# A golf/baseball physics question, and an old thread

It makes the difference that a ball at 50 MPH relative to the ground hit by a bat swung at 1000 MPH relative to the ground will be traveling 950 MPH faster relative to the ground than a ball at 1000 MPH relative to the ground hit by a bat swung at 50 MPH relative to the ground.

Some of the reference frame arguments here are leaving out the fact that we’re interested in the behavior in the ballpark’s reference frame, so you have to transform your results back.

That aside, one can just calculate the final velocity of the baseball assuming a perfectly elastic collision. While the collision is not perfectly elastic in reality, some insight can be gained. The key is the mass difference.

Recall that in billiards, the cue ball has approximately the same mass as the object balls. If you hit an object ball dead-on with a spin-less cue ball, the cue ball stops dead and the object ball takes off. Now increase the density (and thus, mass) of the object ball by a factor of 10. The cue ball will nudge the object ball, but the cue ball will rebound. No initial speed can overcome this behavior. The faster the cue ball hits the extra-heavy object ball, the faster it will bounce back. To be sure: the object ball will move faster, too, but it needn’t move very quickly to balance the momentum of the reversed cue ball, and thus energy conservation can be preserved.

Returning to baseball, let’s assume the bat weighs 1 kg and is moving at 60 mph. It’s only a few lines of algebra to work out the final velocity of the ball assuming an elastic collision. Here’s a graph of the final velocity of the ball under different assumptions about the ball’s mass.

An infinitely heavy baseball clearly keeps plowing right ahead. A 10-kg baseball will be turned around if it’s moving slower than 13.3 mph or so. A ball that weighs the same as the 60-mph bat will rebound at 60 mph regardless of its initial velocity. Finally, a regulation ball rebounds faster and faster as its initial speed is increased.

In reality, the collision is not perfectly elastic and the batter in applying a force on the ball during contact – both very important considerations. Nonetheless, the elastic, unforced example here hopefully clarifies the main question in the thread.

But the bat and ball approach each other at the same speed: 1050 MPH. I can run into a brick wall at 10 MPH or the wall can run into me at 10 MPH and it makes no difference.

Not doubting you, but why would that be?

ETA: my intuition is that it would rebound at the combined speed of the ball and the bat.

Oops, 1/2 of the combined speed.

Thanks, Pasta. I should know better than to give off-the-cuff answers without actually doing the calculations.

The only way to conserve both energy and momentum in that case is for the ball to go forward with the speed the bat originally had, while the bat goes backwards with the speed the ball originally had. The rate at which the ball recedes from the bat is indeed the same as the rate at which they approached each other, but some of that (most of it, if the pitch was more than 60 MPH) is due to the motion of the bat, not of the ball.

What an excellent explanation, thanks.

You have a complete misunderstanding of COR. The COR gives speed at which something will return after colliding, relative to the original speed. Lets say the ball on bat had a COR of 0.5. If the ball hits an immovable bat when the ball is traveling at 50 mph, then the ball speed afterward would be 25 mph in the opposite direction. If you increased the ball speed to 100 mph, the speed afterward would also increase, to 50 mph. Of course the fact that the bat is moving and can change its velocity makes things more complex, but you seem to also be ignoring the force applied on the bat by the batter’s hands, which make simple momentum calculations impossible.

It certainly does make a difference, if what you are interested in is how fast you are moving after the collision. After the collision, that speed differs by 950 MPH between the two cases…

My ignorance has been fought in regard to this question. I was ignoring the energy of the incoming pitch, in the overall equation. I was thinking that some work had to be done on the ball to reverse its direction of flight, and said work would take away from the energy the bat could transfer to the ball.

I do understand COR, though. It’s basically just a “springiness” factorm, as in: what percent of an object’s speed will it maintain–in the opposite direction–after the collision. Both the bat’s COR and the ball’s COR come into play.

With the initial question having been settled–that the pitched ball will travel farther than the teed ball–what about the other question? Can you hit a (stationary, teed) golfball farther with a softball bat or with a golf club? The softball is at 6.8 ounces, and it goes, say, 400 feet at best. The golfball is at 1.6 ounces, and it goes, say, 400 yards at best. Does it not stand to reason that the softball bat would send the golfball farther than the golf club would?

I think I’m an idiot.

You can get a golf club going significantly faster in a single swing than you can a baseball bat. Which works better would be a very complicated question depending on many factors, and would probably be best determined by experiment. And as it happens, the experiment has already been done: If a big hunk of wood worked best for propelling a golf ball, then that’s what golfers would use.

The experiment is ongoing. Not that a big hunk of wood would be better, of course, but that different driver designs may be better. Could more mass in the clubhead increase distance under certain circumstances? Or less? Or more, or less, mass in the club overall (shaft included)?

Assuming that “golfers would play it if that were the case” is tantamount to “the experiment has already been done” is a reach. I’m not ready to give the club manufacturers that much credit.

The comparison to softball is just a thought experiment. I repeat the question: Who is providing more power to the ball? The softball player who propels a 6.8-oz ball 400 feet, or the golfer who propels a 1.6-oz ball 1200 feet?

When hitting with the bat, is the golf ball still on a golf tee, or is it in the baseball strike zone?

With a golf club, a spin is applied to the ball, which provides lift. The golf ball is dimpled to increase that lift. It’s possible that the ball hit with the bat will initially be traveling faster, but that the ball hit by the golf club will travel farther. For the baseball case, we were mostly ignoring the effect of spin (see RealityChuck’s post for the exception).

I’m assuming ideal conditions, meaning the softball is on a perch right in the dead center of the hitting zone as the batter likes it.

I’m interested in the forces and resultant velocities, rather than any exceptions to the rule.

I couldn’t get your link to work, but what your saying does not make sense. A batted ball gets its spin from the interaction with the bat, it does not continue spinning the way the pitcher threw it.
If the ball strikes the top half of the bat, the ball will have backspin. If it hits the bottom half of the bat, it will have topspin.

But the original spin that the pitcher puts on the ball affects the interaction with the bat (if you don’t believe me, get a tennis or ping-pong player to hit a few balls to you with varying top and back spin, and see how they bounce differently). Since bats are moving pretty much horizontally directly against the ball, I’d think the spin from the pitcher could have a significant effect on the spin after being hit by the bat.

This demonstrates my point though. How does a ping pong player put topspin on a ball? By angling the paddle so that it strikes the top of the ball. He creates backspin by angling the paddle so he hits the bottom of the ball.
If you strike a round object off center, the object has little choice but to spin the direction the side that was hit.

I think the force of the spin put on the ball by the pitcher is insignificant when compared to the forces applied to the ball by the bat.
If a baseball is moving toward the plate at 100mph, and the bat only contacts the bottom half of the ball, clearly the ball must end up with backspin. The top of the ball ‘wants’ to continue towards the plate, while the bottom of the ball ‘tries’ to move away from the plate.

cant believe I’ve read thru all of the responses…