Bat Force and Human Lifting Capacity in Softball Pitching

[Mowgli]

Homework Statement

A 0.165kg softball is pitched to you at 20.0m/s. You hit the ball back along the same path, and at the same speed. If the bat was in contact with the ball for 9.90x10^-3 seconds,

a.) what was the average force the bat exerted on the ball?

b.) Could this force lift a 70.0kg person?

Homework Equations

I believe the equation to use is

mass/change in time x (final velocity + initial velocity)

The Attempt at a Solution

a.)

(.165kg (20 + 20 m/s)) / 9.90 x 10^-3
= .165(40) / .0099
= 666.7 N
= 6.70 x 10^3 N

b.)

How do I factor in a 70kg person? Do I use 70(40)/.0099 ?

 

[Doc Al]

What force is required to lift a 70 kg person?

 

[mizzy]

Look at your equation again. Force is equal to the change in momentum over time. Change in momentum = final - initial.

 

[Mowgli]

so I should have .165(20-20)/.0099 and get 16.7 N??

 

[mizzy]

you're hitting the ball back, so the final momentum should be negative.

Can someone confirm this?

 

[Doc Al]

I believe the equation to use is

mass/change in time x (final velocity + initial velocity)

That equation should be: mass/change in time x (final velocity - initial velocity)

But as mizzy says, the sign of the velocity changes since you're hitting the ball in the opposite direction. So 20 - (-20) = 20 + 20.

(All we care about is the magnitude of the change, so it doesn't matter which direction we call negative.)

The Attempt at a Solution

a.)

(.165kg (20 + 20 m/s)) / 9.90 x 10^-3
= .165(40) / .0099
= 666.7 N
= 6.70 x 10^3 N

That's fine, except for the last step. (You have the wrong exponent.)

 

[Mowgli]

Oooh, okay. Yes I noticed it should be ^2

Thank you.

To figure out b.)?? Can this be used to lift a 70kg person? Anyone?

 

[Doc Al]

you're hitting the ball back, so the final momentum should be negative.

Can someone confirm this?

If the initial momentum was positive, then the final would be negative. (But all we really care about is the magnitude of the change.)

 

[Doc Al]

Oooh, okay. Yes I noticed it should be ^2

Make sure you understand the point that mizzy was making.

To figure out b.)?? Can this be used to lift a 70kg person? Anyone?

What force is required to lift a 70 kg person?

 

[Mowgli]

F= 70kg x 9.90x10^-3
= 70x.0099
= .693 N

.693 < 666.7 so less force is required to lift the person?

 

[mizzy]

you multiplied mass by time??

 

[Mowgli]

or by speed? I'm not sure what to use.

70.0kg x 20.0 m/s
= 1400
= 1.4 X 10^3

 

[mizzy]

I'm not too sure. But in your previous answer, mass x time doesn't give you F. Force is in Newtons (kgm/s^2)

 

[Doc Al]

The answer to my question--What force is required to lift a 70 kg person?--has nothing to do with momentum, speed, or time.

How much does the person weigh?

 

[Mowgli]

the person weighs 70kg

 

[Doc Al]

the person weighs 70kg

That's his mass, not his weight. How do you calculate the weight?

 

[Mowgli]

into lbs? multiply by 2.20

so 154 lbs

 

[Doc Al]

into lbs? multiply by 2.20

so 154 lbs

Better calculate the weight in Newtons so you can compare it to the force you calculated in part a.

 

[Mowgli]

oooh! so 154lb x 4.448N
= 684.9N
= 6.80x10^2
and
6.70x10^2 < 6.80x10^2
therefore more force would be needed to life the 70kg person

 

[Doc Al]

oooh! so 154lb x 4.448N
= 684.9N
= 6.80x10^2
and
6.70x10^2 < 6.80x10^2
therefore more force would be needed to life the 70kg person

That's the idea.

But the easy way to calculate the weight is by using W = mg, where g = 9.8 m/s^2.

 

[mizzy]

That's his mass, not his weight. How do you calculate the weight?

weight = mass x accleration of gravity

 

[Mowgli]

got it :) thank you!