Re: How does a bumper car work that are ridden at amusement parks?

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