# Find momentum transfer and force on the head with and without a helmet

paulimerci
Homework Statement:
Let's suppose the head of a young football player has a mass of 4kg and a speed of 10m/s. If he was to collide without a helmet he basically has 1mm(0.001m) of cushion and increases the time the head stops in 0.0005secs. If a helmet provides 20mm(0.02m) of cushion and increases the time the head comes to rest to 0.002 secs. Find the momentum change of his head for both collisions. Find the momentum change of his head for both collisions. Find the force on the head both with and without a helmet.
Relevant Equations:
Impulse - momentum theorem
Without helmet, m = 4kg, ##v_f = 10m/s## ##\Delta t = 0.0005sec##
$$\Delta p = mv_f - mv_i$$
$$\Delta p_{without} = 40 kg m/s$$

$$Impulse_ {without} = F_{net} \cdot \Delta t$$
$$force_{without} = 80000 N$$

With helmet m = 4kg, ##v_f = 10m/s## ##\Delta t = 0.002sec
$$\Delta p = mv_f - mv_i$$
$$\Delta p_{with} = 40 kg m/s$$
$$Impulse_ {with} = F_{net} \cdot \Delta t$$
$$force_{with} = 20000N$$
I'm just confused with the velocity in the question above. Are they stating initial or final? In my derivation above, I took it as the final velocity.
And it looks like the answer to force without and with the helmet are wrong.

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Homework Helper
It's not the impulse but the force that you should find. The impulse is not measured in newtons.
What is the velocity of the head (or anything) when it stops? What is the final velocity of an object that stops its motion?

• paulimerci
paulimerci
It's not the impulse but the force that you should find. The impulse is not measured in newtons.
What is the velocity of the head (or anything) when it stops? What is the final velocity of an object that stops its motion?
Thank you, I corrected that part.
The velocity of the head when it stops becomes zero. An object that stops its motion also becomes zero.

Homework Helper
Then you know what is the final velocity and what is the initial, right?

paulimerci
Then you know what is the final velocity and what is the initial, right?
The initial speed of the head is 10 m/s, and the final velocity is zero because he collides with the ball, which stops its motion. right? Was the answer posted in #1 right then? Except for the change in momentum and forces, both with and without a helmet, everything is in the negative.

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paulimerci
What do you think @ haruspex? Have I done it right?

Homework Helper
What do you think @ haruspex? Have I done it right?
If we assume (as you did) constant force and constant decelleration and take as input the collision duration then I agree with your calculations and the result.

Yes, the momentum of the head is 40 kg m/s in both cases. Since the final momentum is zero, 40 kg m/s must be the impulse delivered to the head.

We divide the collision impulse by the collision duration to get the rate of change of momentum during the collision. The two durations are given as 0.0005 sec and 0.002 sec. The two divisions yield 80,000 N and 20,000 N respsectively.

However something seems off. We have a factor of 20 difference in impact distance but only a factor of 4 difference in collision duration. That does not make sense to me. A quick and dirty energy analysis says that if you increase the distance by a factor of 20 and only decrease the force by a factor of 4 that you must have increased the work absorbed by a factor of 5. But the head has the same initial kinetic energy in both cases. The energy absorbed should be identical in both cases.

So let us back-calculate to see if the collision displacements given in the problem statement are consistent with the given collision durations. We will take the two accelerations we calculated and compute the distance traversed by the head during the two collisions.

Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement.

[Edit: I'd first used head mass = 10 kg. But head mass is actually supposed to be 4 kg. So I had to re-do my calculations. The conclusion is the same -- bad problem statement]

With helmet, we have F = 20,000 N. Divide by mass to get acceleration. a = 5000 m/s2. The collision duration is supposed to be 0.002 second. Distance covered will be ##s = \frac{1}{2}at^2## = 0.01 m. This does not match the displacement given in the problem statement.

There is another calculation that we could have used. Assuming constant decelleration, the average speed during the collision is half of the initial speed. At 10 m/s initial speed, that is 5 m/s average speed. Over a duration of 0.0005 seconds (without helmet) that gives us 0.0025 m (same as the calculation above). Over a duration of 0.002 seconds (with helmet) it gives us 0.01 m (same as the calculation above).

Somebody screwed up. The problem statement in the original post is incorrect.

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• paulimerci and nasu
paulimerci
If we assume (as you did) constant force and constant decelleration and take as input the collision duration then I agree with your calculations and the result.

Yes, the momentum of the head is 40 kg m/s in both cases. Since the final momentum is zero, 40 kg m/s must be the impulse delivered to the head.

