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It seems impossible, yet I'm thinking that maybe because the ball compresses against the bat a bit it acts a little like a spring, and DOES travel faster than the bat?


EDIT: This is just a clarification, and not really part of the question, but I think it may be valuable. For people saying momentum is conserved, I'm not sure what you are imagining, but take a moment to think about the equation you keep on mentioning: $$M_\textrm{bat}V_\textrm{bat} = M_\textrm{ball}V_\textrm{ball}$$ This is saying that the bat somehow transfers ALL of its momentum to the ball.

The only way this can ever happen is if the bat comes to a dead stop when it hits the ball, somehow holds it in place while transferring ALL of its momentum to it (that phrase doesn't even make logical sense), and then the ball flies off at a much larger velocity.

WERE the bat floating through space and struck a ball, the bat would not stop when it hits the ball, and there is no way it makes sense for the ball to go off at THAT much of a larger velocity. Just imagine a spaceship very slowly drifting through space, and an astronaut who suddenly touches it. Conservation of momentum DOES NOT mean $$M_\textrm{ship}V_\textrm{ship} = M_\textrm{astronaut}V_\textrm{astronaut}$$

According to that, the astronaut would shoot off at hundreds (maybe thousands) of kilometers per hour when being touched by a massive spaceship, and the spaceship would come to a stop, but obviously that doesn't happen. TOTAL momentum is conserved, so that $$M_\textrm{ship1}V_\textrm{ship1} + M_\textrm{astronaut1}V_\textrm{astronaut1} = M_\textrm{ship2}V_\textrm{ship2} + M_\textrm{astronaut2}V_\textrm{astronaut2}$$

but even THAT equation doesn't even apply in the case of the baseball strike since there is a human being providing a force EVEN as the bat hits the ball.

I know this is ingrained deeply in the minds of physics students because we have conservation of momentum drilled into our heads as young students in introductory physics, but I encourage everyone to always think intuitively about physics scenarios before applying equations.

Anyways, I hope that was valuable. Cheers!

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    $\begingroup$ It is momentum that is conserved and transferred, not speed. momentum=mv so a large mass hitting a small one and transferring its momentum must give it a larger speed , from conservation law. $\endgroup$ – anna v Sep 19 '18 at 18:46
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    $\begingroup$ When considering the mass of the bat, you'd also have to factor in the mass of the batter. That's part of the platform that is holding the bat. $\endgroup$ – BillDOe Sep 19 '18 at 18:48
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    $\begingroup$ Ever see a batter bunting the ball? $\endgroup$ – Samuel Weir Sep 19 '18 at 18:56
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    $\begingroup$ Can be explained even by logic, no physics needed. Seeing how the bat makes a circular movement, the ball must be faster than the bat. Otherwise, having the ball fly forward-ish would be impossible due to the same-speed bat which "sticks" to it pushing it sideways along its circular path until it rolls off. $\endgroup$ – Damon Sep 20 '18 at 13:28
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    $\begingroup$ Drop a rubber ball onto the pavement. The ball bounces up; the pavement never moved. $\endgroup$ – Carl Witthoft Sep 20 '18 at 15:31

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For an ideal heavy bat, the ball moves faster than its point of contact with the bat. Here's why.

  • Suppose you swing the bat with velocity $+w$ and the ball comes in with velocity $-v$.
  • Work in the reference frame of the bat. In this frame the ball has velocity $-v-w$.
  • Since the bat is much heavier than the ball, and assuming the collision is elastic, the ball simply bounces off the bat as if it were a brick wall, ending up with velocity $v+w$.
  • Transforming back to your frame, the ball ends up with velocity $v+2w$.

This is indeed always greater than the speed of the bat. For example, if you hit the ball from a tee, so $v = 0$, then the baseball ends up going precisely twice as fast as the bat.

This can also be understood from a force perspective. If you think of the bat and ball as squishing during impact like tiny springs, then at the moment they're moving at the same speed $w$, there is a sizable amount of energy stored in the springs. As the collision ends, the springs release this energy, increasing the speed of the ball over that of the bat.

