A bowling ball going 20MPH or a Ping Pong ball going 200MPH?

[Dagenais]

Just wondering, what generates more force?

A bowling ball going 20MPH or a Ping Pong ball going 200MPH?

 

[Integral]

this is a poorly posed question. F=ma in both cases a=0 so F=0.

Now, could you rephrase the question in a meaningful manner?

 

[Dagenais]

I actually read it at another forum, where the same question was asked. And a guy claimed the bowling ball, since he would rather be hit by a 200MPH ping pong ball.

That was all that was given, and I was wondering the same thing.

 

[NSX]

I think the point of the question is so that [itex]m * v[/itex] comes up to the same thing. As well, the conservation of momentum will show interesting results ... how's that?

 

[TALewis]

20 mph = 8.94 m/s
200 mph = 89.4 m/s

Assumptions:

Mass of the bowling ball: 14 lbm = 6.35 kg
Mass of the ping-pong ball: 2.5 gm = 0.0025 kg

Values from shaky google searches.

Momentum of the bowling ball:

6.35 * 8.94 = 56.8 kg-m/s

Momentum of the ping-pong ball:

0.0025*89.4 = 0.224 kg-m/s

I'd say the ping-pong ball hurts a lot less.

 

[CronoSpark]

Well, assuming the bowling ball is more "massive" than the ping pong ball...
Momentum-wise, bowling ball
Force-wise, bowling ball as well

We need to take into account of the strong magnetic force induced by the bowling ball onto another object in this case, assuming that the atomic model is valid

 

[robphy]

Because of the impulse-momentum theorem [tex]F_{avg}\Delta t=\Delta p = \Delta (mv)[/tex], to fully answer the question, one must address the issue of [tex]\Delta t[/tex], the time-interval over which the desired force is to act.

 

[Gokul43201]

Well, assuming the bowling ball is more "massive" than the ping pong ball...
Momentum-wise, bowling ball
Force-wise, bowling ball as well

We need to take into account of the strong magnetic force induced by the bowling ball onto another object in this case, assuming that the atomic model is valid

I don't understand what you're talking about there.

If the bowling ball were only a little more massive than the ping-pong ball, the ping pong ball would have a greater momentum. And the force applied by the decelerating ball depends on the impulse time, which in turn depends on the elastic properties of the ball and what it's hitting.

And what's this "strong magnetic force" have anything to do with being smacked by a ball ? And why does this need THE atomic model to be correct ? And why do you doubt the atomic model ?

I'm sorry if I'm not following the line of logic here...it seems a little fuzzy to me.

 

[CronoSpark]

I don't understand what you're talking about there.

If the bowling ball were only a little more massive than the ping-pong ball, the ping pong ball would have a greater momentum. And the force applied by the decelerating ball depends on the impulse time, which in turn depends on the elastic properties of the ball and what it's hitting.

And what's this "strong magnetic force" have anything to do with being smacked by a ball ? And why does this need THE atomic model to be correct ? And why do you doubt the atomic model ?

I'm sorry if I'm not following the line of logic here...it seems a little fuzzy to me.

Thank you for your inquiries, Goku. To be honest, I am never too sure about my own statements.

Here is how I see the situation.

I am considering an ideal environment, where both the ping pong ball and bowling ball are perfect hollow spheres. Assuming that both of them are traveling at a constant velocity (200 MPH), and also neglecting external forces (for example: air resistance), we analyse which one of them produces more "force".

If we imagine both of the objects in the atomic level, I am imagining that the bowling ball consists of more electrons that are more "closely packed" than that of the ping pong ball. This in turn allows the bowling ball to induce a stronger magnetic repulsion (in other words, force) than that of of the ping pong ball. Of course, we need to also assume that both balls are going to be in "contact" with another object to produce this force.

This explanation can only be used if the atomic model is valid. I have never doubted the model, however, but I feel that I should state it because there is always a possibility that it can be proven false... well at the present time at least.

I am guessing you understand already why the bowling ball should have a higher momentum than the ping pong ball.

That is the way I see it, and it seems to make sense to me. I could be wrong. Once again, thank you for analysing my post, Goku.

 

[Integral]

If you assume [tex] \Delta t [/tex] is the same see TALewis post above. End of discussion.

Chornspark, refrain from idle speculations, please speak form knowledge or ask questions. Do not post questionable knowldedge.

 

[CronoSpark]

If you assume [tex] \Delta t [/tex] is the same see TALewis post above. End of discussion.

Chornspark, refrain from idle speculations, please speak form knowledge or ask questions. Do not post questionable knowldedge.

My speculations did have a basis to it.

However, you are right Integral, it is inappropriate for me to give such an opinion within the General Physics section, so I feel that I should apologize for it.

Robphy and Goku's explanations were correct.

 

[arildno]

Even if you can give the ping-pong ball enough momentum to equal to the momentum of the bowling ball, being hit by the ping-pong ball wil still hurt a lot less.
This is because the material the ping-pong ball is made out of, has far smaller elasticity parameters than the bowling ball
(i.e, the ping-pong ball will more easily deform).
In general, in a collision, the object with the least elasticity parameters will deform the most (think of different spring constants).
The bowling ball will hardly deform when hitting a person, leaving the person to be the one that is deformed (), while the ping-pong ball will probably be smashed soundly in a similar collision.

 

[JohnDubYa]

hurt = pressure. The cross-sectional areas of the two objects are different, so that has to be taken into account as well.