Up to:
$$0.5kx^2 = mg(d_{1}sinθ) + 0.5m(v_{f1})^2$$
your reasoning is correct and the launch speed $v_{f1}$ of the block can be calculated from there.
But after that it becomes a trajectory problem. Define an $x$ and $y$-axis with origin $O$ as shown below. $T$ is the target to hit:
Calculation of the trajectory (and derivation) you find in Wikipedia:
$$y=y_0+x\tan\theta-\frac{gx^2}{2(v\cos\theta)^2}...\text{Eq.1}$$
Where in our case $y_0=0$ and $v$ the speed $v_{f1}$, calculated above. So set up that equation.
In order for the projectile to hit the point $T$, we need to assign coordinates to it. Unfortunately the problem only states '$h$ meter above the ground', which would be:
$$y=h-d_1\sin\theta$$
Unfortunately the problem doesn't state how far away the target is from the incline! If we knew this, we could calculate $v_{f1}$ from $Eq.1$ with $(x,y)$ and then equate it to the value further up. That would then allow to calculate $k$. But not without an $x$ coordinate for the point $T$. So there's information missing for a full solution.
As regards:
- The force the block hit the ground with.
That depends entirely on the ground (assuming the block is hard)! When the block hits the ground it decelerates strongly according:
$$F=ma$$
The deceleration determines the size of $F$ and depends on how hard or soft the ground is.
And:
The maximum height reached by the block.
That can be calculated from energy conservation.
The $y$-component of $v$ is:
$$v_y=v \sin\theta$$
The maximum height $H$ is calculated from:
$$\frac{mv_y^2}{2}=mgH$$
Add the height of the ramp to $H$.
This:
The time between releasing the block and the block hitting the hill.
You'll find the derivation in the link above but it can only be calculated with full coordinates of $T$.
You are right to be suspicious of an answer that contradicts your intuition - but in this case the solution is right
The key (not clearly explained) is that you apply a constant force that is greater than the force needed to move one block: consequently that block will accelerate. As the spring compresses, the first block will accelerate less, and eventually it will decelerate. When it stops, the compression of the spring is maximum - and it applies a greater force to both blocks than the force you used to compress it.
So the solution is correct. If you still have difficulty believing this, think about a hammer: when you swing it, you apply a little bit of force over a long distance; when it hits the nail, it feels a lot of deceleration over a short distance. The force applied to the nail is greater than the force you applied to the hammer. Similar (not same) concept... You do work over a large distance with a little bit of force, to allow you to do the same work over a short distance (with a large force)
Does that help?
Best Answer
The charge on one block not only repels the charge on the other block, but it also repels the $dq$ charge that you are bringing to charge the system.
As you are slowly charging the system and slowly building up the charges, the force $F_e$ as you integrate is not a constant since the charge is increasing slowly. You may also need to account for the self-energy to build the charged system from scratch (a charge of 0 to 2Q).