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Golf Ball Speed
Name: Greg Y.
Status: student
Age: 30s
Location: N/A
Country: N/A
Date: Tuesday, May 21, 2002
Question:
I understand from conservation of momentum that after a
collision, the total momentum of a system will be the same as before the
collision. So, say you have a golf club hitting a stationary golf
ball. It would seem to me that the maximum speed of the golf ball after
the collision would be the speed of the golf club head which just struck
it. In terms of momentum, the maximum momentum of the ball would be
limited by the speed of the club head as follows: max_momentum_of_ball =
Mass_of_ball * Velocity_of_club_head and the final momentum of the club
would simply be whatever is left over thus conserving momentum. Is this
correct, or can the ball actually be given a higher speed than the club
head? If so, then would you please explain why/how?
Replies:
Greg,
Having the same momentum in no way means having the same speed. This is why
hitting a ball with a heavy-headed golf club sends it flying further than
using a light-headed golf club. Another factor in striking a golf club is
the pushing due to your hands. Momentum is lost from the golf club,
transferred to the ball. The club/ball system gets a little extra momentum
from the golf player's hands during the short time the club and ball are in
contact. The majority is due to momentum from club to ball. To find an
approximation of the momentum lost from the club, calculate the loss of club
velocity during the collision. Multiply this by the club-head mass to yield
lost momentum. This momentum transfers to the ball. Now divide by the ball
mass to get an approximation of the ball velocity.
The problem with this calculation is measuring how much velocity the club
loses during the collision. What happens in the "follow-through" does not
directly affect the ball. It is possible that a club loses only 5-10% of
its original velocity while in contact with the ball.
Dr. Ken Mellendorf
Physics Instructor
Illinois Central College
Conservation of momentum does not give you enough information to solve
this problem. You have to consider energy also. Here is how it works:
Before the collision:
p = Mc * V0
T = (Mc * V0 * V0) / 2
where Mc is the mass of the club, and V0 is its speed.
After the collision:
P = Mc * Vc + Mb * Vb
T = (Mc * Vc * Vc + Mb * Vb * Vb) / 2
where Vc is the club speed, Mb and Vb are the mass and speed of the
ball.
These equations can be solved together to give the following expression:
Vg = V0 * 2 * Mc / (Mc + Mb)
So, if the ball is very light compared to the club, it will leave with
nearly twice the original club speed.
Tim Mooney
There are a number of variables here, but in a perfect world the momentum of
the ball plus the momentum of the club would be equal to the momentum of the
club before the collision. Another way of looking at it would be that loss in
momentum of the club due to the collision would equal the momentum of the
ball.
Momentum = mass times velocity
Confounding variables would be friction of the air, increase or decrease of
muscular involvement, and the portion of the energy that is converted to heat
in the collision. Likely there are more. If a robot were playing a round of
golf in a vacuum in an environment where heat remained constant in a
collision.... the problem would be easy.
Larry Krengel
The speed of the ball should be greater than the speed of the club head
at the time of the collision. This is because the ball is elastic.
When it is struck by the club head it will deform slightly and then
rebound to it's original shape, while still in contact with the club.
The return to its original shape imparts a slight additional speed to
the ball.
If the ball deformed but did not return to its original shape the
maximum speed it could attain would be the speed of the club.
You can model this using a basketball, a bread sack, and a book. Move
the book at a constant velocity and strike the basketball. It will
rebound from the book and move faster than the book was moving (try
experimenting with different speeds).
Now try using the bread sack as an air bag between the book and the
basketball and hit the ball with the book moving at the same speeds.
This may take some experimenting, What you want is to have the bread
sack filled with air but 'tied' loosely enough that the air can escape
(but not too quickly) when you compress it. This is a plastic
compression. Part of the energy of the collision goes into deforming
the bag and forcing the air out of it instead of into deforming the
ball. You should find that the ball does not move away as quickly. With
a "perfect" air bag the ball would stay with the book so that you were
pushing it along with the book instead of it bouncing away.
Greg Bradburn
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