Question 8.9: Elastic collision on an air track Let’s look at another coll...
Elastic collision on an air track
Let’s look at another collision between the air-track gliders of Examples 8.4 and 8.6. This time, we equip the gliders with spring bumpers so that the collision will be elastic. As before, glider A has a mass of 0.50 kg, glider B has a mass of 0.30 kg, and each moves with an initial speed of 2.0 m/s as they approach each other. What are the velocities of A and B after the collision?
SET UP We draw “before” and “after” sketches, using the same coordinate system as in Example 8.4 (Figure 8.17). The masses and initial velocities of the gliders are the same as in Examples 8.4 and 8.6, so the initial total kinetic energy (1.6 J) is also the same as in those examples.
SOLVE We can’t use Equations 8.11 and 8.12 because in this instance neither object is at rest before the collision. But we can use conservation of momentum, along with the relative-velocity relationship we’ve just discussed, to obtain two simultaneous equations for the two final velocities. Because we’re dealing with a one-dimensional problem, we’ll omit the x subscripts on the velocities, while keeping in mind that they are all x components of velocities.
From conservation of momentum,
\upsilon _{A,f}=\frac{m_A -m_{B}}{m_A+m_B}\upsilon _{A,i}, (8.11)
\upsilon _{B,f}=\frac{2m_A}{m_A+m_B}\upsilon _{A,i}, (8.12)
m_A\upsilon _{A,i}+m_B\upsilon _{B,i}=m_{A}\upsilon _{A,f}+m_B\upsilon _{A,f}+m_B\upsilon _{B,f},
(0.50 kg)(2.0 m/s)+(0.30 kg)(-2.0 m/s)
=(0.50 kg)\upsilon _{A,f}+(0.30 kg)\upsilon _{B,f},
0.50 \upsilon _{A,f}+0.30\upsilon _{B,f}=0.40 m/s.
(In the last equation we’ve divided through by the unit kg.) From the relative-velocity relationship for an elastic collision,
\upsilon _{B,f} – \upsilon _{A,f}=-(\upsilon _{B,i} – \upsilon _{A,i})
=-(-2.0 m/s -2.0 m/s)=4.0 m/s.
Solving these equations simultaneously, we obtain
\upsilon _{A,f}=-1.0 m/s, \upsilon _{B,f}=3.0 m/s (final x components of velocity)
REFLECT Both gliders reverse their directions of motion; A moves to the left at 1.0 m/s and B moves to the right at 3.0 m/s. This result is different from the result of Example 8.4, but that collision was not an elastic one, so we shouldn’t expect the results to be the same. In this case, the total kinetic energy after the collision is
K_f=\frac{1}{2}(0.50 kg)(-1.0 m/s)^2 + \frac{1}{2}(0.30 kg)(3.0 m/s)^2
= 1.6 J.
This equals the total kinetic energy before the collision, as expected.
Practice Problem: Suppose we interchange the two gliders, making m_A = 0.30 kg and m_B = 0.50 kg. (We also turn them around so that the springs still meet.) If the initial speeds are the same as before, find the two final velocities. Answers: -3.0 m/s, 1.0 m/s.
