Dodge This:

1. Dodge This: Impossible Physics of The

2. In this presentation… Analysis of one scene from The Matrix using: -2 Dimensional Motion -Rotational Motion -Linear and Angular Momentum -Conservation of Linear and Angular Momentum -Energy

4. Vbi 90° COM 1.8m Hagent= 1.8m Magent = 75kg COMagent = 1.017m Vai = 0 m/s Mbullet = .00745kg Vbi = 390m/s Θ = 90°

5. ωf Vf Conservation of Linear and Angular Momentum for completely inelastic collisions, where an object in motion collides with an object at rest tell us that: Ptot = M1V1 = (M1 + M2)Vf Ltot = M1V1RsinΘ = Iωf

6. Using the values stated earlier along with other assumptions we can solve for the final linear and angular velocities. Assumptions: • Agent Brown is approximately a uniform rod of length 1.8m. • Air Resistance is negligible. • The bullet does not exit Agent Brown’s skull (collision is perfectly inelastic). • The bullet travels at a constant velocity from the time it is launched until the time it enters Agent Brown’s skull. • Trinity uses a standard Baretta 92FS with 9mm bullets. • Agent Brown is approximately 6ft tall (1.8m) and is of mass approximately 75kg. • All rotation occurs around the COM. • The human body is approximately as tall as 7.5 times the length of its own head. • The bullet enters Agent Brown’s skull approximately 1/3 the length of one head from the tip of his head.

7. MbVbi= (Mb + Ma)Vf Vf = [M/(Mb + Ma)]*Vbi Vf = .039m/s To obtain the moment of inertia of the Agent Brown-Bullet body we must apply the parallel axis theorem. I = [1/12](Mb + Ma)L^2 + (Mb + Ma)h^2. h = COM – [1/2]L = .117m I = .387(Mb + Ma) = 29.028kg*m^2 If each head is (1/7.5)1.8m = 0.24m. The bullet will enter 0.08m from the top of Agent Brown’s head. If his COM is 1.017m then the radius of the angular momentum is: r = 1.72 – 1.017 = 0.703m. Li = Mb*Vbi*r*sinΘ = 2.0426kg*m^2/s Ltot = Li= Iωf. ωf = L/I = .0704rad/s

8. In the video Agent Brown rotates fast enough so that by the time he hits the ground he has rotated 90 degrees CCW and has moved backwards 2meters. Because his feet never interfere with the falling of his body, we may track the amount of time it would take for his COM to hit the ground as it moves in projectile motion. The time it takes for Agent Brown’s COM to hit the ground is calculated by t = sqrt(2COM/g) = .46seconds We can calculate what angle his body should theoretically rotate through and how far he should move backwards before his body hits the ground. Θ = Θo + ωft = 0.032rad = 1.86 degrees. X = Xo + vft = 0.0180 meters These values are much smaller than the ones seen in the film clip (and are far too different to be due to errors in estimation). To produce the rotational motion shown in the film one would require a bullet with a momentum of 144.108kg*m/s to cause a rotational velocity of 3.49rad/s. (if we keep the radius, masses, and angle constant) And to produce the translational motion shown one would require a bullet with a momentum of 333.03kg*m/s to cause a translational velocity of 4.44m/s. (if we keep the radius, masses, and angle constant) Obviously no bullet can simultaneously have 2 different momentums and it is impossible to produce any bullet that could cause the type of motion displayed in the clip.

9. If you still think this situation is realistic… The initial kinetic energy of the system is equal to the kinetic energy of the bullet which is. [1/2]Mb*Vbi^2 + = 556.60J The total energy of Agent Brown just after he begins rotation: [1/2](Mb + Ma)*Vf^2 + [1/2]Iωf^2 = 0.129J In any inelastic collision the system should lose Kinetic Energy. Using the values of Agent Brown’s motion as seen in the video we can calculate his Kinetic Energy. [1/2](Mb + Ma)*Vf^2 + [1/2]Iωf^2 = 916.086J This value makes no sense because it is larger than the initial Kinetic energy of the system.

10. Fin