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Name: Greg Y.
Status: student
Age: 30s
Location: N/A
Country: N/A
Date: Tuesday, May 21, 2002

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?


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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