Answer
$v=7.09ft/s$
Work Step by Step
We can determine the required speed as follows:
First, we apply the principle of impulse and momentum to figure (1)
$mv_1+\Sigma \int_{0}^{10} F_x dt=mv_2$
$\implies mv_1+A_xt-B_xt=mv_2$
$\implies 0+10A_x-10B_x=\frac{4000}{32.2}v$
$\implies A_x-B_x=12.4224v~~~$[eq(1)]
Now, we apply the principle of impulse and momentum to figure (2)
$I_A\omega_1+\Sigma\int_{0}^{10}M_A dt=mv_2 r+I_A\omega_2~~~$[eq(2)]
We know that
$I_A=I_B=mk^2=(\frac{1500}{32.2})(1.25)^2=72.7873slug\cdot ft$
and $\omega_1=0$
Similarly $\int_{0}^{10} M_Adt=Mt-A_x tr=300(10)-A_x(10)(1.5)=3000-15A_x$
From eq(2)
$0+3000-1.5A_x=(\frac{1500}{32.2})(1.5v)+72.7873(0.6667v)$
This simplifies to:
$A_x=200-7.8935v~~~$[eq(3)]
The principle of impulse and momentum for figure (3) is given as
$I_B\omega_1+\Sigma\int_{0}^{10} M_B dt=mv_2 r+I_B\omega_2$
As $\omega_1=0$
and $\Sigma\int_{0}^{10} M_B dt=B_x tr=B_x(10)(1.5)=15B_x$
$\implies 0+15B_x=(\frac{1500}{32.2})(1.5)+72.7873(0.6667v)$
This simplifies to:
$B_x=7.8935v~~~$[eq(4)]
We plug in the known values from eq(3) and eq(4) into eq(1) to obtain:
$200-7.8935v-7.8935v=12.4224v$
This simplifies to:
$v=7.09ft/s$
