Question

The speed of the block at the point $C$; immediately before it leaves the second incline is

• A. $\sqrt{120}m/s$
• B. $\sqrt{105}m/s$
• C. $\sqrt{90}m/s$
• D. $\sqrt{75}m/s$

Answer 1

The correct option is B $\sqrt{105}m/s$

let the velocity before collision $=v$

let the velocity after the collision $=v_B$

Along the plane velocity before collision can be given by

$v=\sqrt{2gh}$

$v=\sqrt{2\left(10\right)\left(3\right)}m/s$

$v=\sqrt{60}m/s$

Along the place velocity after collision

$v_B=v\cos {30}^o$

$v_B=\sqrt{45}m/s$

the speed of the Block at point $C$,

$v_C=mgh$

$v_C\Rightarrow mg\left(3\right)=\frac{1}{2}m[v_C^2-v_B^2]$

$m\left(10\right)\left(3\right)=\frac{1}{2}m[v_C^2-v_B^2]$

$60=\frac{1}{2}[v_C^2-{\sqrt{\left(45\right)}}^2]$

$60=1[v_C^2-45]$

$v_C^2=60+45m/s$

$v_C^2=105m/s$

$v_C=\sqrt{105}m/s$