How far will a piece from an explosive projectile fly?

[songoku]

Homework Statement

A projectile is fired from a cannon at position X. At its highest point, it explodes into two equal parts. One of the parts falls vertically (as shown in the figure) and lands at distance of 1000 m from X. How far (horizontally) will the second part land from X?
a. 1000 m
b. 1500 m
c. 2000 m
d. 2500 m
e. 3000 m

Relevant Equations

Conservation of momentum

At first, I want to use conservation of momentum when the object explodes at the highest position but I think the projectile is not a closed system (there is weight acting on it). Tried conservation of energy but not enough information provided.

What approach should I take for this question?

Thanks

 

[DaveC426913]

I suspect you are overthinking it. It might make more sense to think of this problem on a billiards table; I'm pretty sure you can ignore the vertical component.

 

[songoku]

I suspect you are overthinking it. It might make more sense to think of this problem on a billiards table; I'm pretty sure you can ignore the vertical component.

Do you mean I can use conservation of momentum at the highest point, even though there is net external force acting on the projectile?

Thanks

 

[hutchphd]

I would break the problem into two pieces. What are the conditions (exact velocities of the two pieces) immediately after the explosion? The result should then be evident.

 

[kuruman]

Horizontal momentum is conserved because there are no horizontal external forces. Can you show that the two pieces take the same amount of time to reach the ground? If so, then where does the center of mass land?

 

[hmmm27]

I'm pretty sure you can ignore the vertical component.

There is no vertical component to the separation, if for no other reason than that answer isn't one of the ones listed.

 

[hutchphd]

There is no vertical component to the separation, if for no other reason than that answer isn't one of the ones listed.

It does say that is the highest point reached (presumably by either fragment ...)

 

[songoku]

I would break the problem into two pieces. What are the conditions (exact velocities of the two pieces) immediately after the explosion? The result should then be evident.

I think I get the result. Since the horizontal momentum is conserved, the second part of the projectile will move to the right because the other part of the projectile undergoes vertical motion with zero initial velocity.

Using conservation of momentum, the speed of the second part of the projectile is twice the initial speed of the projectile before explosion (because the mass is half of initial) so the horizontal distance traveled will also be twice (measured from highest position)

I get the answer is (e)

Horizontal momentum is conserved because there are no horizontal external forces. Can you show that the two pieces take the same amount of time to reach the ground?

Yes

If so, then where does the center of mass land?

Sorry I don't understand this part. Do you mean the center of mass of the two parts of projectile after explosion? Wouldn't it be in the middle of the two fragments?

Thanks

 

[kuruman]

Sorry I don't understand this part. Do you mean the center of mass of the two parts of projectile after explosion? Wouldn't it be in the middle of the two fragments?

What does conservation of linear momentum have to say about the motion of the center of mass when collisions (or explosions) occur? If the projectile did not explode, would it land at a different point from that of the center of mass of the exploded projectile?

 

[DaveC426913]

What does conservation of linear momentum have to say about the motion of the center of mass when collisions (or explosions) occur? If the projectile did not explode, would it land at a different point from that of the center of mass of the exploded projectile?

Ah. This clicked it for me. Once you determine the motion of the CoM, the correct answer follows intuitively.
Elegant solution.

 

[hutchphd]

What does conservation of linear momentum have to say about the motion of the center of mass when collisions (or explosions) occur? If the projectile did not explode, would it land at a different point from that of the center of mass of the exploded projectile?

But one does not know with certainty that this explosion did not impart net momentum to the center of mass of the fragments. The explosive gases leave the system and so this it not obviously true...I believe this is what is rightfully worrying the OP. The intent of the question wants this assumption but it could be more clearly formulated (maybe using an internal spring?)

 

[kuruman]

But one does not know with certainty that this explosion did not impart net momentum to the center of mass of the fragments. The explosive gases leave the system and so this it not obviously true...I believe this is what is rightfully worrying the OP. The intent of the question wants this assumption but it could be more clearly formulated (maybe using an internal spring?)