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  • $\begingroup$ Thanks knzhou, but I'm a little confused by your answer; if the bat is moving, but we switched reference frames to where it is stationary, then when you say the bat is moving with a velocity +w , isn't that just the same thing as saying v is greater. Where did the 2 come from? $\endgroup$ – Joshua Ronis Sep 20 '18 at 10:36
  • $\begingroup$ @JoshuaRonis I edited the answer a bit to try to clarify it, does it make more sense now? $\endgroup$ – knzhou Sep 20 '18 at 10:39
  • $\begingroup$ Ok, I see what you are doing, but I think that what you are forgetting is that if the bat is coming in with speed +w, and we are considering the frame of reference where the bat is stationary, your equation should actually be Vf = -(v+w) + $\endgroup$ – Joshua Ronis Sep 20 '18 at 12:33
  • $\begingroup$ @JoshuaRonis No, that’s accounted for, I’m just doing the transformations in a different order than you are. To do it your way: in that frame, the final speed of the ball is v+w since it just bounces off the bat. So when we transform back to the original frame we get v+2w, the same result. $\endgroup$ – knzhou Sep 20 '18 at 12:40
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    $\begingroup$ If you consider yourself in the reference frame of the bat, the ball is coming towards the bat at a speed -w. Then, the change in the ball's velocity would be +2w, but it would bounce off at a speed +w, not +2w.... OH BUT RELATIVE TO YOU IT IS +2w!!!! Ok, got it now. Thanks, awesome explanation! Its just a little unintuitive...at least to me. I think the only thing I would add is that we are assuming the bat is moving at constant velocity and doesn't loose momentum to the ball. $\endgroup$ – Joshua Ronis Sep 20 '18 at 12:49
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Yes. Consider throwing a ball at a bat which is held stationary: the ball is momentarily stationary but at all other times it is moving faster than the bat.

Now consider sweeping the bat towards an initially stationary ball: if the ball is not to stick to the bat, then it must be moving faster than it when it loses contact with it. (This case is identical to the one above with a different choice of reference frame of course.)

In neither of these cases have I taken proper account of conservation of momentum: the bat must change velocity slightly when it imparts momentum to the ball, so you can't hold it stationary or sweep it at a constant velocity in fact. But this change in velocity of the bat can be made as small as you like by making $m_\text{bat}/m_\text{ball}$ large enough so the argument remains true.


Why we can ignore the person holding the bat

In the comments there has been some discussion about whether the person holding the bat makes a substantial difference. They don't: they certainly can make a difference in detail and obviously are responsible for getting the bat into the right position, but their contribution to the change in velocity of the ball is small. To see this I'll take some numbers from this page (mentioned in the comments).

The ball has a mass of $m = 0.145\,\mathrm{kg}$ and its change in speed $\Delta v \approx 200\,\mathrm{mph}$ or $\Delta v \approx 90\,\mathrm{ms^{-1}}$. This means that the impulse delivered to the ball is

$$I\approx 13\,\mathrm{Ns}$$

Now, let's assume that the person holding the bat exerts a force equivalent to their whole mass on it (they can't do this for any length of time, and in fact they can't do it at all realistically, so this is a safe upper bound). If their mass is $100\,\mathrm{kg}$, then the force they are exerting is $100\,\mathrm{kg}\times 9.8\,\mathrm{ms^{-2}}\approx 981\,\mathrm{N}$. The ball is in contact with the bat for $7\times 10^{-4}\,\mathrm{s}$ ($0.7\,\mathrm{ms}$), so the impulse from the person holding the bat delivered during the collision is

$$\begin{align} I_h &\approx 981\,\mathrm{N}\times 7\times 10^{-4}\,\mathrm{s}\\ &\approx 0.7\,\mathrm{Ns} \end{align}$$

So, the impulse delivered by the human holding the bat, in the best case, is about 5% of total impulse: realistically it will be much less.

This does not show that the human does not affect things like the direction and detailed trajectory of the ball after it is hit: it does show that their contribution to the change in velocity of the ball happens almost entirely before the impact: their job is mostly accelerate the bat and get it into the right place.

It turns out that Dan Russell has a nice summary page, with references on how much the person holding the bat matters. The last two sentences from that page are:

Measurements and computer models show that the collision between bat and ball is over before the bat handle has even begin to vibrate and the ball has left the bat before it even knows the handle exists. Finally, experimental evidence comparing the effect of different grip conditions on resulting batted-ball speed conclusively shows that the manner in which the handle is gripped has no affect on the performance of the bat.

