## College Physics (4th Edition)

The speed of the two acrobats after the collision is $0.64~m/s$ at an angle of $72.6^{\circ}$ above the positive x-axis.
By conservation of momentum, the final momentum of the system is equal to the initial momentum in both the horizontal and vertical directions. We can find the horizontal component of the momentum: $p_x = (60~kg)(3.0~m/s)~cos~10^{\circ}-(80~kg)(2.0~m/s)~cos~20^{\circ}$ $p_x = 26.9~kg~m/s$ We can find the vertical component of the momentum: $p_y = (60~kg)(3.0~m/s)~sin~10^{\circ}+(80~kg)(2.0~m/s)~sin~20^{\circ}$ $p_y = 86.0~kg~m/s$ We can find the magnitude of the momentum: $p = \sqrt{p_x^2+p_y^2}$ $p = \sqrt{(26.9~kg~m/s)^2+(86.0~kg~m/s)^2}$ $p = 90.1~kg~m/s$ We can find the speed of the two acrobats after the collision: $mv = p$ $v = \frac{p}{m}$ $v = \frac{90.1~kg~m/s}{60~kg+80~kg}$ $v = 0.64~m/s$ We can find the angle $\theta$ above the positive x-axis: $tan~\theta = \frac{86.0}{26.9}$ $\theta = tan^{-1}(\frac{86.0}{26.9})$ $\theta = 72.6^{\circ}$ The speed of the two acrobats after the collision is $0.64~m/s$ at an angle of $72.6^{\circ}$ above the positive x-axis.