Question

A block with initial velocity $2\text{m/s}$ is sliding on a horizontal frictionless surface as shown in the figure. What is the value of $\theta$ when the block leaves the surface?

• A. $\theta ={\text{cos}}^{-1}\left(\frac{19}{25}\right)$
• B. $\theta ={\text{cos}}^{-1}\left(\frac{11}{25}\right)$
• C. $\theta ={\text{cos}}^{-1}\left(\frac{13}{25}\right)$
• D. $\theta ={\text{cos}}^{-1}\left(\frac{17}{25}\right)$

Answer 1

The correct option is D $\theta ={\text{cos}}^{-1}\left(\frac{17}{25}\right)$

When the block leaves the surface then the normal force between the block and the surface becomes zero.
$N=0$
At point $P$,
$\Rightarrow \text{mg cos}\theta =\frac{m{v}^2}{R}+N$
$\Rightarrow \text{mg cos}\theta =\frac{m{v}^2}{R}$
[ as $N=0$ at $P$ ]
$\Rightarrow v^2=Rg\cos \theta$ ...............(1)
By using law of conservation of mechanical energy,
Loss in Potential energy $=$ Gain in Kinetic energy
$\text{mgR}(1-\text{cos}\theta )=\frac{m{v}^2}{2}-\frac{m{v}_0^2}{2}$
$\Rightarrow gR(1-\text{cos}\theta )=\frac{v^2}{2}-\frac{v_0^2}{2}$
$\Rightarrow gR(1-\text{cos}\theta )=\frac{Rg\cos \theta }{2}-\frac{v_0^2}{2}$
[ from (1) ]
$\Rightarrow \frac{3gR\cos \theta }{2}=gR+\frac{v_0^2}{2}$
$\Rightarrow \cos \theta =\frac{2}{3gR}(gR+\frac{v_0^2}{2})$
$\Rightarrow \cos \theta =\frac{2}{3\times 10\times 10}(10\times 10+\frac{2^2}{2})$
$\Rightarrow \cos \theta =\frac{17}{25}$
$\Rightarrow \theta ={\text{cos}}^{-1}\left(\frac{17}{25}\right)$
Hence option $\left(d\right)$ is correct.