As an exercise more intended to see if I still have my engineering chops, I am going to solve perhaps the coolest textbook physics problem I have ever seen.
Batman (mass 80.0 kg) stands on a window ledge 5.0 m above the floor (Fig. 8-40). Grabbing a rope attached to a chandelier, he swings down to grapple with The Joker (mass 70.0 kg), who is standing directly under the chandelier. (Assume the stuntman’s center of mass moves down 5.0 m. He releases the rope just as he reaches the villain.)
a) With what speed do the entwined foes start to slide across the floor?
b) If the coefficient of kinetic friction of their bodies with the floor is μ = 0.250, how far do they slide?
To find out how fast Batman and The Joker are going post-POW, we first have to know Batman’s speed. To do this, we transfer all the gravitational potential energy he has–the energy an object has when suspended above the Earth–to kinetic energy–all the energy bound up in his mass at some speed.
The higher you hold on object above the Earth, the more potential energy it has. This is because the longer it falls to Earth, the faster it goes, and this gives the object more energy. The acceleration that causes this potential is (on Earth) 9.81 meters per second per second.
So, at 5 meters above the ground, a 80 kg Batman (no way he is that light, especially with all the armor) has 3,924 Joules of potential energy. This is about the same energy released by one gram of TNT.
But where does all that energy go once Batman swings back to Earth’s surface? All of the potential energy becomes kinetic energy. The equation for kinetic energy is 1/2*mass*velocity^2, and solving for velocity, gives us Batman’s speed when his WHOPs Joker: 9.9 meters per second.
To answer question A, you have to remember that momentum–a product of mass and velocity–is conserved. If Batman and The Joker entangle during their collision, the mass of Batman and the velocity of his hit has to be proportional to the mass and velocity of both Batman and The Joker as one system.
So, the mass and velocity of Batman during his strike equals the mass of both Batman and The Joker multiplied by the velocity at which they skid across the floor. With an 80kg Batman moving at 9.9m/s, the 150kg Batman+Joker pair should begin skidding at 5.3 meters per second (that was a hell of a wallop).
Answer a) The entwined foes start sliding across the floor at 5.3m/s.
But how far does the pair skid?
Question B involves a bit more physics, but stick with me. We know that the floor offers resistance to the sliding duo, and this is in the form of friction. During the skid, friction will apply a force to the pair, and that force is equal to the coefficient of friction of the floor (how “rough” the floor is) multiplied by how heavy the object sliding on it is (e.g., it’s easier to push a 1kg block across a rough floor than it is to push a 100kg block).
So, the floor will resist the pair with 368 Newtons of frictional force (0.25*150kg*9.81m/s^2). And because, as many of you will know, force equals mass times acceleration, this means that the floor can decelerate the pair at a rate of -2.45 meters per second per second (it is negative because it is acting in the opposite direction that the pair is sliding).
Finally, based on equations meant to calculate the stopping distance for cars, we figure out that after SMACKing into The Joker, both he and Batman will slide a significant 5.73 meters.
Answer b) Batman and The Joker slide 5.73m.
If you find any mistakes, be sure to let me know and I’ll correct it promptly.