question_answer

A ball is dropped over an inclined plane of inclination \[45{}^\circ \] If the ball is dropped from a height of 5 m above the inclined plane and collision is elastic, then the length AB over the plane is \[(take,g=10ms)\]

A) \[20\sqrt{2}m\]              

B)  \[10\sqrt{2}m\]

C)  \[5\sqrt{2}m\]              

D) \[15\sqrt{2}m\]

Correct Answer: A

Solution :

Speed of ball with which it strikes the plane is

\[{{\upsilon }_{0}}=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10m{{s}^{-1}}\]

The ball bounces with same speed at an angle \[\theta =45{}^\circ \] with y-axis (shown in above figure), taken perpendicular to plane.

Now, from A to B

\[{{u}_{y}}={{\upsilon }_{0}}\cos \theta =10\times \cos 45{}^\circ =\frac{10}{\sqrt{2}}m{{s}^{-1}}\]

\[{{a}_{y}}=-g\cos \theta =-\frac{10}{\sqrt{2}}m{{s}^{-2}}\]

Now, \[y={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}\]

As \[y=0,\](for entire motion of projectile from A to B)

\[\therefore \,\,\,0={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}\Rightarrow t=\frac{2{{\upsilon }_{0}}\cos \theta }{g\cos \theta }\]

or time of flight, \[t=\frac{2{{\upsilon }_{0}}}{g}\]

\[\Rightarrow \,\,t=\frac{2\times 10}{10}=2s\]

Now, for motion along X-axis,

\[x=L=AB\]

\[{{u}_{x}}={{\upsilon }_{0}}\sin \theta =\frac{10}{\sqrt{2}}ms-1\]

\[{{a}_{x}}=g\sin \theta =\frac{10}{\sqrt{2}}ms-1\]

\[t=2s\]

So, yu\[x=AB={{u}_{x}}t+\frac{1}{2}{{a}_{x}}{{t}^{2}}\]

\[=\frac{10}{\sqrt{2}}\times 2+\frac{1}{2}\times \frac{10}{\sqrt{2}}\times {{\left( 2 \right)}^{2}}\]

\[=\frac{10}{\sqrt{2}}\times 4=20\sqrt{2}m\]