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Hitting a golf ball at a brick wall

  1. Feb 17, 2010 #1
    My friend and I have been discussing this for days only to discover we're completely ignorant of the forces involved:

    If I hit a golf ball at a brick wall that's, say, 5 feet away, will the golf ball take more damage from the wall, or from the club?

    Please, help me settle this lest I lose another night's sleep.
     
  2. jcsd
  3. Feb 18, 2010 #2

    russ_watters

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    Damage? Golf balls are highly elastic over an extremely wide range of deformations. The only damage a golf ball takes is superficial scratching due to the surface roughness of the wall.
     
  4. Feb 18, 2010 #3
    Maybe golf ball wasn't the best example, or maybe damage wasn't the right word or maybe scale everything up.

    My question (rephrased) is does a golf ball take more shock when it gets hit from a resting position, or when it hits a wall hard enough to cause it to bounce in the opposite direction?
     
  5. Feb 18, 2010 #4
    The damage or shock a golf ball gets, depends upon the total impulsive force applied on it. The total force is directly proportional to rate of change of momentum of the ball.
    So, when hit in a wall, the momentum changes from mv to m*-v hence total change is 2mv
    When struck by a bat, the balls velocity changes from 0 to some value say V'. So the total change would be (0 + mV') = mV'. if V' and v were same then the ball would get more damage on striking the wall. That means, had it been made of glass, there is high chance that it breaks when hitting the wall.
     
  6. Feb 18, 2010 #5

    Matterwave

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    The impulse is a good measure of the "shock". But I'd say that the force is even better (after-all, if the impulse is spread over a longer time, there are less forces, this is analogous to breaking your fall by rolling). In this case, you need to know for how long the ball was in contact with the club or with the wall.

    Using your variables, the average force on the ball from the club then is mV'/t' while the force on the ball from the wall is 2mv/t.

    Where t is the time the ball spent in contact with the wall, and t' is the time the ball spent in contact with the gold club.
     
  7. Feb 19, 2010 #6
    If the ball took more shock from the wall when it bounced in the opposite direction, you could place another wall opposite and the ball would continue accelerating and bouncing back and forth untill the ball destroyed it's self or the wall.
    So more force must be applied when it is first hit.
     
  8. Feb 20, 2010 #7
    I'm afraid that's not true, bare in mind, it takes ALOT of force to mereley stop the ball! Never mind reversing it's direction!

    Numerically, if I hit a golf ball at 1m/s over 0.01 s, the force i've applied is mv/t = 100m Newtons, now if the wall reverses it back at me over 0.01 s at 1m/s as well, it's applied 100m newtons to stop the ball, and a further 100m newtons to reverse it's direction of travel. (200m newtons in total)

    So you see, assuming very simplified collisions (they both take the same amount of time in contact) the force of the wall will apply as much force as you have to stop the ball, and more if it comes back off the wall, understand?
     
  9. Feb 20, 2010 #8
    Does the ball therefore come back off the wall with more momentum than when it was first hit.
    If it does, what is there to stop you building a perpetual motion machine, as it would appear you are getting something for nothing.
     
  10. Feb 20, 2010 #9
    No, of course it doesn't.

    Bear in mind we're dealing with velocity here, not speed!

    So as in my earlier example, if i'm a bit more explicit (assuming m=1);

    You've applied a force of 100N to the golf ball. (Assuming the direction towards the wall is positive).

    However the wall applies -100N (i.e, back off the wall) to STOP the ball dead, and if the ball rebounds the wall will have exerted a further -X Newtons. (However of course, as you said, the speed will depend on the value of restitution for the wall, and it cannot rebound with a greater speed than it hit the wall with).

    However you can't ignore the assumptions here, you're assuming the collision with the golf club is exactly the same as the collision with the wall, so for example, the ball hits each for 0.01 seconds. This is paramount and can't really be simplified further.
     
  11. Feb 20, 2010 #10

    rcgldr

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    Assuming that the golf club and wall experience insignifant deformation during the collision with the golf ball, and assuming an elastic golf ball, then ...

    update

    Looking up some actual number, the collision between club and golf ball, results in the golf ball having about 1.5 x speed of club from collision, plus 2400 -> 3000 rpm, a significant increase in angular kinetic energy. The club's mass isn't huge compare to the ball, so it's speed is reduced significantly by the collision. The wall's momentum is huge compared to the ball, so almost all of the velocity change occurs with the ball in wall + ball collision.

    To simplfy matters, you could assume a 100% elastic ball, and compare the collision bewteen a block with the same effective mass as club, sliding on a frictionless surface, colliding with the ball, then the ball later colliding with a wall at the end of the frictionless surface.
     
    Last edited: Feb 20, 2010
  12. Feb 20, 2010 #11
    I am assuming the collision with the golf club takes slightly less time because the ball cannot rebound with a greater speed than it hit the wall with.
    So the maximum amount of force applied for the shortest period of time was when the ball was first hit.
     
  13. Mar 17, 2010 #12
    I think that the wall will take more damage than the golf ball as the golf ball is surrounded by an elastic coating that will reduce the damage to the ball. But there might be few scratches on the ball. It also depends on the http://www.RepairGolfSwing.com" [Broken] and the force at which you are hitting the ball and the force at which the golf ball hits the wall.
     
    Last edited by a moderator: May 4, 2017
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