## College Physics (4th Edition)

The catcher and the base runner slide a distance of $1.04~meters$
We can use conservation of momentum to find the speed of the two players just after the collision: $m_f~v_f= m_0~v_0$ $v_f= \frac{m_0~v_0}{m_f}$ $v_f= \frac{(85~kg)~(8.0~m/s)}{85~kg+95~kg}$ $v_f = 3.78~m/s$ We can find the magnitude of deceleration as the players slide: $m_fa = F_N~\mu_k$ $m_fa = m_f~g~\mu_k$ $a = g~\mu_k$ $a = (9.80~m/s^2)(0.70)$ $a = 6.86~m/s^2$ For this part of the question, we can let $v_0 = 3.78~m/s$. We can find the distance the two players slide: $v_f^2 = v_0^2+2ax$ $x = \frac{v_f^2 - v_0^2}{2a}$ $x = \frac{0 - (3.78~m/s)^2}{(2)(-6.86~m/s^2)}$ $x = 1.04~m$ The catcher and the base runner slide a distance of $1.04~meters$.