Chapter 13.3, Problem 13.150P

A 180-lb man and a 120-lb woman stand at opposite ends of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first.

Fig. P13.150

To determine

Find the velocity of the boat $\left(v_2\right)$ if woman dives first.

Answer to Problem 13.150P

The velocity of the boat $\left(v_2\right)$ if woman dives first is $\underset{\_}{2.80\text{ft}/\text{s}}$.

Explanation of Solution

Given information:

The weight of the man $\left(W_m\right)$ is $180\text{lb}$.

The weight of the woman $\left(W_w\right)$ is $\text{120lb}$.

The weight of the boat $\left(W_b\right)$ is $300\text{lb}$.

The relative velocity of the boat $\left(v_r\right)$ is $16\text{ft/s}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the mass of the man $\left(m_m\right)$ using the formula:

$m_m=\frac{W_m}{g}$

Substitute $180\text{lb}$ for $W_m$.

$m_m=\frac{180\text{lb}}{g}$

Calculate the mass of the woman $\left(m_w\right)$ using the formula:

$m_w=\frac{W_w}{g}$

Substitute $\text{120lb}$ for $W_B$.

$m_w=\frac{120\text{lbs}}{g}$

Calculate the velocity of the boat when woman dives first $\left(v_2\right)$ using the relation:

Show the free body momentum diagram of the boat system after the woman dives as in Figure (1).

Calculate the velocity of the diving of the woman $\left(v_w\right)$ with respect to relative velocity of boat using the relation:

$v_w=v_r-v_1$

Here, $v_1$ is the velocity of the boat after the woman dives.

Substitute $16\text{ft}/\text{s}$ for $v_r$.

$v_w=16\text{ft}/\text{s}-v_1$

Calculate the initial momentum $\left[{\left(P_1\right)}_w\right]$ of the woman before diving using the relation:

${\left(P_1\right)}_w=m_w{v}_w$

Substitute $16-v_1$ for $v_w$ and $\frac{120\text{lbs}}{g}$ for $m_w$.

${\left(P_1\right)}_w=\left(\frac{120\text{lbs}}{g}\right)\left(16-v_1\right)$

The final velocity of the woman after diving is zero, so the final momentum of the woman ${\left(P_2\right)}_w$ is zero.

Calculate the momentum of boat and man after the woman dives using the relation:

${\left(P\right)}_{m+b}=-m_{m+b}{v}_1$

Here, ${\left(P\right)}_{m+b}$ is the momentum of both man and boat and $m_{m+b}$ is the mass of both the man and boat.

Substitute $\frac{300+180\text{lbs}}{g}$ for $m_{m+b}$.

${\left(P\right)}_{m+b}=-\left(\frac{300+180\text{lbs}}{g}\right)v_1$

The expression for principle of conservation of momentum as follows;

${\left(P_1\right)}_w+{\left(P\right)}_{m+b}={\left(P_2\right)}_w$

Substitute $\left(\frac{120\text{lbs}}{g}\right)\left(16-v_1\right)$ for ${\left(P_1\right)}_w$, $\left(\frac{300+180\text{lbs}}{g}\right)v_1$ for ${\left(P_{}\right)}_{m+b}$, and 0 for ${\left(P_2\right)}_w$.

$\begin{array}{l}\left(\frac{120\text{lbs}}{g}\right)\left(16-v_1\right)-\left(\frac{300+180\text{lbs}}{g}\right)v_1=0\\ \left(\frac{120}{g}\right)\left(16-v_1\right)=\left(\frac{300+180}{g}\right)v_1\\ \frac{\left(16-v_1\right)}{v_1}=\frac{480}{120}\end{array}$

$\begin{array}{l}\frac{\left(16-v_1\right)}{v_1}=4\\ 16-v_1=4v_1\\ 5v_1=16\\ v_1=3.20\text{ft}/\text{s}\left(\text{towards the right}\right)\end{array}$

Calculate the velocity of the boat after the man dives $\left(v_2\right)$ using the relation:

Show the free body momentum diagram of the boat after man dives as in Figure (2).

Calculate the velocity of the diving of the man $\left(v_m\right)$ with respect to relative velocity of boat using the relation:

$v_m=v_r-v_2$

Substitute $16\text{ft}/\text{s}$ for $v_r$.

