Work Energy And Power Question 296
Question: A particle fells from a height A on a fixed horizontal plane and rebounds. If e is the coefficient of restitution, the total distance travelled by the particle before it stops rebounding is
Options:
A) $ \frac{h}{2}\frac{[ 1-e^{2} ]}{[ 1+e^{2} ]} $
B) $ \frac{h[ 1-e^{2} ]}{[ 1+e^{2} ]} $
C) $ \frac{h}{2}\frac{[ 1+e^{2} ]}{[ 1-e^{2} ]} $
D) $ \frac{h[ 1+e^{2} ]}{[ 1-e^{2} ]} $
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Answer:
Correct Answer: D
Solution:
[d] The velocity of particle after falling through height h $ u=\sqrt{2gh} $ …(i)
After first rebounding, the velocity of ball is eu and after attaining maximum height it will come to the ground with same velocity eu.
So, after second rebounding its velocity will be $ e^{2}u $ .
Similarly, after third fourth…etc reboundings its velocities will be $ e^{2}u,e^{4}u $ ,… etc.
Since, it first rebounds with velocity eu so if it attains height h then from $ v^{2}=u^{2}-2gh $
$ \therefore 0=e^{2}u^{2}-2gh $
Or $ h_1=\frac{e^{2}u^{2}}{2g}=\frac{e^{2}2gh}{2g}=e^{2}h $ [from Eq. (i)] The same height the ball travels while approaching ground.
Now, it rebounds with velocity $ e^{2}u $ so if it attains a height $ h_2 $ then $ 0 = e^{4}u^{2}- 2gh_2 $ or $ h_2=e^{4}h $
The similar process will follow for further reboundings
Hence, the total distance travelled by the practice before it stops rebounding.
$ =h+2h _{i}+2h_2+…\infty =h+2e^{2}h+2e^{4}h+…\infty $
$ =h+2e^{2}h( 1+e^{2}+e^{4}+…\infty )=h+2e^{2}h( \frac{1}{1-e^{2}} ) $
$ =h( 1+\frac{2e^{2}}{1-e^{2}} )=( \frac{1+e^{2}}{1-e^{2}} )h $
