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highschoolmom2010
 one year ago
If a stationary cart is struck headon by a cart with twice mass of the stationary one and a velocity of 5 m/s, what will be the new velocity of the stationary cart if the collision is inelastic? Show all calculations leading to an answer.
highschoolmom2010
 one year ago
If a stationary cart is struck headon by a cart with twice mass of the stationary one and a velocity of 5 m/s, what will be the new velocity of the stationary cart if the collision is inelastic? Show all calculations leading to an answer.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in general a collision problem can be solved considering the conservation of the total momentum: \(\Delta P_{system}=0\) if this is an "extreme inelastic collision", the colliding objects stick together after the collision, so we can express the momentum of the system before the collision as: \[P_{before}=m_1v_1\] where \(m_1\) is the moving cart, the stationary cart does not appear because \(v_2=0\) the momentum after the collision is: \[P_{after}=(m_1+m_2)v_f\] because the carts now form one mass together. You can now equal \(P_{before}=P_{after}\) and solve to find the final velocity \(v_f\)

highschoolmom2010
 one year ago
Best ResponseYou've already chosen the best response.0i still dont understand this

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large P_{\text{before}}=P_{\text{after}}\] \[\Large m_1*v_1=(m_1+m_2)*v_f\] \[\Large (m_1+m_2)*v_f = m_1*v_1\] \[\Large v_f = \frac{m_1*v_1}{m_1+m_2}\] \[\Large v_f = \frac{m_1*5}{m_1+m_2}\] The only thing wrong here is that we don't know the mass of either cart. So we can't find a numeric value of the final velocity \(\Large v_f\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to make an example, note that if we assume \(m_1=m_2\), both cars have equal mass, we get a value for \(v_f\): \(v_f=\frac{v_1}{2}=2.5m/s\)

highschoolmom2010
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 that is what i didnt understand because we dont know the mass of either cart all is said was the moving cart was 2 times that of the stationary one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@highschoolmom2010 if the moving cart has twice the mass than the stationary one, that means: \(m_1=2m_2\), so we can use that in the expression for the final velocity and get: \(v_f=\frac{m_1v_1}{m_1+m_2}=\frac{2m_2v_1}{3m_2}=\frac{2}{3}v_1=\frac{10}{3}m/s\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@highschoolmom2010 This type of question shows us that we don't necessarily need to know values of every variable in our equation IF INSTEAD we have a way to compare one variable to the other. In this case saying \[m_1 = 2*m_2\] allows us to replace two unknown variables with just one unknown variable. \[v_f = (m_1*v_1)/(m_1+m_2) = (2*m_2*v_1)/(2*m_2+m_2) = (2*m_2*v_1)/(3*m_2)\] This seems counterintuitive at first but if we remember that there are lots of little algebraic math tricks that allow us to cancel out a variable like \[m_2\] then the problem gets a lot simpler. \[v_f = (2*v_1)/3\] Remember this for future problems in physics class there are many times this sort of variable replacement makes our lives easier!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@akoziol4u good comment!
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