Question

The slopes of the windscreen of two cars are ${\alpha }_1={30}^o$ and ${\alpha }_2={15}^o$ respectively. At what ratio $\frac{v_1}{v_2}$ of the velocities of the cars will their drivers see the hailstones bounced back by the windscreen on their cars in the vertical direction? Assume hailstones fall vertically downwards and collisions to be elastic.

• A. $3$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $3$

Here, absolute velocity of hail stones $\overrightarrow{v}$ before colliding with wind screens is vertically downwards and velocity of hail stones with respect to cars after collision ${\overrightarrow{v}}_{HC}^{\prime }$ is vertically upwards. Collision is elastic, hence, velocity of hail stones with respect to cars before collision ${\overrightarrow{v}}_{HC}$ and after collision ${\overrightarrow{v}}_{HC}^{\prime }$ will make equal angles with the normal to the wind screen.
$({\overrightarrow{v}}_{HC}{)}_1=$ velocity of hail stones -velocity of car 1
$=\overrightarrow{v}-{\overrightarrow{v}}_1$
From the figure, we can see that
$\beta +{90}^o-2\beta +{\alpha }_1={90}^o$
or ${\alpha }_1=\beta$
or $2\beta =2{\alpha }_1$
In $\Delta ABC,tan2\beta =tan2{\alpha }_1=\frac{v_1}{v}$ .....(i)
Similarly, we can show that
$tan2{\alpha }_1=\frac{v_2}{v}$ .....(ii)
From Eqs. (i) and (ii), we get
$\frac{v_1}{v_2}=\frac{tan2{\alpha }_1}{tan2{\alpha }_2}=\frac{tan{60}^o}{tan{30}^o}=\frac{\sqrt{3}}{1/\sqrt{3}}=3$
$\therefore \frac{v_1}{v_2}=3$