Question

A ball falls on an inclined plane of inclination $\theta ={30}^{\circ }$ from a height $h=10\text{m}$ above the point of impact and makes a perfectly elastic collision. Find the distance where will it hit the plane again from the initial point of impact. $(\text{Take}g=10{\text{m/s}}^2)$

• A. $20\text{m}$
• B. $40\text{m}$
• C. $10\text{m}$
• D. $80\text{m}$

Answer 1

The correct option is B $40\text{m}$

Speed of the ball just before the collision
$u=\sqrt{2gh}\text{m/s}$
If we rotate the wedge, make face of the wedge as horizontal, we have


Hence now,
acceleration along $x$ direction $=\text{g sin}\theta$
initial speed along $x$ direction $=\text{u sin}\theta$
acceleration along $y$ axis $=-g\text{cos}\theta$
speed along $y$ axis $=\text{u cos}\theta$
As we know time of flight $T=\frac{2u_y}{a_y}$
Here $u_y=\text{u cos}\theta$
$a_y=\text{g cos}\theta$
Since, there is elastic collision hence, velocity of approach is equal to velocity of seperation.
$\Rightarrow T=\frac{2\times \text{u cos}\theta }{\text{g cos}\theta }\Rightarrow (T=\frac{2u}{g})$

Distance travelled in time $T$ in horizontal direction is given by
$S_x=u_{x}T+\frac{1}{2}{a}_x{T}^2$
$\Rightarrow S_x=u\sin \theta \times \frac{2u}{g}+\frac{1}{2}\times {\left(\frac{2u}{g}\right)}^2\times g\sin \theta$
$\Rightarrow S_x=\frac{2u^2\text{sin}\theta }{g}+\frac{2u^2\text{sin}\theta }{g}$
$\Rightarrow S_x=\frac{4u^2\text{sin}\theta }{g}=\frac{4\times (\sqrt{2gh}{)}^2\sin \theta }{g}$
$\Rightarrow S_x=8\times 10\times \text{sin}{30}^{\circ }$
$\Rightarrow S_x=40\text{m}$
Hence, the distance where will it hit the plane again from the initial point of impact is $40\text{m}$