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

The pendulum bob is released from position A, it collides with wall and returns back. When it is at position C, the angular amplitude is `60^@`
In `DeltaOCM, cos 60^@ = (OM)/l implies OM = 1//2`
Let velocity of bob B just before first collision is vB, using conservation of mechanical energy
`DeltaK + DeltaU = 0`
`implies (1/2 m v_B^2 -0) implies v_B = sqrt(2 gl)`
Speed of the ball after rebounding (first time) `v_1 = esqrt((2gl))`
The speed after second rebound is `v_2 = e^2 sqrt((2gl))`
In general after n rebounds, the speed of the bob is `v_n = e^n sqrt((2 gl))`
The velocity of bob `V_B^'` just after collision when angular amplitude is `60^@`. Again using conservation of mechanical energy `Delta K + Delta U = 0`
`implies (0 - 1/2 mv'_B^2) = 0 implies v_B^' = sqrt(gl)`
This means that the velocity of the bob reduces from `sqrt(2gl)` to `sqrt(gl)` due to the collisions with walls.
The final velocity after n collisions is `sqrt(gl) " ":. " " e^n (sqrt(2gl)) = sqrt(gl)`
Where e is coefficient of restitution.
`(2/(sqrt5))^(n) xx sqrt(2gl) = sqrt(2gl) implies (2/(sqrt5))^(n) = 1/(sqrt2)`
Taking `log, n log (2/(sqrt5)) = log 1/(sqrt(2)) implies n = 3.1`
Therefore number of collisions will be 4.