He has a lot of other useful information on the physics of baseball.

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    $\begingroup$ Note that the mass ratio is strictly controlled by the rules of baseball: balls at 0.145 kg, and bats at around 1 kg... $\endgroup$ – DJohnM Sep 19 '18 at 19:15
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    $\begingroup$ @DJohnM the bat is not an isolated system, it is being held by the batter who, in turn, is bracing himself against the earth. Within some reasonable limits one can consider m_bat/m_ball arbitrarily large. With sufficient force applied by the batter (and it's not an infinite force, as the ball has finite mass and compressibility) one can also consider the bat to not change velocity at all. $\endgroup$ – LLlAMnYP Sep 20 '18 at 11:23
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    $\begingroup$ @rob I'm not convinced the bat is isolated. I can guarantee that the ball would not go nearly as far if the batter were to let go in the instant before striking the ball. Also the ball and bat respond drastically different depending on if you hit the ball on a harmonic node (sweet spot) of the bat or not, implying there is some response to its fixed point i.e. the batter. $\endgroup$ – BlackThorn Sep 20 '18 at 15:53
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    $\begingroup$ @BlackThorn: Do you have references to properly-done experiments which demonstrate the first of these claims? There is a lot of myth and lore around this sort of thing. $\endgroup$ – tfb Sep 20 '18 at 16:57
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    $\begingroup$ @tfb So speed of sound in wood is < 5000 m/s. The batter's hands are about 1 meter away. For a shockwave to travel to the batter's hands and back, that is about 2 meters, or > 0.0004 seconds. acs.psu.edu/drussell/bats/impulse.htm claims contact time of 0.0007 s; so mass/grip of human interacting with ball during contact is plausible. $\endgroup$ – Yakk Sep 20 '18 at 19:24
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Consider the trivial case: bat not moving. The ball will bounce off the bat as if the bat were a wall. Obviously, in any form of elastic collision, after the bounce the ball will have a non-zero velocity. This is greater than the 0 velocity of the stationary bat.

Interestingly, the ball-and-bat system is similar to the ball-and-train system used to as an analogy when explaining gravity assists. If you are familiar with the gravity assist, then you can see that after the interface (collision) the projectile (spacecraft, ball) is moving faster than the collider (planet, train, bat).

Ball bouncing off train Image courtesy of NASA.

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    $\begingroup$ What a lovely thing to behold. +1. $\endgroup$ – Wossname Sep 21 '18 at 20:19
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According to newton's third law of motion, both the base ball and the bat experience equal force but unequal acceleration which is because of different masses. If acceleration is different then velocity is also different for both ball and bat. So, ball would travel faster than the bat.

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    $\begingroup$ I don't think this answers the question. This answers a very different question of why the recoil speed of the bat differs from the exit speed of the ball. The OPs question is how the bat can keep accelerating the ball after the ball has reached the bats speed and hence lost contact with it. $\endgroup$ – jacob1729 Sep 20 '18 at 11:16
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Yes, this happens due to the conservation of momentum in a collision.

$$p = mv$$

where:

  • $p$ is momentum
  • $m$ is mass of object
  • $v$ is the velocity of the object

By the Law of Conservation of momentum, momentum before and after the collision should be the same. Which is expressed like this

$$p_{Before} =p_{after} $$ Or in the baseball bat and ball scenario. Assuming all the energy or momentum of the baseball bat is transferred to the ball.

$$p_{Bat} = p_{Ball}$$ $$m_{bat}* v_{bat} = m_{batll}* v_{ball} $$

Assuming $m_{bat} > m_{ball} $

For Law of Conservation of Momentum to be followed

$\therefore v_{ball} > v_{bat} $

Although there are a lot of ways this could go. This is just one way, especially for collision of two objects where there is no guarantee that kinetic energy is conserverd (see inelastic collision). But let's say the collision is elastic. You would use

$$E_{K-Bat} = E_{K-ball}$$ $$\frac 12 m_{bat}* v_{bat}^2 =\frac 12 m_{batll}* v_{ball}^2$$ $$m_{bat}* v_{bat}^2 =m_{batll}* v_{ball}^2$$

