Conservation of Momentum Calculation Woes!

[bruceleereview]

Homework Statement

I am a writer from the UK trying to complete a filler piece about the science of one martial artist kicking another. I have tried online calculators to no avail and tried manually writing out the equation but the result looks incorrect and I lack the grey matter to make sense of it at 11pm on a Saturday night! Assistance please The link to a video setting out the elastic collision is in the solution attempt box.

Relevant Equations

m1v1i + m2v2i = m1v1f + m2v2f

Alleyway kick

The kicker weighs approximately 64kg or 140lb and the gentlemen holding the air shield is approximately 73kg or 160lb.

How far does the kickee travel before hitting the chair and boxes and what speed is he travelling at?

The practice kick prior to the actual kick puts the men approximately a metre apart toe to toe before cutting to a wide shot which the kicker looks to have a larger stride and momentum than before.

 

[haruspex]

the result looks incorrect

What result? Please post your attempt, per forum rules.

How far does the kickee travel before hitting the chair and boxes

Isn't that a matter of estimating the distance from the images?
If you mean to ask how far he might travel before hitting the ground, it depends partly on the angle of the kick. It is clear the kick is somewhat upwards, lifting the target off the ground. It also depends on the posture assumed by the target.. horizontal, vertical, tucked into a ball…

Start by trying to assess the trajectory taken by the target's mass centre. What is the highest point reached, compared with initial height, and at what distance horizontally from launch. From those you can compute the launch velocity.

 

[Lnewqban]

...
The kicker weighs approximately 64kg or 140lb and the gentlemen holding the air shield is approximately 73kg or 160lb.

How far does the kickee travel before hitting the chair and boxes and what speed is he travelling at?

Welcome, @bruceleereview !

If you can, play the video at 25% its original velocity between seconds 0:26 and 0:30.

Your calculation is difficult because energy is added to the kick by different muscles at different times.
Observe how the center of mass location, or hips bone, of Lee acquired certain horizontal velocity during the initial forward jump.

Then a foot is landed ahead of that CM and is used as a pivot and balance point around which the hips transfer the body impulse into the quickly extending kicking leg and foot.

Note how the movement of the hips stops once that energy is fully transferred to the foot.
Also note that the stop point is located directly over the supporting foot, keeping Lee's balance during impact and also during retraction of the kicking leg and balancing arms and head.

At the moment the foot impacts the cushion, it has acquired kinetic energy from the muscles involved in the jump first, and from the muscles extending the leg.

Note that the kinetic energy follows a straight line formed by CM-leg-foot-cushion-upper torso, in such a way that nothing of the pushing force is used for any other thing that providing horizontal impulse to the CM of the person receiving the kick.

As the force of impact is exerted above that person's CM, a sudden moment or torque is applied onto his body, and certain commotion reaches his head, both of which contribute to total loss of balance and little muscular resistance to the subsequent flight and fall.

"I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times." - Bruce Lee

 

[topsquark]

Homework Statement: I am a writer from the UK trying to complete a filler piece about the science of one martial artist kicking another. I have tried online calculators to no avail and tried manually writing out the equation but the result looks incorrect and I lack the grey matter to make sense of it at 11pm on a Saturday night! Assistance please The link to a video setting out the elastic collision is in the solution attempt box.
Relevant Equations: m1v1i + m2v2i = m1v1f + m2v2f

Alleyway kick

The kicker weighs approximately 64kg or 140lb and the gentlemen holding the air shield is approximately 73kg or 160lb.

How far does the kickee travel before hitting the chair and boxes and what speed is he travelling at?

The practice kick prior to the actual kick puts the men approximately a metre apart toe to toe before cutting to a wide shot which the kicker looks to have a larger stride and momentum than before.

Lnewqban has done a very nice job analyzing the general situation, but I'd like to add an obvious comment:

Bruce Lee was an exceptional martial artist but, in the end, a movie is a movie and has to sell. Odds are the reaction the man had after being kicked was out of proportion to the actual force of the kick: it was rigged. It's a kick that can indeed produce an impressive impact (I've done it myself), and note that Bruce's entire forward motion essentially stops after the kick, but I have a hard time believing that even Bruce Lee could kick a man hard enough to send him flying like that.

Think of all of those Western movies where the bad guy gets hit by a bullet and flies backward into the air. Now do the calculation:
bullet mass around 50 mg.
muzzle velocity of a 6-shooter, call it 200 m/s
average mass of a man around 70 kg

If the bullet is shot at standing dead man and is absorbed, the velocity of the final combination will be about 0.147 m/s... just enough to make him fall backward. Not very impressive, so the producers do things.

-Dan

 

[kuruman]

bullet mass around 50 mg.

You mean 50 g(rams). Or did you mean 50 mkg (milli-kilograms)?

I should add that in cheap Western movies you can sometimes see the bad guys flex their knees in preparation for jumping backwards when they're "hit".

 

[bruceleereview]

What result? Please post your attempt, per forum rules.

Apologies. I am self-conscious of sharing as I'm not a scientist or a mathematician. That being said, I plugged the numbers into an online calculator and this is what came out:

https://www.omnicalculator.com/phys...ject1:0!ms,mass_object2:73!kg,v1_object2:0!ms

Isn't that a matter of estimating the distance from the images?

Yes, I took the time to measure the imagery and scale everything appropriately.

