1) Change in momentum of the rock: 138.6 kg m/s
2) Magnitude of change of Earth's velocity: [tex]2.31\cdot 10^{-23} m/s[/tex]
Explanation:
1)
In order to find the change in momentum of the rock through the fall, we have to find its change in velocity first. This can be done by using the law of conservation of energy: in fact, the kinetic energy gained by the rock is equal to the loss in gravitational potential energy. So we can write
[tex]\Delta KE = -\Delta PE[/tex]
where:
[tex]\Delta PE=mg(h_f-h_i)[/tex] is the change in potential energy, where
m = 14 kg is the mass of the rock
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]h_f = 0[/tex] is the final height
[tex]h_i = 5.0 m[/tex] is the initial height
Substituting,
[tex]\Delta PE=(14)(9.8)(0-5)=-686 J[/tex]
So, the rock has gained 686 J of kinetic energy; and since it started from rest, its initial kinetic energy was zero, so its final kinetic energy is 686 J:
[tex]K_f=\frac{1}{2}mv^2=686 J[/tex]
And solving for v, we find the final velocity of the rock:
[tex]v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(686)}{14}}=9.9 m/s[/tex]
So, the change in momentum of the rock is:
[tex]\Delta p = m\Delta v = (14)(9.9-0)=138.6 kg m/s[/tex]
2)
According to the law of conservation of momentum, the total momentum of the rock-Earth system must be conserved:
[tex]p_{rock}+p_{earth}=const.[/tex]
This means that during the fall of the rock, the change in momentum of the rock is opposite to the change in momentum of the Earth:
[tex]\Delta p_{rock} = -\Delta p_{earth}[/tex]
This means that the change in momentum of the Earth is
[tex]\Delta p_{earth}=-138.6 kg m/s[/tex]
And since the mass of the Earth is
[tex]m=6.0\cdot 10^{24}kg[/tex]
Its change in velocity is:
[tex]\Delta v = \frac{\Delta p}{m}=\frac{-138.6}{6.0\cdot 10^{24}}=-2.31\cdot 10^{-23} m/s[/tex]
And the negative sign means the direction of the change in velocity is opposite to that of the rock (so, the Earth accelerates towards the rock).
Learn more about momentum:
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