A TWO-DIMENSIONAL ELASTIC COLLISION
Figure 8.27 shows an elastic collision of two pucks (masses m_A = 0.500 kg and m_B = 0.300 kg) on a frictionless air-hockey table. Puck A has an initial velocity of 4.00 m/s in the positive x-direction and a final velocity of 2.00 m/s in an unknown direction α. Puck B is initially at rest. Find the final speed v_{B2} of puck B and the angles α and β.
IDENTIFY and SET UP:
We’ll use the equations for conservation of energy and conservation of x- and y-momentum. These three equations should be enough to solve for the three target variables.
EXECUTE:
The collision is elastic, so the initial and final kinetic energies of the system are equal:
\begin{aligned}\frac{1}{2} m_{A} v_{A 1}^{2} &=\frac{1}{2} m_{A} v_{A 2}^{2}+\frac{1}{2} m_{B} v_{B 2}^{2} \\v_{B 2}^{2} &=\frac{m_{A} v_{A 1}^{2}-m_{A} v_{A 2}^{2}}{m_{B}} \\&=\frac{(0.500\mathrm{~kg})(4.00 \mathrm{~m} / \mathrm{s})^{2}-(0.500 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})^{2}}{0.300\mathrm{~kg}} \\v_{B 2} &=4.47 \mathrm{~m} / \mathrm{s}\end{aligned}Conservation of the x- and y-components of total momentum gives
\begin{aligned}m_{A} v_{A 1 x}=& m_{A} v_{A 2 x}+m_{B} v_{B 2 x} \\(0.500 \mathrm{~kg})(4.00\mathrm{~m} / \mathrm{s})=&(0.500 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})(\cos \alpha) \\&+(0.300 \mathrm{~kg})(4.47 \mathrm{~m} / \mathrm{s})(\cos \beta) \\0=& m_{A} v_{A 2 y}+m_{B} v_{B2 y} \\0=&(0.500 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})(\sin \alpha) \\&-(0.300\mathrm{~kg})(4.47 \mathrm{~m} / \mathrm{s})(\sin \beta)\end{aligned}These are two simultaneous equations for α and β. You can supply the details of the solution. (Hint: Solve the first equation for cos β and the second for sin β; square each equation and add. Since \sin ^{2} \beta+\cos ^{2} \beta=1, this eliminates β and leaves an equation that you can solve for cos α and hence for α. Substitute this value into either of the two equations and solve for β.) The results are
\alpha=36.9^{\circ} \quad \beta=26.6^{\circ}
EVALUATE: To check the answers we confirm that the y-momentum, which was zero before the collision, is in fact zero after the collision. The y-momenta are
\begin{aligned}p_{A 2 y} &=(0.500 \mathrm{~kg})(2.00 \mathrm{~m} / \mathrm{s})\left(\sin 36.9^{\circ}\right)=+0.600 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \\p_{B 2 y} &=-(0.300 \mathrm{~kg})(4.47\mathrm{~m} / \mathrm{s})\left(\sin 26.6^{\circ}\right)=-0.600 \mathrm{~kg} \cdot \mathrm{m} /\mathrm{s}\end{aligned}and their sum is indeed zero.