Yes, one could be more realistic and start worrying that gases from the explosion carry some of the momentum way. However, the statement of the problem says about the projectile, "At its highest point, it explodes into two equal parts." I think that "equal parts" leaves no room for gases and makes the intent of the problem's author obvious.

 

[kuruman]

Ah. This clicked it for me. Once you determine the motion of the CoM, the correct answer follows intuitively.
Elegant solution.

But it is important to ascertain that the two pieces land at the same time before this solution is applied as is. If, say, the vertically falling piece lands first, one needs to take into account that the CoM keeps moving horizontally until the second piece lands. I remember seeing a problem where this was the case with the second piece going over a cliff.

 

[DaveC426913]

But it is important to ascertain that the two pieces land at the same time for this solution to work. If the vertically falling piece lands first, one needs to take into account that the CoM keeps moving horizontally until the second piece lands. I remember seeing a problem where this was the case.

True, but I'd say that's outside the scope of this question since no such data is provided or implicit.

In its defense, it does say "one part falls vertically", which could give one license to assume initial vertical velocity of zero.

 

[kuruman]

True, but I'd say that's outside the scope of this question since no such data is provided or implicit.

In its defense, it does say "one part falls vertically", which could give one license to assume initial vertical velocity of zero.

I agree 100% with what you say, as this is not a rocket science problem . However, if I were to pick nits, I would say that an object released from rest from a height is "falling vertically" throughout its motion until it stops. "Falling vertically" only establishes that the object has no instantaneous horizontal velocity component and says nothing about the initial vertical component.

A clearer way to formulate the question could have been, "Immediately after the explosion, one of the parts drops vertically while the other part moves horizontally. The first part (as shown in the figure) lands at distance of 1000 m from X. How far (horizontally) will the second part land from X?"

 

[songoku]

What does conservation of linear momentum have to say about the motion of the center of mass when collisions (or explosions) occur? If the projectile did not explode, would it land at a different point from that of the center of mass of the exploded projectile?

The center of mass will move with constant velocity in horizontal direction since no net external force acting on it.

If the projectile did not explode, it would land at same point of the center of mass of exploded projectile.

The center of mass will land at distance 2000 m from X so the second part of exploded projectile should land at distance 3000 m from X.

Is this what you mean?

Thanks

 

[vcsharp2003]

I agree 100% with what you say, as this is not a rocket science problem . However, if I were to pick nits, I

I think we must assume the vertically falling part to have an initial downward velocity ($\ne 0$) since in any explosion parts fly off with certain non-zero velocities due to the explosive forces acting on exploded parts being extremely high in magnitude.

 

[vcsharp2003]

Homework Statement:: A projectile is fired from a cannon at position X. At its highest point, it explodes into two equal parts. One of the parts falls vertically (as shown in the figure) and lands at distance of 1000 m from X. How far (horizontally) will the second part land from X?
a. 1000 m
b. 1500 m
c. 2000 m
d. 2500 m
e. 3000 m
Relevant Equations:: Conservation of momentum

At first, I want to use conservation of momentum when the object explodes at the highest position but I think the projectile is not a closed system (there is weight acting on it).

Conservation of momentum always applies to an explosion if you consider the just before explosion momentum and just after explosion momentum.

This is inspite of the external gravitational force acting on the exploding mass because explosive forces are huge in magnitude compared to gravitational force in typical cases. With above assumption, it means that only internal explosive forces act on exploded parts and there is no external force. With no external force acting, we can therefore say that momentum gets conserved during an explosion.

 

[DaveC426913]

I think we must assume the vertically falling part to have an initial downward velocity ($\ne 0$) since in any explosion parts fly off with certain non-zero velocities due to the explosive forces acting on exploded parts being extremely high in magnitude.

That might be the case in real life, but this is a deliberately simplified homework question, and:

1. we have to assume we are given all the values necessary to solve it.
2. if we were to grant your assumption, then the problem is unsolvable, since we are not given sufficient information.

