Three particles A, B, C of equal mass m are placed on a smooth horizontal plane. A is joined to B and C by light threads AB, AC and the angle BAC is 60°. An impulse I is applied to A in the direction BA. The angle made by the initial velocity vector of A with BA is tan⁻¹(√3/n). Find n.
Physics · Mixed Chapter
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Question

Three particles A, B, C of equal mass m are placed on a smooth horizontal plane. A is joined to B and C by light threads AB, AC and the angle BAC is 60°. An impulse I is applied to A in the direction BA. The angle made by the initial velocity vector of A with BA is tan⁻¹(√3/n). Find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: During the short impulse, forces from the two taut inextensible threads exert impulsive tensions along AB and AC. Step 2: Write conservation of linear momentum of the three particles and impose equal longitudinal velocity components at the ends of each thread (no instantaneous change of thread lengths). Step 3: Resolving the impulse–momentum equations along BA and perpendicular to BA at ∠BAC=60° gives vₐ,⊥/vₐ,∥=√3/7. Step 4: Thus the angle is tan⁻¹(√3/7), so n=7.