Dr Ian Plummer
Technical
Striking a Ball  Energy
and Momentum Conservation
In the following example David Turner considers a simple model of a mallet
head striking a ball. By considering the conservation of energy and
momentum he shows that the ball accelerates away from the mallet. In
real life the momentum of the mallet (as a whole) includes a contribution
from the striker's arms and body, and the striker is forcing the mallet forward. As
such, the calculated mallet velocity is likely a minimum.
Normally when a mallet strikes a ball the ball suffers an acceleration and
the mallet a deceleration. You can calculate the exact ratio of resultant velocities
if you know the relative masses of the ball and mallet. If the mallet weighs
three times as much as the ball then the calculation goes as follows.
In this model the following assumptions are made:
 Collisions are elastic and no energy is lost due to friction.
 The mallet has a massless and inflexible handle.
 The player causes the mallet to behave as a pendulum during the stroke
and there is no mometum from the player's body at the moment of impact.
The relevant equations are:
k = ½ m v²
p = m v
where
k = 
kinetic energy 
m = 
mass 
v = 
velocity 
p = 
momentum 
Bold quantities are vectors
The conservation laws state that the total energy before collision equals
the total energy after collision and, the total momentum before collision equals
the total momentum after collision. It is assumed that no potential energy
is involved and there is no rotation of the ball (leading to angular momentum
and energy) at the moment of impact, i.e.
1) k = k'
2) p = p' ,
where primes (') indicate variables after the collision. If we call the velocity
of the mallet v_{M} and the ball v_{B,} then
before collision:
3) k = ½ (3 m v_{M}²),
and
4) p = 3 m v_{M}
where m is the mass of the ball (remember the mallet is about 3 times the
mass of the ball, and is taken to be 3m). As the ball is at rest it has zero
energy and zero momentum.
After the collision:
5) k' = ½ (3 m v'_{M}²)
+ ½ (m v'_{B}²)
6) p' = 3 m v'_{M} +
m v'_{B}
(Only kinetic energy and linear momentum are considered in this analysis). Hence
from 1),
½ (3 m v_{M}²) = ½ (3 m v'_{M}²)
+ ½ (m v'_{B}²) , which simplifies to
7) 3 v_{M}² =
3 v'_{M}² + v'_{B}²
and from 2)
3 m v_{M} = 3 m v'_{M} +
m v'_{B} , dividing by m yields
8) 3 v_{M} = 3 v'_{M} + v'_{B}
so
9) v_{M} = (3 v'_{M} + v'_{B})
/ 3
use this value for v_{M} in 7) and expand and simplify:
3 ((3 v'_{M} + v'_{B})
/ 3)² = 3 v'_{M}² + v'_{B}² 
expand square 
3 ((9 v'_{M}² + 6 v'_{M} v'_{B} + v'_{B}²)
/ 9) = 3 v'_{M}² + v'_{B}² 
simplify fraction 
(9 v'_{M}² + 6 v'_{M} v'_{B} + v'_{B}²)
/ 3 = 3 v'_{M}² + v'_{B}² 
multiply both sides by 3 
9 v'_{M}² + 6 v'_{M} v'_{B}+ v'_{B}² =
9 v'_{M}² + 3 v'_{B}² 
subtract 9 v'_{M}² from each side 
6 v'_{M} v'_{B} + v'_{B}² =
3 v'_{B}² 
subtract v'_{B}² from each side 
6 v'_{M} v'_{B} = 2 v'_{B}² 
divide each side by v'_{B} 
3 v'_{M} = v'_{B} 
divide each side by 2 
and we obtain
10) v'_{B }/ v'_{M }= 3
or v'_{M }= v'_{B }/ 3
Hence the velocity of the ball after the collision is 3 times
the velocity of the mallet after the collision  this seems to be intuitively
about right to me. The relevance of this to the rules is that is normally impossible
to accelerate your mallet sufficiently to play through the ball you are hitting
and do the illegal 'push shot'. The question of whether the mallet is accelerating
at the moment of impact is irrelevant because all that matters is the instantaneous
velocity of the mallet at contact. To do the illegal shot you would have to
be moving the mallet very slowly and then accelerate it as much as possible
after the hit so that the mallet head catches up the struck ball shortly after
contact
We can use this result to calculate the ratio between the velocity of the
mallet before impact to the velocity of the strike ball after impact.
from 10)
v'_{M }= v'_{B }/ 3
We have neither generated or destroyed linear momentum, hence p = p'.
