Tai? Yes, Randy? I don't think we're going as fast as we were...

Momentum is conserved. (Energy is not conserved - this is an inelastic collision.)

Momentum of man (in the north direction):

$p_1=m v=80\times 8=640$

Momentum of woman (in the east direction):

$p_2=m v=60\times 8=480$

Momentum is a vector. Since north and east are perpendicular to each other, the combined momentum is:

$p=\sqrt{p_1^2+p_2^2}=\sqrt{640^2+480^2}=800$

Since they stick together, we can calculate their resultant velocity with their total mass:

$p=m v=800\\ (80+60) v=800\\ v=\frac{40}{7}=5.71429$

Through conservation of momentum, the initial momentum and final momentum in each direction (east-west and north-south) are conserved independently.

$p=mv$

$m_{final}=60+80=140$

$p_{e-w}=60*8=480$

$v_{e-w}=\frac{480}/{140}=3.429$

$p_{n-s}=80*8=640$

$v_{e-w}=\frac{640}{140}=4.571$

To get total speed we have to add the velocity vectors:

$s=\sqrt{{v_{e-w}}^2+{v_{n-s}}^2}=5.715$

We consider the rightward velocity and the northward velocity.

The rightward speed is $8(\frac{60}{80+60})=\frac{48}{14} \frac{m}{s}$ .

The northward speed is $8(\frac{80}{80+60})=\frac{64}{14} \frac{m}{s}$ .

Hence, by the Pythagorean Theorem, and since their directions are perpendicular to each other, their total speed is $\sqrt{\frac{64}{14}^2+\frac{48}{14}}=\frac{80}{14}=\frac{40}{7} \frac{m}{s}$ , which is about $\boxed{5.714} \frac{m}{s}$ .

since momentum is a vector, we can find thye resultant momentum of the man and the woman by taking the square root of (a^2 + b^2 + abcos90) = square root of (a^2 + b^2)=H. since cos90 =0. this is finding the resultant of vectors using paralellogram method.

here 'a' and 'b' are the momentum of the man and woman.

but we know, (m1+m2)V=H according to the law of coservation of momentum and here m1 and m2 are the masses of the man and woman. thus the final velocity V=H/(m1+m2)

This is just some typical momentum problem. The collision is, obviously, inelastic because the man picks up his partner (the system moves with a constant velocity). From the conservation of momentum we have:

$m_{man}v_{man} + m_{woman}v_{woman} = (m_{man}+m_{woman})v_{system}$

Plugging all our known values, we have: $(80kg)(8\frac{m}{s}\hat{j}+(60kg)(8\frac{m}{s}\hat{i} = (140kg)(v_{system})$ $v_{system} = (3.43\frac{m}{s}\hat{i} + 4.57\frac{m}{s}\hat{j})$

$v_{system} = \sqrt{(3.43)^{2} + (4.57)^{2}}\frac{m}{s}$

$v_{system} = 5.714\frac{m}{s}$

use m1v1 + m2v2 = (m1+m2)v', where v' will be their final speed. (60)(8) + (80)(8i) = (60+80)v', provided that 8i indicates the direction of speed going north. then v' = 3.428571429 East, 4.571428571 North. Then, getting its resultant, v' = 5.714285714, 53.13 degrees N of E

we use conservation of linear momentum in x and y direction to get x and y components of final velocity(assumed to be "v")

$60 \times 8 = 140vx$ (colm in x direction)

$80 \times 8 = 140vy$ (colm in y direction)

and $v = \sqrt {{vx^2 } \times {vy^2}}$ solving we get v = 5.714 m/s