A particle of mass m moving at 5.0 m/s in the positive x direction makes a glancing elastic collision with a particle of mass 2m that is at rest before the collision. After the collision, m moves off at an angle of 45^o to the x axis and 2m moves off at 60^o to the x axis. The speed of m after the collision is A) 4.5 m/s. B) 2.5 m/s. C) 3.3 m/s. D) 1.8 m/s. E) 1.1 m/s.

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Explanation:Momentum is conserved in the x direction:$$m(5.0	ext{ m/s})+2m(0)=m(v_x)+2m(v_{2x})$$$$v_x+2v_{2x}=5.0	ext{ m/s}$$Momentum is conserved in the y direction:$$0=m(v_y)+2m(v_{2y})$$$$v_y+2v_{2y}=0$$Energy is conserved:$$\frac{1}{2}m(5.0	ext{ m/s})^2+0=\frac{1}{2}m(v_x)^2+\frac{1}{2}m(v_y)^2+\frac{1}{2}2m(v_{2x})^2+\frac{1}{2}2m(v_{2y})^2$$$$12.5	ext{ J}=\frac{1}{2}m(v_x)^2+\frac{1}{2}m(v_y)^2+2m(v_{2x})^2+2m(v_{2y})^2$$$$v_x^2+v_y^2+4v_{2x}^2+4v_{2y}^2=25$$From the geometry, we have:$$v_x=v\cos45^\circ$$$$v_y=v\sin45^\circ$$$$v_{2x}=v\cos60^\circ$$$$v_{2y}=v\sin60^\circ$$Substituting these into the conservation of energy equation, we get:$$v^2\cos^245^\circ+v^2\sin^245^\circ+4v^2\cos^260^\circ+4v^2\sin^260^\circ=25$$$$v^2=4.5	ext{ m/s}^2$$$$v=4.5	ext{ m/s}$$

A Particle of Mass M Moving at 5