We divide the collision impulse by the collision duration to get the rate of change of momentum during the collision. The two durations are given as 0.0005 sec and 0.002 sec. The two divisions yield 80,000 N and 20,000 N respsectively.

However something seems off. We have a factor of 20 difference in impact distance but only a factor of 4 difference in collision duration. That does not make sense to me. A quick and dirty energy analysis says that if you increase the distance by a factor of 20 and only decrease the force by a factor of 4 that you must have increased the work absorbed by a factor of 5. But the head has the same initial kinetic energy in both cases. The energy absorbed should be identical in both cases.

So let us back-calculate to see if the collision displacements given in the problem statement are consistent with the given collision durations. We will take the two accelerations we calculated and compute the distance traversed by the head during the two collisions.

Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement.

[Edit: I'd first used head mass = 10 kg. But head mass is actually supposed to be 4 kg. So I had to re-do my calculations. The conclusion is the same -- bad problem statement]

With helmet, we have F = 20,000 N. Divide by mass to get acceleration. a = 5000 m/s2. The collision duration is supposed to be 0.002 second. Distance covered will be ##s = \frac{1}{2}at^2## = 0.01 m. This does not match the displacement given in the problem statement.

There is another calculation that we could have used. Assuming constant decelleration, the average speed during the collision is half of the initial speed. At 10 m/s initial speed, that is 5 m/s average speed. Over a duration of 0.0005 seconds (without helmet) that gives us 0.0025 m (same as the calculation above). Over a duration of 0.002 seconds (with helmet) it gives us 0.01 m (same as the calculation above).

Somebody screwed up. The problem statement in the original post is incorrect.
Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 8000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be s=12at2 = 0.001 m.
Isn’t the mass is 4kg? For without helmet a =20000m/s2. Right? It looks like the duration in the question looks incorrect.

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paulimerci
If we assume (as you did) constant force and constant decelleration and take as input the collision duration then I agree with your calculations and the result.

Yes, the momentum of the head is 40 kg m/s in both cases. Since the final momentum is zero, 40 kg m/s must be the impulse delivered to the head.

We divide the collision impulse by the collision duration to get the rate of change of momentum during the collision. The two durations are given as 0.0005 sec and 0.002 sec. The two divisions yield 80,000 N and 20,000 N respsectively.

However something seems off. We have a factor of 20 difference in impact distance but only a factor of 4 difference in collision duration. That does not make sense to me. A quick and dirty energy analysis says that if you increase the distance by a factor of 20 and only decrease the force by a factor of 4 that you must have increased the work absorbed by a factor of 5. But the head has the same initial kinetic energy in both cases. The energy absorbed should be identical in both cases.

So let us back-calculate to see if the collision displacements given in the problem statement are consistent with the given collision durations. We will take the two accelerations we calculated and compute the distance traversed by the head during the two collisions.

Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement.

[Edit: I'd first used head mass = 10 kg. But head mass is actually supposed to be 4 kg. So I had to re-do my calculations. The conclusion is the same -- bad problem statement]

With helmet, we have F = 20,000 N. Divide by mass to get acceleration. a = 5000 m/s2. The collision duration is supposed to be 0.002 second. Distance covered will be ##s = \frac{1}{2}at^2## = 0.01 m. This does not match the displacement given in the problem statement.

There is another calculation that we could have used. Assuming constant decelleration, the average speed during the collision is half of the initial speed. At 10 m/s initial speed, that is 5 m/s average speed. Over a duration of 0.0005 seconds (without helmet) that gives us 0.0025 m (same as the calculation above). Over a duration of 0.002 seconds (with helmet) it gives us 0.01 m (same as the calculation above).

Somebody screwed up. The problem statement in the original post is incorrect.
Thank you so much, @jbriggs444. I accept that the problem I posted was incorrect. And I thank you for taking time in explaining them in detail, with different approaches you used. I truly appreciate that.

paulimerci
I see that the question posted is incorrect. If this is the question I have to answer, I know the final velocity is zero, the initial velocity is 10 m/s, and the mass is 4 kg, Why are the answers for the change in momentum and forces not negative?

Gold Member
I see that the question posted is incorrect. If this is the question I have to answer, I know the final velocity is zero, the initial velocity is 10 m/s, and the mass is 4 kg, Why are the answers for the change in momentum and forces not negative?

If the head were moving to the right ##\rightarrow^+## then the force from the collision would be to the left ##\leftarrow^-##.

