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.

Two particles, each of m, are connected by a light inextensible string

1.	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» Solution :(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"`