Three particles `A,B,C` of mass `m` each are joined to each other by massless rigid rods to form an equilateral triangle of side `a`. Another particle of mass `m` hits `B` with a velocity `v_(0)` directed along `BC` as shown. The colliding particle stops immediately after impact.

(i) Calculate the time required by the triangle `ABC` to complete half-revolution in its subsequent motion.
(ii) What is the net displacement of point `B` during this interval ?

Answer

(i) `t = (6 a pi)/(sqrt(3) v_(0))` ;
(ii) `s = (a)/(sqrt(3)) sqrt(1 + (2 pi + sqrt(3))^(2)`.

Explanation

Structure will have translational motion of `C.M` and rotational motion about `C.M` From conservation of linear momentum
`mv_(0) = 3 mv_(CM)`
`v_(CM) = (v_(0))/(3)`
Impulse received by particle is `mv_(0)`. Same and opposite impulse will be received by structure so angular impulse `= I omega`
`mv_(0) xx (a)/(2 sqrt(3)) = 3m ((a)/(sqrt(3)))^(2) xx omega`
`omega = (v_(0))/(2 sqrt(3) a)`
(i) time taken to complete half revolution `= (pi)/(omega)`
`t = (pi)/((v_(0)//2 sqrt(3a))) rArr t = (2 sqrt(3) pi a)/(v_(0)) rArr t = (6 pi a)/(sqrt(3) v_(0))`
(ii) linear displacement
=`sqrt((d+a)^(2) + ((a)/(sqrt(3)))^(2)) = sqrt((v_(CM) t + a)^(2) + ((a)/(sqrt(3)))^(2))`
=`sqrt((v_(0)/(3) xx (6 pia)/(sqrt(3) v_(0)) +a)^(2) + ((a)/(sqrt(3)))^(2)) =sqrt((a^(2))/(3) (2 pi + sqrt(3))^(2) + (a^(2))/(3)) = (a)/(sqrt(3)) sqrt((2 pi + sqrt(3))^(2) + 1)`.