3
$\begingroup$

Conservation of momentum: A thought experiment.

A baseball is placed on top of a baseball holder, the kind used to train young batters. A batter hits the stationary ball perfectly horizontal, sending it flying through the air in a relatively straight line, during which time it will eventually fall to the ground due to Earth’s gravitational pull. Ignoring everything but momentum, the total distance traveled by the time it hits ground is, let’s say 200 feet.

Now, a second baseball is instead dropped down in front of the batter from a certain height, straight down, at which time the player again hits the ball squarely on its side, not at an up or down angle, but perfectly horizontal, sending it flying in a relatively straight line through the air.

Will this second ball travel the same distance as the first one, and if so, what happened to the downward momentum the ball obtained before it was hit by the batter? Alternatively, does the initial downward momentum in fact affect the distance traveled?

My friends say that the initial downward momentum is most likely converted into heat by friction with the bat, rendering it incapable of affecting the trajectory of the struck ball. I say the initial downward momentum of the dropped ball must be conserved and that it does in fact make the ball fall faster to the ground, thereby shortening the distance traveled. Any thoughts would be greatly appreciated.

$\endgroup$
  • $\begingroup$ You have it right. $\endgroup$ – garyp Oct 3 '15 at 19:36
  • 1
    $\begingroup$ The "baseball holder" is called a tee, for what it's worth :) $\endgroup$ – hobbs Oct 3 '15 at 22:54
4
$\begingroup$

Provided the bat delivers exactly horizontal momentum impulsively to the second ball, it will not travel as far due to its initial downward velocity, as you say. Dissipating the downward momentum doesn't make much sense in the scenario you described.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Purely semantic, but would it be better to consider the downward momentum rather than velocity of the baseball for our purposes? Just wondering. $\endgroup$ – MildBubby Oct 3 '15 at 19:41
  • $\begingroup$ Any moment of friction between the bat and the falling ball isn't going to stop the ball's downward motion. At most, since it's only on one side of the ball, it'll torque the ball, adding spin, and slightly decrease the downward momentum. $\endgroup$ – DarenW Oct 3 '15 at 19:51
  • $\begingroup$ By "impulsively", you mean "instantaneously"? If so, that's not what happens in the real world. If you watch a slow-mo film of a bat hitting a ball, you'll see that the contact lasts for quite some time and the ball is quite dramatically deformed. (See, for example, the "compression of the ball" section of The Physics of Baseball.) $\endgroup$ – David Richerby Oct 3 '15 at 23:52
  • $\begingroup$ @DavidRicherby I do mean instantaneously. I know a real hit takes some time, but we're already talking about "perfectly horizontal", which only makes sense for a moving ball if contact is instantaneous at a point. Can't always have it both realistic and idealized... I guess the interpretation is "my argument is approximately valid as long as contact is close enough to instantaneous", where how close is close enough depends on how valid you want it to be. $\endgroup$ – Kyle Oman Oct 6 '15 at 2:50
1
$\begingroup$

When you hit the falling ball, friction between bat and ball will momentarily stop the side of the ball that is hit from moving down. However, since this force of friction is not applied at the center of mass of the ball, this will not result in a complete arrest of the vertical motion: instead, the ball will acquire some spin. However, it will not lose all its vertical momentum - although it will lose some.

Let's consider three cases. A - stationary ball is hit; B - ball is dropped and hit; C- ball is dropped. From the above we conclude that ball B will arrive at the ground sooner than ball A; but slower than ball C.

I ignore for the moment the fact that the "top spin" that was imparted on ball B will actually result in an additional force (the Magnus effect) that will pull the ball to the ground more quickly: for the right ratio of ball size (diameter, mass), impact, and air density, it is conceivable that this might make ball B reach the ground before ball C.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Let's try an easier thought experiment. Get a flat bat (since it's easier to visualize, not because it changes anything), and instead of swinging it, hold it still. From a distance above it, drop the ball. Now, the key is to hold it at such an angle that the ball will travel horizontally when it hits the bat. You will see it bounce off with a forwards velocity, and the ball will have conserved momentum.

Now, repeat the experiment but put in a massive swing, but keep the angle you had before. You can intuitively predict that the ball will now not go horizontally, but will go up as well - not what you wanted! (1)

The other side of the spectrum is holding the bat perfectly vertical. You will now see that the ball will start with a bit of a velocity downwards (since there is no upwards force stopping it), and will not travel horizontally but slightly downwards as well. Not what we wanted either! (2)

The key is to hold the bat at an angle somewhere between (1) and (2), such that you actually get the same effect as the 'bounce' at the beginning of this answer. Just like with the 'bounce', the downwards momentum will be converted to forwards momentum, and so all momentum is conserved.

| cite | improve this answer | |
$\endgroup$
-2
$\begingroup$

The batter in both cases is connected to the Earth. Momentum is conserved but you have to add the amount given to the Earth--which is impossible to detect but it is there. If you drop a ball and it lands on the Earth and stops, momentum has not disappeared, it is transferred to the Earth. The answer just above notes that the bat will give some spin to the ball. This is angular momentum. The batter, and the Earth, will get an equal amount of angular momentum in the opposite 'direction'. This will not be detectable. The suggested answers above ignore the Earth and are incorrect.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Well, it's an attempt at an answer, it's just not correct. $\endgroup$ – David Z Oct 4 '15 at 22:22

Not the answer you're looking for? Browse other questions tagged or ask your own question.