An exploding cannon shell

[brotherbobby]

Homework Statement

Consider a cannon shell of mass $M$ moves along a parabolic trajectory in the earths gravitational field. An internal explosion, generating an amount $E$ of mechanical energy, blows the shell into two parts. One part of mass $kM$ where $k<1$, continues moving along the same trajectory with velocity $v'$ while the other part is reduced to rest. Find the velocity of the mass $kM$ immediately after the explosion.

Relevant Equations

1. According to (linear) momentum conservation, if masses $m_1$ and $m_2$ moving with velocities $v_{1i}$ and $v_{2i}$ collide and have velocities $v_{1f}$ and $v_{2f}$ after the collision, then, in the absence of any external forces, $\underline{m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}}$

2. According to energy conservation, if masses $m_1$ and $m_2$ moving with velocities $v_{1i}$ and $v_{2i}$ collide and have velocities $v_{1f}$ and $v_{2f}$ after the collision, then $\underline{\dfrac{1}{2}m_1v^2_{1i}+\dfrac{1}{2}m_2v^2_{2i}\pm W'=\dfrac{1}{2}m_1v^2_{1f}+\dfrac{1}{2}m_2v^2_{2f}}$, where $W'$ is the work done on (or by) the system during the collision.

Problem :

I start by putting the image of the problem to the right.

A cannon of mass $M$ moving with a velocity (say) $v$, splits into two parts of masses $kM$ and $(1-k)M$ following an explosion that generated an amount of energy $E$. The second piece falls straight to the ground while the first piece of mass $kM$ continues in the same direction with a velocity $v'$. What is the velocity ${\color{red}{v' =?}}$

Using momentum conservation, we have $Mv=kMv'\Rightarrow v=kv'\qquad (1).$

Using energy conservation, $\dfrac{1}{2}Mv^2+E=\dfrac{1}{2}kMv'^2$, to which, upon using $(1)$ to replace $v\rightarrow v'$, we get $\dfrac{1}{2}kMv'^2-\dfrac{1}{2}k^2 Mv'^2=E\Rightarrow \boxed{v'=\sqrt{\dfrac{2E}{k(1-k)M}}}\quad\color{DarkGreen}{\Huge\checkmark}$

The answer is correct and matches with that of the text.


Doubt :

Let's have the answer again : $\boxed{v'=\sqrt{\dfrac{2E}{k(1-k)M}}}$, as the velocity of the piece of mass $kM$. Clearly, the value of $k$ has to be between 0 and 1.


(1) What happens (intuitively) when $k\approx 0$?

When $k$ goes to 0, the first piece has neglible mass while (almost) the whole of the mass of the cannon is taken by the second piece which has no kinetic energy. Hence the first piece must take (almost) the entire kinetic energy of the original cannon. Having negligle mass, it means that its velocity $v'\rightarrow \infty$.

Thus, we can say that $\displaystyle{\lim_{k\to 0}\; v'\rightarrow \infty}\qquad (2)$

I plotted the solution for $v'$ in $\verb|Desmos|$ as given above, $v'=\sqrt{\dfrac{2E}{k(1-k)M}}$, taking $E=\dfrac{1}{2}$ and $M=1$. I copy and paste the image of the plot to the right. From the plot, as $k\rightarrow 0$, the velocity $v\rightarrow\infty$.


(2) What happens (intuitively) when $k\approx 1$?

This is where I run into trouble.

Intuitively, as $k\rightarrow 1$, the first piece has almost the entire mass of the cannon while the second piece has a negligible amount and falls down under gravity. Hence the first piece should be moving with a velocity $v'=v+2\dfrac{E}{m}$ using equation $(1)$ above, whatever the value of $v$ was. Putting the values of $E,m$, we get $2E/m=1$

We can capture this as above : $\displaystyle{\lim_{k\to 1}\; v'\rightarrow v+1}\qquad (2)$

But the graph from the plot shows that $v'$ goes to $\infty$ as k goes to 1.

What is going on?


Request : The plot of the function is correct I suppose. I'd like a hint or clue as to where am I going wrong in my reasoning above for the case where $k \approx 1$.

 

[haruspex]

You are overlooking what E/M can be. As k gets closer to 1, less energy will be required to bring the fragment to rest, so E/M is tending to zero.

 

[brotherbobby]

You mean the second fragment with mass $(1-k)M$. Yes its mass tends to 0 as k approaches 1, but is $E$ the reason why its velocity after the explosion is 0?

Either way, even if it so, $\dfrac{E}{M}\rightarrow 0$ would imply $v'\rightarrow 1$ by my reasoning. But it is indefinitely huge as per the plot.

 

[haruspex]

(2) What happens (intuitively) when $k\approx 1$?

You have several variables. You must also specify what is staying constant.

the first piece should be moving with a velocity $v'=v+2\dfrac{E}{m}$ using equation $(1)$ above, whatever the value of $v$ was.

I assume you mean $v'^2=v^2+2\dfrac{E}{m}$.

Putting the values of $E,m$, we get $2E/m=1$

Putting the values of $E,m=1, 2$?

We can capture this as above : $\displaystyle{\lim_{k\to 1}\; v'\rightarrow v+1}\qquad (2)$

But the graph from the plot shows that $v'$ goes to $\infty$ as k goes to 1.

What is going on?

There's no contradiction if your model is also driving $v$ to infinity.
And indeed, you have $v^2=\frac{2E}M\frac k{1-k}$, so it will.

 

[kuruman]

(1) What happens (intuitively) when $k\approx 0$?
(2) What happens (intuitively) when $k\approx 1$?

Intuitively, you can answer these questions using CM considerations without involving energy release. For simplicity, let's say that the explosion occurs when the projectile is at maximum height. Both fragments will hit the ground simultaneously because they have zero vertical velocity before and after the explosion. Of course, the explosion is assumed to be instantaneous.

The fragment $(1-k)M$ will hit the ground at horizontal position $\frac{1}{2}R$ where $R$ is the range of the unexploded projectile and also the distance where the CM of the exploded projectile will land. We need to find the horizontal distance $x$ from the point of launch where the fragment $kM$ lands. After cancelling the masses, $$X_{cm}=(1-k)\frac{1}{2}R+kx=R \implies x=\frac{R}{2}\left (\frac{1}{k}+1\right ).$$ This says that

• when $k=1$, there is no explosion $\left((1-k)=0\right)~$ and the projectile lands as expected at $R$.
• when $k\rightarrow 0$, an increasing amount of mass drops straight down to $\frac{1}{2}R$ whilst an increasingly smaller leftover mass must land increasingly farther away from the point of launch to keep the CM nailed at $R$.

A similar reasoning would apply in the general case of an explosion at an arbitrary point along the trajectory but the math would be more complicated.