First slide

Connected bodies

Question

In fig., the blocks are of equal mass. The pulley
is fixed. In the position shown, A moves with a speed u and the speed of B is v. Then

Moderate

A

v=u
B

$v = u / \cos θ$
C

$v = u / \sin θ$
D

$v = u \sin θ$

Solution

See fig.
From figure
$\begin{array}{l} x^{2} + y^{2} = l^{2} \\ ∴ 2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 2 l \frac{dl}{dt} \end{array}$
There is not vertical motion of the block. Hence
$\frac{dy}{dt} = 0, \frac{dx}{dt} = v_{x} and \frac{dl}{dt} = v_{A} = u$
$Hence 2 \times v_{x} = 2 lu or {xv}_{x} = lu$
$∴ v_{x} = \frac{l}{x} u = \frac{u}{\cos θ}$ .

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