So that would look like:

$$\vec{F_{avg}} \Delta t = 0 - m \vec{v_o} \implies \vec{F_{avg}} = -\frac{m \vec{v_o} }{\Delta t}$$

If ##\vec{v_o} = v_o \boldsymbol{u}## is positive (1D) the average Force is to the left (negative), you would find:

$$\vec{F_{avg}} = -\frac{m v_o \boldsymbol{u} }{\Delta t}$$

Where ##\boldsymbol{u}## is just a unit vector for the positive direction.

However, you could just as easily choose the right as positive and have the head moving to the left. In which case you would find after substituting ##\vec{v_o} = -v_o \boldsymbol{u}##:

$$\vec{F_{avg}} \Delta t = 0 - m \vec{v_o} \implies \vec{F_{avg}} = \frac{m v_o }{\Delta t} \boldsymbol{u}$$

Basically, the directions are arbitrary here. The magnitudes of the forces aren't.

That took me a few edits once I looked at what I was saying. I hope I didn't confuse you along the way.

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• paulimerci
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@paulimerci, a few points in addition to what has already been said...

The question as stated in Post #1 is incomplete. It is good practice to state any assumptions needed for your answer.

I’d say the key assumptions you should include in your answer are:

1. The (unknown) mass of the helmet is negligible.

2. The (unknown) force exerted by the player’s neck on his head is negligible.

3. The player's initial velocity is taken to be in the positive direction.

4. Since the figures given are inconsistent with constant accelerations, assume each force is variable and only the average force (for each case) is required. (This means you don't use the distances, only the times.)

Also, note that ##\Delta p = mv_f - mv_i## gives
##\Delta p = (4kg*0) – (4kg*10m/s) =## a negative value.
(And you could have factored-out the mass.)

• paulimerci
paulimerci
If the head were moving to the right ##\rightarrow^+## then the force from the collision would be to the left ##\leftarrow^-##.

So that would look like:

$$\vec{F_{avg}} \Delta t = 0 - m \vec{v_o} \implies \vec{F_{avg}} = -\frac{m \vec{v_o} }{\Delta t}$$

If ##\vec{v_o} = v_o \boldsymbol{u}## is positive (1D) the average Force is to the left (negative), you would find:

$$\vec{F_{avg}} = -\frac{m v_o \boldsymbol{u} }{\Delta t}$$

Where ##\boldsymbol{u}## is just a unit vector for the positive direction.

However, you could just as easily choose the right as positive and have the head moving to the left. In which case you would find after substituting ##\vec{v_o} = -v_o \boldsymbol{u}##:

$$\vec{F_{avg}} \Delta t = 0 - m \vec{v_o} \implies \vec{F_{avg}} = \frac{m v_o }{\Delta t} \boldsymbol{u}$$

Basically, the directions are arbitrary here. The magnitudes of the forces aren't.

That took me a few edits once I looked at what I was saying. I hope I didn't confuse you along the way.
No, you didn't. As Steve4Physics mentioned, I should have made assumptions before starting to sort out the problem. I'll try to do that next time. Thank you @erobz @Steve4Physics.

• erobz and Steve4Physics
paulimerci
If we assume (as you did) constant force and constant decelleration and take as input the collision duration then I agree with your calculations and the result.

Yes, the momentum of the head is 40 kg m/s in both cases. Since the final momentum is zero, 40 kg m/s must be the impulse delivered to the head.

We divide the collision impulse by the collision duration to get the rate of change of momentum during the collision. The two durations are given as 0.0005 sec and 0.002 sec. The two divisions yield 80,000 N and 20,000 N respsectively.

However something seems off. We have a factor of 20 difference in impact distance but only a factor of 4 difference in collision duration. That does not make sense to me. A quick and dirty energy analysis says that if you increase the distance by a factor of 20 and only decrease the force by a factor of 4 that you must have increased the work absorbed by a factor of 5. But the head has the same initial kinetic energy in both cases. The energy absorbed should be identical in both cases.

So let us back-calculate to see if the collision displacements given in the problem statement are consistent with the given collision durations. We will take the two accelerations we calculated and compute the distance traversed by the head during the two collisions.

Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement.

[Edit: I'd first used head mass = 10 kg. But head mass is actually supposed to be 4 kg. So I had to re-do my calculations. The conclusion is the same -- bad problem statement]

With helmet, we have F = 20,000 N. Divide by mass to get acceleration. a = 5000 m/s2. The collision duration is supposed to be 0.002 second. Distance covered will be ##s = \frac{1}{2}at^2## = 0.01 m. This does not match the displacement given in the problem statement.

