Velocities of four masses | Conservation of Momentum

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SUMMARY

The discussion centers on the application of conservation of momentum to a scenario involving four masses, specifically three deer and one additional mass. The calculations presented lead to a final velocity of the additional mass, denoted as v_{f,R}, being (-1.47\hat{i}+1.75\hat{j}) m/s. However, the textbook states the correct answer is (-0.21\hat{i} + 0.25\hat{j}) m/s, prompting questions about the assumptions made in the calculations. Participants emphasize the importance of using the correct symbol for momentum, which is p, rather than \rho.

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Homework Statement
Three 70-kg deer are standing on a flat 200-kg rock that is on an ice-covered pond. A gunshot goes off and the deer scatter, with deer A running at ##(15\hat{i} + 5\hat{j})\frac{m}{s}##, deer B running at ##(-12\hat{i} + 8\hat{j})\frac{m}{s}##, and deer C running at ##(1.2\hat{i} - 18\hat{j})\frac{m}{s}##. What is the velocity of the rock on which they were standing?
Relevant Equations
##\rho_i=\rho_f##
##\vec{\rho_{D1,i}}+\vec{\rho_{D2,i}}+\vec{\rho_{D3,i}}+\vec{\rho_{R,i}} = \vec{\rho_{D1,f}} +\vec{\rho_{D2,f}} +\vec{\rho_{D3,f}} +\vec{\rho_{R,f}}##
##\vec{0} = \vec{\rho_{D1,f}} +\vec{\rho_{D2,f}} +\vec{\rho_{D3,f}} +\vec{\rho_{R,f}}##
##\vec{0} = m_D(15\hat{i} + 5\hat{j}) + m_D(-12\hat{i} + 8\hat{j}) + m_D(1.2\hat{i} - 18\hat{j}) + m_R\vec{v_{f,R}} ##
##\vec{0} = m_D((15\hat{i} + 5\hat{j}) + (-12\hat{i} + 8\hat{j}) + (1.2\hat{i} - 18\hat{j})) + m_R\vec{v_{f,R}}##
##\vec{0} = m_D(4.2\hat{i}-5\hat{j}) + m_R\vec{v_{f,R}}##
##\vec{0} = (294\hat{i}-350\hat{j}) + 200\vec{v_{f,R}}##
##(-294\hat{i}+350\hat{j}) = 200\vec{v_{f,R}}##
##\vec{v_{f,R}} = (-1.47\hat{i}+1.75\hat{j})\frac{m}{s}##

Textbook says the correct answer is ## (-0.21\hat{i} + 0.25\hat{j})\frac{m}{s}##. Are my assumptions with respect to how conservation of momentum works in this case wrong? Any help is appreciated.
 
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Hmm... The "correct answer" would work for three 10-kg deer.
 
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Hill said:
Hmm... The "correct answer" would work for three 10-kg deer.
Ah thank you. Kept thinking I was making an algebra error or something. The answer key's probably wrong. Wouldn't be the first time.
 
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I_Try_Math said:
Relevant Equations: ##\rho_i=\rho_f##
For future reference, the conventional symbol for momentum is the Latin ##p## (pee) not the Greek ##\rho## (rho). Please use the correct symbol to avoid confusion.
 
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kuruman said:
For future reference, the conventional symbol for momentum is the Latin ##p## (pee) not the Greek ##\rho## (rho). Please use the correct symbol to avoid confusion.
I was not aware of that. Thank you for pointing that out.
 
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Whoever made up this question has been stuck in an office for too long!
 
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PeroK said:
Whoever made up this question has been stuck in an office for too long!
Armchair deer hunter? At least the question wasn't "Find the direction the gunshot came from", the intended answer being a unit vector in the direction of the velocity of the CM of the deer.
 
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kuruman said:
Armchair deer hunter?
Na, this is someone who is completely imagining what it's like to hunt deer. To your comment, he shot at the middle of them... :woot:.

Clearly the (wild) gun shot trigged a landslide. The deer rode the rock down the mountain side landing them in the middle of a frozen pond...stunned. When the ice cracked underneath them, they jumped off the boulder.
 
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