Object slides from a surface's frictionless part to a rough part. How long does it take to stop?

Question

A rectangular prism of length $L$ is sliding on a flat surface with speed $v_0$, without friction. It slides onto a rough section of the surface where the coefficient of kinetic friction is $\mu$. Determine the amount of time it takes the object to stop, starting from the time it first touched the rough section. ($L = 0.49\ \mathrm{m}, v_0 = 1.96\ \mathrm{m/s}, \mu = 0.20, g = 10\ \mathrm{m/s^2}$)

(As a sidenote, the force of friction is $F_k = \mu F_n$, where $F_n$ is the normal force, in this case, the object's weight)

This was problem 3.B. for 11th grade in the 2nd phase of the 2020 Vermes Miklós physics contest, organized in Romania. At this level we aren't required to know calculus but I didn't see a way to solve the general problem without it.

Answer 1

I will assume that mass is uniformly distributed along the volume of the prism..ie, that constant volumes have constant masses which means that the density is constant. So, $M(t)=\rho Ax(t)$ where $\rho$ is the density, $A$ is the cross sectional area and $x(t)$ is the distance travelled measured from the point where the rough and smooth surfaces meet. This gives us the following differential equation $$\frac{d^{2}x}{dt^{2}}+sx=0\ ,\ s=\frac{uAg \rho}{m}\ $$ Its solution is $$x\left(t\right)=\frac{v_{0}}{\sqrt{s}}\sin\left(t\sqrt{s}\right) \tag{1}$$ and differentiating gives the speed as a function of time which is $$v\left(t\right)=v_{0}\cos\left(t\sqrt{s}\right) \tag{2}$$ These equations describe the motion of the prism as long as the force acting on it is variable, which is as long as some part of the prism is still on the smooth surface. When the prism travels a distance more than $L$, the force acting on it becomes constant and these equations stop describing this motion. It turns out for your case that it does travel a distance more than $L$. So to calculate the time it takes to stop we need two different times, the time it takes to travel the distance $L$ and the time it takes after that to stop under the effect of the constant force acting on it from this point, this force is $F=-umg$ which gives $a=-ug$ where $a$ is the acceleration. Here is what we are going to do, First I set $x(t)=L$ in $(1)$ to find the time $t_1$ it takes to travel the distance $L$. I plug this value of $t_1$ in $(2)$ to find the velocity $v_1$ it has at this point (which, from now on, will be altered by constant acceleration only). I then plug this $v_1$ in the equation $v=v_1+at$ and set $v=0$ to find the time $t_2$ it needs to stop from this point. And finally the time we need is $t=t_1+t_2$. This is quite straight forward but a bit tedious, anyway the final solution is this

$$t=\sqrt{\frac{L}{ug}}\sin^{-1}\left(\frac{\sqrt{Lug}}{v_{0}}\right)+\frac{\sqrt{v_{0}^{2}-Lug}}{ug}$$

I think this solution is far too complicated to be arrived at without calculus and I do not know how one can get that $sin^{-1}$ term just from the usual equations of motion. Having said that, I think there must be a way to justify dealing with this using constant deceleration, in other words, my solution must be wrong if a high school student is to be able to solve this problem.