Any help appreciated with this problem.
A train consists of a railway engine of mass $M$ tonne coupled to two trucks, each of mass $m$ tonne. Initially the train is at rest and each coupling is slack to the extent of $a$ meters. A constant force $P$ tonne wt is then applied to the engine to drive the train forwards. Neglecting frictional resistance, find the time that elapses before the second truck starts to move. $\bigg[Ans : \sqrt{\dfrac{2a(2M + m)}{Pg}} \bigg]$.
My attempt.
Using Newton's 2nd Law of motion ($F = ma$) and $s = ut + \frac{1}{2} at^2$ I found the time when the first couple becomes taut. $t_1 = \sqrt{\dfrac{2aM}{Pg}}$.
Using the the same logic I found the time taken for the second couple to become taut. In this case the mass is $(M + m)$. This gives $t_2 = \sqrt{\dfrac{2a(M +m)}{Pg}}$.
So I just added $t_1 + t_2 = \sqrt{\dfrac{2aM}{Pg}} + \sqrt{\dfrac{2a(M +m)}{Pg}}$.
Which obviously isn't the answer given.
Thanks in advance.
I'll call the constant force $F$. The time until the first coupling is taut is; as you say
$$ t_1=\sqrt{\frac{2aM}{F}} $$
The velocity of the engine at $t_1$ is $v_1=\frac{F}{M}t_1$.
At $t=t_1$, suppose the engine and first truck have a momentum conserving collision, and they move at the same velocity $V$ just after the collision. Then
$$ p_\text{initial}=Mv_1:=p_\text{final}=(M+m)V\\ \therefore V=\frac{M}{M+m}v_1=\frac{\sqrt{2aMF}}{M+m} $$
After the collision, we have a new initial value problem $F=(M+m)\ddot{x}$. I'll define a new $t=0$, $x=0$ here, so the initial conditions are $x(0)=0$, $\dot{x}(0)=V$. The solution is
$$ x(t)=Vt+\frac{Ft^2}{2(M+m)} $$
The second truck will move when $x(t_2)=a$. Solving the quadratic equation for $t_2$ (picking the positive root, and simplifying) yields
$$ t_2=-\sqrt{\frac{2aM}{F}}+\sqrt{\frac{2a(2M+m)}{F}} $$
The total time is $t_1+t_2=\sqrt{\frac{2a(2M+m)}{F}}$