$v_m=16\text{ft}/\text{s}-v_2$

Calculate the initial momentum $\left[{\left(P_1\right)}_m\right]$ of the man before diving using the relation:

${\left(P_1\right)}_m=m_m{v}_m$

Substitute $16-v_2$ for $v_w$ and $\frac{180\text{lbs}}{g}$ for $m_w$.

${\left(P_1\right)}_m=\left(\frac{180\text{lbs}}{g}\right)\left(16-v_2\right)$

Calculate the momentum of the boat after the woman has dived using the relation:

${\left(P\right)}_{m+b}=\left(\frac{300+180\text{lbs}}{g}\right)v_1$

Calculate the final momentum of the boat after the man has dived $\left[{\left(P\right)}_b\right]$ using the relation:

${\left(P\right)}_b=-m_b{v}_2$

Substitute $\frac{300\text{lbs}}{g}$ for $m_b$.

${\left(P\right)}_b=-\left(\frac{300}{g}\right)v_2$

The expression for the principle of conservation of momentum to the man and boat as follows;

${\left(P\right)}_{m+b}={\left(P_2\right)}_b+{\left(P_1\right)}_m$

Substitute $\left(\frac{300+180\text{lbs}}{g}\right)v_1$ for ${\left(P\right)}_{m+b}$, $-\left(\frac{300}{g}\right)v_2$ for ${\left(P_2\right)}_b$, and $\left(\frac{180\text{lbs}}{g}\right)\left(16-v_2\right)$ for ${\left(P_1\right)}_m$.

$\begin{array}{l}\left(\frac{300+180\text{lbs}}{g}\right)v_1=-\left(\frac{300}{g}\right)v_2+\left(\frac{180\text{lbs}}{g}\right)\left(16-v_2\right)\\ 480v_1=2,880-180v_2-300v_2\\ 480v_2=2,880-480v_1\end{array}$

Substitute $3.20\text{ft}/\text{s}$ for $v_1$.

$\begin{array}{l}480v_2=2,880-480\left(3.20\right)\\ 480v_2=2,880-1,536\\ v_2=\frac{1,344}{480}\\ v_2=2.80\text{ft}/\text{s}\end{array}$

Therefore, the velocity of the boat $\left(v_2\right)$ if woman dives first is $\underset{\_}{2.80\text{ft}/\text{s}}$.

To determine

Find the velocity of the boat $\left(v_2^{\text{'}}\right)$ if man dives first.

Answer to Problem 13.150P

The velocity of the boat $\left(v_2^{\text{'}}\right)$ if man dives first is $\underset{\_}{0.229\text{ft}/\text{s}}$.

Explanation of Solution

Given information:

The weight of the man $\left(W_m\right)$ is $180\text{lb}$.

The weight of the woman $\left(W_w\right)$ is $\text{120lb}$.

The weight of the boat $\left(W_b\right)$ is $300\text{lb}$.

The relative velocity of the boat $\left(v_r\right)$ is $16\text{ft/s}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Show the free body momentum diagram of the boat system after the man dives as in Figure (3).

Calculate the velocity of the diving of the man $\left(v_m\right)$ with respect to relative velocity of boat using the relation:

$v_m=v_r-{v^{\prime }}_1$

Here, ${v^{\prime }}_1$ is the velocity of the boat after the man dives.

Substitute $16\text{ft}/\text{s}$ for $v_r$.

$v_m=16\text{ft}/\text{s}-{v^{\prime }}_1$

Calculate the initial momentum $\left[{\left(P_1\right)}_m\right]$ of the man before diving using the relation:

${\left(P_1\right)}_m=m_m{v}_m$

Substitute $16-{v^{\prime }}_1$ for $v_m$ and $\frac{180\text{lbs}}{g}$ for $m_m$.

${\left(P_1\right)}_m=\left(\frac{180\text{lbs}}{g}\right)\left(16-{v^{\prime }}_1\right)$

The final velocity of the man after diving is zero, so the final momentum ${\left(P_2\right)}_m$ of the man is zero.

Calculate the momentum of boat and woman after the man dives using the relation:

${\left(P\right)}_{w+b}=-m_{w+b}{v^{\prime }}_1$

Substitute $\frac{300+120\text{lbs}}{g}$ for $m_{w+b}$.

