(adsbygoogle = window.adsbygoogle || []).push({}); Using conservation of momentum: A particle A of mass 4m moves with speed [itex]\frac{3}{2}u[/itex] collides directly with a particle B of mass m which is at rest on a smooth horizontal table. The coefficient of restitution between the particles is [itex]e[/itex]. Given that there are no further collisions, find the range of possible values for e.

[itex] 4.\frac{3}{2} + 0 = 4v_A + v_B [/itex]

[itex] 6 = 4v_A + v_B [/itex]

Coefficient of restitution:

[itex] \frac{2(v_B - v_A)}{3} = e [/itex]

[itex] 2v_B - 3e = 2v_A [/itex]

[itex] 6 = 2(2v_B - 3e) + v_B [/itex]

Where do I go from here...? Thanks.

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# Homework Help: Coefficient of Restitution, Conservation of Momentum

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