Question 13.158: Two disks sliding on a frictionless horizontal plane with op...

Total momentum conserved:

Relative velocities:

\begin{aligned}\nu_{B}^{\prime}-\nu_{A}^{\prime} & =e\left(\nu_{A}-\nu_{B}\right) \\\nu^{\prime} & =2 e \nu_{0}\quad\quad\text{(2)}\end{aligned}

Subtracting Eq. (2) from Eq. (1) and dividing by \nu_{0},

\frac{m_{A}}{m_{B}}-1-2 e=0 \quad \frac{m_{A}}{m_{B}}=1+2 e \quad m_{B}=\frac{m_{A}}{1+2 e}

Since weight is proportional to mass, \quad\quad \quad W_{B}=\frac{W_{A}}{1+2e}(3)

(a) With W_{A}=6 \mathrm{\ lb} and e = 0.5,

W_{B}=\frac{6}{1+(2)(0.5)}\quad\quad\quad W_{B}=3.00 \mathrm{\ lb}\blacktriangleleft 

(b) With W_{A}=6 \mathrm{\ lb} and e = 1,

W_{B}=\frac{6}{1+(2)(1)}=2 \mathrm{\ lb}

With W_{A}=6 \mathrm{\ lb} and e = 0,

W_{B}=\frac{6}{1+(2)(0)}=6 \mathrm{\ lb}

Range: \quad\quad \quad\quad 2.00 \mathrm{\ lb} \leq W_{B} \leq 6.00 \mathrm{\ lb}\blacktriangleleft