It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with 3 g bullets at the rate of 100 bullets/min, and the speed of each bullet is 500 m's.
Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets? Why and how did you get your answer?
2006-10-12 03:56:52 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Well, actually we need to know either a) the time it takes for the bouncing or b) how much superman's chest goes in for this rebounding.
The momentum change of each bullet is of course 0,003 * 500 * 2 (complete change in direction) = 3kg m / s.
Let's say superman's chest has contact with each bullet for 0,005 seconds, it means the total force from one bullet is 600 N (like a the weight of a 60 kg person). in 60 seconds, you get 100 bullets which is 1000 N per second, which I guess is the (average force) you are talking about...
The answer above mine is wrong, because you can't multiply mass with speed and get Force. mass times speed (actually velocity) equals momentum (unit = kg.m/s) Force is Newton (kg.m/s^2)
2006-10-12 04:06:50 · answer #1 · answered by hilmic 2 · 0⤊ 1⤋
So, if you want an average force, you just have to figure out the average change in momentum per unit time.
Start out calculating the amount of momentum each bullet has-- .003 kg*500m/s=1.5kg*m/s.
Now, we know that the reflected momentum is the same as the initial-- just in the opposite direction. So, the change in momentum for each bullet is just 2*the initial-- 3kg*m/s.
Knowing that the bullets come at the rate of 1.67 bullets/second (100 bullets/min), we know that superman's chest has to give 3kg*m/s of momentum to 1.67 bullets every second. So, the average force should just be the product of the two-- 5kg*m/s^2.
Note that this is the average force-- if you looked at the real force, there would be lots of time when the force was zero, and short times when the force is extremely high. Since we don't care about that, we can just take the change in momentum over the change in time...
Hope this made sense (it did to me)
2006-10-12 06:46:53 · answer #2 · answered by wherearethetacos 3 · 1⤊ 0⤋
3g = 0.003 kg. Each bullet changes velocity from 500 m/s forward to 500 m/s backward for a total velocity change of 1000 m/s. Now the bullet changes speed in a fraction of a second which is a very high acceleration, and the rest of the 1 second there is no acceleration, but I will average it out by saying that the bullet takes a full second to bounce off so the average acceleration over one second is 1000 m/s/s. Using F=MA, I get 0.003 kg times 1000 m/s/s = 3 Newtons for each bullet that hits. 100 a minute is 100/60 or 1.66.. per second. 3 times 1.66... is 5 Newtons. That is a little over one pound.
2006-10-12 04:05:57 · answer #3 · answered by campbelp2002 7 · 0⤊ 1⤋
Hi The "Man of Steel" would only get hit with 1 bullet because the first bullet, traveling straight back, would plug up the gun.
2006-10-12 04:00:09 · answer #4 · answered by Cirric 7 · 0⤊ 1⤋
there is no one called superman
2006-10-12 04:43:44 · answer #5 · answered by thiru k 2 · 0⤊ 1⤋