Answer
The catcher and the base runner slide a distance of $1.04~meters$
Work Step by Step
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 $1.04~meters$.
