When you hit a baseball, does the ball ever travel faster than the bat? 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!
 A: 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.
A: 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.
A: 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.
A: 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 $v_{bat}$
A: 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).

Image courtesy of NASA.
A: 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.
A: 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.
A: 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).
A: 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.
A: 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 $m_{bat} >> m_{ball}$
