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: $8.493(10)^{-3} m/s$

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 $V_{f}$:

$V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}}$ (5)

$V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}$

Finally:

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