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Dr Ian Plummer

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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:-

  1. Collisions are elastic and no energy is lost due to friction.
  2. The mallet has a massless and inflexible handle.
  3. 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 vM and the ball vB, then before collision:

3)     k = ½ (3 m vM²), and
4)     p = 3 m vM

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'M²) + ½ (m v'B²)
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 vM²) = ½ (3 m v'M²) + ½ (m v'B²) , which simplifies to

7)     3 vM² = 3 v'M² + v'B²

and from 2)

3 m vM = 3 m v'M + m v'B , dividing by m yields
8)     3 vM = 3 v'M + v'B

so

9)     vM = (3 v'M + v'B) / 3

use this value for vM in 7) and expand and simplify:-

 
3 ((3 v'M + v'B) / 3)² = 3 v'M² + v'B² expand square
3 ((9 v'M² + 6 v'M v'B + v'B²) / 9) = 3 v'M² + v'B²  simplify fraction
(9 v'M² + 6 v'M v'B + v'B²) / 3 = 3 v'M² + v'B² multiply both sides by 3
9 v'M² + 6 v'M v'B+ v'B² = 9 v'M² + 3 v'B²  subtract 9 v'M² from each side
6 v'M v'B + v'B² = 3 v'B² subtract v'B² from each side
6 v'M v'B = 2 v'B² 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 vM = 3m v'M + m v'B divide by m
3 vM = 3 v'M + v'B subst 3 v'M= v' from 10)
3 vMv'B + v'B = 2 v'B

Hence

11)    1.5 vMv'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 non-rotating 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 re-acquires 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


Updated 28.i.16
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