 

[vcsharp2003]

if we were to grant your assumption, then the problem is unsolvable, since we are not given sufficient information.

It is solvable. I have done it before posting my thoughts.

We simply start with all unkowns like $u$ as initial projected velocity at $\theta$ with ground. We also assume that second part after explosion fires of at $v_2$ at $\alpha$ angle with horizontal. The first part fires off at $v_1$ in a vertically downward direction.

Then we simply apply law of conservation of momentum to explosion and also the horizontal range formula to projectile.

It will all finally work out. I can show the complete solution, but it's prohibited in this forum to show full solution. I can give more hints if you want to solve this problem using these unknowns.

EDIT: I found even my solution was using an implicit assumption as explained in my post#23. So, without making some assumption, you're right about the problem being unsolvable.

 

[DaveC426913]

It is solvable. I have done it before posting my thoughts.

We simply start with all unkowns like $u$ as initial projected velocity at $\theta$ with ground. We also assume that second part after explosion fires of at $v_2$ at $\alpha$ angle with horizontal. The first part fires off at $v_1$ in a vertically downward direction.

Then we simply apply law of conservation of momentum to explosion and also the horizontal range formula to projectile.

It will all finally work out. I can show the complete solution, but it's prohibited in this forum to show full solution. I can give more hints if you want to solve this problem using these unknowns.

If the initial velocity of the vertically falling piece is unknown but not equal to zero, it could be virtually any value.

You're telling us that - even given the pieces have unknown trajectories - you can provide a unique solution (one of the given answers in this multiple choice question) ?

I am skeptical.

(I have asked if there is a way to split this side-discussion off from the homework problem so you can be free to post your logic.)

 

[vcsharp2003]

If the initial velocity of the vertically falling piece is unknown but not equal to zero, it could be virtually any value.

You're telling us that - even given the pieces have unknown trajectories - you can provide a unique solution (one of the given answers in this multiple choice question) ?

I am skeptical.

(I have asked if there is a way to split this side-discussion off from the homework problem so you can be free to post your logic.)

Sorry, let me review my approach again and then I'll get back.

 

[vcsharp2003]

If the initial velocity of the vertically falling piece is unknown but not equal to zero, it could be virtually any value.

You're telling us that - even given the pieces have unknown trajectories - you can provide a unique solution (one of the given answers in this multiple choice question) ?

I am skeptical.

(I have asked if there is a way to split this side-discussion off from the homework problem so you can be free to post your logic.)

I have reviewed my solution with all unknowns that I mentioned in post#20.

However, I found that I was also making an implicit assumption that the non-vertical exoloded part will take the same time to reach the ground as the initial projectile from ground level took to reach the highest point. I think this assumption will only be true if second part explodes off horizontally. Then, it would logically follow that vertical part explodes off with zero initial velocity.
So, your assumption is as good as mine. Without an assumption, this problem is indeed unsolvable as pointed out by you. I'm sorry for creating false expectations. I'll add an edit to my post#20.

 

[kuruman]

The way I see it, if piece A drops vertically down, the horizontal velocity of piece B will be the same regardless of how much momentum the two pieces acquire in the vertical direction. The more momentum piece B has in the vertical direction, the longer it stays in the air and the more distance it travels in the horizontal direction from the point of the explosion. Without the vertical velocity, one cannot find that distance.

Edited in response to the removal of post #24.

 

[DaveC426913]

However, I found that I was also making an implicit assumption that the non-vertical exoloded part will take the same time to reach the ground as the initial projectile from ground level took to reach the highest point.

I make the same assumption but for a different reason: the wording of the question says "one of the parts falls vertically". Since no initial v is given, we must assume it is zero, else the question is unanswerable.

Without an assumption, this problem is indeed unsolvable

Agree. Al though in the context of the question, I'd say it is a very reasonable assumption.

I'm sorry for creating false expectations. I'll add an edit to my post#20.

