Ball hitting the bat

Classical Mechanics Level 2

A ball approaches a batsman horizontally at speed $v_{\text{ball}}$ , and the batsman swings the bat, hitting the ball back along the same path. If the speed of the bat is $v_{\text{bat}}$ when it hits the ball, then what is the recoil speed of the ball?

Details and Assumptions:

The collision is perfectly elastic.
For all practical purposes, the speed of the bat is constant throughout the collision.

v_{\text{ball}}

v_{\text{ball}} - v_{\text{bat}}

v_{\text{ball}} + v_{\text{bat}}

v_{\text{ball}} + 2 v_{\text{bat}}

2 solutions

Andrew Dickson
Jul 16, 2017

Relevant wiki: Analyzing Elastic Collisions

There's an easy way to think of this intuitively.

If the bat was stationary, and the collision was elastic, the ball would bounce away with the exact opposite of its original speed.

In reality the bat is moving, but we can take the perspective of the bat in which the ball is moving towards it with speed $v_{ball}+v_{bat}.$

From our earlier logic, we then have the ball moving away from the bat with the same speed, $v_{ball}+v_{bat}.$

However, this is only with reference to the bat. Since the bat is moving with speed $v_{bat}$ in our original reference frame, we have to add on its velocity to get our answer, $v_{ball}+2v_{bat}.$

For this sytem to behave as stated in the initial conditions for the question (i.e. the bat maintains a constant velocity, either the bat has infinite mass, or their is additional force applied from something external to the ball/bat system. These answers are either inconsistent with expectations of energy conservation or other physical laws, because the initial question is fundamentally flawed.

Chris Sekely - 3 years, 10 months ago

See the considerations in my comment above.

Josh Silverman Staff - 3 years, 10 months ago

Something is "hinky" here, or the question is misleading making this question bad or useless for testing or demonstrating a physical principle. Questions should reinforce intuition not screw it up.

By definition, I believe, that an elastic collision is a collision in which both momentum and kinetic energy are conserved.

Since we do not know, but can intuitively imply, that momentum is not conserved, because the momentum of the ball changes but the momentum of the bat does not - one of the givens ( The collision is perfectly elastic ) is untrue.

There is an unaccounted for force hiding somewhere in there.

bp kline - 3 years, 10 months ago

The two assumptions are that the collision is elastic and that the bat's speed is practically constant.

If we were to calculate this question like a normal two body collision with conservation of energy and momentum, we'd find the final velocity of the ball and bat to be

$\begin{aligned} v_\textrm{bat}^f &= \frac{v^i_\textrm{bat}\left(M_\textrm{bat} - M_\textrm{ball}\right) + 2 M_\textrm{ball}v_\textrm{ball}^i}{M_\textrm{ball} + M_\textrm{bat}}\\ v_\textrm{ball}^f &= \frac{v^i_\textrm{ball}\left(M_\textrm{ball} - M_\textrm{bat}\right) + 2 M_\textrm{bat}v_\textrm{bat}^i}{M_\textrm{bat} + M_\textrm{ball}}. \end{aligned}$

Now, let's analyze the expression for $v_\textrm{ball}^f.$ The mass of a bat is roughly $\SI{34}{oz}$ plus the weight of the batter's arms (for a typical adult) which are about $\SI{8}{lb}$ each or $\SI{256}{oz}$ altogether. So $M_\textrm{bat}\approx\SI{290}{oz}.$ The weight of a baseball is comparatively light, only $\SI{5}{oz}.$

Thus, we have

$v_\textrm{ball}^f = \frac{\left(\SI{5}{oz} - \SI{290}{oz}\right) \times v_\textrm{ball} + 2 \times \SI{290}{oz} \times v_\textrm{bat}}{\SI{295}{oz}} \approx -0.96 v^i_\textrm{ball} + 2 \times 0.98 v^i_\textrm{bat}$

which is practically $-v_\textrm{ball}^i + 2v^i_\textrm{bat}.$ Again, this is assuming that the bat/arm simply moves as a free body and collides with the ball. In reality, the bat is driven by the batter's rigid body, so the situation is even closer to the simplifying approximation of infinite mass, which will bring our answer even closer to $-v_\textrm{ball}^i + 2v^i_\textrm{bat}.$

Is momentum conserved?

