Question 6.MF.1.1: In a one-dimensional collision, suppose the first object has...
In a one-dimensional collision, suppose the first object has mass m_{1}=1 kg and initial velocity v_{1 i}=3 m / s , whereas the second object has mass m_{2}=2 kg and initial velocity v_{2 i}=-3 m / s . Find the final velocities for the two objects. (For clarity, significant figure conventions are not observed here.)
Substitute the given values and X=v_{1 f} and Y=v_{2 f} into Equations 6.10 and 6.11, respectively, and simplify, obtaining
m_{1} v_{1 i}+m_{2} v_{2 i}=m_{1} v_{1 f}+m_{2} v_{2 f} [6.10]
\frac{1}{2} m_{1} v_{1} i^{2}+\frac{1}{2} m_{2} v_{2 i}{ }^{2}=\frac{1}{2} m_{1} v_{1} f^{2}+\frac{1}{2} m_{2} v_{2} f^{2} [6.11]
\begin{gathered}-3=X+2 Y (1)\\27=X^{2}+2 Y^{2} (2)\end{gathered}
Equation (1) is that of a straight line, whereas Equation (2) describes an ellipse. Solve Equation (1) for X and substitute into Equation (2), obtaining 27 = (-3-2 Y)^{2}+2 Y^{2} , which can be simplified to
Y^{2}+2 Y-3=0
In general, the quadratic formula must now be applied, but this equation factors, giving Y=v_{2 f}= 1 m/s or -3 m/s. Only the first answer, 1 m/s, makes sense. Substituting it into Equation (1) yields
X=v_{1 f}=-5 m / s
It is also possible to use Equation 6.10 together with the derived Equation 6.14. This situation, illustrated in Example 6.5, is equivalent to finding the intersection of two straight lines. It’s easier to remember the equation for the conservation of energy than the special Equation 6.14, so it’s a good idea to be able to solve such problems both ways.
v_{1 i}-v_{2 i}=-\left(v_{1 f}-v_{2 f}\right) [6.14]