If you mean to ask how far he might travel before hitting the ground, it depends partly on the angle of the kick. It is clear the kick is somewhat upwards, lifting the target off the ground. It also depends on the posture assumed by the target.. horizontal, vertical, tucked into a ball.

I'm going to keep it plain and the elevation level, although the kick isn't and the trajectory of the recipient isn't either. Looking for an estimation as you say. Given the small measurements I came out with, it looks increasingly like a parlour trick with the short distance of the setup and placement of target, chair and boxes designed to enhance the effect of the impact and energy transfer.

 

[haruspex]

Apologies. I am self-conscious of sharing as I'm not a scientist or a mathematician. That being said, I plugged the numbers into an online calculator and this is what came out:

https://www.omnicalculator.com/phys...ject1:0!ms,mass_object2:73!kg,v1_object2:0!msYes, I took the time to measure the imagery and scale everything appropriately.

I'm going to keep it plain and the elevation level, although the kick isn't and the trajectory of the recipient isn't either. Looking for an estimation as you say. Given the small measurements I came out with, it looks increasingly like a parlour trick with the short distance of the setup and placement of target, chair and boxes designed to enhance the effect of the impact and energy transfer.

I see you took the initial velocity as 1.02m/s. Where did that come from?
In a jump, an athlete's mass centre can rise about 1m. Using $v^2=2as$, that equates to a launch speed of about 4.5m/s. So 2-3m/s might be nearer the mark.

Look at the target's left leg. At first, it stays on the ground, and the knee bends a little. At this point he is merely falling over, which does not require any great impulse. If you can study individual frames, you might be able to estimate the initial speed, after the kick, of his mass centre.
Then the leg straightens and the foot does leave the ground. In effect, he jumps backwards into the boxes.

 

[bruceleereview]

I see you took the initial velocity as 1.02m/s. Where did that come from?
In a jump, an athlete's mass centre can rise about 1m. Using $v^2=2as$, that equates to a launch speed of about 4.5m/s. So 2-3m/s might be nearer the mark.

Look at the target's left leg. At first, it stays on the ground, and the knee bends a little. At this point he is merely falling over, which does not require any great impulse. If you can study individual frames, you might be able to estimate the initial speed, after the kick, of his mass centre.
Then the leg straightens and the foot does leave the ground. In effect, he jumps backwards into the boxes.

Hi, I took a series of screen shots. Knowing Lee's height, I calculated the scale factor and then utilised that to figure out his distance - once he is settled and ready to kick - from the gentleman to the left of frame who stays in a fixed position and his distance from the gentleman holding the air shield who is also fixed. I then did the same for the next frame where he has bridged the gap and his kicking leg is still chambered and about to extend and push the man holding the air shield off his fixed point. I then watched the footage back with stopwatch in hand and noted how long it took him to get between these two points. That's how I arrived at how fast he travelled. I didn't take any body mechanics into account. I treated the movement like generic object A striking object B on a 2D plane. In the calculator I changed it to a partial elastic collision to account for the elevation and off-centre push.

Presumably, albeit with some loss, object B's distance travelled before he clips the chair and the boxes is equidistant to that travelled by Lee including the leg extension?

This is not my area at all, but I am trying to make sense of it in broad, basic terms.

 

[haruspex]

Hi, I took a series of screen shots. Knowing Lee's height, I calculated the scale factor and then utilised that to figure out his distance - once he is settled and ready to kick - from the gentleman to the left of frame who stays in a fixed position and his distance from the gentleman holding the air shield who is also fixed. I then did the same for the next frame where he has bridged the gap and his kicking leg is still chambered and about to extend and push the man holding the air shield off his fixed point. I then watched the footage back with stopwatch in hand and noted how long it took him to get between these two points. That's how I arrived at how fast he travelled. I didn't take any body mechanics into account. I treated the movement like generic object A striking object B on a 2D plane. In the calculator I changed it to a partial elastic collision to account for the elevation and off-centre push.

Presumably, albeit with some loss, object B's distance travelled before he clips the chair and the boxes is equidistant to that travelled by Lee including the leg extension?

This is not my area at all, but I am trying to make sense of it in broad, basic terms.

Ok, but as I noted, the movement of the stuntman into the boxes has little to do with how hard he was hit. It's only the mass centre velocity right after the impact that relates to Mr Lee's assault.

 

[bruceleereview]

Ok, but as I noted, the movement of the stuntman into the boxes has little to do with how hard he was hit. It's only the mass centre velocity right after the impact that relates to Mr Lee's assault.

Indeed, I think that's what I meant in my layman's terms. He was travelling from the point of interception to the boxes at the same speed Lee was moving at the point of impact. Lee comes to a full stop once his leg is fully extended.

So how best then to articulate this scenario in basic scientific terms for the non-scientist?

 

[haruspex]

He was travelling from the point of interception to the boxes at the same speed Lee was moving at the point of impact.

Only initially. After that he falls over towards the boxes, then appears even to jump towards them, so he gains speed.

how best then to articulate this scenario in basic scientific terms for the non-scientist?

In basic scientific terms it is conservation of momentum, the equation you quoted.

 

[wrobel]

Oh nice topic!. I am a former karateka. How many times I unsuccessfully tried to construct mathematical models of punches. Biomechanics is a hard science.

 

[topsquark]

Oh nice topic!. I am a former karateka. How many times I unsuccessfully tried to construct mathematical models of punches. Biomechanics is a hard science.

I loved my Jiu-Jitsu classes. It's all about Physics!

-Dan