Question

Rod AB is placed against a block which is moving towards right with a speed of $1m/s$. If at an instant when the rod makes an angle ${60}^0$ with the horizontal, end A is sliding left with a speed of $1m/s$. Then the speed of the point of contact 'B' of the rod is $\sqrt{\frac{7}{x}}$ $m/s$. Find $x=?$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

From FBD, $1\cos 60=v\cos 30-1\cos 60$
$\Rightarrow \frac{1}{2}=\frac{\sqrt{3}}{2}v-\frac{1}{2}$
$\Rightarrow v=\frac{2}{\sqrt{3}}$
$\therefore v_{net}=\sqrt{1^2+(\frac{2}{\sqrt{3}}{)}^2}=\sqrt{\frac{7}{3}}$
therefore, $x=3$ is the correct option.