Question

During a maneuver in space, a space craft separates into two pieces, each of mass m. Before the separation, the spacecraft was moving with a speed v. If one of the pieces is at rest after the separation, which one of the following statements concerning this maneuver is true?

a) This maneuver conserves kinetic energy.b) The maneuver does not conserve total energy.c) This maneuver does not conserve momentum.d) If one piece is at rest, the other is moving with a speed 2v.e) One piece cannot be at rest. The must both be moving with a speed v/2.

Answer 1

Answer: d) If one piece is at rest, the other is moving with a speed 2v.

Step-by-step explanation:

The momentum
`p` is given by the following equation:

`p=m.V` (1)

Where:

`m` is the mass of the object

`V` is the velocity of the object

Now, the total momentum must be conserved. According to this:

`p_(i)=p_(f)` (2)

Where
`p_(i)` is the initial momentum (before the maneuver) and
`p_(f)` the final momentum (after the maneuver).

Being:

`p_(i)=2m.V` (3)

`p_(f)=m.V_(1)+m.V_(2)` (4)

Then:

`2m.V=m.V_(1)+m.V_(2)` (5) Conservation of momentum

If we are told
`V_(1)=0` because one of the pieces of mass
`m` is at rest, in order to fullfil the conservation of momentum
`V_(2)=2V`.

In this way:

`2m.V=m(2V)` (6)

`2m.V=2m.V`