 Solution

During ejection the mitral valve is closed, so $Q_v=0$. Adding the arterial and ventricular balances cancels aortic flow:where the last approximation assumes $p_v\ll p_a$.

Model the rapid contraction as a switch to approximately constant systolic elastance $E_S$, with $C_s=1/E_S$ the systolic ventricular compliance. A sufficiently small aortic-valve resistance makes $p_{LV}\simeq p_a$ after a short pressure-equalization transient. The systolic pressure must be high enough to open the valve, $E_S(V^{+}-V_0)>p_a^{+}$. During that transient neglect systemic runoff. Conservation of combined arterial and stressed ventricular volume givessoThis is an approximation across a rapid transient, not a discontinuity at finite valve resistance. The subsequent common pressure obeys $R_c(C_a+C_s){\dot{p}}_a\simeq -p_a$, givingThe source's $R_C$ in this expression denotes the same systemic resistance $R_c$.

At the end of firing, the remaining ventricular volume is $V_{\mathrm{end}}\simeq V_0+C_s{p}_a(\Delta t_F)$. The stroke volume is $V^{+}-V_{\mathrm{end}}$, henceThe assumptions are a closed mitral valve throughout ejection, negligible venous pressure, rapid valve equalization relative to systemic runoff, and nearly constant systolic elastance during the interval measured by $\Delta t_F$. These give the ventricular ejection pressure and stroke-volume approximation.