Question

An astronaut with a mass of `70` `kg` is floating in space. If the astronaut throws an object with a mass of `14` `kg` at a speed of `7/6` `ms^-1`, how much will his speed change by?

Answer 1

The total momentum before the throw is `0` `kgms^-1`, and since momentum is conserved this will also be the total momentum after the throw: `v_1=-(m_2v_2)/m_1=-(14*7/6)/70=-0.23` #ms^-1. The minus sign shows it is in the opposite direction to the thrown object.

Explanation:

Momentum is given by:

`p=mv`

Momentum is conserved - the total momentum after an event is equal to the total momentum before.

In this case, if we consider the astronaut and object together to be at rest in space, `v=0 to p=mv=0`, so the momentum before the throw is `0` `kgms^-1`.

After the throw, the total momentum will still be zero. If we call the astronaut '1' and the thrown object '2', it will be made up as follows:

`p_"total"=m_1v_1+m_2v_2=0`

Rearranging,

`m_1v_1=-m_2v_2`

`v_1=-(m_2v_2)/m_1`

The minus sign just indicates that the astronaut's motion is in the opposite direction to that of the thrown object, as Newton's Third Law would suggest.

I've substituted in the mass of the thrown object, its speed and the mass of the astronaut in the answer above.

(It's worth noting that we didn't need to assume that the astronaut was stationary, since we are asked only how much his speed changes... an existing velocity could be included but would cancel out to give the same answer because motion is relative.)