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

[javacola]

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?

 

[Dick]

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.

 

[javacola]

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.

 

[Dick]

The initial vx=50m/s. Because it's traveling right. Final vx=(-49.15)m/s. That's MINUS because it's traveling left.

 

[javacola]

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?

 

[Dick]

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.

 

[javacola]

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)

 

[javacola]

Nothing?

 

[Dick]

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.

 

[m0llym0nster]

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.