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Here is the situation:

You are attacking someone with a wooden pole (such as a pole arm or tree branch). You either (1) hit as hard as you can and the pole breaks into two pieces on impact OR (2) hit quite hard but the pole remains in tact. Assume that you are hitting the same spot with the same angle and everything, so the only difference is how much force was applied.

Which would cause more damage?

My gut instinct is that a weapon that breaks would transfer less of the impact to the person it hits, thereby causing less damage. However my boyfriend pointed out that if you are hitting someone hard enough that the pole breaks, you have used the maximum amount of force that the pole can withstand, so the most force you can transfer with a single blow has been done.

[Edit: I don't have an education in physics beyond high school, but it looks like this question has never been answered on this site before. That could be because I didn't know the physics terms to search for though! Any pointers in the right direction are appreciated.]

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  • $\begingroup$ In baseball, when a batter hits a ball hard enough to break the bat, the ball typically doesn't go very far. $\endgroup$ – mmesser314 Dec 4 '19 at 3:38
  • $\begingroup$ You know you shouldn't hit at the wrong part of the bat, energy and money are wasted! $\endgroup$ – user6760 Dec 4 '19 at 4:15
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Typically the broken weapon does less damage.

There's two major reasons. The first is that the breaking of the weapon consumes energy. This is energy that could have been imparted to the object that was hit. If you don't break the weapon, you get to transfer it all.

The second major reason is that, in slow motions, impacts are messy. The energy and momentum transfers occur over time, not instantaneously. If your weapon breaks, that ends your ability to transfer energy and momentum from your body into the weapon and into the target. Indeed, one of the side effects of bad form when striking is that this connection breaks. (sometimes you want that break. In particular hitting hard things with sledge hammers is best done when your hands are relaxed so that they do not transmit vibrations back to you)

Another argument would be to point out that the weapon breaking should have similar effects to the target breaking. It is well known in martial arts communities that it hurts far worse to fail to break a board or a brick than it hurts to go through it. If the board retains integrity, it has to stop your hand completely. If it breaks, your hand gets to go through.

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  • $\begingroup$ The board is the not the weapon, your hand is the weapon in the last case. However breaking your hand will make future impulses pretty impossible, but the one where it happened is not hindered. $\endgroup$ – user192234 Dec 4 '19 at 3:55
  • $\begingroup$ @user192234 Yes, the board breaking case is the dual to the situation confetti was looking at. However, thanks to the laws of relative motion, the effect of a board breaking or not breaking on a hand should correspond to the effect of a weapon breaking or not breaking on a target. In both cases, it is the thing following the words "or not breaking on a..." which is receiving the damage we are interested in. $\endgroup$ – Cort Ammon Dec 4 '19 at 3:58
  • $\begingroup$ You proved you wrong by negation. But I don't want to get into arguments. I'm wrong you are right. :) $\endgroup$ – user192234 Dec 4 '19 at 4:03
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The force applied on an object is proportional to the change in momentum:

$F = m (v-u)/t = dp/dt$

Momentum is a vector, so direction matters. The largest change in momentum happens when the object you're swinging reverses direction. For example, if you're swinging a stick at a pillar, if it bounces backwards after contact, that exerts the largest force.

If the stick breaks, however, the final momentum is no longer negative, but zero (or close to it). Therefore you exert less force and inflict less damage.

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  • $\begingroup$ Less damage than, what? $\endgroup$ – user192234 Dec 4 '19 at 3:45
  • $\begingroup$ @user192234 that if it didn't break. $\endgroup$ – Allure Dec 4 '19 at 4:00
  • $\begingroup$ Why didn't it? :) $\endgroup$ – user192234 Dec 4 '19 at 4:02
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My take on this is that the answer involves clarifying the question.

You do not transfer force you apply it. You transfer momentum - force times time. The amount that you push back your opponent is the momentum transfer.

If the force profile was literally made greater in magnitude but otherwise identical, then by definition you must be transferring more momentum.

The pole will break, however, on the maximum force applied rather than on the momentum transferred. So, if you attempt to apply the pole in exactly the same way, but hard enough, then the pole will break some time through the application of force, and the force actually applied will drop to near to nothing. As a result, less damage is done.

That is - the actual successful application of more force will cause more damage regardless of the breaking of the pole. But, the breaking of the pole, in practice, will be accompanied by a failure to apply the intended force - and so potentially by less damage.

The details would depend on the force profile. If the profile was a long period of steady force with a peak of force at the end, then increasing the force would result in more momentum transferred (before the pole breaks) followed by the pole breaking right at the end, when it no longer matters.

Having said all that, however, brittle objects break in a different manner to tough objects. One object can break because of a sharp rap, while it might withstand a sustained effort at a higher force. The other might withstand many sharp raps, but bend and break under a sustained effort (consider the sagging of a particle board bookshelf after many years)

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You don't want to break the weapon you want to break another object with the weapon. It's a game theoretic thing to not lose your weapon in a fight :D.

Physically it is harder on the enemy than if the weapon had not broken. Though :]

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