practice problem 1
Now, the bears I live with average, the males, eight to twelve hundred pounds [360 to 540 kg]. They're the largest bears in the world…. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach.
- Compute the speed of a grizzly bear using Mr. Treadwell's hundred meter statement.
- Compute the kinetic energy of a grizzly bear using the speed you calculated in part a. and the average mass stated by Mr. Treadwell.
- How fast would a 250 lb man have to run to have the same kinetic energy you calculated in part b? (Do not use a calculator to compute your answer.)
- How fast would a 4000 lb car have to drive to have the same kinetic energy you calculated in part b? (Do not use a calculator to compute your answer.)
Speed is distance over time.
v = ∆s ∆t v = 100 m 5.85 s v = 17 m/s
Kinetic energy is the product of mass and speed squared. Let's use a mass in the middle of the range stated by Mr. Treadwell.
K = ½mv2 K = ½(450 kg)(17 m/s)2 K = 65,700 J
When kinetic energy is constant, mass inversely proportional to the square of speed. Mass goes down when we replace the 1,000 pound grizzly bear with a 250 pound man. To keep the kinetic energy constant, the man will have to run faster. Since our hypothetical man has ¼ the mass of a grizzly, he needs to run 2 times faster to have the same kinetic energy.
K = ½mv2 ⇒ K = ½(¼m)(2v)2
That's 34 m/s in International units or 82 mph in Anglo-American units. When it comes to kinetic energy, a bear at top speed is like a man running as fast as a speeding car.
vman = 2vbear vman = 2(17 m/s) = 34 m/s vman = 2(41 mph) = 82 mph
Use reasoning similar to part c. Mass goes up when we replace the 1,000 pound grizzly bear with a 4,000 pound car. Four times the mass needs ½ the speed to have the same kinetic energy.
K = ½mv2 ⇒ K = ½(4m)(½v)2
That's 8.5 m/s or 20.5 mph. A bear at top speed is like a car driving through a school zone.
vcar = ½vbear vcar = ½(17 m/s) = 8.5 m/s vcar = ½(41 mph) = 20.5 mph
practice problem 2
The Space Shuttle Columbia disintegrated during reentry on the morning of 1 February 2003. The cause of the accident was determined months later. A review of video footage taken during the launch 16 days earlier showed a large piece of foam insulation falling off the external fuel tank shortly after liftoff then striking the leading edge of the orbiter's left wing. This compromised the thermal protection system at the point of impact and allowed the superheated gases generated on reentry to melt the aluminum frame there. The left wing snapped off first, the orbiter tumbled and broke apart, scattering pieces across eastern Texas. All seven crew onboard were killed
Eighty-two seconds into STS 107 [the mission number], a sizeable piece of debris struck the left wing of the Columbia. Visual evidence and other sensor data established that the debris came from the bipod ramp area and impacted the wing on the wing leading edge. At this time Columbia was traveling at a speed of about 2300 feet/second (fps) through an altitude of about 65,900 feet. Based on a combination of image analysis and advanced computational methods, the Board determined that a foam projectile with a total weight of 1.67 lb and impact velocity of 775 fps would best represent the debris strike….
Just prior to separating from the External Tank (ET), the foam was traveling with the orbiter at about 2300 fps. The visual evidence shows that the debris impacted the wing approximately 0.161 seconds after separating from the ET. In that time, the debris slowed down from 2300 fps to about 1500 fps, so it hit the orbiter with a relative velocity of about 800 fps. In essence, the debris slowed down and the Orbiter did not, so that the Orbiter ran into the debris.
Show that a piece of rigid foam insulation like the one that struck the Space Shuttle Columbia possesses a considerable amount of kinetic energy despite being "just a piece of foam".
- Determine the kinetic energy of the foam debris that struck Columbia in 2003.
- How fast would a 10 lb sledge hammer have to travel in order to have the same kinetic energy as the foam? State your answer in miles per hour or kilometers per hour as you prefer.
- How massive would a defensive tackle of American or Canadian football have to be if he ran as fast as a world class sprinter and had the same kinetic energy as the foam debris? State your answer in pounds or kilograms as you prefer.
The kinetic energy of the foam is calculated directly from the kinetic energy formula.
K = ½mv2
K = ½(1.67 lbs)(775 ft/s)2
K = something in archaic units
I wouldn't accept whatever answer a regular calculator gave you. Say hello to Google calculator
Google calculator does not understand significant digits. The answer should really be stated as 21.1 kJ. Is this a big number? Let's put it into context.
The speed of a sledgehammer with equivalent energy can be found by rearranging the kinetic energy formula to make speed the subject.
v = ⎛
2(21134.187 J) ⎞½
m 10 lbs v = something equally unusable
We should again resort to Google calculator and ask it to calculate our answers in the appropriate units.
Sledgehammers are normally considered quite destructive. They are the tool of choice for home demolition work. In order for a small sledgehammer (like the kind used to tear down drywall or plasterboard) to have the same energy as the foam debris it would have to be traveling as fast as race car: about 216 mph or 348 km/h.
The mass of a defensive tackle with equivalent energy can also be found by rearranging the kinetic energy formula. This time make mass the subject. For the sake of argument we'll assume that a particularly fast tackle could run could run 100 m in 10 s. These units work out nicely.
m = 2K = 2(21134.187 J) v2 (100 m/10 s)2 m = 422.68374… kg
That was so easy I was able to calculate the solution in my head. Multiplying by 2 and then shifting the decimal two places is not a chore. You can verify the metric answer with Google calculator if you wish and then modify the procedure so as to get an answer in english pounds.
Again, Google does not do significant figures. The answers are 422 kg or 932 lbs. This is like getting hit by four exceptionally fast tackles all at once. It's harder to imagine four big fast men causing damage to the the wing on a space shuttle. This might not be the best analogy to use for comparison.
One last word about the Columbia accident. In an experiment done five months later, a piece of foam similar to a bipod ramp was fired from a pneumatic cannon at a replica shuttle wing. The resulting impact punched a hole in the replica wing's leading edge that was larger than a human head. This level of structural damage is sufficient to explain the breakup of Columbia.
|Pneumatic gun used to fire the simulated foam bipod ramp||Shuttle wing mockup before impact||Shuttle wing mockup after impact|
practice problem 3
practice problem 4