Finding velocities of two people on skates

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Homework Help Overview

The problem involves two individuals on skates, where one throws a snowball to the other. The first person has a mass of 66.0 kg and is moving at a speed of 2.50 m/s while throwing a 0.0430 kg snowball at 31.0 m/s. The second person, with a mass of 56.0 kg, is initially at rest and catches the snowball. The objective is to determine their velocities after the snowball is exchanged, disregarding friction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Some participants suggest using conservation of linear momentum as a starting point for the problem. Others question the clarity of the initial conditions and the definitions of the states before, during, and after the snowball is thrown and caught.

Discussion Status

The discussion is ongoing, with participants exploring the correct application of momentum conservation. There is recognition of the need to establish distinct time frames in the problem, and some guidance has been offered regarding the formulation of momentum equations.

Contextual Notes

Participants note the importance of accurately identifying what is moving at specific speeds and the implications of the initial conditions on the momentum calculations. There is also a mention of potential confusion regarding the high velocities calculated in initial attempts.

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Homework Statement


A 66.0 kg person throws a 0.0430 kg snowball forward with a ground speed of 31.0 m/s. A second person, with a mass of 56.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.50 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.
a. thrower
b. catcher

The Attempt at a Solution


I do not even know where to begin
 
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As with all of the three problems that you posted consecutively, begin by conserving linear momentum.
 
mv + mv = mv + mv
(66.0 kg)(31.0 m/s) + (56.0 kg)(0 m/s) = (66.0 kg)v1f + (56.0 kg)v2f

2046 m/s = 66v1f + 56v2f
(1924 m/s = -(v1f - v2f) )66

2046 m/s = 66v1f + 56v2f
126984 m/s = -66v1f + 66v2f

129030 m/s = 122v2f
v2f = 1057.6 m/s

1924 m/s = - v1f + 1057.6 m/s
v1f= -866.4 m/s

these numbers seem very high
 
mandy9008 said:
mv + mv = mv + mv
(66.0 kg)(31.0 m/s) + (56.0 kg)(0 m/s) = (66.0 kg)v1f + (56.0 kg)v2f
What are you saying here? What is moving at 31.0 m/s, the 66-kg person or the snowball? There are three distinct times here.

1. Time "before": Both persons are at rest, one is holding the snowball.
2. Time "in-between": The snowball is flying through the air, the person who threw it is moving, the other person is not.
3. Time "after": The snowball is caught and both persons are moving.

You need to write two momentum conservation equations, one linking "before" and "in-between" and one linking "in-between" and "after". Your system is two persons and one snowball.
 
kuruman said:
What are you saying here? What is moving at 31.0 m/s, the 66-kg person or the snowball? There are three distinct times here.

1. Time "before": Both persons are at rest, one is holding the snowball.
Per the given information, the first person is moving at 2.5 m/s.
kuruman said:
2. Time "in-between": The snowball is flying through the air, the person who threw it is moving, the other person is not.
The thrower is still moving.
kuruman said:
3. Time "after": The snowball is caught and both persons are moving.

You need to write two momentum conservation equations, one linking "before" and "in-between" and one linking "in-between" and "after". Your system is two persons and one snowball.
 
Mark44 said:
Per the given information, the first person is moving at 2.5 m/s. The thrower is still moving.
Sorry, I missed that part, bu tit does not change the strategy for answering the question.
 

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