Question

An astronaut with a mass of `80 kg` is floating in space. If the astronaut throws an object with a mass of `16 kg` at a speed of `3/8 m/s`, how much will his speed change by?

Answer 1

The speed of the astronaut changes by `0.075m/s` in the opposite direction of the rock.

Explanation:

This is a problem of momentum conservation, specifically an explosion. In an explosion, an internal impulse acts in order to propel the parts of a system into a variety of directions. This is often a single object, but can be, such as in this case, multiple objects which were initially at rest together. For collisions/explosions occurring in an isolated system, momentum is always conserved-no exceptions.

The Law of Conservation of Momentum:

`vecP_f=vecP_i`

For multiple objects,

`vecP=vecp_(t ot)=sumvecp=vecp_1+vecp_2+...vecp_n`

In our case, we have the momentum of the thrown object and the astronaut. We'll call the mass of astronaut `m_1` and the mass of the object `m_2`. So, we're given that `m_1=80kg`, `v_(1i)=0`, `m_2=16kg`, `v_(2i)=0m/s`, and `v_(2f)=3/8m/s`. We want to find `v_(1f)`, the final velocity of the astronaut, and compare this to his original value. We can set up an equation for momentum conservation:

`m_1v_(1i)+m_2v_(2i)=m_1v_(1f)+m_2v_(2f)`

However, both the thrown object and the astronaut are initially at rest, so the total momentum before the "explosion" is `0`.

`0=m_1v_(1f)+m_2v_(2f)`

We can manipulate the equation to solve for `v_(1f)`

`v_(f1)=-(m_2v_(f2))/m_1`

Using our known values:

`v_(f1)=-((16kg)(3/8m/s))/(80kg)`

`=>v_(f1)=-3/40m/s=-0.075m/s`

In the above equations we've defined the direction that the object is thrown in as positive, so a negative answer tells us that astronaut moves in the opposite direction of the object. Because his initial velocity was `0`, this is also how much his velocity changes by: `-0.075m/s` or `0.075m/s` in the opposite direction of the rock.

Hope this helps!