There is another calculation that we could have used. Assuming constant decelleration, the average speed during the collision is half of the initial speed. At 10 m/s initial speed, that is 5 m/s average speed. Over a duration of 0.0005 seconds (without helmet) that gives us 0.0025 m (same as the calculation above). Over a duration of 0.002 seconds (with helmet) it gives us 0.01 m (same as the calculation above).

Somebody screwed up. The problem statement in the original post is incorrect.
"Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement."

@jbriggs Why is v_i zero in the above kinematic equation you provided? or did you conduct a quick test? Even if I add v_i = 10 m/s, this still doesn’t match with displacement provided in the problem statement. Just want to know!

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"Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement."
You have assumed a constant force. Although the question asks you to find "the force", you can only find the average force.
##\int F.dt=m\Delta v=F_{avg}\Delta t##.
For the displacement, ##\int F.dx=\Delta(\frac 12mv^2)##.
Is there a force profile, ##F=F(t)##, that satisfies both?

Picture the velocity- time graph. It starts at some velocity v and declines to zero over a period t. The average acceleration is -v/t, but the displacement is the area under the graph. How can you make the displacement small while keeping -v/t fixed?

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Homework Helper
You have a factor of 1/2 missing in the work-energy expression. The work is change in kinetic energy.

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You have a factor of 1/2 missing in the work-energy expression. The work is change in kinetic energy.
Whoops. fixed, thanks.

Homework Helper
"Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement."
Note that the correct attribution would be to myself (@jbriggs444) as author of the quoted passage:
Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement.
When responding to short passage within a post, it is best if you can avoid quoting the whole thing and only quote the relevant passage, leaving that passage alone within the quote tags that are automatically supplied.

One way to do that is to mouse over the relevant passage in the original post, selecting that passage and then clicking "quote" on the resulting pop-up. That will save the passage, making it available to use with the "insert quotes" button on the editting panel.

One flaw in this technique is that ##\LaTeX## in the quotated passage will not be properly captured.

Another way is to quote the whole thing (as you have been doing) and then erase all but the desired passage. This approach leaves ##\LaTeX## intact.

Yet another variation on this latter technique is to turn off BB code processing on the editting panel. Then you can see how the quote is formatted behind the scenes. e.g.

[QUOTE="jbriggs444, post: 6852126, member: 422467"]
[...]Without helmet, we have F = 80,000 N. Divide by mass to get acceleration. a = 20,000 m/s2. The collision duration is supposed to be 0.0005 second. Distance covered will be ##s=\frac{1}{2}at^2## = 0.0025 m. This does not match the displacement given in the problem statement.
[...]
[/QUOTE]

You can see here what the undedrlying BB code looks like. You can chop up the original quoted text into pieces and insert your own BB code tags so that you can quote multiple passages and insert your own text in between. Do not forget to use the preview feature in case you butcher the result :-)

Now, with all that said, let us look at your actual question:
@jbriggs Why is v_i zero in the above kinematic equation you provided? or did you conduct a quick test? Even if I add v_i = 10 m/s, this still doesn’t match with displacement provided in the problem statement. Just want to know!
Your concern is that I used ##s=\frac{1}{2}at^2##. This equation is supposed to work only when ##v_i = 0##.

The answer is that I am contemplating the collision in time reversed fashion, adopting the frame in which the head is at rest after the collision. In this frame the exterior of the crushed helmet is at rest and the head accelerates away.

The amount by which the helmet (or head) is crushed is an "invariant". That means that we will get the same result no matter what frame we use to do the calculation. So I automatically chose a frame that makes the result easy to calculate.

I apologize for not making that clear. Shifting frames like that is something that I do automatically without even thinking about it sometimes.

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• paulimerci
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You have assumed a constant force.
Yes. I am the author of the quoted passage. I did indeed assume a constant force and had indicated that explicitly upthread.

I am aware of the difficulties in the casual use of the unqualified term "average" when vacillating between a time-weighted and a displacement-weighted average.

• paulimerci
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Yes. I am the author of the quoted passage. I did indeed assume a constant force and had indicated that explicitly upthread.

I am aware of the difficulties in the casual use of the unqualified term "average" when vacillating between a time-weighted and a displacement-weighted average.
I should have noted the quote marks.
We have a factor of 20 difference in impact distance but only a factor of 4 difference in collision duration.
Not impossible. With no protection, the impact duration is mostly skull deformation. Large forces and accelerations. The added protection of the helmet may deform rather more readily, so at first the deceleration is much slower. The v-t graph declines gently at first, then more sharply. That could produce a much greater average velocity.
The implication is that the helmet is too soft, doing very little to reduce the peak force.

• jbriggs444