Chapter 13.3, Problem 13.149P

To determine

The initial speed of the bullet $v_0$.

Answer to Problem 13.149P

The initial speed of the bullet is $\underset{\_}{\underset{\_}{v_0=769.3ft s^{-1}}}$.

Explanation of Solution

Given information:

Weight of the bullet, $m_B=0.5 oz$

Weight of $A$ and $C$, $m_A=m_C=3 lb$

Coefficient of friction between the blocks and the plane is ${\mu }_k=\mathrm{0.25.}$

General impulse-momentum principal,

$m{v}_1+\sum Im{p}_{1\to 2}=m{v}_2$

Calculation:

Consider the motion of a block under friction,

Applying, $F=ma$

$\uparrow R=mg$

$\to -F=ma$

Since, friction force, $F=\mu R$ ,

$\begin{array}{l}-\mu R=ma\\ -\mu mg=ma\end{array}$

$a=-\mu g$

Applying $\to v^2=u^2+2aS$ ,

$0^2=u^2+2\times (-\mu g)S$

$u=\sqrt{2\mu gS}\mathrm{...........}(1)$

Applying $\to v=u+at$

$0=\sqrt{2\mu gS}+(-\mu g)t$

$t=\sqrt{\frac{2S}{\mu g}}\mathrm{............}(2)$

Consider block $A$ and bullet,

Impulse −momentum diagram,

Apply equation (1) considering the motion of $A$ block,

$\begin{array}{l}{v}_A=\sqrt{2{\mu }_{k}g{S}_1}\\ v_A=\sqrt{2\times 0.25\times 32.174ft s^{-2}\times 0.5 ft}\\ v_A=2.84ft s^{-1}\end{array}$

Apply equation (2) considering the motion of $A$ block,

$\Delta t_1=\sqrt{\frac{2S_1}{{\mu }_{k}g}}$

$\begin{array}{l}\Delta t_1=\sqrt{\frac{2\times 0.5ft}{0.25\times 32.174ft s^{-2}}}\\ \Delta t_1=0.35 s\end{array}$

$m_B=0.5 oz=0.5 oz\times \frac{0.0625 lb o{z}^{-1}}{32.174ft s^{-2}}=9.71\times {10}^{-4} lb s^{2}f{t}^{-1}$

$m_A=\frac{3 lb }{32.174ft s^{-2}}=0.0932 lb s^{2}f{t}^{-1}$

Applying general impulse-momentum principal,

$m{v}_1+\sum Im{p}_{1\to 2}=m{v}_2$

$m_B{v}_0-F_1\Delta t_1=m_B{v}_1+m_A{v}_A$

Since, $F_1={\mu }_k{m}_{A}g$ ,

$m_B{v}_0-{\mu }_k{m}_{A}g\Delta t_1=m_B{v}_1+m_A{v}_A$

$v_0=\frac{m_B{v}_1+m_A{v}_A+{\mu }_k{m}_{A}g\Delta t_1}{m_B}\mathrm{..........}(3)$

Consider block $C$ and bullet,

Impulse −momentum diagram,

Apply equation (1) considering the motion of $A$ block,

$\begin{array}{l}{v}_C=\sqrt{2{\mu }_{k}g{S}_1}\\ v_C=\sqrt{2\times 0.25\times 32.174ft s^{-2}\times \frac{4}{12} ft}\\ v_C=2.32ft s^{-1}\end{array}$

Apply equation (2) considering the motion of $A$ block,

$\Delta t_2=\sqrt{\frac{2S_2}{{\mu }_{k}g}}$

$\begin{array}{l}\Delta t_2=\sqrt{\frac{2\times \frac{4}{12} ft}{0.25\times 32.174ft s^{-2}}}\\ \Delta t_2=0.29 s\end{array}$

$m_C=\frac{3 lb }{32.174ft s^{-2}}=0.0932 lb s^{2}f{t}^{-1}$

Applying general impulse-momentum principal,

$m{v}_1+\sum Im{p}_{1\to 2}=m{v}_2$

$m_B{v}_1-F_2\Delta t_2=0$

$m_B{v}_1-{\mu }_k(m_C+m_B)g\Delta t_2=0$

$v_1=\frac{{\mu }_k(m_C+m_B)g\Delta t_2}{m_B}$

$v_1=\frac{0.25\times (3 lb +0.03215 lb )\times 0.29 s}{9.71\times {10}^{-4} lb s^{2}f{t}^{-1}}$

$v_1=226.4ft s^{-1}$

From equation (3),

$v_0=\frac{m_B{v}_1+m_A{v}_A+{\mu }_k{m}_{A}g\Delta t_1}{m_B}$

$v_0=\frac{(9.71\times {10}^{-4} lb s^{2}f{t}^{-1}\times 226.4ft s^{-1})+(0.0932 lb s^{2}f{t}^{-1}\times 2.84ft s^{-1})+(0.25\times 3lb\times 0.35 s)}{9.71\times {10}^{-4} lb s^{2}f{t}^{-1}}$

$v_0=769.3ft s^{-1}$

Conclusion:

Thus, the initial speed of the bullet is $\underset{\_}{\underset{\_}{v_0=769.3ft s^{-1}}}.$