## College Physics (4th Edition)

The magnitude of object B's velocity after the collision is $5.39~m/s$
By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the horizontal component $v_x$ of the final velocity of object B: $M~v_x +M~(1.0~m/s) = M~(6.0~m/s)$ $v_x = (6.0~m/s)-(1.0~m/s)$ $v_x = 5.0~m/s$ We can find the vertical component $v_y$ of the final velocity of object B: $M~v_y +M~(2.0~m/s) = M~(0)$ $v_y = -2.0~m/s$ We can find the magnitude of object B's velocity after the collision: $\sqrt{v_x^2+v_y^2} = \sqrt{(5.0~m/s)^2+(-2.0~m/s)^2} = 5.39~m/s$ The magnitude of object B's velocity after the collision is $5.39~m/s$.