Momentum - Baseball Homework: Average Force, Direction

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Homework Help Overview

The problem involves a baseball with a mass of 0.145 kg that is pitched horizontally at a speed of 39.0 m/s and then strikes a bat, causing it to be popped straight up to a height of 37.0 m. The questions focus on calculating the average force exerted during the contact time of 1.0 ms and determining the direction of that force.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculating momentum using the formula p=mv and applying it to find the average force using F=p/t. There are questions about how to incorporate the height into the calculations and what it implies for the velocity after leaving the bat.
  • Some participants explore using kinematic equations and conservation of energy to find the initial velocity after the ball leaves the bat, leading to varied results and confusion over the calculations.
  • There are suggestions to consider the change in velocity as a vector and to use the root sum of squares to find the magnitude of the change in velocity for the average force calculation.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and results. Some have provided guidance on considering the vector nature of velocity and the relationship between the horizontal and vertical components. There is a recognition of the complexity of the problem, and participants are exploring different interpretations and methods without reaching a consensus.

Contextual Notes

Participants express frustration with the problem, indicating that they are struggling to understand the concepts involved. There are mentions of limited attempts remaining for submitting answers, which adds pressure to the discussion.

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Homework Statement


A 0.145 kg baseball pitched horizontally at 39.0 m/s strikes a bat and is popped straight up to a height of 37.0 m. (a)If the contact time between bat and ball is 1.0 ms, calculate the magnitude of the average force between the ball and bat during contact. (b)Find the direction of the average force on the ball.

Homework Equations


p=mv
F=p/t


The Attempt at a Solution


This seemed like a really simple problem to me, but Mastering Physics said my answer was wrong. I tried to find momentum using p=mv, p= 0.145 kg(39.0 m/s) = 5.655
Then I plugged this into F=p/t and got F=5.655/0.001 s =5655N, which it said was wrong. Am I missing something? What do I do with the height?
Also I don't know what to do with part b at all.
 
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tiggrulz13 said:

Homework Statement


A 0.145 kg baseball pitched horizontally at 39.0 m/s strikes a bat and is popped straight up to a height of 37.0 m. (a)If the contact time between bat and ball is 1.0 ms, calculate the magnitude of the average force between the ball and bat during contact. (b)Find the direction of the average force on the ball.

Homework Equations


p=mv
F=p/t


The Attempt at a Solution


This seemed like a really simple problem to me, but Mastering Physics said my answer was wrong. I tried to find momentum using p=mv, p= 0.145 kg(39.0 m/s) = 5.655
Then I plugged this into F=p/t and got F=5.655/0.001 s =5655N, which it said was wrong. Am I missing something? What do I do with the height?
Also I don't know what to do with part b at all.

The height should give you the initial velocity after leaving the bat shouldn't it?
 
How would you do that?
 
Ok, I tried using x=xo + volt + 1/2at^2 to find the initial velocity after leaving the bat, but I got a huge number:
37=0 + Vo(0.001) + 1/2 (39/.001)(.001)^2
Vo=36980.5 m/s
I'm sorry, for some reason I'm just not getting it tonight.
I also thought about using conservation of energy to find V and got:
1/2mv^2 +mgh = 1/2mv^2 + mgh
1/2(.145)(39)^2 + 0 = 1/2(.145)V^2 + (.145)(9.8)(37)
and i got v=28.2099 m/s
 
tiggrulz13 said:
Ok, I tried using x=xo + volt + 1/2at^2 to find the initial velocity after leaving the bat, but I got a huge number:
37=0 + Vo(0.001) + 1/2 (39/.001)(.001)^2
Vo=36980.5 m/s
I'm sorry, for some reason I'm just not getting it tonight.
I also thought about using conservation of energy to find V and got:
1/2mv^2 +mgh = 1/2mv^2 + mgh
1/2(.145)(39)^2 + 0 = 1/2(.145)V^2 + (.145)(9.8)(37)
and i got v=28.2099 m/s

Consider using the V2 = 2gh relationship to figure the vertical initial velocity. (At its height its V is 0.)

Now you have a system that had MV in the positive x direction get changed to MV in the positive up direction in .001 second.
 
I used the v^2 = 2gh equation to get v=26.9295
I then plugged this into the F=p/t equation:
F = 0.145(39-26.9295) / 0.001
F=1750.22
and that was wrong - and I don't have anymore attempts left on that part, apparently the correct answer was 6900 N.

For the second part I need to find the direction (theta) of the average force. How do I go about doing that?

(I'm sorry if I'm taking a lot of your time, thank you for trying to help me, for some reason physics is just not clicking with me tonight)
 
tiggrulz13 said:
I used the v^2 = 2gh equation to get v=26.9295
I then plugged this into the F=p/t equation:
F = 0.145(39-26.9295) / 0.001
F=1750.22
and that was wrong - and I don't have anymore attempts left on that part, apparently the correct answer was 6900 N.

For the second part I need to find the direction (theta) of the average force. How do I go about doing that?

(I'm sorry if I'm taking a lot of your time, thank you for trying to help me, for some reason physics is just not clicking with me tonight)

This is incorrect. Velocity is a vector.

This means that the change in velocity is a right triangle (horizontal pitch to vertical pop fly). The magnitude of the change in velocity then is given by the root sum of the squares of the two.
(26.932 + 392)1/2

Using that change in speed times the mass divided by the time is the average force applied.

As to angle then, it is the angle of the formed by the vectors of velocity - but against the direction of the initial pitch. That would then be the Tan of Vup/Vpitch above the direction the pitch arrived from.
 

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