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4-1.Newton's Laws of Motion
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Two particles $A$ and $B$ are connected by rigid rod $A B$. The rod slides along perpendicular rails as shown here. The velocity of $A$ to the left is $10\; m / s$. What is the velocity of $B$(in $m/s$) when angle $\alpha=60^{\circ}$?
A
$10$
B
$9.8$
C
$17.3$
D
$5.8$
(AIPMT-1998)
Solution
$L ^{2}=x^{2}+y^{2}$
On differentiating we get,
$0=2 x \frac{ dx }{ dt }+2 y \frac{ dy }{ dt }$
The Velocity of $A =\frac{ dx }{ dt }=-10 m / s$
The velocity of particle $B=\frac{ d y}{ dt }$
$\frac{ d y}{ dt }=\frac{- x }{ y } \frac{ dx }{ dt }=\cot 60^{0} \times 10=\frac{1}{\sqrt{3}} \times 10=5.8 m / s$
Standard 11
Physics
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