A particle falling vertically from a height hits a plane surface … - Vidyadip

Question

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle $\theta$ with speed $\text{v}_0$ and rebounds elastically. Find the distance along the plane where if will hit second time.

(Hint:

After rebound, particle still has speed Vo to start.
Work out angle particle speed has with horizontal after it rebounds.
Rest is similar to if particle is projected up the incline.

✓

Answer

Particle rebounces from P so it will be an elastic collision. As it strikes plane inclined at $\text{v}_0$ speed so speed of particle after rebounces will be $\text{v}_0$

Again consider the new axia X'OX and YOY' axis at P as origin 'O'. The componenets of g and $\text{v}_0$ in new OX and OY axis are:

$\text{v}_\text{x}=\text{v}_0\sin\theta\text{ and v}_\text{y}=\text{v}_0\cos\theta$

$\text{g}_\text{x}=\text{g}\cos\theta,\text{g}_\text{y}=\text{g}\sin\theta$ acting vertically downwords

Consider the motion of particle from O to A in new YOY' axis.

$\text{s}_\text{y}=\text{u}_\text{y}\text{t}+\frac{1}{2}\text{a}_\text{y}\text{t}^2$

$\text{s}_\text{y}=0\ \text{v}_\text{y}=\text{v}_0\cos\theta\ \text{a}_\text{y}=-\text{g}\sin\theta $ (upward)

$\therefore\text{t=T}$ (time of flight)

$0=\text{T}\Big[\text{v}_0\cos\theta-\frac{1}{2}\text{g}\text{ sin}\theta\text{ T}\Big]$

This means either $\text{T}=0\text{ or v}_0\cos\theta-\frac{\text{g}\cos\theta(\text{T})}{2}=0$

T cannot be zero $\Rightarrow\text{T}=\frac{2\text{v}_0\cos\theta}{\text{g}\cos\theta}$

$\text{T}=\frac{2\text{v}_0}{\text{g}}$

Now consider the motion along OX axis.

$\text{S}_\text{x}=\text{L, u}_\text{x}=\upsilon_0\sin\theta,\text{a}_\text{x}=\text{g}\sin\theta,\text{t}=\text{T}=\frac{2\upsilon_0}{\text{g}}$

$\text{S}_\text{x}=\text{u}_\text{x}\text{t}+\frac{1}{2}\text{a}_\text{x}\text{t}^2$

$\text{L}=\Big[\frac{2\text{v}_\text{0}}{\text{g}}\Big]\text{v}_0\sin\theta+\frac{1}{2}\text{g}\sin\theta\Big[\frac{2\text{v}_0}{\text{g}}\Big]^2$

$\text{L}=\frac{2\text{v}^2_0}{\text{g}}\sin\theta+\frac{1}{2}\text{g}\sin\theta.\frac{4\text{v}^2_0}{\text{g}^2}$

$=\frac{2\text{v}^2_0}{\text{g}}[\sin\theta+\sin\theta]=\frac{2\text{v}^2_0}{\text{g}}2\sin\theta$

$\Rightarrow\text{L}=\frac{4\text{v}^2_0}{\text{g}}\sin\theta.$

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