Three identical particles A, B and C (each of mass m) lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ABC is 1350 as shown in diagram. An impulse J is applied to the particle C in the direction BC for a very short time interval. Then just after applying impulse J,The speed of the particle ‘A’ is nJ7m. The value of n is

Impulse =ΔP Along X-axis: J2cos45= mv1+mv1  J22=2mv1--(1)Along Y-axis: J2sin45= mv2 J22=mv2--(2)(1)=(2) v2=2v1---(3) v3=v1+v22J-J2=mv3 J-22mv1=mv1+v22 J=7mv12 v1=vA=J27m----(4) v2=2v1=2J27mfrom (3)---(5)vB=v12+v22=5v12=10J7mvC=v1+v22=3v12=32J27m=3J7mv1y=22J7m;J1=3Jmand  J2=2J7II method    The external impulse applied to C causes both strings to jerk exerting internal impulses J1 and J2V2=v1x−−−−−−−(1)J1cos⁡45∘−J2=mv1x−−−−−−−−−(3)J1cos⁡45∘=mv1y−−−−−−−(4)J−J1=mv−….....−(5)Also velocities of B&C along BC are are equal i.e.,  v1ycos450+v1xcos450=v−−−−−6After solving we get,  v2=v1x=2J7mv=3J7m;v1y=22J7m;VA=2J7m,VB=10J7m,J1=3Jmand  J2=2J7