Conservation of Momentum and Energy
I have been learning about Momentum and Kinetic Energy in
science lessons at school, but there is one thing I really cannot
get my head around. I have been taught that if an object with a
certain momentum (mv) hits another object and all the momentum is
conserved, then the new object will have the same momentum, so if
the new object had half the mass of the first it would travel at
double the speed, e.g. 100kg object travelling at 10m/s hits a 50kg
object, so the 50kg object will go at 20m/s (50 * 20 = 100 * 10).
What confuses me is when I try to work out the Kinetic Energy of the
two objects. If the 100kg object is travelling at 10m/s then it has
kinetic energy of 5000J (100/2 * 10^2), but if it hits a 50kg object
and all the momentum is conserved then the new object will have
kinetic energy of: 50/2 * 20^2 = 10000J. I was under the impression
that energy could not be created or destroyed, so how does this
100kg object travelling through a vacuum with 5000J of energy
suddenly hit this 50kg object and produce 10000J of energy, where
does the extra energy come from?
You are right, the kinetic energy is conserved,
and your example is scientifically considered impossible.
The collision would not happen like that.
The combined energy of the two objects cannot be larger after than before.
It can, however, be less, if there is inelasticity (dull bounce).
The combined momentum cannot be larger or smaller.
It must be exactly the same before and after.
But it is the sum that is conserved, the net momentum.
Notice that usually there is momentum in both objects after a collision,
and you need to find and use both velocities
to add up the net momentum.
And direction matters.
In X, Y, and Z directions there can be + or - momentum for each object.
Momentum like velocity is a vector business.
For kinetic energy, direction does not matter.
The kinetic energy always has a "plus" sign.
Squaring the velocity sees to that.
Energy is a scalar quantity.
The usual example of when all momentum of A is transferred to B, is:
when the mass of A and B are equal to each other and the bounce is perfectly elastic.
In that case A can be stopped after starting B in motion.
This 100% momentum transfer can also happen for just the right combinations of:
A lighter than B, and a corresponding percentage of inelastic energy loss.
Both of these are contradicted by your A=100kg & B=50kg.
(If A is heavier than B, then A always keeps going forwards, slower than B, after the collision.)
You need to find the velocities of both using different formulas than you used.
The usual method:
a) find the velocity of a flying observer who would see net zero net momentum before the collision.
From this point of view, the two objects come in from opposite sides,
do a head-to-head collision in front of you,
then bounce back away in exactly the direction they came from.
b) put yourself in that observer's shoes: subtract that observer velocity from vA and vB
c) now momentum of A = -1 times momentum of B. Check that.
d) find the sum of kinetic energy now (it's less, it's minimized)
e) multiply the kinetic energy sum by the elastic efficiency,
i.e., 53% (0.53), or 100%(1.00), or some such.
You need to have a presumed elastic efficiency to calculate the results.
f) now both velocities are reversed in sign
and multiplied by the square-root of the efficiency (taken as 0.0-1.0)
g) still momentum of A = -1 times momentum of B, but both are smaller than before.
h) return to your original observation point,
by adding the observer velocity in (a) back into vA and vB.
That is not exactly a mathematical formula.
Hope it is not too hand-wavy for you.
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Update: June 2012