Question 9.122: A man (weighing 915 N) stands on a long railroad flatcar (we......
A man (weighing 915 N) stands on a long railroad flatcar (weighing 2415 N) as it rolls at 18.2 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 4.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?
Denoting the new speed of the car as v, then the new speed of the man relative to the ground is v – v_{rel}. Conservation of momentum requires
\left\lceil\frac{W}{g}+\frac{w}{g}| v_0=\left\lceil\frac{W}{g}| v+\left\lceil\frac{w}{g}\mid b v-v_{rel}g \right. \text{.}\right.\right.
Consequently, the change of velocity is
\Delta \vec{v}=v-v_0=\frac{w v_{rel}}{W+w}=\frac{(915 \,N )(4.00 \,m / s )}{(2415 \,N )+(915\, N )}=1.10\ m / s.