Now you're right that some velocity is imparted to the bat/arms by the ball. But the mass of the ball is so small compared to that of the bat/arms that it makes a very small contribution, and thus only needs to pick up a small backward velocity to conserve the system's momentum. To see this, we can plug in some numbers as before, and we see that $v^f_\textrm{bat} \approx 0.96 v_\textrm{bat}^i + 2\times 0.02 v^i_\textrm{ball}.$

Again, in reality the bat is driven by the rigid body and so the system moves closer to the approximation of $M_\textrm{bat} \gg M_\textrm{ball}.$ But even if we consider it as a simple two body collision, the answer is already practically given by what we get with the approximations given in the problem's assumptions.

Note that the question asked about speed (not velocity), so keep the conventions in mind here.

Josh Silverman Staff - 3 years, 10 months ago

I think you are making unwarranted assumptions. You are over analyzing in a confusing way, this should be simple, these problems when you have the data are all very simple, provided sufficient information is given.

bp kline - 3 years, 9 months ago

@Bp Kline – What is the unwarranted assumption you find? My response above is addressing a few of the assumptions and explaining the extent to which they should hold.

Josh Silverman Staff - 3 years, 9 months ago

I'm not a physicist but I thought that the conservation of momentum for this system would apply. Given that the bat's momentum has not changed then, by definition, the momentum of the ball must be the same. Given that the ball's mass did not change then the velocity must be the same. What am I missing?

Glenn Wallace - 3 years, 10 months ago

What about the force that the batsman apply to keep the velocity of the bat same?

Rohit Gupta - 3 years, 10 months ago

A batsman cannot modulate his force on a bat in the split second of a collision, in that time period everything is instantaneous, and constant. When the ball strikes the bat it has a certain momentum and angular momentum presumably, and it has a certain properties of elasticity. The beginning and the ending do not add up in my book.

bp kline - 3 years, 9 months ago

I believe the "official" solution (and your solution) is actually wrong, but please check my reasoning. First, in your explanation, you are assuming the absolute value of the ball's velocity relative to the bat remains constant. Instead, I propose that the total energy (kinetic energy in this case) of the system remains constant. In fact, this is how an elastic colission is defined (https://en.m.wikipedia.org/wiki/Elastic_collision).

Some basics first... I will indicate subscripts using square brackets: [ ]. Also, I will establish the initial conditions as follows:

V[bat] is constant and positive from left to right. V[ball-initial]=V[i] is negative from left to right V[ball-final]=V[f] is positive from right to left Total kinetic energy is conserved relative to any fixed frame of reference, noted as E[k] I will define all above variables as positive and indicate appropriate signs (positive or negative explicitly in the formulas) Masses of bat and ball are M[bat], M[ball] So, starting from a reference frame where both bat and ball are moving:

Before colission: E[k1]=.5×M[bat]×(V[bat])^2+ .5×.M[ball]×(-V[i])^2
After colission: E[k2]=.5×M[bat]×(V[bat])^2+ .5×.M[ball]×(V[f])^2 Let E[k1]=E[k2] After reducing and simplifying, we get V[f] = +-V[i]. We would then choose the positive answer.

Now, onto the frame of reference where the bat remains stationary:

E[k1] = .5×M[ball]×(-V[i]+V[bat])^2 E[k2] = .5×M[ball]×(V[f] - V[bat])^2

After setting both equal and reducing a little, we have: (-V[i]+V[bat])^2 = (V[f] - V[bat])^2 V[i]^2-2×V[i]×V[bat]+V[bat]^2 = V[f]^2-2×V[i]×V[bat]+V[bat]^2 So, again... V[f]^2 = V[i]^2 V[f] = +-V[i] Again, we choose the positive one.

This explanation, is somewhat long, but thanks for following it through. I hope I communicated clearly enough and didn't make any mistakes.

Chris Sekely - 3 years, 10 months ago

Don't quote me on this, but I think conservation of energy isn't applicable in this situation because someone has to be applying a force to the bat to keep it from slowing down. Energy is being added to the system.

Alternatively, the bat has infinite mass, in which case there's infinite energy anyways, and I'm not really sure how conservation of energy works.

Andrew Dickson - 3 years, 10 months ago

I completely agree with you. The striker must apply a force to keep the bat moving at constant speed.

Rohit Gupta - 3 years, 10 months ago

This question models reality very badly. For things to happen as described, a movement of the bat is required that moves the bat at constant velocity, no matter what force is encountered in the process. Like a slowly moving rigid and stiff robot would do, or a sweeping rake of a bowling alley. That would mean that after you missed the ball, bat speed and movement would still exactly be the same as after hitting it.

A human player does exactly the opposite: he gives velocity and thus momentum to a bat with mass, and at the moment of collision, almost no extra force is applied (such that total momentum of bat and ball is preserved). The bat will considerably slow down.

