Calculating Speed in a Tug-of-War on Ice

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In the tug-of-war scenario on ice, James and Ramon are initially 20.0 m apart, with masses of 90.0 kg and 60.0 kg, respectively. Ramon pulls with a speed of 0.70 m/s, prompting a discussion about whether to apply conservation of momentum or center of mass principles. It is clarified that the system should be considered as a whole, leading to the conclusion that the center of mass velocity is not simply 0.70 m/s. The correct approach involves using the equation for the center of mass to solve for James's speed, which is ultimately linked to the forces exerted by each participant. The discussion concludes with a successful resolution of the problem.
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James and Ramon are standing 20.0 m apart on the slippery surface of a frozen pond. Ramon has mass 60.0 kg and James has mass 90.0 kg. Ramon pulls on the rope to give himself a speed of 0.70 m/s. What is James's speed?

From what I understand, since there is an external force due to Ramon's pull, momentum is not conserved, and I can't use the conservation of momentum equation. So, I'm thinking this is a center of mass problem, with V_cm = 0.70 m/s, and James's speed is 0.70 m/s. However, that's not the correct answer. Can anyone help me with this?
 
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Well, there's certainly an "external" force on James, if you consider James and Ramon as separate systems. But what if you consider them as parts of the same system?

Alternatively, consider Newton's 3rd law: Whatever force that Ramon exerts on James must equal the force that James exerts on Ramon.

You can view this as a center of mass problem (note that this is equivalent to using conservation of momentum). But what makes you think that the V_cm = 0.70 m/s? That's James's speed after the pull, not the velocity of the cm.
 
It turns out that I have to use center of mass for this problem.
What I have so far:
(m_J)(v_J)+(m_R)(v_R)=(m_J + m_R)V_cm
The thing is, I have two unknowns here: v_J and V_cm. I suppose v_J is what I'm solving for. I'm stuck.
 
first solve this prob to get a better understanding...

Two people of equal mass, 6 meters apart, attempt a tug of war on frictionless ice. If they pull on opposite ends of the rope with equal forces, each slides 3 meters to a point midway between them.Suppose instead that only one person pulls and the other fastens the rope around his or her waist. How for does each person slide? (Neglect any effects of the rope's mass.)
 
iwonde said:
It turns out that I have to use center of mass for this problem.
What I have so far:
(m_J)(v_J)+(m_R)(v_R)=(m_J + m_R)V_cm
The thing is, I have two unknowns here: v_J and V_cm. I suppose v_J is what I'm solving for. I'm stuck.
What you're using is conservation of momentum. And you are given V_cm: They start from rest. :wink:
 
Thanks, guys! I got it! =)
 

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