# Dynamics: Impulse and Linear momentum (impulsive force exerted on catchers glove)

1. Feb 29, 2012

### jaredogden

1. The problem statement, all variables and given/known data

A baseball player catching a ball can soften the impact by pulling
his hand back. Assuming that a 5-oz ball reaches his glove at
90 mi/h and that the player pulls his hand back during the impact
at an average speed of 30 ft/s over a distance of 6 in., bringing the
ball to a stop, determine the average impulsive force exerted on
the player’s hand.

2. Relevant equations

mv1 + Imp1→2 = mv2
v2 = v02 + 2a(x - x0
x = x0 + v0t + 1/2at2

3. The attempt at a solution

(o ft/s)2 = (132 ft/s)2 + 2a(6/12 ft)
a = -17421 ft/s2

(6/12 ft) = 0 ft + 132 ft/st + 1/2(-17424 ft/s2)t2
t = 0.007575s

0.3125 lbs (132 ft/s) + F(0.007575s) = 0.3125 lbs (0 ft/s)
F = 54446 lbs

The answer is wayyyy less as it should be..
76.9 lbs

I just can't figure out how to get there..

2. Feb 29, 2012

### Staff: Mentor

This is a quadratic; not correctly solved.

3. Feb 29, 2012

### jaredogden

oh sorry

0 = -8712 ft/s2t2 + (135 ft/s)t + 0.5 ft

t = [-(135 ft/s) + sqrt((135 ft/s)2) - (4*(-8712 ft/s2)*(0.5 ft))]/(2*(-8712 ft/s2))

t = -0.003088 s

t = [-(135 ft/s) - sqrt((135 ft/s)2) - (4*(-8712 ft/s2)*(0.5 ft))]/(2*(-8712 ft/s2))

t = 0.0185841 s

mv1 + Imp1→2 = mv2
0.3125 lbs (132 ft/s) + F(0.0185841 s) = 0.3125 lbs (0 ft/s)
F = 2219.639
F = 2220 lbs

I'm still getting the wrong answer.

4. Feb 29, 2012

### Staff: Mentor

I believe one of these terms has the wrong sign.

It is always helpful, when you know the correct answer, to provide it so others know their working is right.

5. Mar 1, 2012

### jaredogden

Oh I put = + 0.5 ft.. in my calculator I did -0.5 ft

the answer for t is still the same.

I'm obviously missing something in my approach..

6. Mar 1, 2012

### Staff: Mentor

I've almost forgotten how to work in imperial measures.
Isn't there a factor of 32.2 because F is in poundals https://www.physicsforums.com/images/icons/icon5.gif [Broken]

Last edited by a moderator: May 5, 2017
7. Mar 1, 2012

### jaredogden

Ah of course... I'm so stupid. Yeah there is, mass in English units is in slugs not lbs. I should be dividing .3125 lbs by 32.2 ft/s2

I'm closer... but still off by about 6.4 lbs.. I must be forgetting something..

mv1 + Imp1→2 = mv2
(0.3125 lbs/(32.2 ft/s2)) (132 ft/s) + F(0.0185841 s) = 0.3125 lbs (0 ft/s)
F = 70.499 lbs
F = 70.5 lbs

This is frustrating, I feel like an idiot.

8. Mar 1, 2012

### jaredogden

I just realized in the second time I did the quadratic formula I was using 135 ft/s instead of 132 ft/s.

Also used that for the linear momentum.. I'm redoing everything right now.

9. Mar 1, 2012

### Staff: Mentor

Yes, I noticed your 135 mistake. Still doesn't quite get me to 76 though. I used 0.0183 sec