Magnitude of the average force during a collision

[Jay232323]

Homework Statement

A 0.170-kg baseball pitched horizontally at 36.0 m/s strikes a bat and is popped straight up to a height of 36.0 m. If the contact time between bat and ball is 2.0 ms, calculate the magnitude of the average force between the ball and bat during contact.

$m_b=0.17kg$
$v=36.0m/s$
$h=36.0m$
$t=0.002s$

Homework Equations

$F_{avg}=\frac {\Delta p} {\Delta t}$

$KE_i=PE_f$

The Attempt at a Solution

Working with the first equation to get a form that uses the variables I've got.

$F_{avg}=\frac {\Delta p} {\Delta t}=\frac {p_f-p_i} {\Delta t}=\frac {m_bv'-m_bv} {\Delta t}=\frac {m_b(v'-v)} {\Delta t}$

Doing the same with the second equation, I can find an expression for the velocity of the ball after the collision.

$KE_i=PE_f$

$\frac 1 2 m_b(v')^2=m_bgh$

$v'=\sqrt {2gh}$

Plugging that back into the equation for the average force, I get

$F_{avg}=\frac {m_b(\sqrt {2gh}-v)} {\Delta t}$

Hey, I know all of those values!

$F_{avg}=\frac {(0.17)(\sqrt {2*9.81*36}-36)} {0.002}=-800.98N$

And since we're only concerned about the magnitude

$F_{avg}=800.98N$

The answer that I should be getting is

$F_{avg}=3800N$

 

[TSny]

Hint: Vectors!

(Welcome to PF)

 

[Let'sthink]

Note that momentum final initial and force are all vector quantities. So be careful in calculating change of momentum its direction and magnitude.