Question

Two particles $A$ and $B$ are connected by a rigid rod $AB$. The rod slides along perpendicular rails as shown here. The velocity of $A$ to the left is $10m/s$. What is the velocity of $B$ when angle $\alpha ={60}^o$?

• A. $9.8$ $m/s$
• B. $10$ $m/s$
• C. $5.8$ $m/s$
• D. $17.3$ $m/s$

Answer 1

Answer: (D)

$17.3$ $m/s$

Let us consider the upward velocity of rod B to be $vm/s$
Now, Velocity of rod $A=10m/s$
We know that, Slope of a velocity graph is given by, $\tan \alpha =\frac{v_y}{v_x}$
Here, $v_y=vm/sand{v}_x=10m/s$
$\therefore \tan {60}^{\circ }=\frac{v}{10}\Rightarrow v=10\times \tan {60}^{\circ }$
$\therefore v=17.3m/s$