Two particles, each of `m`, are connected by a light inextensible string of length 2l. Initially they lie on a smooth horizontal table at points `A` and `B` distant `l` apart. The particle at `A` is projected across the table with velocity `u`. Find the speed with which the second particle begins to move if the direction of `u` is , (a) along BA, (b) at an angle of `120^(@)` with AB, (c) perpendicular to AB. In each case calculate (in terms of `m` and `u`) the impulse of tension in the string.

Answer

(a)u/2,mu/2; (b) `usqrt(13)//8,m usqrt(13)//8;(c) usqrt3//4,m usqrt3//4`

Explanation

(a) ` (##BSL_PHY_CMMC_E01_098_S01.png" width="80%">
By conservation of momentum `rArrmv_(1)+mv_(1)=0+m u`
`rArr2mv_(1)=m u`
`rArrv_(1)=(u)/(2)`
Impulse of tension=change in momentum of particle `B`
`=mv_(1)-0=(m u)/(2)`
(b) (just before taut)
(just after taut)
By sine rule `rArr(l)/(sintheta)=(2l)/(sin120^(@))rArrsintheta=(sqrt3)/(4)rArrcostheta=(sqrt(13))/(4)`
By conservation of momentum along the string
`mv_(2)+mv_(2)=m(ucostheta)+m(0)`
`rArr" "v_(2)=(1)/(2)ucostheta=((1)/(2)u)((sqrt(13))/(4))=(sqrt(13))/(8)u`
Impulse=change in momentum of mass `B`
`=m((sqrt(13))/(8)u)-0=(sqrt(13))/(8)m u`
(c)

By conervation of momentum along the string
`mv_(3)+mv_(3)=m(ucos30^(@))+0`
`rArrv_(3)=(1)/(2)ucos30^(@)=u(sqrt3)/(4)m u`
Impulse=change in momentum of mass `B`
`=m(u(sqrt3)/(4))-0=(sqrt3)/(4)"mu"`