Question #4192

Particle [i]A[/i] of mass '[b]m[/b]' has initial velocity [b]v[/b]. After colliding with a particle [i]B[/i] of mass & '[b]2m[/b]' the particles have the following paths [i]B[/i] makes an angle [b]45[/b] degree while [i]A[/i] makes an angle [b]X[/b]. & First, & [i]A[/i] was moving straight. & Find [i][b]X[/b][/i].

Expert's answer

Projections of momentum

mv=2mv'cos45 + mv'' cosx

v=2v'cos45 + v''cosx

0=2v'sin45 - v''sinx → 2v'=(v''sinx)/sin45

v=(v''sinx)/sin45 cos45 + v''cosx →

v=v'' (sinx+cosx)→ v''(1-cos^2 x + cosx)=v

→ -cos^{2} x + cosx + 1 = v/v''

→ cos^{2} x - cosx - (1+v/v'')=0

→ cosx=1/2 - √(1+4(1+v/v')/2

**x=arccos 1/2 - √(1+4(1+v/v'') )/2.**

mv=2mv'cos45 + mv'' cosx

v=2v'cos45 + v''cosx

0=2v'sin45 - v''sinx → 2v'=(v''sinx)/sin45

v=(v''sinx)/sin45 cos45 + v''cosx →

v=v'' (sinx+cosx)→ v''(1-cos^2 x + cosx)=v

→ -cos

→ cos

→ cosx=1/2 - √(1+4(1+v/v')/2

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