Question

A ball falls on an inclined plane of inclination $\theta ={30}^o$ from a height $h=10\text{m}$ above the point of impact and makes a perfectly elastic collision. Find the maximum height attained by the ball, perpendicuar to wedge surface. (take $g=10{\text{m/s}}^2$)

• A. $5\sqrt{3}\text{m}$
• B. $5\text{m}$
• C. $10\text{m}$
• D. $10\sqrt{3}\text{m}$

Answer 1

The correct option is A $5\sqrt{3}\text{m}$

Speed of the ball just before the collision $u=\sqrt{2gh}=\sqrt{2\times 10\times 10}=10\sqrt{2}\text{m/s}$

If we rotate the wedge make face of the wedge as horizontal.


Hence after collision,
Acceleration along x direction $=g\sin \theta$
Initial speed along x direction $=u\sin \theta$
Acceleration along y axis $=-g\cos \theta$
Speed along y axis $=u\cos \theta$


At maximum Height $v_y=0.$
Hence, from $v^2=u^2+2aS$
$0=(u\cos \theta {)}^2+2\times (-g\cos \theta )S_y$
$S_y=\frac{u^2\cos \theta }{2g}$
$S_y=\frac{(10\sqrt{2}{)}^2\times \cos {30}^o}{2\times 10}$
$S_y=\frac{100\times 2\times \frac{\sqrt{3}}{2}}{2\times 10}$
$S_y=5\sqrt{3}\text{m}.$