## College Physics (4th Edition)

The magnitude of the target body's momentum after the collision is $8.7~kg~m/s$
Let's assume the projectile is initially moving in the +x-direction and then gets deflected $60.0^{\circ}$ in the +y-direction.. By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the horizontal component $p_x$ of the final momentum of the object that was stationary: $p_x +(2.0~kg)~(3.0~m/s)~cos~60.0^{\circ} = (2.0~kg)~(5.0~m/s)$ $p_x = (2.0~kg)[(5.0~m/s)-(3.0~m/s)~cos~60.0^{\circ}]$ $p_x = 7.0~kg~m/s$ We can find the vertical component $p_y$ of the final momentum of the object that was stationary: $p_y +(2.0~kg)~(3.0~m/s)~sin~60.0^{\circ} = (2.0~kg)~(0)$ $p_y = 0-(2.0~kg)~(3.0~m/s)~sin~60.0^{\circ}$ $p_y = -5.20~kg~m/s$ We can find the magnitude of the target body's momentum after the collision: $\sqrt{p_x^2+p_y^2} = \sqrt{(7.0~kg~m/s)^2+(-5.20~kg~m/s)^2} = 8.7~kg~m/s$ The magnitude of the target body's momentum after the collision is $8.7~kg~m/s$.