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Centre of Mass, Linear Momentum, Collision — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsCentre of Mass, Linear Momentum, Collision5 Marks
Question
A projectile is fired with a speed u at an angle $\theta$ above a horizontal field. The coefficient of restitution of collision between the projectile and the field is e. How far from the starting point, does the projectile makes its second collision with the field?
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Answer

The projected velocity $= u$.
The angle of projection $=\theta.$
When the projectile hits the ground for the $1^{st}$ time, the velocity would be the same i.e. $u$.
Here, the component of velocity parallel to ground, $\text{u}\cos\theta$ should remain constant.
But the vertical component of the projectile undergoes a change after the collision.
$\Rightarrow\text{e}=\frac{\text{u}\sin\theta}{\text{v}}$
$\Rightarrow\text{v}=\text{eu}\sin\theta.$
Now for the $2^{nd}$ projectile motion $, U = $ velocity of projection $=\sqrt{(\text{u}\cos\theta)^2+(\text{eu}\sin\theta)^2}$ and Angle of projection $=\alpha=\tan^{-1}\Big(\frac{\text{eu}\sin\theta}{\text{u}\cos\theta}\Big)=\tan^{-1}(\text{e}\tan\theta)$ or $\tan\alpha=\text{e}\tan\theta\ \dots(2)$
Because, $\text{y}=\text{x}\tan\alpha-\frac{\text{gx}^2\sec^2\alpha}{\text{2u}^2}\ \dots(3)$
Here, $\text{y}=0,\ \tan\alpha=\text{e}\tan\theta,\ \sec^2\alpha=1+\text{e}^2\tan^2\theta$ And $\text{u}^2=\text{u}^2\cos^2\theta+\text{e}^2\sin^2\theta$
Putting the above values in the equation $(3),$ $\text{xe}\tan\theta=\frac{\text{gx}^2(1+\text{e}^2\tan^2\theta)}{\text{2u}^2(\cos^2\theta+\text{e}^2\sin^2\theta)}$
$\Rightarrow\text{x}=\frac{2\text{eu}^2\tan\theta(\cos^2\theta+\text{e}^2\sin^2\theta)}{\text{g}(1+\text{e}^2\tan^2\theta)}$
$\Rightarrow\text{x}=\frac{2\text{eu}^2\tan\theta-\cos^2\theta}{\text{g}}=\frac{\text{eu}^2\sin2\theta}{\text{g}}$
$\Rightarrow$ So, from the starting point $O,$ it will fall at a distance $=\frac{\text{u}^2\sin2\theta}{\text{g}}+\frac{\text{eu}^2\sin2\theta}{\text{g}}=\frac{\text{u}^2\sin2\theta}{\text{g}}(1+\text{e})$

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