If a baseball bat is heavier does the ball go farther? Why?

Answer 1

Up to a point. Sorry if this answer gets long-winded but I can't explain it any more briefly. If you make certain simplifying assumptions, you find that there is an optimum ratio of bat mass to ball mass. This ratio depends on the relative energies of the bat and ball before impact. Let's assume the batter can put the same energy into his swing regardless of bat mass (although this is obviously not perfectly true), and that the impact point is the center of mass of the bat, which is not rotating but just moving at right angles to the length of the bat and opposite to the ball's path.
Some definitions:
Define v1 and v2 as, respectively, pre-impact bat and ball speeds, e1 and e2 as their pre-impact energies, v1a and v2a as post-impact speeds, and m1 and m2 as their masses. Positive speed is taken to be in the direction of the swing, negative in the direction of the pitch. We are seeking an m1/m2 ratio that maximizes v2a. Define k as the elastic coefficient (the ratio of collision velocity after and before the impact; for the ball dropped onto a rigid surface, this is equivalent to the square root of the ratio of the bounce height divided by the initial height and is between 0 and 1. A rough estimate for a hardball is 0.5.). Assume a perfectly rigid bat.
Define v12 as the speed of the system comprising m1 & m2, which you can obtain by dividing the total momentum by the total mass, i.e. v12 = (m1 * v1 + m2 * v2) / (m1 + m2); define dv1 and dv2 as the speeds of bat and ball << relative to v12 >> before impact, and dv1a and dv2a as their v12-relative speeds after impact.
Collisions:
From conservation of momentum, any collision, elastic or inelastic, doesn't change v12. The kinetic energy of the collision is the sum of the bat and ball energies, 0.5*m1*dv1^2 + 0.5*m2*dv2^2. After an inelastic collision (k=0), and at the moment of peak deformation in an elastic collision (0 Inelastic collision:
For an inelastic collision, assuming e1=e2=m2=1, we have a fairly simple expression to find the m1 that maximizes v2a and thus v12; we maximize (m1*SQRT(2/m1)-SQRT(2))/(m1+1) (from the definition of v12 and the fact that v=SQRT(2*e/m)). Rather than resorting to calculus, repeated evaluation of the function shows a maximum at m1=5.83 (approx.).
Elastic collision:
Introducing elasticity complicates the equation, but interestingly does not affect the optimum m1/m2. Before a collision, the ratio of the bat and ball speeds relative to the system speed, dv1 and dv2, is the inverse of the ratio of the corresponding masses, that is, dv1/dv2=-m2/m1. This falls out of the definition of v12. Elastic deformation of the ball changes some or all of the kinetic energy to potential energy. At the instant of full deformation, the bat and ball have the same speed, v12. Some or all of the potential energy is then returned as kinetic energy, specifically, speeds relative to v12 of the bat and ball (dv1a and dv2a). These speeds, like dv1 and dv2, are in inverse ratio to the masses. The net result is that the final speed differences are the initial speed differences multiplied by -k (the minus sign reflects the sign change due to rebound). The individual speeds are then computed by adding dv1a and dv2a to v12. The necessary equations are:
v1 = SQRT(2 * e1 / m1); v2 = -SQRT(2 * e2 / m2)
v12 = (m1 * v1 + m2 * v2) / (m1 + m2)
dv1 = v1 - v12; dv2 = v2 - v12
dv1a = -k * dv1; dv2a = -k * dv2
v1a = v12 + dv1a; v2a = v12 + dv2a

When coded and executed with e1=1, e2=1, m2=1 and variable m1, a maximum value of v2a is found when m1 = 5.83. What is interesting is that this value provides a maximum for any value of k. Below are best bat/ball mass (m1/m2) ratios for a range of e1/e2 ratios:
e1/e2......m1/m2
0.5........9.90
1..........5.83
2..........3.73
3..........3.00

Answer 2

Not exactly. What you are referring to is conservation of momentum. Momentum is a quanity mass * velocity. Now if you have a bat of mass 5 and swing it at velocity 5, assuming perfect conditions, a stationary ball will gain a momentum of 25.

Now, if you have a heavier bat and swing it with the same velocity, say mass of 10 and velocity of 5, then a momentum of 50 is transfered, and the ball should go faster and further.

However, take in to account that the bat is heavier, then the velocity of the swing will decrease say to 4. So then the mass*velocity will be (5*4=20) and the ball will go slower than its original 25.

Basically, if you swing a heavier bat with the same velocity as the lighter bat, only then will the ball go further.

Answer 3

Momentum, p imparted to the ball will be greater. P=m*v. So as the bat comes back the ball goes away. But at the same time if the time of impact is more then the ball will move slower. So you can take a heavy bat and hit or take a light bat and hit really hard. See if you take an egg and drop it on a sidewalk, it will break but if you drop it just as hard on a pillow, it won't because the time of impact is longer. So the force = momentum/time=mass*velocity/time.
When either time is decreased or mass is increased or velocity is increased, then the ball will go more distance. The amount of time the bat spends touching the ball is just as important as the velocity with which the ball is met by the bat.

Answer 4

It all depends distance is equal to velocity times time. d=v*t if the vleocity and the speed are faster then the object will go farther. Since The bat is heavier you are swinging slower therefor it will not go as fast as if you swing a lighter bat with more speed.

Answer 5

assuming that you can swing the bat at the same speed, and it is equally as hard as the lighter bat then yes. The heavier bat has more momentum behind it, so it doesnt slow as much when the baseball hits it.

Answer 6

No, the ball doesn't go farther just because the bat is heavy, it all depends on the speed of the pitch and how fast you swing the bat

Answer 7

The heavier the bat, the more mass and density it has. The more mass, the more momentum it will have. When you hit the ball, the momentum energy (kinetic) is transferred to the ball and will push the ball farther the more energy there is.

Answer 8

Yes if the bat is heavier the ball will go farther!!
You gotta put more wood behind it!!!

Answer 9

the opposite is true. You can swing a lighter bat faster, and then the ball goes further. This is why "corking" your bat is considered cheating. It's when wood is removed from the center of the bat and replaced with light cork. This explains why modern bats and golf clubs are made from titanium and aluminum, because it's lighter

Answer 10

Let us recall acceleration*mass or force the is rate of change of momentum. So if the base ball bat is heavier its mass is higher. Hence the momentum imparted to the ball will be higher.(because as per law of conservation of momentum the momentum of the bat will have to equal momentum imparted to ball minus losses). Hence since rate of change of momentum is higher it will accelerate more and hence it will cover larger distance.(s=ut+gt^2/2) where s=dist,t=time,g=acceleration,u=intial velocity