The **aim of this question** is to find the **change in speed of two bodies** after collision by utilizing the concept of **elastic collisions.**

Whenever two bodies collide, their **momentum and energy remain constant** as per the **energy and momentum conservation laws**. Based on these laws we derive the concept of** elastic collisions** where the **friction is ignored.**

During **elastic collisions** the speed of two bodies after the collision can be **determined by the following formula**:

\[ v’_B \ = \dfrac{ 2m_A }{ m_A + m_B } v_A –Â \dfrac{ m_A – m_B }{ m_A + m_B } v_B \]

\[ v’_AÂ \ = \dfrac{ m_A – m_B }{ m_A + m_B } v_A +Â \dfrac{ 2 m_B }{ m_A + m_B } v_B \]

Where $ v’_A $ and $ v’_B $ are the **final speeds after c****ollision**, $ v_A $ and $ v_B $ are the **speeds before collision,** and $ m_A $ and $ m_B $ are the **masses** of the colliding bodies.

If we **consider a special case of elastic collision** such that both the bodies have **equal mass** ( i.e. $ m_A \ = \ m_B \ = \ m), the above** equations reduce to:**

\[ v’_B \ = \dfrac{ 2m }{ m + m } v_A –Â \dfrac{ m – m }{ m + m } v_B \]

\[ v’_A \ = \dfrac{ m – m }{ m_A + m_B } v_A +Â \dfrac{ 2 m }{ m + m } v_B \]

The above **equations further reduce to:**

\[ v’_B \ = v_A \]

\[ v’_A \ = v_B \]

Which means that whenever two equally massed bodies collide, they **exchange their speeds.**

## Expert Answer

**Given:**

\[ m \ = \ 0.5 \ lb \ = \ 0.5 \times 0.453592 \ kg \ = \ 0.23 \ kg \]

**Part (a) – Downward movement of mass A.**

**Total Energy of mass A at the top:**

\[ TE_{top} \ = \ KE_A + PE_A \]

\[ TE_{top} \ = \ \dfrac{ 1 }{ 2 } m v_A^2 + m g h \]

\[ TE_{top} \ = \ \dfrac{ 1 }{ 2 } (0.23) (0)^2 + (0.23) (9.8) (3) \]

\[ TE_{top} \ = \ 6.762 \]

**Total Energy of mass A at the bottom:**

\[ TE_{bottom} \ = \ KE_A + PE_A \]

\[ TE_{bottom} \ = \ \dfrac{ 1 }{ 2 } m v_A^2 + m g h \]

\[ TE_{bottom} \ = \ \dfrac{ 1 }{ 2 } (0.23) v_A^2 + (0.23) (9.8) (0) \]

\[ TE_{bottom} \ = \ 0.115 v_A^2 \]

**From the energy conservation law:**

\[ TE_{bottom} \ = \ TE_{top} \]

\[ 0.115 v_A^2 \ = \ 6.762 \]

\[ v_A^2 \ = \dfrac{ 6.762 }{ 0.115 } \]

\[ v_A^2 \ = 58.8 \]

\[ v_A \ = 7.67 \ m/s \]

**Part (b) – Collision of mass A with mass B.**

Speeds before collision:

\[ v_A \ = 7.67 \ m/s \]

\[ v_B \ = 0 \ m/s \]

Speeds after collision (as derived above):

\[ v’_B \ = v_A \]

\[ v’_A \ = v_B \]

Substituting values:

\[ v’_B \ =Â 7.67 \ m/s \]

\[ v’_A \ = 0 \ m/s \]

**Part (c) – Collision of mass B with mass C.**

Speeds before collision:

\[ v_B \ = 7.67 \ m/s \]

\[ v_C \ = 0 \ m/s \]

Speeds after collision (similar to part b):

\[ v’_C \ = v_B \]

\[ v’_B \ = v_C \]

Substituting values:

\[ v’_C \ =Â 7.67 \ m/s \]

\[ v’_B \ = 0 \ m/s \]

## Numerical Result

After the second collision:

\[ v’_A \ = 0 \ m/s \]

\[ v’_B \ = 0 \ m/s \]

\[ v’_C \ =Â 7.67 \ m/s \]

## Example

Suppose** two bodies of mass 2kg and 4 kg** have **speeds of 1 m/s and 2 m/s**. If they collide, what will be **their final speeds after the collision**.

**Speed of first body:**

\[ v’_AÂ \ = \dfrac{ m_A – m_B }{ m_A + m_B } v_A +Â \dfrac{ 2 m_B }{ m_A + m_B } v_B \]

\[ v’_A \ = \dfrac{ 2 – 4 }{ 2 + 4 } ( 1 ) +Â \dfrac{ 2 ( 4 ) }{ 2 + 4 } ( 2 ) \]

\[ v’_A \ = \dfrac{ -2 }{ 6 } +Â \dfrac{ 16 }{ 6 } \]

\[ v’_A \ = 2.33 \ m/s \]

**Similarly:**

\[ v’_B \ = \dfrac{ 2m_A }{ m_A + m_B } v_A –Â \dfrac{ m_A – m_B }{ m_A + m_B } v_B \]

\[ v’_B \ = \dfrac{ 2 ( 2 ) }{ 2 + 4 } ( 1 ) –Â \dfrac{ 2 – 4 }{ 2 + 4 } ( 2 ) \]

\[ v’_B \ = \dfrac{ 4 }{ 6 } + \dfrac{ 4 }{ 6 } \]

\[ v’_B \ = 1.33 \ m/s \]