Try playing baseball with a bat without mass. Jimmy Connors attached lead strips to the top of his tennis rackets. It's all about momentum.

Manuel Ruiz - 3 years ago

This problem is inherently "hinky" because we're asked to assume a counterfactual: that it's possible for the velocity of exactly one object in a two-object collision to remain constant. This implies other counterfactuals like infinite mass.

So we need to treat the problem as an impossible but limiting case of a more general problem: As the bat's ∆v approaches 0, what is the limit of the ball's final velocity?

We can't get a sensible answer using the bat's momentum, so let's use its kinetic energy E, which is proportional to v^2. Any change in E is proportional to ∆(v^2) which = 2v∆v + (∆v)^2.

To use this to find the limit of the ball's final velocity, we want ∆E to approach 0 as quickly as possible, as ∆v -> 0. So it's easiest to use the reference frame of the bat, where v=0.

Given these assumptions, it's clear that in order to conserve total kinetic energy, the ball's velocity relative to the bat must just flip its sign, producing the "correct" answer given.

Wes Pinchot - 3 years, 8 months ago

Tapas Mazumdar
Jul 10, 2017

Relevant wiki: Analyzing Elastic Collisions

Since the collision is elastic, the coefficient of restitution ( $e$ ) for the collision would be equal to $1$ and so by definition

$e = \dfrac{|\text{Relative velocity of separation}|}{|\text{Relative velocity of approach}|} \implies 1 = \dfrac{v_{\text{recoil}} - v_{\text{bat}}}{v_{\text{ball}} + v_{\text{bat}}} \implies v_{\text{recoil}} = v_{\text{ball}} + 2 v_{\text{bat}}$

Note that we've assumed $v_{\text{recoil}} > v_{\text{bat}}$ because generally during a play of bat and ball the ball should travel with a faster velocity than that of the bat for the batsman to make a proper hit.

Am I confused in thinking that (vball + 2vbat) describes the relative speed of the ball, to the bat, after collision, but not the absolute speed, i.e., to the field of play? I can't visualize a calculation for speed of the ball that doesn't take conservation of momentum into account.

Brian Egedy - 3 years, 10 months ago

No, all speeds are with respect to the ground.
If you look at speeds relative to the bat, then the ball approaches the bat with speed $v_\text{ball} + v_{bat}$ and returns with the same speed.
With respect to ground, the ball approaches with $v_\text{ball}$ and returns with $v_\text{ball} + 2v_\text{bat}$

Pranshu Gaba - 3 years, 10 months ago

I am often accused of being pedantic, however in teaching this is a benefit as it is easier for students to follow and understand each step of a calculation and learn from it.
I originally set out to disprove the answer, but in doing so actually proved it. In answering my own weaknesses, perhaps others will find this pedantic proof to be helpful.

<<Bat and Ball problem>> Defining terms: Coefficient of Restitution, e... The Coefficient of Restitution is the ratio of final to initial relative velocities in the normal direction. NOTE: The “normal direction” for the ball is the positive direction (left to right in my calculation) and that of the rebounded ball is negative. This must be taken into account to obtain a positive value for e…

e = (relative velocity after collision) / -(relative velocity after collision) <= 1 * Problems of this nature are a constant headache and source of error especially as a simplified approach, that the value of e is simply the modulus of the quotient, is often used for convenience as this means that an answer can be arrived at without due diligence to the signs of vectors. This is alright when calculating e, since the +/- is simply ignored, but when determining the value for a velocity using e, the sign is critical.

Since we are told that this is a perfectly elastic collision, then we know that e = 1

We are told that the change in momentum of the bat remains constant for the purposes of calculation. This is an approximation to simplify the calculation but is assumed to give a negligible error especially as many other approximations are made for this calculation, such as... * the collision occurs in a single dimension and that all impacts and velocities take place along the same straight line (normal direction). That is, the ball is hit back along the line of its approach, * the curve of the swing of the bat is negligible at the instant of the calculation, * there is no spin involved on either the bat or ball, * friction between ball and bat is zero, * wind resistance is zero, * energy is conserved (no losses due to sound, heat, deformation of materials or any other work done by deforming the bat or ball...)

We need to refer to velocity (a vector) rather than speed (a scalar) when determining relative velocities, so care is needed regarding + and – values of vectors since they denote direction. For this calculation, I take the positive direction to be left to right. This is arbitrary but makes no difference to the calculation providing it is used consistently throughout the calculation.

