A simple pendalum is suspended from a peg on a verticle wall . The pendulum is pulled away from the well is a horizental position (see fig) and released . The bell his the well the coefficient of resitution being `(2)/(sqrt(5)`

what is the miximum number of colision after which the amplitube of secillections between less that `60` digree ?

Answer

4

Explanation

When simple pendulum released from position A strikes the wall with velocity `v` then by conservation of mechanical energy.
`mgL+0=(1)/(2)mv^(2)+0, i.e.v=sqrt(2 gL)`
Now as coefficient of restitution is `e` so, speed of pendulum after first collision will be `v_(1)=ev=esqrt(2gL)`
Now after completing Oscillation in accordance with conservation of mechanical energy it will strike the wall with same velocity and so its velocity after second collision will be
`v_(2)=ev_(1)=e(esqrt(2gL))=e^(2)sqrt(2gL)`
So the velocity of the pendulum after `n` collision will be `v_(n)=e^(n)v=e^(n)sqrt(2gL)`
Now if it rises to a height `h`, by conservation of mechanical energy
`(1)/(2)m(V_(n))^(2)=mgh, i.e., e^(2n)2gL=gh`

`e^(2n)=(h)/(L)=(L(1-costheta))/(L)`
or
`((2)/(sqrt(5)))^(2n)=1-cos theta[as e=(2)/(sqrt(2))]or ((4)/(5))^(n)=1-cos theta`
Now for `theta` to be lesser than `60, cos theta gt(1//2)`
i.e., `1-cos theta lt(1)/(2)`
So, `((4)/(5))^(n)lt(1)/(2)or ((5)/(4))^(n)gt2` or
`n(log 10-3log2)gtlog2`
or `ngt(0.301)/(0.097)[as log 10 =1 and log2=0.3]`
or `gt3.1`
As `n` (no.of collisions) must be integer so for `theta lt60^(@),n=4`.