Chapter 13.4, Problem 13.188P

When the rope is at an angle of α = 30°, the 1-lb sphere A has a speed v₀ = 4 ft/s. The coefficient of restitution between A and the 2-lb wedge B is 0.7 and the length of rope l = 2.6 ft. The spring constant has a value of 2 lb/in. and θ = 20°. Determine (a) the velocities of A and B immediately after the impact, (b) the maximum deflection of the spring, assuming A does not strike B again before this point.

To determine

Find the velocity of A $\left({v^{\prime }}_A\right)$ and B $\left({v^{\prime }}_B\right)$ immediately after impact.

Answer to Problem 13.188P

The velocity of A $\left({v^{\prime }}_A\right)$ and B $\left({v^{\prime }}_B\right)$ immediately after impact are $\underset{\_}{2.36\text{ft}/\text{s}\left(83.8°\right)}$ and $\underset{\_}{3.23\text{ft}/\text{s}(\to )}$ respectively.

Explanation of Solution

Given information:

The angle of the rope $\left(\alpha \right)$ is $30°$.

The weight of the sphere A $\left(W_A\right)$ is $1\text{lb}$.

The weight of the wedge B $\left(W_B\right)$ is $2\text{lb}$.

The speed of the sphere A $\left(v_A\right)$ is $\text{4ft/s}$.

The coefficient of restitution between A and wedge (e) is 0.7.

The length of the rope (l) is $2.6\text{ft}$.

The spring constant (k) is $\text{2lb/in}\text{.}$

The angle $\left(\theta \right)$ is $20°$.

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

Calculation:

Calculate the mass of sphere A $\left(m_A\right)$ using the relation:

$m_A=\frac{W_A}{g}$

Substitute $1\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_A=\frac{1\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Calculate the mass of wedge B $\left(m_B\right)$ using the relation:

$m_B=\frac{W_B}{g}$

Substitute $2\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_B=\frac{2\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =0.06211\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Calculate the initial altitude of sphere $\left(h_0\right)$ using the relation:

$h_0=l\left(1-\cos \alpha \right)$

Substitute $2.6\text{ft}$ for l and $30°$ for $\alpha$.

$\begin{array}{c}{h}_0=\left(2.6\text{ft}\right)\left(1-\cos 30°\right)\\ =0.3483\text{ft}\end{array}$

Calculate the initial potential energy of sphere $\left(V_0\right)$ using the formula:

$V_0=m_{A}g{h}_0$

Substitute $0.031\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_A$, $32.2\text{ft}/{\text{s}}^2$ for g, and $0.3483\text{ft}$ for $h_0$.

$\begin{array}{c}{V}_0=\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(0.3483\text{ft}\right)\\ =0.3483\text{lb}\cdot \text{ft}\end{array}$

Calculate the initial kinetic energy of sphere $\left(T_0\right)$ using the formula:

$T_0=\frac{1}{2}{m}_A{v}_0^2$

Here, $v_0$ is the initial velocity of sphere.

Substitute $0.031\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_A$ and $4\text{ft}/\text{s}$ for $v_0$.

$\begin{array}{c}{T}_0=\frac{1}{2}\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left(4\text{ft}/\text{s}\right)}^2\\ =0.248\text{lb}\cdot \text{ft}\end{array}$

Calculate the altitude of sphere just before impact $\left(h_1\right)$ using the relation:

$h_1=l\left(1-\cos \alpha \right)$

Substitute $2.6\text{ft}$ for l and $0°$ for $\alpha$.

$\begin{array}{c}{h}_1=\left(2.6\text{ft}\right)\left(1-\cos 0°\right)\\ =\left(2.6\text{ft}\right)\left(1-1\right)\\ =0\end{array}$

Calculate the initial potential energy of sphere just before impact $\left(V_1\right)$ using the relation:

$V_1=m_{A}g{h}_1$

Substitute $0.031\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_A$, $32.2\text{ft}/{\text{s}}^2$ for g, and 0 for $h_1$.

$\begin{array}{c}{V}_1=\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\left(0\right)\\ =0\end{array}$

Calculate the kinetic energy of sphere just before impact $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}{m}_A{v}_A^2$

Here, $v_A$ is the final velocity of sphere.

Substitute $0.031\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_A$.

$\begin{array}{c}{T}_1=\frac{1}{2}\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left(v_A\right)}^2\\ =0.0155v_A^2\text{lb}\cdot \text{ft}\end{array}$

The expression for the principle of conservation of energy between initial and final stage of sphere as follows:

$T_0+V_0=T_1+V_1$

Substitute $0.248\text{lb}\cdot \text{ft}$ for $T_0$, $0.3483\text{lb}\cdot \text{ft}$ for $V_0$, $0.0155v_A^2\text{lb}\cdot \text{ft}$ for $T_1$, and 0 for $V_1$.

