Calculating Velocity of Ice Skater After Throwing Snowball

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In summary, the skater, with a mass of 63 kg, throws a 4.06 kg snowball with a velocity of (-14.3 i + 17.2 j) m/s while at rest on frictionless ice. The subsequent velocity of the skater can be found using the conservation of momentum equation, (m1+m2)( -14.3 i + 17.2 j ) = m1v1+m2v2, where m1 is the mass of the skater, m2 is the mass of the snowball, and v1 and v2 are the velocities before and after the throw, respectively.
  • #1
PierceJ
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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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  • #2
What is your question? Where are the units? What about significant digits?
 
  • #3
The question is there "Find the subsequent velocity of the skater."
The units are also there, m/s
4 significant digits
 
  • #4
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:
  • #5
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)?
 
  • #6
p = Mw + mv?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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.
 

1. How does the ball affect the skater's motion?

The ball exerts a force on the skater upon impact, causing a change in the skater's velocity and direction of motion.

2. Will the skater's motion always be affected by the ball?

Yes, the skater's motion will always be affected by the ball due to the transfer of energy and momentum between the two objects.

3. In what direction will the skater move after being hit by the ball?

The direction of the skater's motion will depend on the direction and magnitude of the force applied by the ball. If the force is applied in the same direction as the skater's motion, the skater will continue moving in that direction but with a change in velocity. If the force is applied in a different direction, the skater's motion will change accordingly.

4. How does the mass of the ball affect the skater's motion?

The mass of the ball will affect the amount of force it exerts on the skater and therefore the magnitude of the change in the skater's motion. A heavier ball will exert a greater force and cause a larger change in the skater's motion compared to a lighter ball.

5. Will the skater's motion always be predictable after being hit by the ball?

No, the skater's motion may not always be predictable after being hit by the ball as it depends on various factors such as the force applied, the direction of the force, and the initial velocity of the skater. Other external factors such as friction and air resistance may also affect the skater's motion and make it less predictable.

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