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\[x=AB={{u}_{x}}t+\frac{1}{2}{{a}_{x}}{{t}^{2}}\] |
| \[=\frac{10}{\sqrt{2}}\times 2+\frac{1}{2}\times \frac{10}{\sqrt{2}}\times {{\left( 2 \right)}^{2}}\] |
| \[=\frac{10}{\sqrt{2}}\times 4=20\sqrt{2}m\] |
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