If the ball is stopped in 0.03 s, what is the average force of the ball on the catcher's hand?

Question

At a ball game, consider a baseball of mass m=0.15 kg that falls directly downward at a speed v=40 m/s into the hands of a fan. What impulse Ft must be supplied to bring the ball to rest? If the ball is stopped in 0.03 s, what is the average force of the ball on the catcher's hand?

Answer 1

Force is a change of momentum with respect to time
F=m dv/dt

or

F= m v/t= 0.15 x 40 / 0.03= 200N

Please note that impulse is not force.
Impulse if application of force averaged over time. For example if one applies 200 N of force for .03 sec the ipulse
I=F t= 200 x .03 =6 N s or 6 kg m/s

I is also a change in momentum
I=p2 -p1=m(v2-v1) or

I =.15(0 - 40)=-6 kg m/s
have fun!

Answer 2

change in momentum / time = force
mV - mu / t = Force
0.15*40 - 0 / 0.03 = Force
= 6 / 0.03
Force = 200 N

Answer 3

m = 0.15 kg
u = 40 m/s
v = 0 m/s

Impulse = Change in momentum = m(v - u) = -6 kg m/s

Force = Impulse / time = 200 N

Hope this helps.

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Answer 4

First compute the acceleration (or deceleration). The ball changes velocity from 40 m/s to 0 m/s in .03 sec, so the acceleration is 40/.03 or 1333.3 (m/s)/s.

Force is computed as mass times acceleration (F = ma), Newton's 2nd Law

Force in metric systems is measured in newtons
So, F = .15kg * 1333.3m/s/s = 200 newtons

To convert to pounds, newtons = 0.225 lbs.
So the force is 200 newtons = 45 lbs.

Answer 5

let's figure out the deacceleration
from vf = vi + a*t
a = -vi/t

F = m*a = .15 kg * -40 m/s / 0.03 s = -200 N

That is the force the hand exerts on the ball.

The ball would exert and equal and opposite force on the hand.

F = 200N

Answer 6

they are in a place to take transport of out yet often sort new balls are used for that. assorted the time the used balls in the interest are paced in the communities prepare bucket and used for prepare.

Answer 7

impulse = (average force)×(time) = (change in momentum)

How much is the change in momentum? (Hint: the initial momentum is (0.15kg)(40m/s), and the final momentum is (0.15kg)(0m/s).)

Now you know the change in momentum; and they tell you the "time". Plug into the above equation and solve for "average force".