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Model Paper 2 — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 25 Marks
Question
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle with speed $v_0$ and rebounds elastically as shown in the figure. Find the distance along the plane where it will hit the second time.
Image
Hint:
i. After rebound, particle still has speed $V_0$ to start.
ii. Work out angle particle speed has with horizontal after it rebounds.
iii. Rest is similar to if particle is projected up the incline.]
✓

Answer

From the figure resolving the components of $v_0$ and $g$, we get
Image
$v_x=v_0 \sin \theta$ and $v_y=v_0 \cos \theta$
$g_x=g \cos \theta, g_y=g \sin \theta$ acting vertically downwards
Consider the motion of particle from O to A in new YOY' axis.
$
y=u_y t+\frac{1}{2} a_y t^2
$
Where, $z=0, \quad v_y=v_0 \cos \theta, \quad a_y=-g \sin \theta$
$
\begin{aligned}
& \therefore t=T \quad(\text { time of flight }), y=0 \\
& \Rightarrow 0=v_0 \cos \theta T-\frac{1}{2} g \sin \theta T^2 \\
& \Rightarrow T=\frac{2 v_0 \cos \theta}{g \cos \theta} \\
& T=\frac{2 v_0}{g}
\end{aligned}
$
Now consider the motion along OX axis.
$
\begin{aligned}
& x=L, u_x=v_0 \sin \theta, a_x=g \sin \theta, t=T=\frac{2 v_0}{g} \\
& x=u_x t+\frac{1}{2} a_x t^2 \\
& L=\left[\frac{2 v_5}{g}\right] v_0 \sin \theta+\frac{1}{2} g \sin \theta\left[\frac{2 v_0}{g}\right]^2 \\
& L=\frac{2 v_0^2}{g} \sin \theta+\frac{1}{2} g \sin \theta \cdot \frac{4 v_0^2}{g^2} \\
& =\frac{2 v_0^2}{g}[\sin \theta+\sin \theta]=\frac{4 v_0^2}{g} \sin \theta \\
& \Rightarrow L=\frac{4 v_0^2}{g} \sin \theta
\end{aligned}
$
Hence the value of L is $\frac{4 v_0^2}{g} \sin \theta$.

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