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```        Please solve the questions given in the photograph from FIITJEE AITS.
```
6 years ago

510 Points
```										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)
```
6 years ago
510 Points
```										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 )
```
6 years ago
510 Points
```										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
```
6 years ago
510 Points
```										approve if u like my ans
```
6 years ago
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