## College Physics (4th Edition)

(a) The boy's change in momentum is $-751.2~kg~m/s$ (b) The impulse on the net is $986.4~kg~m/s$ (c) The average force on the net due to the boy is $2466~N$
(a) We can find the speed after falling 8.0 meters: $v_f^2 = v_0^2+2ay$ $v_f = \sqrt{v_0^2+2ay}$ $v_f = \sqrt{0+(2)(9.80~m/s^2)(8.0~m)}$ $v_f = 12.52~m/s$ We can find the change in momentum: $\Delta p =m~\Delta v = (60.0~kg)(-12.52~m/s) = -751.2~kg~m/s$ The boy's change in momentum is $-751.2~kg~m/s$ (b) The impulse on the net is the sum of the boy's initial momentum and the gravitational force during the collision: $J = p_0+mg~t$ $J = (751.2~kg~m/s)+(60.0~kg)(9.80~m/s^2)(0.40~s)$ $J = 986.4~kg~m/s$ The impulse on the net is $986.4~kg~m/s$ (c) We can find the average force on the net due to the boy: $F~t = 986.4~kg~m/s$ $F = \frac{986.4~kg~m/s}{t}$ $F = \frac{986.4~kg~m/s}{0.40~s}$ $F = 2466~N$ The average force on the net due to the boy is $2466~N$