 

[vcsharp2003]

But it is important to ascertain that the two pieces land at the same time before this solution is applied as is. If, say, the vertically falling piece lands first, one needs to take into account that the CoM keeps moving horizontally until the second piece lands. I remember seeing a problem where this was the case with the second piece going over a cliff.

If we know that com will take the projectile path as if the explosion hadn't occurred, then can we not say with certainty that when com hits ground, then both exploded parts must also hit it otherwise the com would be above ground level due to the line joining the two parts being slanted to the ground?

For example, the com would be above ground along the slanted line $m_1m_2$, if both parts didn't touch ground at the same time as com touching the ground.

 

[jbriggs444]

can we not say with certainty that when com hits ground, then both exploded parts must also hit it

[If we allow a non-horizontal explosion then] One part can hit the ground first. At which point the center of mass will no longer be in free fall since an external force acts on part of the system.

 

[hutchphd]

I think the problem could have been better stated. But a reasonable interpretation of

"A projectile is fired from a cannon at position X. At its highest point, it explodes into two equal parts. One of the parts falls vertically (as shown in the figure) and lands at distance of 1000 m from X. How far (horizontally) will the second part land from X?"

is that no piece of the projectile exceeds the highest point. Ambiguity gone.

/

 

[vcsharp2003]

[If we allow a non-horizontal explosion then] One part can hit the ground first. At which point the center of mass will no longer be in free fall since an external force acts on part of the system.

Was my logic for saying that com should touch the ground at the same time as it's components correct?

 

[jbriggs444]

Was my logic for saying that com should touch the ground at the same time as it's components correct?

If I understand the logic, it is that:

Neither part can tunnel under the ground. If the center of mass is at the ground then both pieces must be on the ground. Proof by contradiction -- if one mass were above ground and the other no lower than the ground then the center of mass would be above ground.

That logic is reasonable. But it provides no assurance that the center of mass is in free fall all the way to the ground.

 

[Steve4Physics]

But it provides no assurance that the center of mass is in free fall all the way to the ground.

We could change the problem slightly. Suppose there is a very deep hole in the ground, so that the vertically falling fragment reaches ground-level but then continues (free-falling) down the hole.

While both fragments are in free-fall, their CoM is in free-fall all the way down to the ground (and potentially through the ground!)

 

[kuruman]

If we know that com will take the projectile path as if the explosion hadn't occurred, then can we not say with certainty that when com hits ground, then both exploded parts must also hit it otherwise the com would be above ground level due to the line joining the two parts being slanted to the ground?

Yes, the fact that the CoM hits the ground is no guarantee that the two fragments will hit the ground at the same time. Here is the mathematical proof that compares the unexploded trajectory with the trajectories of the to equal fragment.
Let $V_{0x},~V_{0y}$ be the components of the initial velocity. The maximum height where the explosion occurs is $H=\dfrac{V_{0y}^2}{2g}$. The time of flight for the unexploded projectile is $T_{\!f}=\dfrac{2V_{0y}}{g}$ and the time elapsed until the explosion occurs is $\frac{1 }{2}T_{\!f} = \dfrac{V_{0y}}{g}$.
Now assume that when the explosion breaks up the projectile into two equal pieces, piece A falls vertically to the ground it has initial velocity $\vec v_A=\{0,-v_y\}$ while piece B has velocity $\vec v_B=\{2V_{0x},v_y\}$ in accordance with momentum conservation. The vertical position of the CoM is $$Y_{\text{cm}}=\frac{\frac{m}{2}y_A+\frac{m}{2}y_B}{m}=\frac{1}{2}\left(H-v_yt-\frac{1}{2}gt^2\right)+\frac{1}{2}\left(H+v_yt-\frac{1}{2}gt^2\right).$$With $H=\frac{V_{0y}^2}{2g}$, this becomes, $Y_{\text{cm}}=\frac{V_{0y}^2}{2g}-\frac{1}{2}gt^2$ which gives the time of flight for the CoM after the explosion, $T_{\text{cm}}=\frac{V_{0y}}{g}=\frac{1}{2}T_{\!f}$ as expected. The CoM will land horizontally at the same spot as the unexploded projectile as long as the fragments are in free fall for at least time $\frac{1}{2}T_{\!f}.$