It will still follow the relationship that I said about $v_{ball}$ always being larger than $v_{bat}$

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  • $\begingroup$ Hmm, the momentum of the ball would only equal the momentum of the bat if the bat suddenly stopped. It's not like the bat is flying to space and hits a ball and comes to a dead stop while the ball flies off with all of the bat's previous momentum. $\endgroup$ – Joshua Ronis Sep 20 '18 at 9:46
  • $\begingroup$ Well in that case, if all of the momentum is not transferred you could also say $p_{after} = m_{ball} * v_{ball} +m_{bat} * v_{bat-after} . Which means that the bat didnt fully stop $\endgroup$ – Zirc Sep 20 '18 at 10:50
  • $\begingroup$ In that case, the velocity of the ball is not guaranteed to be larger than the velocity of the bat. $\endgroup$ – Zirc Sep 20 '18 at 10:51
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    $\begingroup$ So for a normal ball and an unobtainum-neutronium alloy bat, by swinging the bat as slowly as a very strong robot can manage, the ball should exit the solar system. $\endgroup$ – Stian Yttervik Sep 20 '18 at 10:56
  • $\begingroup$ Exactly BUT ONLY IF ALL THE MOMENTUM is transferred. In my answer i have omitted something which is the momentum of the bat after the collision. Did the bat stop? Maybe, maybe not. Then again it all goes back to the formula p-before= p-after. If the momentum is not fully transfer, you have to calculate the momentum before and after of each object. $\endgroup$ – Zirc Sep 20 '18 at 11:01
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From what I understood, the ball and the bat are in contact for a very short period of time.Even if the bat is being accelerated by an external force , during impact(very shor period of time) its velocity would not have changed much due to the acceleration(there wont be enough time for the acceleration or force to change the bats velocity).So conservation of momentum can be applied(i.e. the external force can be ignored).After impact, the bat would not move much(due to the collision) because of the external force by the person.But actually there will change in velocity of the bat due to collission.If you hang a bat by a rope and throw a ball at it, I think the bat will move significantly.

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  • $\begingroup$ The duration of the impact is not really important: it's the integral $\int F\cdot dt$ that determines the change in momentum. If the ball picks up momentum, the bat loses an equivalent momentum. But since the ball is lighter than the bat, the bat gains more speed than the bat loses. $m_1 \Delta v_1 = -m_2 \Delta v_2$. $\endgroup$ – Floris Sep 21 '18 at 21:34
  • $\begingroup$ The bat is experiencing an impulse due to the collision (as you have mentioned).The force I have mentioned is the force applied by the person holding the bat on the bat.This is an external force.To apply momentum consersvation, there should not be any external forces to the system of bodies.Here since the collision time is very small, the effect of the external force can be neglected.So conservation can be applied.The effect of the external force on the bat before,afteer the collission is immaterial(except for the velocity attained just before collission). $\endgroup$ – Mohan Sep 22 '18 at 1:18
  • $\begingroup$ Ah OK - I misinterpreted what you said. Yes, the time during which the externally applied force is acting is small enough (and the force small enough) that it is unlikely to affect the motion of the bat relative to the ball. $\endgroup$ – Floris Sep 22 '18 at 2:17
  • $\begingroup$ @Floris your analysis is flawed, you can hit a heavier ball with a lighter bat and still accelerate the ball to a speed faster than the bat. Why? Presumably because you're not just swinging the bat and leaving it, you're constantly accelerating it through the swing and during the contact and deformation (of bat and ball - watch slomos it's quite surprising how long the contact is and how much deformation there is, even of a "rigid" bat). Now, if you throw the bat, you're right you can just look at momentum of the bat vs ball. $\endgroup$ – pbhj Sep 22 '18 at 21:08
  • $\begingroup$ @pbhj - no, if a heavier object hits a lighter object with an elastic collision, the light object ends up going faster than the heavy object. “External force” (hand holding bat) is not needed. $\endgroup$ – Floris Sep 23 '18 at 16:47
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As others have pointed out, YES the ball can/will have greater speed.

In order to understand this, you need to consider the compressibility of the objects.

  • You swing the bat towards the ball,
  • the bat bends on impact,
  • the ball deforms.

