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highschoolmom2010

  • one year ago

If a stationary cart is struck head-on 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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  1. anonymous
    • one year ago
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    in 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 \(v_f\)

  2. highschoolmom2010
    • one year ago
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    i still dont understand this

  3. jim_thompson5910
    • one year ago
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    \[\Large P_{\text{before}}=P_{\text{after}}\] \[\Large m_1*v_1=(m_1+m_2)*v_f\] \[\Large (m_1+m_2)*v_f = m_1*v_1\] \[\Large v_f = \frac{m_1*v_1}{m_1+m_2}\] \[\Large v_f = \frac{m_1*5}{m_1+m_2}\] 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 \(\Large v_f\)

  4. anonymous
    • one year ago
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    to make an example, note that if we assume \(m_1=m_2\), both cars have equal mass, we get a value for \(v_f\): \(v_f=\frac{v_1}{2}=2.5m/s\)

  5. highschoolmom2010
    • one year ago
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    @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.

  6. anonymous
    • one year ago
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    @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: \(v_f=\frac{m_1v_1}{m_1+m_2}=\frac{2m_2v_1}{3m_2}=\frac{2}{3}v_1=\frac{10}{3}m/s\)

  7. anonymous
    • one year ago
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    @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!

  8. anonymous
    • one year ago
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    @akoziol4u good comment!

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