## College Physics (4th Edition)

We can rank the fragments in order of the magnitude of their momentum, from smallest to largest: $d = e \lt a = b \lt c$
(a) We can find the speed: $\frac{1}{2}mv^2 = 400~J$ $v = \sqrt{\frac{(2)(400~J)}{m}}$ $v = \sqrt{\frac{(2)(400~J)}{8~kg}}$ $v = 10~m/s$ We can find the momentum: $p = mv = (8~kg)(10~m/s) = 80~kg~m/s$ (b) We can find the speed: $\frac{1}{2}mv^2 = 1600~J$ $v = \sqrt{\frac{(2)(1600~J)}{m}}$ $v = \sqrt{\frac{(2)(1600~J)}{2~kg}}$ $v = 40~m/s$ We can find the momentum: $p = mv = (2~kg)(40~m/s) = 80~kg~m/s$ (c) We can find the speed: $\frac{1}{2}mv^2 = 1600~J$ $v = \sqrt{\frac{(2)(1600~J)}{m}}$ $v = \sqrt{\frac{(2)(1600~J)}{4~kg}}$ $v = 28.28~m/s$ We can find the momentum: $p = mv = (4~kg)(28.28~m/s) = 113~kg~m/s$ (d) We can find the speed: $\frac{1}{2}mv^2 = 100~J$ $v = \sqrt{\frac{(2)(100~J)}{m}}$ $v = \sqrt{\frac{(2)(100~J)}{16~kg}}$ $v = 3.54~m/s$ We can find the momentum: $p = mv = (16~kg)(3.54~m/s) = 57~kg~m/s$ (e) We can find the speed: $\frac{1}{2}mv^2 = 1600~J$ $v = \sqrt{\frac{(2)(1600~J)}{m}}$ $v = \sqrt{\frac{(2)(1600~J)}{1~kg}}$ $v = 56.57~m/s$ We can find the momentum: $p = mv = (1~kg)(56.57~m/s) = 57~kg~m/s$ We can rank the fragments in order of the magnitude of their momentum, from smallest to largest: $d = e \lt a = b \lt c$