Question

A plane mirror is inclined at an angle $\theta ={37}^{\circ }$ with horizontal surface. A particle is projected from point $P$ (see figure) at $t=0$ with a velocity $v=20\text{m/s}$ at an angle $\alpha ={53}^{\circ }$ with horizontal. Assuming the particle does not collide with the mirror. If in time $t$ (in $\text{s}$), image will be momentarily at rest with respect to particle, Find $10t$.

• A. 7
• B. 7.00
• C. 7.0

Answer 1

Answer: (A), (B), (C)

The image will come momentarily to rest with respect to the particle when it moves parallel to the mirror.
Let at the time $t$ the particle moves parallel to the mirror.

$v^{\prime }\sin \theta =v\sin \alpha –gt$ .........$\left(i\right)$

and $v^{\prime }\cos \theta =v\cos \alpha$
$\Rightarrow v^{\prime }=\frac{v\cos \alpha }{\cos \theta }$ .........$\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$,

$\frac{v\cos \alpha }{\cos \theta }\times \sin \theta =v\sin \alpha –gt$

$\Rightarrow t=\frac{v\sin \alpha \cos \theta -v\cos \alpha \sin \theta }{g\cos \theta }$

$\Rightarrow t=\frac{20{\left(\frac{4}{5}\right)}^2-20{\left(\frac{3}{5}\right)}^2}{10\left(\frac{4}{5}\right)}$
$\Rightarrow t=\frac{20\left(7\right)}{25\left(8\right)}=0.7$
$t=0.7\text{s}$

Therefore, the value of $10t$ is $7$.