# How do airsoft spring guns work

## Conservation of momentum and shocks

We consider the following energy balance table:

position | I. | II | III |
---|---|---|---|

\ ({E _ {{\ rm {pot}}}} \) | - | \ (m \ cdot g \ cdot y_2 \) | \ (m \ cdot g \ cdot y_3 \) |

\ ({E _ {{\ rm {kin}}}} \) | - | \ (\ frac {1} {2} \ cdot m \ cdot {v _ {{\ rm {II}}}} ^ 2 \) | - |

\ ({E _ {{\ rm {Spann}}}} \) | \ (\ frac {1} {2} \ cdot D \ cdot {y_1} ^ 2 \) | \ (\ frac {1} {2} \ cdot D \ cdot {y} ^ 2 \) | - |

\ ({E _ {{\ rm {ges}}}} \) | \ (\ frac {1} {2} \ cdot D \ cdot {y_1} ^ 2 \) | \ (m \ cdot g \ cdot y_2 + \ frac {1} {2} \ cdot m \ cdot {v _ {{\ rm {II}}}} ^ 2 + \ frac {1} {2} \ cdot D \ cdot {y} ^ 2 \) | \ (m \ cdot g \ cdot y_3 \) |

Since, according to the law of conservation of energy, the total energy of the system must be the same at all times, i.e. especially at times I and II, the following applies

\ [\ frac {1} {2} \ cdot D \ cdot {y_1} ^ 2 = m \ cdot g \ cdot {y_2} + \ frac {1} {2} \ cdot m \ cdot {v _ {{\ rm {II}}}} ^ 2 + \ frac {1} {2} \ cdot D \ cdot {y ^ 2} \]

Solving this equation for \ ({v _ {{\ rm {II}}}} \) gives

\ [{v _ {{\ rm {II}}}} = \ sqrt {\ frac {{\ frac {1} {2} \ cdot D \ cdot {y_1} ^ 2 - \ frac {1} {2} \ cdot D \ cdot {y ^ 2} - m \ cdot g \ cdot {y_2}}} {{\ frac {1} {2} \ cdot m}}} \]

Substituting the given values yields

\ [{v _ {{\ rm {II}}}} = \ sqrt {\ frac {{\ frac {1} {2} \ cdot 90.0 \ frac {{\ rm {N}}} {{\ rm {m}}} \ cdot {{\ left ({0.160 {\ rm {m}}} \ right)} ^ 2} - \ frac {1} {2} \ cdot 90.0 \ frac {{\ rm { N}}} {{\ rm {m}}} \ cdot {{\ left ({0.011 {\ rm {m}}} \ right)} ^ 2} - 0.100 {\ rm {kg}} \ cdot 9, 81 \ frac {{\ rm {m}}} {{{{\ rm {s}} ^ {\ rm {2}}}}} \ cdot 0.149 {\ rm {m}}}} {{\ frac { 1} {2} \ cdot 0.100 {\ rm {kg}}}}} = 4.47 \ frac {{\ rm {m}}} {{\ rm {s}}} \]

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