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

Date: Wed May 3 09:13:23 2000
Area of science: Engineering
ID: 956108895.Eg
Message:
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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
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
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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