10} = \frac{4}{5}$

(i) At an instant angle of inclined string with horizontal is $θ$
Velocity of block $A$ is, $u$ horizontal
Velocity of block $B$ is, $v$ vertical
Above case is shown in below figure,

From string constraint relation, we have
$v = u \cos θ . . . (1)$

(ii) Since there is not any external force and frictional force, therefore from conservation of mechanical energy, we have
$- ∆ U = ∆ K$
$\Rightarrow m g \times 5 = \frac{1}{2} m v^{2} + \frac{1}{2} m u^{2}$
$\Rightarrow m g \times 5 = \frac{1}{2} m {(u \cos θ)}^{2} + \frac{1}{2} m u^{2}$ , from $(1)$
$\Rightarrow u = \frac{50}{\sqrt{41}} m s^{- 1}$
$\Rightarrow v = u \cos θ = \frac{50}{\sqrt{41}} \times \frac{4}{5} = \frac{40}{\sqrt{41}} m s^{- 1}$

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