Motion In A Plane Ques 1
1. Two particles $A$ and $B$ are connected by a rigid rod $\mathrm{AB}$. The rod slides along perpendicular rails as shown here. The velocity of $A$ to the left is $10 \mathrm{m} / \mathrm{s}$. What is the velocity of $B$ when angle $\alpha=60^{\circ}$ ?
[1998]
(a) $5.8 \mathrm{m} / \mathrm{s}$
(b) $9.8 \mathrm{m} / \mathrm{s}$
(c) $10 \mathrm{m} / \mathrm{s}$
(d) $17.3 \mathrm{m} / \mathrm{s}$
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Answer:
Correct Answer: 1.(a)
Solution: (a) Given, $v_A=10 \mathrm{m} / \mathrm{sec}$
$ \alpha=60^{\circ} $
let length of the rod $=\mathrm{L}$
From figure,
$ \Rightarrow x^2+y^2=\mathrm{L} $
differentiation with respect to time ’ $t$ '
$ \begin{aligned} & \Rightarrow 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0 \ & \Rightarrow x \frac{d x}{d t}+y \frac{d y}{d t}=0 \end{aligned} $
where, $\frac{d x}{d t}=v_A$ and $\frac{d y}{d t}=v_B$
So, $x v_A+y v_B=0$
$ v_B=\frac{-x}{y} v_A $
where $\frac{x}{y}=\cot \alpha$
So, $v_B=-v_A \cot \alpha$
$ \begin{aligned} & v_B=\frac{-10}{\sqrt{3}}=-5.773 \simeq 5.8 \mathrm{m} / \mathrm{sec} \end{aligned} $
BETA
