Calculating Impulsive Force on Baseball Player's Hand

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Discussion Overview

The discussion revolves around calculating the average impulsive force exerted on a baseball player's hand when catching a ball, considering the player's technique of pulling his hand back to soften the impact. The scope includes mathematical reasoning and problem-solving related to physics concepts, specifically in the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the problem statement and initial calculations, leading to an impulsive force of 54446 lbs, which they believe is incorrect.
  • Another participant points out a potential error in solving a quadratic equation related to time.
  • A subsequent reply corrects the quadratic formula application but still arrives at an incorrect answer.
  • Participants discuss the importance of sign conventions in their calculations, with one suggesting a sign error in the equation.
  • One participant realizes the need to convert mass from pounds to slugs, indicating a misunderstanding of units in English measurements.
  • Another participant acknowledges a mistake in using the wrong initial velocity in their calculations, which affects the results.
  • Despite various corrections and adjustments, participants express frustration over not reaching the expected answer of 76.9 lbs.

Areas of Agreement / Disagreement

Participants express uncertainty and disagreement regarding the correct calculations and approaches to solve the problem. There is no consensus on the final answer, and multiple competing views on the methodology remain present.

Contextual Notes

Participants note potential limitations in their calculations, including unresolved mathematical steps, unit conversions, and the dependence on correct initial values. There is also mention of the need for clarity on sign conventions in their equations.

jaredogden
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Homework Statement



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.

Homework Equations



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

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..
 
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jaredogden said:
(6/12 ft) = 0 ft + 132 ft/st + 1/2(-17424 ft/s2)t2
t = 0.007575s
This is a quadratic; not correctly solved.
 
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.
 
jaredogden said:
0 = -8712 ft/s2t2 + (135 ft/s)t + 0.5 ft
I believe one of these terms has the wrong sign.

I'm still getting the wrong answer.
It is always helpful, when you know the correct answer, to provide it so others know their working is right.
 
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..

jaredogden said:
The answer is wayyyy less as it should be..
76.9 lbs
 
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
 
Last edited by a moderator:
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

The answer is 76.9

This is frustrating, I feel like an idiot.
 
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.
 
Yes, I noticed your 135 mistake. Still doesn't quite get me to 76 though. I used 0.0183 sec
 

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