Kinetic energy from ball and bat is absorbed into this deformed ball/bent bat duo system. The ball is accelerated to the bat speed, and the bat has slowed down a tiny amount - here is where "idealized" momentum has played its' role. Now it is time for elastic relaxation to do its' job.

  • As the ball reforms itself...
  • the bat bends back
  • an impulse is imparted to both, and this impulse is (has to be) like and opposite.

This impulse speeds the ball up and slows the bat down. Equal shares of momentum to both. This affects the ball's speed a lot more than the bat's. The bat and the ball are NOT behaving "ideally" - as a solid steel bat hitting a solid steel ball would behave very differently, as would hitting a rubber ball with the same steel bat. The last case would go out the park with much less trouble than the first.

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    $\begingroup$ "and the bat has slowed down a tiny amount" - not necessarily. You can hold a bat so it falls stationary when the ball hits, but in a normal swing-through, as in baseball or cricket, you're applying force to accelerate the bat, so the acceleration will reduce, but not necessarily the speed. $\endgroup$ – pbhj Sep 22 '18 at 21:03
  • $\begingroup$ @pbhj You are right, I did not consider continuous acceleration, it was not necessary for explaining. I am unsure whether it improves the explanation. If your comment gets more upvotes than just mine, I'll make the effort :) $\endgroup$ – Stian Yttervik Sep 23 '18 at 5:21
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Yes, if the ball is traveling at less than its terminal velocity after being hit and if you hit the ball off a sufficiently tall cliff. The ball will be effected by gravity and will accelerate toward the earth eventually reaching its terminal velocity (the point where air resistance equals the force applied by gravity).

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If the ball would not at any point of time travel faster than the bat, the bat would fly at least as far as the ball. So you need to better qualify what you mean by "ever" here. Obviously you are only thinking about a limited time span.

Or you mean something like "does the ball attain some speed faster than what the bat reaches at any point of time?"

Which is trickier since a baseball does not appear all that elastic: for an elastic collision, obviously the smallest object will achieve the largest speed due to conversation of momentum. For a completely inelastic collision, the speed of all objects after collision will be equal and less than what the hitting object had before the collision.

My guess would be that a baseball is elastic enough to fly further than the bat would even assuming you let go of it with the same inclination as a perfectly hit ball.

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You may want to consider the bat hitting the ball as a free collision between two objects. You have (1) momentum conservation: $$ \vec{p}_{bat}^{before} + \vec{p}_{ball}^{before} = \vec{p}_{bat}^{after} + \vec{p}_{ball}^{after} $$ And (2) energy conservation. $$ E_{bat}^{before} + E_{ball}^{before} = \lambda \left(E_{bat}^{after} + E_{ball}^{after}\right) $$ where $\lambda$ refers to the elasticity of the collision. If $\lambda = 1$ you have a perfectly elastic collision, if $\lambda = 0$ the collision is perfectly inelastic. (Please note that the equation above is only valid in the center of gravity frame of reference.) You might even have superelastic collisions with $\lambda > 1$ if you have some kind of an energysource like an explosive charge or something.

The astronaut that gets hit by a heavy spaceship is a collision of $\lambda \sim 0$, hence the astronaut moves at the same velocity as the spaceship after the collision.

In this picture the bat hitting the ball is a collision with $ 0 < \lambda < 1 $. This may result in a speed of the ball higher than the speed of the bat.

Most likely the momentum however is NOT conserved in the case of the bat hitting the ball. Momentum conservation is a result of the "homogeneity of space". This homogeneity of space however is violated by a person standing at a specific spot in space holding a bat. Or in other words: the bat is connected to a person who is connected to the Earth. Earths mass $ M >> m_{ball} $. Hence the momentum of the bat is practically infinite. In this picture the ball gets repelled by the bat with an energy content of $\lambda E_{impact}$ If we assume $\lambda = 1$ then $$ \vec{p}_{ball}^{after} = -\vec{p}_{ball}^{before} $$ or $$ \vec{v}_{ball}^{after} = -\vec{v}_{ball}^{before} $$ if we do not consider the rest-frame of the bat but the stadiums rest-frame the resulting speed of the ball is $$ {v}_{ball}^{after} = {v}_{ball}^{before} + 2 {v}_{bat}^{before} $$

Please note that you get the same result for a free collision where $m_{bat} >> m_{ball}$

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