A particle A suffers an oblique elastic collision with a particle B that is at rest initially. If their masses are the same, then, after the collision

detailed solution

Correct option is C

c)       a)By law of conservation of momentumi) along x axismu1+0=mv1+cosϕ+mv2cosθ ……………(i)ii)along y axis0=mv1sinϕ−mv2sinϕ ……………………..(ii)b)By law of conservation of energy12mu12=12mv12+12mv22⇒u12=v12+v22..............(iii) From eq (i) & (ii) we getv12=v12+v22+2v1v2cos(ϕ+θ) Now use eq (iii) in eq (iv) we getcos(ϕ+θ)=0→θ+ϕ=900 i.e, after collision they will move in mutually perpendicular directions.