This **question aims** to develop the basic understanding of** elastic collisions** for the case of **two bodies.**

Whenever two bodies have a collision, they have to obey **momentum and energy conservation laws**. An **elastic collision** is a type of collision where these two laws hold but the**Â effects** such as the **friction are ignored**.

The speed of two bodies after an **elasticÂ collision**Â can beÂ **calculated by using the following equations**:

\[ v’_1 \ = \dfrac{ m_1 – m_2 }{ m_1 + m_2 } v_1 +Â \dfrac{ 2 m_2 }{ m_1 + m_2 } v_2 \]

\[ v’_2 \ = \dfrac{ 2m_1 }{ m_1 + m_2 } v_1 –Â \dfrac{ m_1 – m_2 }{ m_1 + m_2 } v_2 \]

Where $ v’_1 $ and $ v’_2 $ are the **final speeds after c****ollision**, $ v_1 $ and $ v_2 $ are the **speeds beforeÂ collision,** and $ m_1 $ and $ m_2 $ are the **masses**Â of the colliding bodies.

## Expert Answer

**Given:**

\[ m_{ 1 } \ = \ 20.0 \ g \ =\ 0.02 \ kg \]

\[ v_{ 1 } \ = \ 0.2 \ m/s \]

\[ m_{ 2 } \ = \ 30.0 \ g \ =\ 0.03 \ kg \]

\[ v_{ 2 } \ = \ 0.3 \ m/s \]

Speed of first body after an **elasticÂ collision**Â can beÂ **calculated by using the following equation**:

\[ v’_1 \ = \dfrac{ m_1 – m_2 }{ m_1 + m_2 } v_1 \ + \ \dfrac{ 2 m_2 }{ m_1 + m_1 } v_2 \]

**Substituting values:**

\[ v’_1 \ = \dfrac{ ( 0.02 ) – ( 0.03 ) }{ ( 0.02 ) + ( 0.03 ) } ( 0.2 ) \ + \Â \dfrac{ 2 ( 0.03 ) }{ ( 0.02 ) + ( 0.03 ) } ( 0.3 ) \]

\[ v’_1 \ = \dfrac{ -0.01 }{ 0.05 } ( 0.2 ) \ + \Â \dfrac{ 0.06 }{ 0.05 } ( 0.3 ) \]

\[ v’_1 \ = -0.04 \ + \Â 0.36 \]

\[ v’_1 \ = 0.32 \ m/s \]

Speed of second body after an **elasticÂ collision**Â can beÂ **calculated by using the following equation**:

\[ v’_2 \ = \dfrac{ 2m_1 }{ m_1 + m_2 } v_1 \ – \ \dfrac{ m_1 – m_2 }{ m_1 + m_2 } v_2 \]

**Substituting values:**

\[ v’_2 \ = \dfrac{ 2 ( 0.02 ) }{ ( 0.02 ) + ( 0.03 ) } ( 0.2 ) \ – \ \dfrac{ ( 0.02 ) – ( 0.03 ) }{ ( 0.02 ) + ( 0.03 ) } ( 0.3 ) \]

\[ v’_2 \ = \dfrac{ 0.04 }{ 0.05 } ( 0.2 ) \ – \ \dfrac{ -0.01 }{ 0.05 } ( 0.3 ) \]

\[ v’_2 \ = 0.16 \ + \ 0.06 \]

\[ v’_2 \ = 0.22 \ m/s \]

## Numerical Results

After the collision:

\[ v’_1 \ = 0.32 \ m/s \]

\[ v’_2 \ = 0.22 \ m/s \]

## Example

**Find the speed of the bodies if their initial speeds are reduced by a factor of 2**.

In this case, the **formulae** suggest that** reducing the speeds by a factor of 2** will also **reduce the speeds after collision by the same factor**. So:

\[ v’_1 \ = 2 \times 0.32 \ m/s \]

\[ v’_1 \ = 0.64 \ m/s \]

**And:**

\[ v’_2 \ = 2 \times 0.22 \ m/s \]

\[ v’_2 \ = 0.44 \ m/s \]