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 and end A is sliding left with a speed of 1m/s. Then the speed of the point of contact ${}^{{}^{\prime }}{B}^{{}^{\prime }}$ of the rod is :

• A. $1m/s$
• B. $\frac{\sqrt{3}}{2}m/s$
• C. $\frac{1}{2}m/s$
• D. $\sqrt{\frac{7}{3}}m/s$

Answer 1

Answer: (D)

$\sqrt{\frac{7}{3}}m/s$

$\text{After breaking the velocity into components we get}$

$⟹1cos60=vcos30-1cos60$

$⟹v=\frac{2}{\sqrt{3}}$

$v_{net}=\sqrt{{\left(\frac{2}{\sqrt{3}}\right)}^2+1^2}=\sqrt{\frac{7}{3}}m/s.$