Question

A 62.7 kg astronaut is holding a 6.5 kg tool pack while drifting at 12.3 m/s to the right. She throws the tool pack away so thatit is drifting at 2.4 m/s to the left. How fast is she moving after the throw? Assume that they are out in deep space with noother forces acting on them.

Answer 1

`v_(new)=9.9ms^(-1)`

Step-by-step explanation: Assuming that it took one second to throw the tool away, then we have to use the final velocity formula to get the new velocity:

`v(t)=v_o+at`

One second would imply that the deceleration in the initial astronaut velocity is:

`\begin{gathered} a=-2.4ms^(-2) \\ \end{gathered}`

Note the sign convention!

The new velocity of the astronaut after the throw would be now:

`\begin{gathered} t=1s \\ \therefore\Rightarrow \\ v(1s)=12.3ms^(-1)+(-2.4ms^(-2))(1s)=(12.3-2.4)ms^(-1) \\ v_(new)=9.9ms^(-1) \end{gathered}`