$\begin{array}{l}0.248\text{lb}\cdot \text{ft}+0.3483\text{lb}\cdot \text{ft}=0.0155v_A^2\text{lb}\cdot \text{ft}+0\\ 0.0155v_A^2=0.5963\\ v_A={\left(\frac{0.5963}{0.0155}\right)}^{0.5}\\ v_A=6.20\text{ft}/\text{s}\end{array}$

Show the impulse-momentum diagram for sphere as in Figure (1).

The expression for the momentum in tangential direction as follows:

$m_A{v}_A\sin \theta =m_A{\left({v^{\prime }}_A\right)}_t$

Here, ${\left({v^{\prime }}_A\right)}_t$ is the tangential component of velocity of sphere at the end of restitution period.

Substitute $0.031\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_A$, $20°$ for $\theta$, and $6.20\text{ft}/\text{s}$ for $v_A$.

$\begin{array}{l}\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left(6.20\text{ft}/\text{s}\right)\sin 20°=\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left({v^{\prime }}_A\right)}_t\\ 0.031{\left({v^{\prime }}_A\right)}_t=0.0657\\ {\left({v^{\prime }}_A\right)}_t=\left(2.1203\text{ft}/\text{s}\right)\nearrow 70°\end{array}$

Show the impulse-momentum diagram of wedge as in Figure (2).

The expression for the momentum in x-direction as follows:

$m_A{v}_A+0=m_A{\left({v^{\prime }}_A\right)}_n\cos \theta +m_A{\left({v^{\prime }}_A\right)}_t\sin \theta +m_B{v^{\prime }}_B$

Here, ${\left({v^{\prime }}_A\right)}_n$ is the normal component of velocity of sphere at the end of restitution period.

Substitute $0.031\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_A$, $0.06211\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_B$, $6.20\text{ft}/\text{s}$ for $v_A$, $\left(2.1203\text{ft}/\text{s}\right)$ for ${\left({v^{\prime }}_A\right)}_t$, and 20  for $\theta$.

$\begin{array}{l}\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\left(6.20\text{ft}/\text{s}\right)+0=\left\{\begin{array}{l}\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left({v^{\prime }}_A\right)}_n\cos 20°\\ +\left[\begin{array}{l}\left(0.031\text{lb}\cdot {\text{s}}^2/\text{ft}\right)\\ \left(2.1203\text{ft}/\text{s}\right)\sin 20°\end{array}\right]\\ +\left(0.06211\text{lb}\cdot {\text{s}}^2/\text{ft}\right){v^{\prime }}_B\end{array}\right\}\\ 0.1921-0.02278=0.0291{\left({v^{\prime }}_A\right)}_n+0.0621{v^{\prime }}_B\\ 0.0291{\left({v^{\prime }}_A\right)}_n+0.0621{v^{\prime }}_B=0.17\end{array}$ (1)

Calculate the coefficient of restitution (e) using the formula:

$e=\frac{{\left({v^{\prime }}_B\right)}_n-{\left({v^{\prime }}_A\right)}_n}{{\left(v_A\right)}_n-{\left(v_B\right)}_n}$

Substitute 0.7 for e, $v_A\cos \theta$ for ${\left(v_A\right)}_n$, $6.20\text{ft}/\text{s}$ for $v_A$, 0 for ${\left(v_B\right)}_n$, and ${v^{\prime }}_B\cos \theta$ for ${\left({v^{\prime }}_B\right)}_n$.

$\begin{array}{l}0.7=\frac{{v^{\prime }}_B\cos \theta -{\left({v^{\prime }}_A\right)}_n}{\left(6.1994\text{ft}/\text{s}\right)\cos \theta -0}\\ \left(6.1994\text{ft}/\text{s}\right)\cos 20°\left(0.7\right)={v^{\prime }}_B\cos 20-{\left({v^{\prime }}_A\right)}_n\\ 0.9396{v^{\prime }}_B-{\left({v^{\prime }}_A\right)}_n=4.0778\\ {\left({v^{\prime }}_A\right)}_n=0.9396{v^{\prime }}_B-4.0778\end{array}$

Find the velocity of sphere B immediately after the impact:

Substitute $0.9396{v^{\prime }}_B-4.0778$ for ${\left({v^{\prime }}_A\right)}_n$ in equation (1).

$\begin{array}{l}0.0291\left({\left({v^{\prime }}_A\right)}_n\right)+0.0621{v^{\prime }}_B=0.17\\ 0.0291\left(0.9396{v^{\prime }}_B-4.0778\right)+0.0621{v^{\prime }}_B=0.17\\ 0.0273{v^{\prime }}_B-0.1186+0.0621{v^{\prime }}_B=0.17\\ 0.0894{v^{\prime }}_B=0.2886\\ {v^{\prime }}_B=\frac{0.2886}{0.0894}\\ {v^{\prime }}_B=3.23\text{ft}/\text{s}\end{array}$

Find the normal component of velocity of sphere:

Substitute $3.2279\text{ft}/\text{s}$ for ${v^{\prime }}_B$ in equation (1).