${\left(P\right)}_{w+b}=-\left(\frac{300+120\text{lbs}}{g}\right){v^{\prime }}_1$

The expression for principle of conservation of momentum to the man as follows;

${\left(P_1\right)}_m+{\left(P\right)}_{w+b}={\left(P_2\right)}_m$

Substitute $\left(\frac{180\text{lbs}}{g}\right)\left(16-{v^{\prime }}_1\right)$ for ${\left(P_1\right)}_m$, $-\left(\frac{300+120\text{lbs}}{g}\right){v^{\prime }}_1$ for ${\left(P\right)}_{w+b}$, and 0 for ${\left(P_2\right)}_m$.

$\begin{array}{l}\left(\frac{180\text{lbs}}{g}\right)\left(16-{v^{\prime }}_1\right)-\left(\frac{300+120\text{lbs}}{g}\right){v^{\prime }}_1=0\\ 180\left(16-{v^{\prime }}_1\right)=420{v^{\prime }}_1\\ 2,880-180{v^{\prime }}_1=420{v^{\prime }}_1\\ 600{v^{\prime }}_1=2,880\\ {v^{\prime }}_1=4.8\text{ft}/\text{s}\left(\text{towards left}\right)\end{array}$

Show the free body momentum diagram of the boat after man dives as in Figure (4).

Calculate the velocity of the diving of the woman $\left(v_w\right)$ with respect to relative velocity of boat using the relation:

$v_w=v_r-{v^{\prime }}_2$

Here, ${v^{\prime }}_2$ is the velocity of the boat after the woman dives.

Substitute $16\text{ft}/\text{s}$ for $v_r$.

$v_w=16\text{ft}/\text{s}-{v^{\prime }}_2$

Calculate the initial momentum $\left[{\left(P_1\right)}_w\right]$ of the woman before diving using the relation:

${\left(P_1\right)}_w=m_w{v}_w$

Substitute $16-{v^{\prime }}_2$ for $v_w$ and $\frac{120\text{lbs}}{g}$ for $m_w$.

${\left(P_1\right)}_w=\left(\frac{120\text{lbs}}{g}\right)\left(16-{v^{\prime }}_2\right)$

Calculate the final momentum of the boat after the woman has dived $\left[{\left(P\right)}_b\right]$ using the relation:

${\left(P\right)}_b=m_b{v^{\prime }}_2$

Substitute $\frac{300\text{lbs}}{g}$ for $m_b$.

${\left(P\right)}_b=\left(\frac{300}{g}\right){v^{\prime }}_2$

Calculate the velocity of the boat after the woman dives $\left({v^{\prime }}_2\right)$:

${\left(P\right)}_{w+b}={\left(P_2\right)}_b+{\left(P_1\right)}_w$

Substitute $-\left(\frac{300+120\text{lbs}}{g}\right){v^{\prime }}_1$ for ${\left(P\right)}_{w+b}$, $\left(\frac{300}{g}\right){v^{\prime }}_2$ for ${\left(P\right)}_b$, and $-\left(\frac{120\text{lbs}}{g}\right)\left(16-{v^{\prime }}_2\right)$ for ${\left(P_1\right)}_w$.

$\begin{array}{l}-\left(\frac{300+120\text{lbs}}{g}\right){v^{\prime }}_1=\left(\frac{300}{g}\right){v^{\prime }}_2-\left(\frac{120\text{lbs}}{g}\right)\left(16-{v^{\prime }}_2\right)\\ -420{v^{\prime }}_1=300{v^{\prime }}_2-1,920+120{v^{\prime }}_2\\ -420{v^{\prime }}_1=420{v^{\prime }}_2-1,920\end{array}$

Substitute $4.80\text{ft}/\text{s}$ for ${v^{\prime }}_1$.

$\begin{array}{l}-420\left(4.80\text{ft}/\text{s}\right)=420{v^{\prime }}_2-1,920\\ 420{v^{\prime }}_2-1,920=-2016\\ 420{v^{\prime }}_2=-96\\ {v^{\prime }}_2=\frac{-96}{420}\\ {v^{\prime }}_2=0.229\text{ft}/\text{s}\end{array}$

Therefore, the velocity of the boat $\left(v_2^{\text{'}}\right)$ if man dives first is $\underset{\_}{0.229\text{ft}/\text{s}}$.