Terminal Trouble

Let us assume that before collision with a new roller, velocity of plank is $v'$ , and after collision, velocity is $v$ . Let $n = \frac{L}{d} >>1$ be the number of cylinders in contact with the plank before the collision.

By conservation of angular momentum, about center of new cylinder, (draw impulse diagram and notice that there is no net external angular impulse on the system of plank and cylinders about this point.)

$\displaystyle M v'r + \frac{m (n-1) v' r}{2} = M vr + \frac{nm vr}{2}$

$\displaystyle \Rightarrow v' - v = \frac{mv}{2M + (n-1) m} \approx \frac{mv}{2M + nm}, \text{ as } n >>1$ .

Now, before this collision, there was a similar collision, initial velocity was $v$ and then when this collision velocity was about to occur, velocity was $v'$ . Applying conservation of energy in period in between,

$\displaystyle \frac{1}{2} Mv^2 + \frac{1}{4} nm v^2 + M g d \sin \alpha= \frac{1}{2} M v'^2 + \frac{1}{4} nm v'^2$

$\displaystyle \Rightarrow \frac{2M+nm}{4} (v^2-v'^2) = Mgd \sin \alpha$

$\displaystyle \Rightarrow \frac{2M+nm}{4} (v+v')(v-v') = Mgd \sin \alpha$

$\displaystyle \Rightarrow \frac{2M+nm}{4} (2v)\frac{mv}{2M + nm} \approx Mgd \sin \alpha$

$\displaystyle \Rightarrow v \approx \sqrt{2\frac{M}{m} gd \sin \alpha} = 14 m/s$

Imgur

the energy acquired by the block M during passing a length $l$ will be loss by making the cylinder rotate with the same terminal velocity so : the potential energy gained is $\Delta E =Mgsin(a)$ and the kinetic energy gained by a cylinder is $\Delta Ek= \frac{ m v^2}{4}$ and its obvious that $\Delta E= \Delta Ek$ so that $Mgsin(a) = \frac{1}{4}m v^2$ then $v= \sqrt{9.8*20} = 14 m/s$