Question

A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the asteroid stick together, becoming a single object with a single velocity. If the mass of the comet is 1.71 ✕ 1014 kg and the mass of the asteroid is 6.06 ✕ 1020 kg, what is the final velocity of their combination? (Assume the comet's initial direction is positive. Indicate the direction with the sign of your answer.)

Answer 1

Answer:
`8.493(10)^(-3) m/s`

Step-by-step explanation:

According to the conservation of linear momentum principle, the initial momentum
`p_(i)` (before the collision) must be equal to the final momentum
`p_(f)` (after the collision):

`p_(i)=p_(f)` (1)

In addition, the initial momentum is:

`p_(i)=m_(1)V_(1)+m_(2)V_(2)` (2)

Where:

`m_(1)=1.71(10)^(14) kg` is the mass of the comet

`m_(2)=6.06(10)^(20) kg` is the mass of the asteroid

`V_(1)=3.01(10)^(4) m/s` is the velocity of the comet, which is positive

`V_(2)=0 m/s` is the velocity of the asteroid, since it is at rest

And the final momentum is:

`p_(f)=(m_(1)+m_(2))V_(f)` (3)

Where:

`V_(f)` is the final velocity

Then :

`m_(1)V_(1)+m_(2)V_(2)=(m_(1)+m_(2))V_(f)` (4)

Isolating
`V_(f)`:

`V_(f)=(m_(1)V_(1))/(m_(1)+m_(2))` (5)

`V_(f)=((1.71(10)^(14) kg)(3.01(10)^(4) m/s))/(1.71(10)^(14) kg+6.06(10)^(20) kg)`

Finally:

`V_(f)=8.493(10)^(-3) m/s` This is the final velocity, which is also in the positive direction.