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Please solve the questions given in the photograph from FIITJEE AITS.
just after first collision let velociy of bob is v then
applying momentam conservation
initial momantam of system(Pi) = (m/2) (u)
u = (2gL)1/2 so Pi = m(2gL)1/2/2
final momemtam (Pf) = mv
mv = m(2gL)1/2 /2 (from momentam conservation)
v = (gL/2)1/2 ................1
KE of bob is mv2/2
KE = mgL/4 ...............2
when bob is at max height then its KE is converted into PE
mgL/4 = mgHmax
L/4 = Hmax
from floor height of bob is L+Hmax = L/4+L = 5L/4 (Q15 => option c)
let tension in string just after collision is T then
T - mg = mv2/r (centripital force) { tension is in upward direction , weight is in downward direction }
T = mg + mv2/L
v = (gL/2)1/2 so
T = 3mg/2 (ans14 =>c )
just after second collision , let velocity of bob is v1 & tension in string is T ...
after collision this system becomes conical pendulam so verticle component of tension is balanced by weight &
centripital force is provided by horizontal component of tension ...
let string makes an angle of @ with verticle then
Tcos@ = weight = 2mg ............1 (weight of bob + particle)
Tsin@ = 2mv12 /r .............2 (r is radius of its path)
r = Lsin@ .........3
from 1,2 & 3
v12 = {gLsin@tan@} ..................4
time period of this system = T = 2pir/v1
= 2pi(Lsin@)/{gLsin@tan@}1/2 ........................5
now , @ can be calculated by using trignometry ... height of this cone = L - L/4 = 3L/4
slant height = L
cos@ = 3/4
so , time perion will be T = 2pi(3L/4g)1/2
approve if u like my ans
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