Superman Physics ^.^ help..!?

Question

It is well known that bullets fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with bullets of mass m = 2.9 g at a rate of R = 115 bullets/min. The speed of each bullet is v = 560 m/s. Suppose the bullets rebound with no change in speed. What is the average force exerted by the stream of bullets on Superman's chest?

Answer 1

Do Over (x2) (thanks to the two following contributors):

Average force is:

(change in momentum)/time

Assuming he's being shot straight on - and that his chest is flat (glancing blows would change the results):

transferred momentum for each bullet is mv = 2x(0.0029kgx560m/s) = 3.248 kgm/s

total transferred momentum each minute = 115(3.248)/min = 373.52 (kgm/s)/min

Change this to kgm/s^2 by dividing by 60 sec/min gives:
6.22 kgm/s^2 or 6.22 Newtons average force exerted by stream of bullets.

As to average not being useful as suggested below, this kind of averaging is exactly what statistical thermodynamics uses in dealing with pressures exerted against a wall - a gas under pressure in a room has many individual atoms - like bullets - which bounce off the walls - exerting an average force on the wall. It is true that in physiological applications - if the high speed or small size puncture the skin - that is most important and the boundary conditions change - and the body starts to bleed - but that is why the example uses Superman - it is a more fun way to introduce the average force statistical thermodynamics concepts early - without talking about atoms hitting walls (yawn).

Answer 2

OK, Luthor, here's the straight skinny on the big S.

I agree with the earlier answer. But we do know the following for each bullet. f = ma = m dv/dt; and by your assumption of incident speed = rebound speed, the change in velocity (dv) is 2*560 mps.

And we can also guess this, since the two speeds are the same value, the big S's chest is perfectly elastic. So the time to rebound will be very very short. And the time to make an impression (s) on his chest is t = s/v; where v = 560 mps. Thus, the turn around time (time in and time out) is dt = 2t.

As a WAG, not knowing what dt really is, we can set dt = 1 sec realizing dt <<< 1 sec is more likely, In which case we can say f > m (dv/dt) = .0029*(2*560)/1 ~ 3.2 kg-m/sec^2 (Newtons) for each bullet. That's about .6 pounds of force, which is not very much. And if we assume the bullets arrive at the big S chest in sequence rather than some together, that's the max force he'll feel.

My guess is that dt ~ .001 seconds; so the force per bullet would be more like 600 pounds or 3,200 Newtons. But given that Mr. S hides around as Clark Kent, we can't test the rebound time through experiment.

I just had an epiphany. The kinetic energy of the rebound off his chest is equal to the work done by his chest during the rebound. That is, WE = F s = 1/2 mv^2 = KE; where s is the depression of his chest and its expansion. Thus, F = 1/2 mv^2/s = 1/2 mv^2/(vt) = 1/2 mv/t; where t is the time the bullet takes to make a depression s. But as the man of steel has no compression when that bullet hits his chest, s = 0 = vt; and t = 0. Therefore, ta da, each bullet hits with infinite force. Clearly, in the real world s <> 0, so the force would be less than infinity. But, hey, Superman does not live in the real world.

But this does point out an interesting relationship. That is, the shot person receives more force when there is a small depression s than when s is large. That results, because the turn around time (time in and time out) dt = 2t is also very very small when s is small.

By the way, browse "bullet impact force" on the Internet and you'll find a lot of interesting stuff on impact forces...including the observation there is controversy on how to work this kind of problems.

PS: Dr. R is correct. I answered on a per bullet basis. Check this out to see why. Average typically means Sum(x)/N(x), which is the sum of all x's divided by the total number of x's. You can find this in any beginning stat text. If f = the force of one bullet, then your average force is F avg = Sum(f)/N(f), but the Sum(f) = nf and N(f) = n; so F = nf/n = f and the average force IS the force of one bullet.

The deal is this, with one bullet hitting big S's chest at a time, there is never more than f, the force of one bullet hitting him over that minute. Force is not accumulated like drops of rain in a bucket. Once the velocity is turned around and headed in the other direction, the force is gone. Tissue and bone damage is accumulated with each bullet, but in the man of steel even that fails to happen.

Answer 3

Seems to me that there's not enough information to calculate the "force" on Superman's chest.

You could calculate the energy per minute (or per bullet) exerted by his chest, but not the actual force. (energy = 1/2 * mass * velocity^2)

You'd need to know how fast the bullets are rebounding (i.e. the time it takes from going full speed forward to full speed backward). Without that, you can't answer the force question (Force = mass * acceleration).

It has to do with the time (and distance) it takes to reverse the bullets energy. You could get by with knowing how far in superman's chest collapses with each bullet - you could then determine how fast a bullet went from full speed to a dead stop. The further in his chest contracts, the lower the force exerted by his chest (and the longer it takes to reverse the bullet). The less his chest collapses, the more force exterted, and the less time it takes to reverse a bullet. (If his chest doesn't collapse at all, then the force is "infinite" and the equations blow up.)

To take it to an extreme, consider gravity as the force acting to reverse a bullet's path. The force of gravity on a bullet is very low (low mass on the bullet), therefore, it would take a long distance to stop a bullet fired straight up into the air (maybe hundreds of feet).

If you fired the bullet up in the air on Jupiter (more gravitational "force"), the bullet wouldn't go nearly as high. More force means a faster return time and less distance traveled during the turn-around.


*Later Edit*

I'd go with Tipper Bore's answer below and skip this one...

Answer 4

Tipper is right about *average* force being (change in momentum)/time, but the change in momentum is 2mv, where v is the initial bullet speed, since the change in velocity is 2v due to the elastic rebound. As it stands, the answer would be correct if superman stopped the bullets.
Edit: Oh, I see he caught the factor of two. Good.

sdsstrat is answering a different question - force during each bullet's impact.

Answer 5

Ignoring the above solutions, my answer could be: Superman and villian type a device. while the two are table certain(i.e. villian continues to be in superman's hand), the linear momentum of the device is 0. As superman throws a villian forward, the linear momentum of villian will improve. Now if superman and villian are to follows regulations of conservation of momentum, superman's momentum ought to cut back(i.e. he ought to acquire a destructive momentum or get a destructive velocity). So Superman shouldnot proceed to be table certain, he ought to circulate backwards as he throws the villian forward.