Calculating Velocity of Ice Skater After Throwing Snowball

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A 63 kg ice skater at rest throws a 4.06 kg snowball with a specified velocity, prompting the need to calculate the skater's subsequent velocity. The total momentum of the system is calculated, leading to the equation ( -58.058i + 69.832j ) = 67.06v, resulting in the skater's velocity of ( -0.8658i + 1.041j). Participants discuss the importance of correctly interpreting the problem and the conservation of momentum principles. Clarifications about the direction of the skater's movement and the significance of units and significant digits are also raised. The conversation emphasizes understanding momentum before and after the throw to solve the problem accurately.
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Homework Statement


A 63 kg ice skater, at rest on frictionless ice, tosses a 4.06 kg snowball with velocity given by v = ( -14.3 i + 17.2 j ) m/s, where the x and y axes are both in the horizontal plane. Find the subsequent velocity of the skater.

Homework Equations


p = mv
p = m1v1+...

The Attempt at a Solution


I started by finding the total momentum of the system by multiplying 4.06 by each part of the vector and I got:
( -58.058i + 69.832j )
Then I set that equal to the two masses added together times v.
( -58.058i + 69.832j ) = 67.06v
v = ( -0.8658i + 1.041j)
 
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What is your question? Where are the units? What about significant digits?
 
The question is there "Find the subsequent velocity of the skater."
The units are also there, m/s
4 significant digits
 
PierceJ said:

Homework Statement


A 63 kg ice skater, at rest on frictionless ice, tosses a 4.06 kg snowball with velocity given by v = ( -14.3 i + 17.2 j ) m/s, where the x and y axes are both in the horizontal plane. Find the subsequent velocity of the skater.

Homework Equations


p = mv
p = m1v1+...

The Attempt at a Solution


I started by finding the total momentum of the system by multiplying 4.06 by each part of the vector and I got:
( -58.058i + 69.832j )
Then I set that equal to the two masses added together times v.
( -58.058i + 69.832j ) = 67.06v
v = ( -0.8658i + 1.041j)

Is the skater moving in the same direction as the ball he has just thrown?

Or was the ball thrown at him, as the title of your post suggests?
 
Last edited:
If the skater moves off with velocity w,and the masses are m (snowball) and M, what is the conservation equation (in symbols, not numbers)?
 
p = Mw + mv?
 
PierceJ said:
The question is there "Find the subsequent velocity of the skater."
I meant what is your question, why did you post? I now realize that I misread the question, as I had in mind what you wrote for the title of the thread. But is the problem really "Ball thrown at skater," as PeroK also asked?

PierceJ said:
The units are also there, m/s
I meant in your answer.

QUOTE="PierceJ, post: 5030654, member: 546115"]
4 significant digits[/QUOTE]
Why 4? Is this compatible with the data from the problem?

But you should start by answering haruspex's question correctly.
 
I realize now that I have read the problem wrong, but even so I still don't know how to answer haruspex's question. I know this forum is designed for me to figure out with only a little help but I'm completely lost here.
 
PierceJ said:
I realize now that I have read the problem wrong, but even so I still don't know how to answer haruspex's question. I know this forum is designed for me to figure out with only a little help but I'm completely lost here.
The question is about conservation. What can you write that relates before the throw to after the throw?
 
  • #10
(m1+m2)( -14.3 i + 17.2 j ) = m1v1+m2v2?
 
  • #11
Start from the momentum.
 
  • #12
PierceJ said:
(m1+m2)( -14.3 i + 17.2 j ) = m1v1+m2v2?
It says the skater throws, not catches, the snowball. What is the momentum of the system before the skater does so? What is the momentum of the snowball after the skater has thrown it?
 
  • #13
The momentum before is 0? Cause the guy isn't moving?
Momentum of the snowball after is its mass times the velocity given in the problem.
 
  • #14
PierceJ said:
The momentum before is 0? Cause the guy isn't moving?
Momentum of the snowball after is its mass times the velocity given in the problem.
Right. So have another go at the equation.
 
  • #15
Ah I see, I got it now. Thank you for putting up with me haha.
 
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