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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.

A 0.145 Kg Baseball Pitched At

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.

Expert Answer

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:

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

\[ ( 1000 ) \ = \ 100 \times v_f  \]

\[ v_f \ = \ \dfrac{ 1000 }{ ( 100 ) } \]

\[ v_f \ = \ 10 \ m/s \]

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