 # Please solve the questions given in the photograph from FIITJEE AITS. 12 years ago

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)

12 years ago

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 )

12 years ago

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

12 years ago

approve if u like my ans