# A 20.0 g marble slides to the left with a velocity of magnitude 0.200 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on elastic collision with a larger 30.0 g marble sliding to the right with a velocity of magnitude 0.300 m/s. Find the magnitude of the velocity of 30.0 g marble after the collision.

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 effects such as the friction are ignored.

The speed of two bodies after an elastic collision can be calculated by using the following equations:

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

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

Where $v’_1$ and $v’_2$ are the final speeds after collision, $v_1$ and $v_2$ are the speeds before collision, and $m_1$ and $m_2$ are the masses of the colliding bodies.

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 elastic collision can be 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:

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

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

$v’_1 \ = -0.04 \ + \ 0.36$

$v’_1 \ = 0.32 \ m/s$

Speed of second body after an elastic collision can be 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:

$v’_2 \ = 2 \times 0.22 \ m/s$

$v’_2 \ = 0.44 \ m/s$