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Dear No Name Entered, There are two major physics actions that occur with bumper cars. One is how they react when they strike another car and the other is how they are powered in the first place. I'm not sure which you are looking for so I will try and give insight to both. (And I will try to be brief as I tend to be long winded!) First, one of the interesting things about bumper cars is how they get power. For one thing, they are one of only a few rides that have three full degrees of freedom. This means they can move in three directions independently; Front and back, left and right, and rotate. Try to think of any other rides (bumper boats is one, but kind of like cars) that you have full control of three different movements at once! This makes it difficult to get power to the motor for your power. A tether (a rope with an electric cord) can get tangled with other cars. Following an electric track doesn't give the driver independent control. Batteries is an option but heavy, expensive, and frequently need recharged. So what is done for most (and you have probably seen this) is have a long metal tail on the car that pushes next to the roof of the bumper car area. The roof is electrified and the end of the tail usually is a spring to maintain contact with the roof. This is where your propulsion (your movement) power comes from. What is interesting is that there is also electricity at the floor of the bumper cars room as well. (That is why most are metal). You might wonder why you don't get shocked when you walk across to the cars? For one, they usually cut power when people leave and get on. But even if you fall out when moving you won't get shocked. The reason is the floor has electric flow but no potential. Potential means the ability to do work. A rock held above your head has potential. It can smash a can, break your toe, splash water. All of those are examples of work. Water can have potential as well. Water held above a water wheel has the ability to turn the wheel when falling to a lower pool. Electricity works much the same way. Electric potential is called voltage. Electricity can do work (turn a motor, light a bulb) when going from a higher voltage to a lower one. It is convenient (and safer) to go from a high voltage to a voltage of zero, or ground. This is the potential of the earth, which is the lowest voltage or potential. Kind of like the ocean is for water. The floor of the bumper cars has the same potential as the earth (ground) and the same potential as you. That is why you don't get shocked! However, there are electrons (electricity) flowing under you from the car motor to the earth. If the didn't your car wouldn't run. It is like having a water wheel that dumps into a tank. Soon the tank fills up and the water has no where to flow. No flow, no wheel movement, no work gets done. So like the water flowing off in the river, the electricity flows from the high voltage ceiling to the motor then to the floor and off to the earth. The only way you could get shocked from the floor is to provide an easier path from the floor to the earth (unlikely since you have high resistance compared to metal) or if you provided the ONLY path (If the ground line to the earth broke)! That would be bad! (Reference Me on the electric part. Most is off the top of my head). The other fun thing with bumper cars is the collisions. This has to do with Newton's Second Law of Motion. Translated, it states that "the rate of change of momentum of a particle is proportional to the net force applied to the particle". What this means is the famous F=ma equation, where force is equal to the mass of a particle times acceleration. Acceleration is the rate of change of velocity, so momentum is mass times velocity. In straight line motion, momentum P is equal to mass of the particle times its speed, P=mv. If there are no external forces, then the change of momentum is zero, therefore momentum is constant. This is helpful in dealing with two particles that collide with little or no external forces. This is similar to the bumper cars if we ignore the motor thrust during impact (which is reasonable since they have a clutch system, or every time you hit someone you could break the motor). The equation that governs the conservation of momentum is as follows: m1v1 + m2v2 = m1v'1 + m2v'2 where: m1 is mass of first particle m2 is mass of second particle v1 is velocity of first particle before collision v2 is velocity of second particle before collision v'1 is velocity of first particle after collision v'2 is velocity of second particle after collision Remember that Velocity is a vector, so direction matters (your math and physics teacher can help you with vectors if you need additional help). We will stick to straight line collisions so we won't worry about direction. Also, you can only use this equation for certain collisions as there are usually two unknowns (v1' and v2') but only one equation. Another equation we can use is the conservation of energy. Ignoring losses from impact to sound and heat, the energy two objects have before an elastic (where the objects bounce) impact is the same as the energy after impact. This equation is: ½ m1v1² + ½ m2v2² = ½ m1v'1² + ½ m1v'2² Where the variables (m1, m2, etc…) are the same as the first equation. Now it should be obvious there would be quite a bit of algebra to figure things out, but it is doable. If you know all of the initial conditions (mass and speed of both bodies before collision), solve the first equation for one of the final velocities in terms of the other final velocity, then plug that equation into the second equation. A little complicated and a teacher can help you with it. I will give you an example of some special conditions though. Say you are in a bumper car that weighs mass m1 and you are traveling at velocity v1. You directly strike (as opposed to a glancing blow) a bumper car of mass m2 and isn't moving. Ignoring friction and other losses (wheel direction, etc..) and using both the above equations, the equation for the velocity of both cars after impact works out to be: v'2 = v1(2m1 / (m1+m2)) v'1 = v1((m1 - m2) / (m1+m2)) Say your car weighs 150kg and the other car is 100kg and you are traveling at 0.5 m/s. (Your units need to be consistent!). Then the second cars final velocity is then v'2 = 0.5(2*150 / (150 + 100) = 0.6 m/s I'll let you solve for v'1! The following website has some info and graphics on bumper car collisions. Not a great site but kinda cool for showing the physics of amusement park rides. http://www.learner.org/ exhibits/parkphysics/ I hope this wasn't too complicated and gave you some of the info you needed. If you are struggling with the math or have other questions on collisions, feel free to e-mail me privately at bradk@jymis.com. BK Momentum reference: Physics for Scientists and Engineers: 2 Edition by Douglas C. Giancoli

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