Question

If a 1 kilogram rock and a 6 kilogram rock were dropped from the same height above the moon's surface at the same time, they would both strike the moon's surface at the same time. The gravitational force with which the moon pulls on the 6 kg rock is 6 times greater than on the 1 kg rock. Why then do the two rocks strike the moon's surface at the same time?

Answer 1

Answer:

• Because the mass is also 6 times greater, so the acceleration is the same.

Explanation:

Force is mass multiplied by acceleration. This is (in one dimension):

$F = m a$

Now, we can see what acceleration will every rock feel:

Lets call $F_1$ the force over the first rock, that has a mass $m_1$, and lets call  $F_2$ the force over the second rock, that has a mass $m_2$. We can write the following equations:

$F_1 = m_1 * a_1$

and

$F_2 = m_2 * a_2$.

We also know that:

$F_2 = 6* F_1$, so:

$6 * F_1 = m_2 * a_2$.

But the mass is also six times greater.

$m_2 = 6* m_1$

so...

$6 * F_1 = 6 * m_1 * a_2$.

Now, lets obtain the acceleration. For the first rock we got:

$a_1 = \frac{F_1}{m_1}$

and for the second rock

$a_2 = \frac{6 * F_1}{ 6 * m_1}$

$a_2 = \frac{ F_1}{ m_1}$

But this is the same acceleration that the first rock has! So, the kinematics will be the same.