Applying conservation of energy on a springs/projectile motion problem

Question

My physics teacher proposed a hypothetical problem to our class as we are nearing the end of the Work-Energy Unit (that will in all likelihood show up on the test). I will probably have various question throughout (that I'll put in blocks), and I apologize in advance for a long read. The problem goes something like this:

An inclined plane at angle $θ$ has a spring placed parallel to the incline. A block of mass $m$ is held against the spring, such that it compresses a distance of $x$ meters. Once the block is released, it moves along the incline for $d_{1}$ meters, then moves in the air as a projectile until it hits a hill that is $h$ meters above the ground.

Our teacher then said he could ask for various things, such as:

1. The spring constant.
2. The maximum height reached by the block.
3. The force the block hit the ground with.
4. The time between releasing the block and the block hitting the hill.

Our physics book contains the equation: $$PE_{gi} + PE_{ei} + KE_{i} + W_{net,ext} = PE_{gf} + PE_{ef} + KE_f$$

$PE_{gi}:$ Potential Energy due to gravity initially

$PE_{ei}:$ Potential Energy due to spring elasticity initially

$KE_{i}:$ Initial Kinetic Energy

$W_{net,ext}:$ Net external work (external, so excluding gravity or work done by the spring)

$PE_{gf}:$ Potential Energy due to gravity finally

$PE_{ef}:$ Potential Energy due to spring elasticity initially

$KE_f:$ Final Kinetic Energy

Here's how I propose to solve the problem:

To find the spring constant, we need Hooke's Law: $0.5kx^2$. Since the block starts at ground level, there would be no $PE_{gi}$, and since the block is not moving initially, there is no $KE_{i}.$ There are also no external forces acting on the block, so there is no $W_{net,ext},$ nor is the spring acting on the block at the end of the incline, so we can remove $PE_{ei}$. Since the block lands on the hill, it possesses no final velocity, so we can also remove $KE_{f}$.

Therefore, our textbook's equation becomes $$PE_{ei} = PE_{gf} \rightarrow 0.5kx^2 = mgh,$$ in which we can then solve for $k$.

Is the above allowed? If solved using $h$ as the height of the hill, wouldn't this solve for $k$ as the spring constant necessary to move a block $h$ meters up the incline, not the spring constant necessary to fire a block up and off an incline?

Assuming $k$ was calculated correctly:

We have to find the final velocity reached by the block on the incline. This final velocity, which I will call $v_{f1}$, will then become the initial velocity, $v_{i}$, of a projectile motion problem. To find $v_{f1}:$ $$PE_{ei} = PE_{gf} + KE_f$$ which equals $$0.5kx^2 = mg(d_{1}sinθ) + 0.5m(v_{f1})^2$$

We can solve for $v_{f1}$ using simple algebra. This becomes a projectile motion problem from here, which I am fairly confident in.

Assuming $k$ was calculated correctly, is the above line of thinking correct?

Someone in class proposed a solution, which I don't think is correct, but I'm not sure why. He said the whole thing could be solved using Conservation of Energy, without having to split the problem into Conservation of Energy to find the spring constant then do a projectile motion problem. Since the block ends up resting on a hill, there is no final velocity, therefore he removed $KE_{f}$ such that $$PE_{ei} = PE_{gf} \rightarrow 0.5kx^2 = mgh$$.

Can you assume that there is no velocity when the block is on the hill? Isn't there a final velocity of the block in every projectile motion problem, thereby disallowing you from removing $KE_{f}$?

Just some final questions, again sorry for the long-winded post.

What would be a better way to go about solving the question? Is it better to split the problem into two parts, or is it possible to do the whole thing using Conservation of Energy? Our teacher stressed that we could solve almost every problem using the one formula $PE_{gi} + PE_{ei} + KE_{i} + W_{net,ext} = PE_{gf} + PE_{ef} + KE_f$.

Any help is appreciated, thank you!

Answer 1

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:

3. 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 $T$.