# What is the change in velocity of the international space station?

## During a docking maneuver, a supply pod of 4000kg accidentally collides with the international space station of mass 100000kg with a relative velocity of $2 m {s}^{-} 1$. Assuming the pod attaches to the station, what is the change in velocity of the international space station?

Mar 11, 2017

$0.077 m {s}^{-} 1$, rounded to two decimal places.

#### Explanation:

Let $u$ be the velocity of International space station before accident.

Therefore, velocity of supply pod $= \left(u + 2\right) m {s}^{-} 1$

Initial momentum of both bodies $= 100000 \times u + 4000 \times \left(u + 2\right)$
$= {10}^{5} u + 4 \times {10}^{3} u + 8 \times {10}^{3}$

After the accident let $v$ be velocity of the combined mass
Final momentum$= \left({10}^{5} + 4 \times {10}^{3}\right) v$

From Law of conservation of momentum we have

$\left({10}^{5} + 4 \times {10}^{3}\right) u + 8 \times {10}^{3} = \left({10}^{5} + 4 \times {10}^{3}\right) v$
$\implies \left({10}^{5} + 4 \times {10}^{3}\right) v - \left({10}^{5} + 4 \times {10}^{3}\right) u = 8 \times {10}^{3}$
$\implies \left({10}^{5} + 4 \times {10}^{3}\right) \left(v - u\right) = 8 \times {10}^{3}$
$\implies \left(v - u\right) = \frac{8 \times {10}^{3}}{\left({10}^{5} + 4 \times {10}^{3}\right)} = \frac{8}{104} = \frac{1}{13} = 0.077 m {s}^{-} 1$