Impulse-Momentum: Find Magnitude of Earth's Impulse

No extra output.In summary, the conversation discusses the concept of conservation of momentum in the context of a person standing on a frictionless ice and throwing a rock in order to get off the ice. The question at hand is what is the magnitude of the impulse on the Earth as a result of the throw. The conversation goes on to consider external forces and the change in momentum of the Earth as a result of the throw, eventually arriving at a final answer of 90.99 for the magnitude of the impulse on the Earth.
  • #1
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Homework Statement



You are standing on a large sheet of frictionless ice and are holding a large rock. In order to get off the ice, you throw the rock so it has velocity relative to the Earth of 11.3 m/s at an angle of 34 above the horizontal. Your mass is 67.0 kg and the rock's mass is 14.4 kg.

What is the magnitude of the impulse on the Earth as a result of this throw?

Homework Equations



p=mv

J = p_2 - p_1

The Attempt at a Solution



The first part of the question had me find the speed of me after the throw.

mmevme = mrockvrock

67(v) = 14.4(11.3cos34)
v= 2.01 m/s

Now i am asked to find the impulse as stated above.

I thought the impulse would just be zero since momentum of this is conserved but i was wrong.

Can anyone help me with this please?

Thank you :)
 
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  • #2
Momentum is conserved only in the absence of external forces. Are there external forces in the x direction? In the y direction?
 
  • #3
If momentum is conserved only in the absence of external forces then i would say there are no external forces because momentum was conserved from finding the first question.

But from you asking me if there are external forces in the x and y direction i want to say i am probably wrong about what i just said...
 
  • #4
Ok you can see that the rock gains a velocity in the positive y direction after the throw.

For momentum to be conserved the Earth needs to gain momentum in the negative y direction.

Assume the Earth has no momentum prior to the rock being thrown and solve for the change in momentum.
 
  • #5
Ok i get what you are saying.

so if the Earth has no momentum at first then it would be zero and so the change in momentum will just be the momentum of the y velocity of the rock

J = (14.4)11.3sin(34) - 0

J = 90.99

Does this seem right?

Thanks for the help
 
  • #6
mybrohshi5 said:
Ok i get what you are saying.

so if the Earth has no momentum at first then it would be zero and so the change in momentum will just be the momentum of the y velocity of the rock

J = (14.4)11.3sin(34) - 0

J = 90.99

Does this seem right?

Ok that would be the momentum for the rock in the y direction. Now for momentum to be conserved you need to find the appropriate speed the Earth gains to conserve momentum (it shouldn't be the same).
 
  • #7
sorry i just edited what i calculated.

does my edit seem right? i factored in mass
 
  • #8
so the speed of the Earth will be

v_earth = 1.523 E^-23

Im a little confused now.

so if impulse is change in momentum, yet change in momentum is conserved then how is the impulse not zero?
 
  • #9
mybrohshi5 said:
so the speed of the Earth will be

v_earth = 1.523 E^-23

Im a little confused now.

so if impulse is change in momentum, yet change in momentum is conserved then how is the impulse not zero?

Ok good that seems correct. Now you have to look at what the question asks. It wants impulse on the the Earth (not on the total system which would be, as you stated, zero).

So the initial momentum of the Earth is zero and the final momentum of the Earth is V_earth*M_earth the change in momentum would be the difference of the two.

I believe the answer will come out to be 90.99 (because the momentum of the rock equals the momentum of the Earth). Sorry for the extra superfluous step there but hopefully it explains things a bit more.
 
  • #10
It was 90.99

It did help explain things more. I get it now :)

Thank you for the help. I appreciate it.
 
  • #11
mybrohshi5 said:
It was 90.99

It did help explain things more. I get it now :)

Thank you for the help. I appreciate it.

No problem :)
 

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