Question 460418
Assuming no frictional losses, the total momentum within the system will remain unchanged, and it will remain zero (zero because we are accounting for velocities in different directions, not the actual speeds).


The masses of the boy and girl can be found by taking 740 N and 490 N and dividing each by g. Hence, we can find the momentum of each person:


*[tex \LARGE p_{boy} = \frac{740 N}{g}(5 \frac{m}{s}) = \frac{3700}{g} \frac{kg*m}{s}]


The momentum of the girl is


*[tex \LARGE p_{girl} = \frac{490 N}{g}(v_{girl}) = \frac{490v_{girl}}{g} \frac{kg*m}{s}]


The two momenta add up to zero so we can solve for v_girl:


*[tex \LARGE 490v_{girl} + 3700 = 0]


*[tex \LARGE v_{girl} \approx -7.55 \frac{m}{s}] (negative because the velocity is in the opposite direction of the boy's)


(Excuse my weird notation, but I decided not to divide 740, 490 by 9.81 simply because it is not needed and g will cancel out anyway. I just left it in there along with the units since the units have to be consistent.)