How far will a piece from an explosive projectile fly?

[Steve4Physics]

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

Yes, that's correct.

And if that happened, then of course both A and B would have non-zero initial vertical velocities due to conservation of momentum in the vertical direction.

 

[vcsharp2003]

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.

This problem really doesn't need any com formula or com concepts to solve it. Just projectile concepts and law of conservation of momentum would yield the correct answer. We can simply assume that the fragments are point masses and not bother about how the com of fragments moves after explosion.

 

[Steve4Physics]

This problem really doesn't need any com formula or com concepts to solve it. Just projectile concepts and law of conservation of momentum would yield the correct answer. We can simply assume that the fragments are point masses and not bother about how the com of fragments moves after explosion.

There is often more than one way to tackle the same problem. And it's worth considering which is the best option.

The original question almost certainly intended for there to be zero initial vertical velocity components from the explosion. One way to solve the problem is then:

1. Realise that fragments (A and B) must land at the same time.

2. Realise that the CoM lands where the unexploded projectile would have landed, 1000m from A, therefore 2000m from X.

3. Realise that the CoM is midway between A and B. Therefore B lands a distance (2000m+1000m =) 3000m from X. Answer ‘e’.

Personally, that's my approach of choice.

 

[vcsharp2003]

There is often more than one way to tackle the same problem. And it's worth considering which is the best option.

The original question almost certainly intended for there to be zero initial vertical velocity components from the explosion. One way to solve the problem is then:

1. Realise that fragments (A and B) must land at the same time.

2. Realise that the CoM lands where the unexploded projectile would have landed, 1000m from A, therefore 2000m from X.

3. Realise that the CoM is midway between A and B. Therefore B lands a distance (2000m+1000m =) 3000m from X. Answer ‘e’.

Personally, that's my approach of choice.

I think the com approach is shorter and therefore a shortcut, versus the approach using projectile concepts + law of conservation of momentum.

But the shortcut approach would not work when the first fragment fires of with non-zero vertical velocity. Then, only the other approach would work. But, I guess this state of affairs could only happen if explosion occurred during the projectile's upward ascent rather than its highest point.

 

[jbriggs444]

But the shortcut approach would not work when the first fragment fires of with non-zero vertical velocity.

No, that is not the correct criterion.

The shortcut approach would not work when the first fragment fires off with a non-zero vertical velocity relative to the second fragment.

It does not matter whether this happens during the upward part of the trajectory or the downward part or right in the middle. As long as the fragments have equal vertical velocities, they will land simultaneously.

 

[vcsharp2003]

The shortcut approach would not work when the first fragment fires off with a non-zero vertical velocity relative to the second fragment.

And why would the shortcut not be applicable in the special case you mentioned? Is it because the two fragments would land at different instants of time?

 

[jbriggs444]

And why would the shortcut not be applicable in the special case you mentioned? Is it because the two fragments would land at different instants of time?

The shortcut is applicable if the fragments land simultaneously.

 

[kuruman]

And why would the shortcut not be applicable in the special case you mentioned? Is it because the two fragments would land at different instants of time?

Please read post #33. It shows that the necessary and sufficient condition for the two fragments to land at equal distances from the CoM is that their post-explosion time of flight be the same.

Also, one can do this problem in one's head as indicated #33. It is, of course, applicable only if there is no transfer of vertical momentum between the fragments.

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.

Personally, I think I have said all there is to be said in this thread and say no more.

 

[DrClaude]

The OP has been answered, so we can close this thread.

Thanks to all that have contributed.