## College Physics (4th Edition)

The velocity of the birds immediately after the collision is $20.1~m/s$
By conservation of momentum, the final momentum of the system is equal to the initial momentum in both the horizontal and vertical directions. Initially, the east and west component of momentum is zero. Let west be the positive direction. We can find the west component of the birds' velocity after the collision: $p_{fx} = p_{0x}$ $(0.270~kg+0.220~kg)~v_x +(0.80~kg)(13~m/s)~sin~10^{\circ}-(0.70~kg)(14~m/s)~sin~30^{\circ} = 0$ $v_x = \frac{(0.70~kg)(14~m/s)~sin~30^{\circ} -(0.80~kg)(13~m/s)~sin~10^{\circ}}{0.270~kg+0.220~kg}$ $v_x = 6.31~m/s$ Let north be the positive direction. We can find the north component of the birds' velocity after the collision: $p_{fy} = p_{0y}$ $(0.270~kg+0.220~kg)~v_y -(0.80~kg)(13~m/s)~cos~10^{\circ}+(0.70~kg)(14~m/s)~cos~30^{\circ} = (0.270~kg+0.80~kg)(20~m/s)-(0.220~kg+0.70~kg)(15~m/s)$ $(0.490~kg)~v_y-1.755~kg~m/s = 7.60~kg~m/s$ $v_y = \frac{7.60~kg~m/s +1.755~kg~m/s}{0.490~kg}$ $v_y = 19.1~m/s$ We can find the velocity of the two birds: $v = \sqrt{v_x^2+v_y^2}$ $v = \sqrt{(6.31~m/s)^2+(19.1~m/s)^2}$ $v = 20.1~m/s$ The velocity of the birds immediately after the collision is $20.1~m/s$.