What is the average force on a baseball after being hit by a bat?

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    Baseball Force
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The discussion focuses on calculating the average force on a baseball after being struck by a bat, with specific attention to both horizontal and vertical components of force. Participants initially struggle with determining the correct components of velocity and acceleration, emphasizing the need to split these into their respective components before applying formulas. The correct approach involves using momentum and impulse equations rather than treating acceleration as a scalar. There is a consensus that gravity does not significantly affect the calculations due to the brief contact time of 1.65 milliseconds. Ultimately, the thread highlights the importance of correctly identifying initial velocities and applying the right equations for accurate results.
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Homework Statement


A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s, and it leaves the bat traveling to the left at an angle of 35 degrees above horizontal with a speed of 60.0 m/s. The ball and bat are in contact for 1.65 milliseconds.

A) Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right

B) Find the vertical component of the average force on the ball.

Express answers with two significant figures.

Homework Equations


F= ma
V= Vo + at


The Attempt at a Solution


First I plugged into the kinematics equation:
60 m/s= 50 m/s + a (0.00165 s)
a= 6060.61 m/s^2

Then I simply used sin and cos to get the components:
6060.61sin(35) = 3476.2 = 3500 (sig figs) = Fy
6060.61cos(35) = 4694.56 = 4700 (sig figs) = Fx

Both of these are wrong. What am I doing wrong?
 
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You need to split the velocity into components first. THEN find the components of acceleration. You can't treat acceleration as a scalar and then split into components.
 
Dick said:
You need to split the velocity into components first. THEN find the components of acceleration. You can't treat acceleration as a scalar and then split into components.

Ok, did that and still got it wrong.

Spitting 60 into components I get: Fx= 49.15 and Fy= 34.4146

Now plugging in: 49.15 = 50 + a(0.00165)
a= -515.15

Fx= (.145)(-515.15)
Fx= 74.69 = 75 (sig figs)

Wrong still.
 
The initial vx=50m/s. Because it's traveling right. Final vx=(-49.15)m/s. That's MINUS because it's traveling left.
 
Thanks, I got the first part.

But the same thing isn't working for the second. It isn't negative since it's going up. Do I need to factor in gravity somehow?
 
javacola said:
Thanks, I got the first part.

But the same thing isn't working for the second. It isn't negative since it's going up. Do I need to factor in gravity?

What did you do and what did you get for the second part? I doubt you have to deal with gravity. It's not going to do much in 1.65 milliseconds.
 
Same thing basically.

34.4146 (vertical component of velocity) = 50 + a (0.00165)
a= -9445.69697

Fy= (-9445.69697)(.145)
Fy= -1369.626= -1400 (sig figs)
 
Nothing?
 
javacola said:
Same thing basically.

34.4146 (vertical component of velocity) = 50 + a (0.00165)
a= -9445.69697

Fy= (-9445.69697)(.145)
Fy= -1369.626= -1400 (sig figs)

Is 50 really the initial vertical velocity? I don't think so.
 
  • #10
Thought I would update this one.


Baseball force problem

--------------------------------------------------------------------------------
1. Homework Statement
A bat strikes a 0.145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 50.0 m/s, and it leaves the bat traveling to the left at an angle of 35 degrees above horizontal with a speed of 60.0 m/s. The ball and bat are in contact for 1.65 milliseconds.

A) Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right

B) Find the vertical component of the average force on the ball.

Express answers with two significant figures.

2. Homework Equations

sigma F = (p2-p1)/(t2-t1).

I did this problem via momentum/impulse


FX = (mass of ball * v2cos(angle) - mass of ball * v1) / time in seconds

(.145 * 50cos(35) - .145*60)/.00165 =


FY = (mass of ball * v2sin(angle) - 0 (has no initial velocity in y direction, so .145*0=0) /time
(.145*50cos(35)-.145*60)/.00165 =


my problem had different numbers in it, but I figure this could help someone else looking for an answer. I haven't seen anyone do it this way yet.


also: http://people.physics.tamu.edu/mahapatra/teaching/ch8_supl_sols.pdf is what I checked my answers with.
 
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