Oct 6, 2011
Batman and the Joker helping to teach physics
From Young and Freedman’s University Physics, pages 263-264…
The problem…
A movie stuntman (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 movie’s villain (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?
(via Reddit)
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Abraham –
I believe the page you are looking for is this guy:
http://www.physicsofsuperheroes.com/
Particularly, the videos:
http://www.physicsofsuperheroes.com/videos.php
He is actually a really good speaker, came and spoke at my undergrad.
…can’t resist…
*does some math*
a) 7.23 (m / s)
b) 10.7 (m)
I was hoping someone would do this lol! I forgot pretty much all my high school physics…
I Think Stephen has too.
V = sqrt(2gh)*Mb/(Mb+Mj) where
g=9.8
h = 5
Mb = 80
Mj = 70
V=5.28
I see we both used the conservation of energy to solve the problem:
K = U
U = ghMb
K = 0.5(Mb + Mj) v^2
ghMb = 0.5(Mb+Mj)v^2
But I solved for v to get:
v = sqrt[2ghMb/(Mb+Mj)]
thus my different answer.
BTW, this is on page 280 of the 12th edition of Y&F. It has a less appealing picture.
Stephen did the problem as if it were a elastic collision, when it’s actually a perfectly inelastic collision. An elastic collision is one in which total kinetic energy is conserved, and it involves objects that are rigid (e.g. billiard balls or marbles). An elastic collision is one in which kinetic energy is not conserved, as some energy is spent deforming the objects (in this case, maybe cracking a few Joker ribs). A perfectly inelastic collision is an inelastic collision in which the objects stick together.
So, because energy is not conserved, we must instead use momentum conservation for the collision.
Batman’s velocity as he hits the Joker is calculated by converting his gravitational potential energy into kinetic energy.
Gravitational potential energy:
U = mgh = 80 kg * 9.8 m/s^2 * 5 m = 3920 J
Kinetic energy:
T = 3920 J = 1/2 * m * v^2
v = sqrt(2T/m) = sqrt(2 * 3920 J / 80 kg) = 7.23 m/s
And now we must use conservation of momentum. m_1 is Batman’s mass and m_2 is the mass of Batman and Joker combined.
m_1 * v_1 = m_2 * v_2
v_2 = m_1 * v_1 / m_2 = 80 kg * 7.23 m/s / (80 kg + 70 kg) = 5.28 m/s
I’ll explain part b in a separate comment.
I think this is from the 9th or 10th edition. Not sure.
Actually, looking at it again, it looks like Stephen forgot Joker’s mass altogether.
Okay, so μ is the coefficient of friction. It tells you the ratio of the magnitude of the force of friction to the magnitude of the force that an object exerts on the ground.
So we’ll consider Batman and Joker a single object. Their combined mass is 150 kg. Their force on the ground is equal to their mass times the acceleration due to gravity. This force is equal in magnitude to the force of the ground on the object, which is called the normal force, so I’ll call it F_norm.
F_norm = 150 kg * 9.8 m/s = 1470 Newtons
Then the force of friction is
f_fric = μ * F_norm = 0.250 * 1470 N = 367.5 N
The acceleration is found using Newton’s second law, Force = mass * acceleration.
F = m * a
a = F / m = 367.5 N / 150 kg = 2.45 m/s^2
We’ll consider this negative, as it’s opposite the direction of motion.
Using the equation of motion
v_f^2 = v_i^2 + 2ax
(with v_f = final velocity, v_1 = initial velocity, and x = distance)
0 = (5.28 m/s)^2 + 2 * (-2.45 m/s^2) * x
x = 5.7 m
So they slide 5.7m after impact.
Both of these answers are checked with the key.
Yes, I know, crazy long posts. It’s been a while since I’ve gotten to do baby physics, so maybe I’m a bit too excited. I’m currently a college senior in physics, and I miss the easy stuff.
// sqrt(2 * 3920 J / 80 kg) = 7.23 m/s //
wrong, it’s 9.89 m/s
Bravo but unfortunately you’re failing to factor in Hooke’s Law for the rope or the force of air on the stuntman. I’ll let you get away with not using Hooke’s Law since no constant is given though but shame on you for not considering drag.
This is the best comment thread on 22 Words ever.
I agree.
Dear Abraham,
I you post stuff like this, you’ll flush all the geeks like me out of the woodwork. Works every time; you’ll see.
The speed, friction, and all that other stuff is secondary. The main question is, “How many takes did the stuntman need to make the scene work?”
Can you tell that I have never taken a physics class?
MJ McNamara got the problem right. About 19 feet. Good job! Mr. McNamara, I hope you have a great career in physics – I’ve loved it.
My dear wife, knowing that this post was shark chum for geeks, sent me this link last night, and I saw it at 11 PM. Of course, I had to solve it!
conservation of energy till the big bang
mgh = 1/2 m v*v
v solves to 9.89 m/s
conservation of momentum of system just after impact
(m1+m2)v2 = m1*v
v2 solves to 5.0596 m/s
acceleration after impact: mu*g = -2.45 m per sec sqr
Eqn of motion: v sqr – u sqr = 2*a*s
u = v2 = 5.0596 m/s
v = 0
s solves to 5.22 m
It’s been a while since I took Mechanics, but wouldn’t batman’s motion essentially be that of a pendulum? I’d think you’d need to crunch the second order ODE to then get batman’s velocity at impact with the Joker, and then depending on the type of collision use the appropriate formula. Or would the derived formula for batman’s velocity spit out the same result as the conservation of energy crunching that other commenters have used?
I find it highly unlikely batman, being a strong fit man, with addition of his many pieces of armour, could possibly weigh 80kg.
Not enough information given. You have to account for MOE being center or degree of off-center and the kinetic energy absorption and deflection by material in Batman’s boots and Joker’s jacket, not to mention whether Batman’s knees are bent or not.
Haha, I wish I had those kind of tasks when I was going to school :D I like physics but this certainly adds some humorous as well.
wow, one of the few threads where people can disagree and there is no name calling, no flaming, and its all arguing with facts and not emotions… I LOVE MY FELLOW GEEKS :-)If you all enjoyed this I think you should read “Physics of Superheros”