We can now find where the pieces are vertically when the CoM lands.
$$\begin{align} & y_A=\frac{V_{0y}^2}{2g}-v_y \frac{V_{0y}}{g}-\frac{1}{2}g\left(\frac{V_{0y}}{g}\right)^2=-v_y \frac{V_{0y}}{g}\nonumber \\ & y_B=\frac{V_{0y}^2}{2g}+v_y \frac{V_{0y}}{g}-\frac{1}{2}g\left(\frac{V_{0y}}{g}\right)^2=+v_y \frac{V_{0y}}{g}\nonumber \end{align}.$$This shows that if we know that the CoM hits the ground at the same spot as the unexploded projectile, then the two fragments will hit the ground at the same time only if the explosion imparts no vertical momentum to them ($v_y=0$). It also shows that for the CoM to land at the same spot as the unexploded projectile when $v_y\neq 0$, fragment A must fall through a hole in the ground as @Steve4Physics has argued.

Now consider the hole-in-the-ground suggestion by @Steve4Physics and assume that fragment A keeps falling through it until the fragment B lands. We know that the CoM lands at the same spot as the unexploded projectile at horizonttal distance $\Delta X_{\text{cm}}=V_{0x}(\frac{1}{2}T_{\!f})=\dfrac{V_{0x}V_{0y}}{g}$ from the point of the explosion. To find where fragment B lands, which is the goal of this problem, first we find its time of flight, $$0=\frac{V_{0y}^2}{2g}+v_y t_B-\frac{1}{2}gt_B^2\implies t_B=\frac{v_y+\sqrt{V_{0y}^2+v_y^2 }}{g}.$$Then we can find where the second fragment lands horizontally from the point of the explosion, $$\Delta X_B=2V_{0x}t_B=\frac{2V_{0x}\left(v_y+\sqrt{V_{0y}^2+v_y^2 }\right)}{g}.$$It should be clear that unless we know $v_y$, we cannot predict where the second fragment will land. If we are not given $v_y$ explicitly we can infer that it is zero with statements such as
(a) Both fragments land at the same time;
(b) Fragment A drops straight down with zero initial vertical velocity;
(c) Fragment B has zero vertical initial velocity.
Lacking such statements, as is the case here, we have to assume that $v_y=0$, otherwise we cannot answer the question.

Summary
This question can be answered directly by observing that fragment B takes the same amount amount of time to come down as it takes to go up to max height. Since its speed is doubled at max height, it must cover twice the horizontal distance during the down trip.

On edit: Corrected subscript typos and added the summary.

 

[vcsharp2003]

Yes, the fact that the CoM hits the ground is no guarantee that the two fragments will hit the ground at the same time.

So, what was wrong in my logic presented in post#27? I am still trying to figure out where in my logic I overlooked some other facts.

I am unable to come up with a scenario/example where they do not hit the ground simultaneously.

From my logic in post#27, it would automatically follow that the vertical part starts with 0 initial velocity and the other part starts with horizontal velocity, since only then would the fragments reach the ground at the same time as com of whole mass.

 

[Steve4Physics]

So, what was wrong in my logic presented in post#27? I am still trying to figure out where in my logic I overlooked some other facts.

A = vertically falling fragment (say with non-zero intial vertical velocity).
B = other fragment.

I think the fact that you overlooked in Post #27 is that if A hits the ground first, a new external force acts on A. As a result, the CoM no longer follows its original trajectory.

For the CoM to follow its original trajectory, we need (for example) a deep hole in the ground - so A continues (in free fall) down the hole. Then no new extenal force acts on A while B is also in free fall.

 

[vcsharp2003]

I think the fact that you overlooked in Post #27 is that if A hits the ground first, a new external force acts on A. As a result, the CoM no longer follows its original trajectory.

And above would happen only if fragment A would explode off with a non-zero vertical velocity. Is that correct?