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WannabeNewton

Science Advisor

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## Homework Statement

A plank of length [itex]2l[/itex] and mass M lies on a friction-less plane. A ball of mass m and speed [itex]v_{0}[/itex] strikes its end as shown. Find the final velocity [itex]v_{f}[/itex] of the ball assuming mechanical energy is conserved and [itex]v_{f}[/itex] is along the original line of motion. Ans. clue. For m = M, [itex]v_{f} = \frac{3}{5}v_{0}[/itex].

## Homework Equations

## The Attempt at a Solution

Ok so because [itex]\sum F = \partial _{t}P = 0[/itex] and [itex]\sum \tau = \partial _{t}L = 0[/itex] (internal collision forces cancel as per Newton's 3rd law) we can use conservation of both linear and angular momentum. We are free to choose a viable coordinate system so any coordinate system with its origin fixed on the line of motion of the point particle of mass m will do the trick. In this frame, [itex]L_{0} = 0[/itex] because the plank is initially at rest and the particle's momentum is directed parallel to its position vector with respect to the origin in this frame. Since the question states that even after the collision the particle moves along the same line of motion, the angular momentum after comes entirely from the plank which is now rotating as well as translating in some form so [itex]L_{f} = I_{com}\omega(-\hat{k}) + lMV(\hat{k}) = (lMV - \frac{1}{3}Ml^{2}\omega)\hat{k} = 0 [/itex] and we get that [itex]\omega = \frac{3V}{l}[/itex]. We also have conservation of linear momentum and here is where I got confused. I'll do as I did on my note book and state my concern after: [itex]mv_{0} = -mv_{f} + MV[/itex] so [itex]V^{2} = \frac{m^{2}}{M^{2}}(v_{0} + v_{f})^{2}[/itex] and using this together with the expression for the angular velocity of the plank after the collision I was able to simply the conservation of mechanical energy equation, [itex]\frac{1}{2}mv_{0}^{2} = \frac{1}{2}mv_{f}^{2} + \frac{1}{2}MV^{2} + \frac{1}{2}I_{com}w^{2}[/itex], down to [itex](1 + \frac{4m}{M})v_{f}^{2} + 8\frac{m}{M}v_{0}v_{f} + (\frac{4m}{M} - 1)v_{0}^{2} = 0[/itex] which gives the solutions [itex]v_{f} = -{v_{0}\frac{(\frac{4m}{M} \pm 1)}{(\frac{4m}{M} + 1)}}[/itex]. If I use m = M for the Ans. clue. then I get [itex]v_{f} = -\frac{3}{5}v_{0}[/itex] or [itex]v_{f} = -v_{0}[/itex]. My questions are these: I assumed at the beggining that the point particle would shoot backwards after colliding with the plank hence the negative sign but if I assume it keeps going in the same direction and do everything up till now the exact same way I get the right answer [itex]v_{f} = \frac{3}{5}v_{0}[/itex] but it makes intuitive sense to me that the particle would bounce back after hitting the plank thus causing the plank to rotate, and translate in the opposite direction; why would the particle keep moving forward? Secondly, there is the case v0 = vf and I don't see a justification for why this solution can be eliminated other than the fact that it would be a trivial solution where nothing at all happens to the plank. Thank you.