Problem source

In this question take $g = 10 ms^{-2}.$
The point $A$ lies on a fixed rough plane inclined at $30^{\circ}$ to the horizontal and $\ell$ is the line of greatest slope through $A$. A particle $P$ is projected up $\ell$ from $A$ with initial  speed $6$ms$^{-1}$. A time $T$ seconds later, a particle $Q$ is projected from $A$ up $\ell$, also  with speed $6$ms$^{-1}$.  The coefficient of friction between 
each particle and the plane is $1/(5\sqrt{3})\,$ and the mass of each particle is $4$kg.
\begin{questionparts}
\item Given that $T<1+\sqrt{3/2}$, show 
that the particles collide at a time $(3-\sqrt6)T+1$ seconds after $P$ is projected.
\item In the case $T=1+\sqrt{2/3}\,$, 
determine the energy lost due to friction from the instant at which $P$ is projected to the time of the collision.
\end{questionparts}

Solution source

Since the particles are identical and are projected with the same speed, the only way they can reach the same point $x$ at the same time, is if $A$ has reached it's apex and started descending.

Considering $P$, we must have (setting the level of $A$ to be the $0$ G.P.E. level), suppose it travels a distance $x$ before becoming stationary:


\begin{align*}
\text{N2}(\nwarrow): && R - 4g \cos(30^\circ) &= 0 \\
\Rightarrow && R &= 20\sqrt{3} \\
\Rightarrow && \mu R &= \frac1{5 \sqrt{3}} (20 \sqrt{3}) \\
&&&= 4 \\
\end{align*}

Therefore in the two phases of the journey the particle is being accelerated down the slope by either $6$ or $4$.

$v^2 = u^2 + 2as \Rightarrow 0 = 36 - 12s \Rightarrow s = 3$. $v = u + at \Rightarrow t = 1$. Therefore after $1$ second $P$ reaches its highest point having travelled $3$ metres. It will pass back to the start in $s = ut + \frac12 a t^2 \Rightarrow 3 = \frac12 4 t^2 \Rightarrow t = \sqrt{3/2}$ seconds, ie the constraint is that the particle hasn't already past $Q$ before the collision.

The collision will occur when $s = 6t - \frac12 6 t^2$ and $s =3 - \frac12 4 (t+T-1)^2$ coincide, ie:

\begin{align*}
&& 6t - 3t^2 &= 3 - 2(t+T-1)^2 \\
&& 0 &= 3 -2(T-1)^2 -(4(T-1)+6)t + t^2 \\
&& 0 &= 3 -2(T-1)^2 -(4T+2)t + t^2 \\
\Rightarrow && t &= \frac{4T+2 \pm \sqrt{(4T+2)^2 - 4(3-2(T-1)^2)}}{2} \\
&&&=  \frac{4T+2 \pm \sqrt{24T^2}}{2} \\
&&&= 2T + 1 \pm \sqrt{6} T \\
&&&= (2 \pm \sqrt{6})T + 1
\end{align*}

we must take the smaller root, ie $(2-\sqrt{6})T + 1$.

In the case the collision occurs exactly at the start, the particle $P$ has traveled $6$ meters, against a force of $4$ newtons of friction, ie work done is $24$ Joules.