A 14 kg rock starting from rest free falls through a distance of 5.0 m with no air resistance. Find the momentum change of the rock caused by its fall and the resulting change in the magnitude of earth’s velocity. Earth’s mass is 6.0 × 1024 kg. Show all your work, assuming the rock-earth system is closed.

Question

irishboy123 · Answer

start by posting your own thoughts/workings/calculations.

anonymous · Answer

all i got so far is p=mv p= 14kg(0m)  p=o kg/m and then im stuck @IrishBoy123

matt101 · Answer

That's a good start! But that's only the INITIAL momentum at the beginning of the situation when the velocity of the rock is 0, before the it started to fall.

But we're looking for the CHANGE in momentum (the impulse). That means we need to know the FINAL velocity of the rock as well, in order to calculate the final momentum. Remember, the change in momentum is final momentum minus initial momentum.

What is the final velocity of the rock? And what is the final momentum?

anonymous · Answer

how do I find those?

matt101 · Answer

Let's focus on velocity for now. You know the rock fell 5 m. You know the initial velocity of the rock was 0 m/s. You also know, even though it doesn't say in the question, that the acceleration the rock was experiencing was 9.8 m/s^2.

You have:
d = 5 m
v(i) = 0 m/s
a = 9.8 m/s^2

And you want to find v(f). Do you know an equation that will let you find v(f)?

anonymous · Answer

i dont  know

anonymous · Answer

@matt101

matt101 · Answer

The equation is:

$$v_f^2=v_i^2+2ad$$
I'm assuming that if you're already learning about momentum, you must have learned this equation when you were doing kinematics/motion. Use this equation to solve for v(f) and let me know what you get!

anonymous · Answer

960.4 m/s^2

matt101 · Answer

Can you show me your work?

anonymous · Answer

0m/s^2 + 2(9.8m/s^2) (5)  
0m/s + 2 (192.08) (5)
960.04

matt101 · Answer

Ok remember the units for acceleration are m/s^2 - we're not talking about squaring the 9.8. So the equation is:

$$v_f^2=0^2+2(9.8)(5)$$$$v_f^2=98$$$$v_f \approx 10$$
So the final velocity is about 10 m/s, That makes the final momentum...?

anonymous · Answer

15 m/s?

matt101 · Answer

It's calculated the exact same way you did it the first time: p=mv.

What's the final momentum?

anonymous · Answer

50?

matt101 · Answer

Just plug it in to p=mv! You have the mass of the rock and you have the velocity of the rock so:

$$p=mv$$$$p=(14)(10)$$$$p=140$$
The final momentum is 140 kg m/s. The CHANGE in momentum is final momentum minus initial momentum: 140-0=140. The change in momentum happens to be the same as the final momentum, because you started from rest (0 momentum).

Does this make sense so far? Any questions?

anonymous · Answer

yes that makes sense. how did you go from 98 to 10

matt101 · Answer

The equation uses the velocities SQUARED, so I had to take the square root to solve for the final velocity. The square root of 98 is about 10 (it's closer to 9.9, but I'm just using 10 for easier numbers).

anonymous · Answer

oh okay thanks

matt101 · Answer

Now the second part of the question is asking about the change in the Earth's velocity.

By conservation of momentum, you know that whatever momentum you start with is equal to whatever momentum you finish with. So the initial momentum of the rock and the earth is equal to the final momentum of the rock and the Earth. So:

$$mv_i+MV_i=mv_f+MV_f$$
The lower case letters are for the rock, the upper case letters are for the Earth. Do you think you can finish the question now?

anonymous · Answer

what would be the initial velocity and mass of earth?

matt101 · Answer

From our frame of reference, the Earth isn't moving at the beginning. The mass of the earth is given in the question, 6.0 × 10^24 kg.

anonymous · Answer

so it would be 14(0) + 6.0x10^24 (0) = 14(0) + 6.0x10^24 (0)

anonymous · Answer

the first side equals 0

matt101 · Answer

Oh I should clarify something! When we're looking at the rock hitting the Earth, the initial situation is just before the rock hits the Earth, and the final situation is just after the rock hits the Earth.

That means the initial velocity of the rock IN THIS CASE is really the final velocity of the rock from the first question: 10 m/s. We know that when the rock hits the Earth it stops moving, so the final velocity of the rock IN THIS CASE is 0 m/s. It's the final velocity of the Earth that we now want to find, so V(f) is our unknown that we're solving for!

anonymous · Answer

the first side of the equation is 0 .

for the other side i dont know

matt101 · Answer

We have to change some numbers from what you wrote above, because there must be some momentum to start with if the Earth experiences a change in its own momentum.

The initial velocity of the rock in this case is 10 m/s, and the initial velocity of the Earth is 0 m/s as I mentioned above. The final velocity of the rock in this case is 0 m/s, and it's the final velocity of the Earth that we want to find. So to update your equation:

14(10) + 6.0x10^24 (0) = 14(0) + 6.0x10^24 (V(f))

Solve for V(f)!

anonymous · Answer

woulnt it be 0

matt101 · Answer

Nope! Let's look at it step by step:

$$(14)(10)+(6 \times 10^{24})(0)=(14)(0)+(6 \times 10^{24})V_f$$$$140+0=0+(6 \times 10^{24})V_f$$$$140=(6 \times 10^{24})V_f$$$$V_f=2.3 \times 10^{-23}$$
Does that make sense?

anonymous · Answer

yea i think so

matt101 · Answer

If you have any other questions let me know!

anonymous · Answer

thank you