## College Physics (4th Edition)

After the collision, the speed of the combination is $4.8~m/s$
By conservation of momentum, the final momentum of the two objects is equal to the initial momentum of the 3.0-kg object. We can find the final speed $v_f$ of the two objects: $m_f~v_f = m_0~v_0$ $v_f = \frac{m_0~v_0}{m_f}$ $v_f = \frac{(3.0~kg)(8.0~m/s)}{3.0~kg+2.0~kg}$ $v_f = 4.8~m/s$ After the collision, the speed of the combination is $4.8~m/s$.