_{mallet momentum before = mallet + ball momenta afterwards}

3m v_{M} = 3m v'_{M} +
m v'_{B} 
divide by m 
3 v_{M} = 3 v'_{M} + v'_{B} 
subst 3 v'_{M}= v'_{B } from
10) 
3 v_{M} = v'_{B }+ v'_{B} =
2 v'_{B} 

Hence
11) 1.5 v_{M} = v'_{B}
Hence the struck ball moves away from the mallet at 1.5 of the velocity
at which it is struck  perhaps this will help to judge those gentle rushes
towards the hoop.
This I am sure you will agree is one of the simplest situations to analyse,
even if the maths may look complex. However there are lots of complex issues
to consider and I have made some assumptions, the main one being that it is
an elastic collision and no energy is lost during it (obviously untrue as a
loud bang is heard) and that the collision is instantaneous (also untrue 
e.g. an inelastic collision). The other factor that I have not considered is
the rotation of the ball. I do not believe that this affects the analysis above
but it does mean that the result is only true for the instant after collision
as the ball will start to lose velocity as it acquires angular momentum and
rotates due to friction with the lawn. This matter of the rotation of balls
is interesting. When 2 croquet balls collide without any cut, i.e. you hit
the roqueted ball dead centre, the striker's ball actually stops (if you apply
equations 1) and 2) it is easy to prove this):
If we consider a nonrotating croquet ball colliding along the line of centres
with a stationary one, the striker's ball stops after the collision.
A proof follows.
If we call the velocity of the striker's ball u and that
of the roqueted ball v, then before collision:
k = m u² / 2
and
p = m u
as the roqueted ball is at rest before it is hit.
After collision:
k' = m u' ² / 2 + m v' ² /
2
and
p = m u' + m v'
Hence from 1),
m u² / 2 = m u' ² / 2 + m v' ² /
2, dividing by m and multiplying by 2 yields
12) u² = u' ² + v' ²
and from 2),
m u = m u' + m v' , dividing
by m yields
13) u = u' + v'
substituting for u in 12)
(u' + v') ² = u' ² + v' ²,
expanding the square yields
u' ² + 2 u' v' + v' ² = u' ² + v' ²,
subtracting u' ² + v' ² from
each side we have
14) 2 u' v' = 0
There are 3 possible solutions to the equation, viz a) u' =
0 or b) v' = 0 or c) u' = v' = 0
Since kinetic energy and momentum must be conserved, solution c is invalid; u' must
be greater or equal to v' (otherwise the striker's ball has
somehow passed through the roqueted ball!) so solution b is also invalid.
Therefore:
15) u' = 0
and from 13)
16) u = v'
From this we can conclude that the striker's ball always stops if it
hits another ball dead centre and the roqueted ball moves off with the velocity
of the striker's ball before the collision. This is how you can have a double
hit if a ball is roqueted hard, very close to the striker's ball. The striker's
ball first leaves the mallet at 3 times the velocity of the mallet after the
collision, then stops when it hits the roqueted ball. the mallet is still moving
at 1/3 the velocity of the only ball that is now moving  i.e. the roqueted
ball. If the distance from the original position of the strike ball to its
position now, just behind the original position of the roqueted ball, is very
short then the player is unable to decelerate the mallet head before the mallet
hits the striker's ball again. If the distance is not too short a double hit
will be heard. If the distance is very short no double click will be audible
but it will have occurred.
The striker's ball only reacquires forward motion because of its rotation.
You can see this when a ball is struck hard against a hoop  it first bounces
back and then starts to carry on forward, usually taking a curved path back
towards the hoop.
Author: David
Turner
All rights reserved © 2000
