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Thread: Mechanics Problem; Any help would be appreciated

  1. #1
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    Question Mechanics Problem; Any help would be appreciated

    The details of the question are below;

    Identical Cars A, B and C, are initially at rest in a straight line on a horizontal surface with their brakes off. Car A is pushed towards the others so that it hits Car B with speed $\displaystyle {2ms^-1}$.
    Given that the coefficient of restitution between any two cars is 0.75, find the speeds of the cars after all collisions have finished.
    The answers are below;
    Spoiler:
    A=0.223, B=0.246, C=1.53


    Please bear in mind that this is college work (16-18) and so try to keep the explanations limited to that as much as possible.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by osmosis786 View Post
    Identical Cars A, B and C, are initially at rest in a straight line on a horizontal surface with their brakes off. Car A is pushed towards the others so that it hits Car B with speed $\displaystyle 2ms^{-1}$.
    Given that the coefficient of restitution between any two cars is 0.75, find the speeds of the cars after all collisions have finished.
    Suppose that X and Y are two objects with masses $\displaystyle M_X$ and $\displaystyle M_Y$, moving along a line with velocities $\displaystyle u_X$ and $\displaystyle u_Y$. When one of them shunts into the other, they separate, moving at new speeds $\displaystyle v_X$ and $\displaystyle v_Y$. You need two equations to work out the new speeds after the collision.

    The first equation is given by conservation of momentum: $\displaystyle M_Xv_X + M_Yv_Y = M_Xu_X + M_Yv_Y$. If both objects have the same mass (as in the case of these cars) then the M's all cancel, and you are left with the conservation of momentum equation $\displaystyle v_X+v_Y = u_X+u_Y.$

    The second equation is the restitution equation, which says that the coefficient of restitution is equal to $\displaystyle \dfrac{v_Y-v_X}{u_X-u_Y}$. In this problem, the coefficient is 0.75, so the equation becomes $\displaystyle v_Y-v_X = 0.75(u_X-u_Y).$

    Now, what happens in these car pile-ups? The first collision occurs when car A, travelling at 2m/s, hits car B. If the velocities after this collision are $\displaystyle v_A$ and $\displaystyle v_B$, then the momentum equation says that $\displaystyle v_A+v_B = 2+0 = 2$. The restitution equation says that $\displaystyle v_B-v_A = 0.75(2-0) = 1.5.$ That gives you two simultaneous equations for $\displaystyle v_A$ and $\displaystyle v_B$. Solve them. You should find that $\displaystyle v_A = 0.25$ and $\displaystyle v_B = 1.75.$

    The next thing that happens is that car B, travelling at 1.75m/s, hits the stationary car C. You can go through the same procedure as for the first collision, solving two simultaneous equations to find the velocities $\displaystyle v_B'$ and $\displaystyle v_C$ after the collision. I made them $\displaystyle v_B' = 0.219$ and $\displaystyle v_C = 1.531$.

    But now car A, still moving at 0.25m/s, is going faster than car B, moving at 0.219m/s. So car A will catch up with car B and there will be another collision. Once again, go through the whole procedure of solving two simultaneous equations to find the velocities $\displaystyle v_A'$ and $\displaystyle v_B''$ after this collision. You should get $\displaystyle v_A' = 0.223$ and $\displaystyle v_B'' = 0.246.$ At this stage, each car is going slower than the one in front of it, so there will be no more collisions and the calculation is complete.
    Last edited by Opalg; Mar 26th 2011 at 11:40 PM.
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  3. #3
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    Thanks a lot. I see where I was going wrong now.

    Kudos to you.
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