# A 0.145 kg baseball pitched at 40 m/s is hit on a horizontal line drive straight back toward the pitcher at 50 m/s. If the contact time between bat and ball is 1 ms, calculate the average force between the bat and ball during contest.

This question aims to introduce the concept of Newton’s second law of motion.

According to Newton’s 2nd law of motion, whenever a body experiences a change in its velocity, there is a moving agent called the force that acts upon it in accordance with its mass. Mathematically:

$F \ = \ m a$

The acceleration of a body is further defined as the rate of change in velocity. Mathematically:

$a \ = \ \dfrac{ \delta v }{ \delta t } \ = \ \dfrac{ v_f \ – \ v_i }{ t_2 \ – \ t_1 }$

In above equations, $v_f$ is the final velocity, $v_i$ is the initial velocity, $t_2$ is the final timestamp, $t_1$ is the initial timestamp, $F$ is the force, $a$ is the acceleration, and $m$ is the mass of the body.

According to the 2nd law of motion:

$F \ = \ m a$

$F \ = \ m \dfrac{ \delta v }{ \delta t }$

$F \ = \ m \dfrac{ v_f \ – \ v_i }{ t_2 \ – \ t_1 } \ … \ … \ … \ ( 1 )$

Since $v_f \ = \ 40 \ m/s$, $v_i \ = \ 50 \ m/s$, $t_2 \ – \ t_1 \ = \ 1 \ ms \ = \ 0.001 \ s$, and $m \ = \ 0.145 \ kg$:

$F \ = \ ( 0.145 \ kg ) \dfrac{ ( 50 \ m/s ) \ – \ ( – \ 40 \ m/s ) }{ ( 0.001 \ s ) }$

$F \ = \ ( 0.145 \ kg ) \dfrac{ ( 50 \ m/s \ + \ 40 \ m/s ) }{ ( 0.001 \ s ) }$

$F \ = \ ( 0.145 \ kg ) \dfrac{ ( 90 \ m/s ) }{ ( 0.001 \ s ) }$

$F \ = \ ( 0.145 \ kg ) ( 90000 \ m/s^2 )$

$F \ = \ 13050 \ kg m/s^2$

$F \ = \ 13050 \ N$

## Numerical Result

$F \ = \ 13050 \ N$

## Example

Imagine a striker hits a stationary soccer ball of mass 0.1 kg with a force of 1000 N. If the contact time between the striker’s foot and the ball was 0.001 seconds, what will be the speed of the ball?

Recall equation (1):

$F \ = \ m \dfrac{ v_f \ – \ v_i }{ t_2 \ – \ t_1 }$

Substituting values:

$( 1000 ) \ = \ ( 0.1 ) \dfrac{ ( v_f ) \ – \ ( 0 ) }{ ( 0.001 ) }$

$( 1000 ) \ = \ 100 \times v_f$

$v_f \ = \ \dfrac{ 1000 }{ ( 100 ) }$

$v_f \ = \ 10 \ m/s$