To simplify the calculation, we use the frame of reference such that the bat is not moving relative to itself. The (true) velocity of the bat must, therefore, be incorporated into the relative velocity of the ball…

Initial true velocity of ball = v[ball] (positive since moving left to right) Initial true velocity of bat = -v[bat] (negative since moving right to left)

Relative velocity of object a to object b = (velocity of b) – (velocity of a) NOTE: The order is important if we are to obtain a correct sign for the relative velocity.

Initial relative velocity of ball (to bat) = (-v[bat]) - v[ball] = v[ball] + v[bat] Initial relative velocity of bat (to itself) = (-v[bat]) - (-v[bat]) = 0

Final relative velocity of the ball = v[final] (assuming, for now, that the direction of motion is left to right, a negative value will denote movement right to left).

Final relative velocity of the bat = 0 by definition of the bat being the frame of reference and velocity does not (for calculation purposes) change due to the impact.

Calculation on the ball in the frame reference of the bat… Coefficient of restitution, e = (v[bat] + v[ball]) / -v[final] = 1 (From * ) Final relative velocity of ball, -v[final] = v[bat] + v[ball]

However, this value is in the reference frame of the bat which is moving at a velocity of -v(bat) relative the ground or real world. Therefore to determine the final velocity of the ball as measured by a spectator (assumed to be stationary), we must be add the velocity of the bat to the final value. NOTE: This is the stage where it is critical that the signs of the velocities are all correct.

Final true velocity of ball = -v[final] + (-v[bat]) = -(v[bat] + v[ball]) - v[bat] = -v[bat] - v[ball] - v[bat] = -2v[bat] - v[ball] = -(2v[bat] + v[ball]) = -(v[ball] + 2v[bat]) (rearranging to show the expression as it appears in the question.)

This tells us that after being struck the ball, as described in the question, has a velocity of -(v[ball] + 2v[bat]) or, in other words, a speed of (v[ball] + 2v[bat]) to the left.

QED

To those still having doubt as to why the speed is increased by a value of 2v[bat] and not simply v[bat], I offer another conundrum...

If a bicycle is travelling in a straight line on a flat* road with a horizontal velocity, v... a - what is the velocity of the tyre on the ground? b - what is the velocity of the axle (centre of the wheel)? c - what is the velocity of the highest point on the tyre? d - what is the velocity of the tyre at the same height as the axle? ...all directions horizontal.

Wandering off topic... There are some interesting teaching points here about logical thought and the meaning of data. For teachers wishing to make students think about which data is relevant, lots of irrelevant data could be included such as the gear used, up/down a hill, the bicycle is braking, the size/width of the wheel, wind resistance etc. Or you could give the wheel diameter and rotational speed and ask them to calculate each of the above values. Interesting version could be conjured by making 2 cyclists riding side by side, one has a cycle with a larger diameter wheel than the other. Which wheel will have the larger values for the velocities asked above?

*Flat road is necessary to define that the point of contact with the road and highest point of the wheel are diametrically opposite, though you could introduce a rough track for a mountain bike and resolve velocities in a horizontal direction...

There are many variations - have fun.

Martin Carr - 3 years, 10 months ago

Greetings, Newton's second law,even with perfect elasticity, and reviewing fully physics.org section of bat and ball derivitive tri equation vs presented binary equation fully disagree. Moreover, exact formal needs derived and discussed. Known statistics for Relative pitched balls 100mph, average batters speed 65 to 70 mph, sum average speed of out of the park ball 110 mph, apply perfect elasticity in presented data results in a hit ball speed of 240 mph.....third portion this equation needs to be resolved. Answer "C" is closer to all known actual yields'. Regards.

Craig Stevenson - 3 years, 10 months ago

Thank you for this comment. Reality takes over here and the presented equations are unrelatable.

Brock Walsh - 3 years, 10 months ago

See my comment in the thread for @Andrew Dickson 's solution.

Josh Silverman Staff - 3 years, 10 months ago

Your math makes sense and is very clear. Please read my response to Andrew (above) and then consider the following: In a closed system (bat and ball only) this answer can't be correct because the final kinetic energy of the system is greater than the total initial kinetic energy. Am I missing something here? I wonder, because my math makes sense, but does yours and Andrew's and we're getting different answers.

Chris Sekely - 3 years, 10 months ago

I think what you are missing is the the system is not closed. The hands of the person holding the bat are part of the system. If the system was just the ball and bat then it would not be possible for the ball to change direction and thus momentum without the bat's momentum and thus speed also changing. Because the bat continues to travel at the same speed there must be another force applied to it which would be from the batter's hands.

