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:**

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

\[ ( 1000 ) \ = \ 100 \times v_f \]

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

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