JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Rebounding of Ball After Collision With Ground

If a ball is dropped from a height h on a horizontal floor, then it strikes with the floor with a speed.

\[{{v}_{0}}=\sqrt{2g{{h}_{0}}}\]                   [From \[{{v}^{2}}={{u}^{2}}+2gh]\]

and it rebounds from the floor with a speed

\[{{v}_{1}}=e\,{{v}_{0}}\]\[=e\sqrt{2g{{h}_{0}}}\]     \[\left[ \text{As }e=\frac{\text{velocity after collision}}{\text{velocity before collision}} \right]\]

(1) First height of rebound : \[{{h}_{1}}=\frac{v_{1}^{2}}{2g}={{e}^{2}}{{h}_{0}}\]

\[\therefore \,\,\,\,{{h}_{1}}={{e}^{2}}{{h}_{0}}\]

(2) Height of the ball after nth rebound : Obviously, the velocity of ball after nth rebound will be

\[{{v}_{n}}={{e}^{n}}{{v}_{0}}\]

Therefore the height after nth rebound will be

\[{{h}_{n}}=\frac{v_{n}^{2}}{2g}={{e}^{2n}}{{h}_{0}}\]

\[\therefore \] \[{{h}_{n}}={{e}^{2n}}{{h}_{0}}\]

(3) Total distance travelled by the ball before it stops bouncing

\[H={{h}_{0}}+2{{h}_{1}}+2{{h}_{2}}+2{{h}_{3}}+...\]\[={{h}_{0}}+2{{e}^{2}}{{h}_{0}}+2{{e}^{4}}{{h}_{0}}+2{{e}^{6}}{{h}_{0}}+...\]

\[H={{h}_{0}}[1+2{{e}^{2}}(1+{{e}^{2}}+{{e}^{4}}+{{e}^{6}}....)]\]

\[={{h}_{0}}\left[ 1+2{{e}^{2}}\left( \frac{1}{1-{{e}^{2}}} \right) \right]\]      

\[\left[ \text{As}\,\,\,1+{{e}^{2}}+{{e}^{4}}+....=\frac{1}{1-{{e}^{2}}} \right]\]

\[\therefore \] \[H={{h}_{0}}\left[ \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right]\]

(4) Total time taken  by the ball to stop bouncing

\[T={{t}_{0}}+2{{t}_{1}}+2{{t}_{2}}+2{{t}_{3}}+..\]\[=\sqrt{\frac{2{{h}_{0}}}{g}}+2\sqrt{\frac{2{{h}_{1}}}{g}}+2\sqrt{\frac{2{{h}_{2}}}{g}}+..\]

\[=\sqrt{\frac{2{{h}_{0}}}{g}}\,\,\,\,[1+2e+2{{e}^{2}}+......]\]  [As \[{{h}_{1}}={{e}^{2}}{{h}_{0}}\]; \[{{h}_{2}}={{e}^{4}}{{h}_{0}}\]]

\[=\sqrt{\frac{2{{h}_{0}}}{g}}\,\,\,\,[1+2e(1+e+{{e}^{2}}+{{e}^{3}}+......)]\]

\[=\sqrt{\frac{2{{h}_{0}}}{g}}\,\left( \frac{1+e}{1-e} \right)\]

\[\therefore \] \[T=\left( \frac{1+e}{1-e} \right)\,\sqrt{\frac{2{{h}_{0}}}{g}}\]