Justin Roughley - 3 years, 10 months ago

Yes, I agree. I also recognize now that I slightly misread the problem. I feel a little dumb for that, but otherwise, I suppose my reasoning was sound. :-|

Chris Sekely - 3 years, 10 months ago

You may also consider an infinite mass for the bat or at least a huge difference in masses, so the mass of the ball is negligible comparing to the bat (the ball hits a train - the train won't feel the ball.)

Horia Tudosie - 3 years, 10 months ago

if the bat is a wall and not movable then perfect elasticity would result in the ball moving away from the wall at minus the intial velocity, but only if the wall is not movable. otherwise the ball would be traveling slower by some amount depending on the mass difference between it and the wall. If the ball is not moving and the wall is then PE would result in the ball moving away from the wall at some velocity given to it by the wall. the wall would be traveling slower by the same amount of energy transferred to the ball but depending on the mass difference between the ball and the wall how much slower dependent on the mass of the two objects. If the mass were the same then they would be travelling in opposite directions.l.

If both the wall and the ball are moving the velocity of either still depends on the mass difference. I don't see why the ball would necessarily gain 2x the velocity of the Wall. The 240 mph velocity of the ball mentioned in another remark below is roughly 352 feet per second.

I think if this analysis was true there would be a whole lot more home runs in the game of baseball. Presuming a 45 degree angle upwards flight at roughly 352 feet per second at the start the force of gravity at 32'/s² to bring it's upward flight to zero or roughly 4.5 seconds. and the same length of time to fall back to the park or 9 seconds or about 1,600 feet of travel. that's 1/3rd of a mile. I don't think ball parks are that large.

R leever - 3 years, 10 months ago

I don't see why the ball would necessarily gain 2x the velocity of the Wall.

You could either do out the full calculation (the results of which are in my comment in the other thread), or by using a shift in reference frames to extend your argument to the moving case (like is done in the other solution).

I think if this analysis was true there would be a whole lot more home runs in the game of baseball.

A key difference between this problem and baseball is that baseball doesn't have perfectly elastic collisions. Quite a bit of energy is lost to dissipation in the deformation of the ball and the bat, not to mention drag.

Josh Silverman Staff - 3 years, 10 months ago

Some real life numbers from http://www.fangraphs.com/blogs/batted-ball-velocity-adrian-beltre-and-xander-bogaerts/

Batted balls are usually at roughly the same velocity as pitchers, or lower. The main reason would be (my guess) that most contact is not with the bat's center of mass. When the bat hits the ball squarely, the velocity is higher than that of the pitch, and the probability of a home run goes up. Also, the launch angle is almost never as high as 45 degrees. A 20-30 degree launch angle is far more common. A typical major league pitch is down near the bottom of the strike zone, at the knees. Good pitchers do not like getting pitches up higher in the strike zone, where it is easier for a batter to get under the pitch. (Of course there is some game theory involved here, as pitchers with particularly good fastballs will mix in high heat which is hard for a batter to adjust to when he's anticipating a lower pitch.)
Oh, here are the numbers for this season: https://baseballsavant.mlb.com/statcast_leaderboard

Shouldn't be a surprise to baseball fans that Aaron Judge has hit the fastest batted ball this season at 121 mph.
Some other numbers: average bat speed is about 70 mph. Also, it's rare for a batter to be able to maintain that kind of speed on contact.
This Quora answer https://www.quora.com/What-is-the-average-bat-speed-of-hitters-in-MLB

suggests that, one a very well-hit ball, about 50% of the bat's energy will be transferred to the ball.
All of these factors explain why home runs are not happening constantly.

Richard Desper - 3 years, 10 months ago

@Richard Desper – And these are as important to consider as the deformation and drag. I think a cleaner example may have been a steel ball and the front of a train 😂 .

Josh Silverman Staff - 3 years, 10 months ago

The answer would be correct if speed of the ball and bat was described. However because velocity is a vector not a scalar none of the answers are correct.

Charles Ivie - 3 years, 10 months ago

The question is asking about the speed of the ball.

Josh Silverman Staff - 3 years, 10 months ago

The way the question was rephrased it is now correct.

Charles Ivie - 3 years, 10 months ago

The question is ambiguous as it does ask what the recoil speed is relative too, the bat or the ground

chris cronin - 3 years, 10 months ago

All the speeds in the problem are relative to the ground.

Pranshu Gaba - 3 years, 10 months ago

Ball hitting the bat

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