Answer
(a) They start to slide across the floor with a speed of 5.28 m/s.
(b) They slide a distance of 5.69 meters.
Work Step by Step
(a) We can find the stuntman's speed when they collide.
$K = PE$
$\frac{1}{2}mv_1^2 = mgh$
$v_1 = \sqrt{2gh} = \sqrt{(2)(9.80~m/s^2)(5.0~m)}$
$v_1 = 9.90~m/s$
We can use conservation of momentum to find their speed just after the collision.
$m_2v_2 = m_1v_1$
$v_2 = \frac{m_1v_1}{m_2}$
$v_2 = \frac{(80.0~kg)(9.90~m/s)}{150.0~kg}$
$v_2 = 5.28~m/s$
They start to slide across the floor with a speed of 5.28 m/s.
(b) We can find the magnitude of deceleration.
$ma = F_f$
$ma = mg~\mu_k$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.250)$
$a = 2.45~m/s^2$
We can find the distance $x$ that they slide.
$x = \frac{v^2-v_0^2}{2a}$
$x = \frac{0-(5.28~m/s)^2}{(2)(-2.45~m/s^2)}$
$x = 5.69~m$
They slide a distance of 5.69 meters.
