Question 8.10: An elastic straight-line collision We repeat the air-track c......
An elastic straight-line collision
We repeat the air-track collision of Example 8.5 (Section 8.2), but now we add ideal spring bumpers to the gliders so that the collision is elastic. What are the final velocities of the gliders?
IDENTIFY and SET UP The net external force on the system is zero, so the momentum of the system is conserved. Figure 8.25 shows our sketch. We’ll find our target variables, v_{A 2 x} and v_{B 2 x}, by using Eq. (8.27),
v_{B 2 x}-v_{A 2 x}=-\left(v_{B 1 x}-v_{A 1 x}\right) (8.27)
the relative-velocity relationship for an elastic collision, and the momentum-conservation equation.
EXECUTE From Eq. (8.27),
\begin{aligned} v_{B 2 x}-v_{A 2 x} & =-\left(v_{B 1 x}-v_{A 1 x}\right) \\ & =-(-2.0 \mathrm{~m} / \mathrm{s}-2.0 \mathrm{~m} / \mathrm{s})=4.0 \mathrm{~m} / \mathrm{s} \end{aligned}
From conservation of momentum,
\begin{aligned} m_{A} v_{A 1 x}+m_{B} v_{B 1 x} & =m_{A} v_{A 2 x}+m_{B} v_{B 2 x} \\ (0.50 \mathrm{~kg})(2.0 \mathrm{~m} / \mathrm{s})+(0.30 \mathrm{~kg})(-2.0 \mathrm{~m} / \mathrm{s}) & \\ & =(0.50 \mathrm{~kg}) v_{A 2 x}+(0.30 \mathrm{~kg}) v_{B 2 x} \\ 0.50 v_{A 2 x}+0.30 v_{B 2 x} & =0.40 \mathrm{~m} / \mathrm{s} \end{aligned}
(To get the last equation we divided both sides of the equation just above it by 1 \mathrm{~kg}. This makes the units the same as in the first equation.) Solving these equations simultaneously, we find
v_{A 2 x}=-1.0 \mathrm{~m} / \mathrm{s} \quad v_{B 2 x}=3.0 \mathrm{~m} / \mathrm{s}
EVALUATE Both objects reverse their direction of motion; A moves to the left at 1.0 \mathrm{~m} / \mathrm{s} and B moves to the right at 3.0 \mathrm{~m} / \mathrm{s}. This is unlike the result of Example 8.5 because that collision was not elastic. The more massive glider A slows down in the collision and so loses kinetic energy.
The less massive glider B speeds up and gains kinetic energy. The total kinetic energy before the collision (which we calculated in Example 8.7) is 1.6 \mathrm{~J}. The total kinetic energy after the collision is
\frac{1}{2}(0.50 \mathrm{~kg})(-1.0 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2}(0.30 \mathrm{~kg})(3.0 \mathrm{~m} / \mathrm{s})^{2}=1.6 \mathrm{~J}
The kinetic energies before and after this elastic collision are equal. Kinetic energy is transferred from A to B, but none of it is lost.
CAUTION Be careful with the elastic collision equations You could not have solved this problem by using Eqs. (8.24) and (8.25),
\begin{array}{c}{{m_{B}(v_{A1x}+v_{A2x})=m_{A}(v_{A1x}-v_{A2x})}}\\ {{v_{A2x}={\frac{m_{A}-m_{B}}{m_{A}+m_{B}}}v_{A1x}}}\end{array} (8.24)
v_{B2x}={\frac{2m_{A}}{m_{A}+m_{B}}}v_{A14} (8.25)
which apply only if object B is initially at rest. Always be sure that you use equations that are applicable! II
KEYCONCEPT In an elastic collision both total momentum and total kinetic energy are conserved. The relative velocity of the two colliding objects has the same magnitude after the collision as before, but in the opposite direction.