$\begin{array}{l}0.0291{\left({v^{\prime }}_A\right)}_n+0.0621\left(3.2279\text{ft}/\text{s}\right)=0.17\\ 0.0291{\left({v^{\prime }}_A\right)}_n=-0.0304\\ {\left({v^{\prime }}_A\right)}_n=-1.0446\text{ft}/\text{s}\end{array}$

Calculate the resultant velocity of sphere A $\left({v^{\prime }}_A\right)$ using the relation:

${v^{\prime }}_A=\sqrt{{\left({v^{\prime }}_A\right)}_n^2+{\left({v^{\prime }}_A\right)}_t^2}$

Substitute $-1.0446\text{ft}/\text{s}$ for ${\left({v^{\prime }}_A\right)}_n$ and $2.1203\text{ft}/\text{s}$ for ${\left({v^{\prime }}_A\right)}_t$.

$\begin{array}{c}{v^{\prime }}_A=\sqrt{{\left(-1.0446\text{ft}/\text{s}\right)}^2+{\left(2.1203\text{ft}/\text{s}\right)}^2}\\ =2.36\text{ft}/\text{s}\end{array}$

Calculate the angle for $\left(\beta \right)$ using the relation:

$\beta ={\tan }^{-1}\left(\frac{{\left({v^{\prime }}_A\right)}_t}{-{\left({v^{\prime }}_A\right)}_n}\right)$

Substitute $-1.0446\text{ft}/\text{s}$ for ${\left({v^{\prime }}_A\right)}_n$ and $2.1203\text{ft}/\text{s}$ for ${\left({v^{\prime }}_A\right)}_t$.

$\begin{array}{c}\beta ={\tan }^{-1}\left(\frac{2.1203\text{ft}/\text{s}}{1.0446\text{ft}/\text{s}}\right)\\ =63.77°\end{array}$

Calculate the resultant angle of velocity of sphere A $\left(\varphi \right)$ using the relation:

$\varphi =\beta +\theta$

Substitute 63.77  for $\beta$ and $20°$ for $\theta$.

$\begin{array}{c}\varphi =63.77°+20°\\ =83.77°\end{array}$

Therefore, the velocity of A $\left({v^{\prime }}_A\right)$ and B $\left({v^{\prime }}_B\right)$ immediately after impact are $\underset{\_}{2.36\text{ft}/\text{s}\left(83.8°\right)}$ and $\underset{\_}{3.23\text{ft}/\text{s}(\to )}$ respectively.

To determine

Find the maximum deflection of the spring assuming A does not strike B again before this point.

Answer to Problem 13.188P

The maximum deflection of the spring $\left(\Delta x\right)$ is $\underset{\_}{1.97\text{in}\text{.}}$.

Explanation of Solution

Given information:

The angle of the rope $\left(\alpha \right)$ is $30°$.

The weight of the sphere A $\left(W_A\right)$ is $1\text{lb}$.

The weight of the wedge B $\left(W_B\right)$ is $2\text{lb}$.

The speed of the sphere A $\left(v_A\right)$ is $\text{4ft/s}$.

The coefficient of restitution between A and wedge (e) is 0.7.

The length of the rope (l) is $2.6\text{ft}$.

The spring constant (k) is $\text{2lb/in}\text{.}$

The angle $\left(\theta \right)$ is $20°$.

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

Calculation:

Calculate the kinetic energy of wedge block just before impact $\left(T_{B1}\right)$ using the formula:

$T_{B1}=\frac{1}{2}{m}_B{v^{\prime }}_B^2$

Here, ${v^{\prime }}_B$ is the final velocity of wedge.

Substitute $0.06211\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m_B$ and $3.23\text{ft}/\text{s}$ for ${v^{\prime }}_B$.

$\begin{array}{c}{T}_{B1}=\frac{1}{2}\left(0.06211\text{lb}\cdot {\text{s}}^2/\text{ft}\right){\left(3.23\text{ft}/\text{s}\right)}^2\\ =0.3239\text{lb}\cdot \text{ft}\end{array}$

The expression for the potential energy of spring at the end of impact $\left(V_{B2}\right)$ as follows:

$V_{B2}=\frac{1}{2}k{\left(\Delta x\right)}^2$

The expression for the principle of conservation of energy for wedge block as follows:

$T_{B1}+V_{B1}=T_{B2}+V_{B2}$

Here, $V_{B1}$ is the potential energy of spring just before impact and $T_{B2}$ is the kinetic energy of spring at the end of impact.

Substitute $0.3239\text{lb}\cdot \text{ft}$ for $T_{B1}$, 0 for $V_{B1}$, 0 for $T_{B2}$, and $\frac{1}{2}k{\left(\Delta x\right)}^2$ for $V_{B2}$.

$\begin{array}{l}0.3239\text{lb}\cdot \text{ft}+0=0+\frac{1}{2}k{\left(\Delta x\right)}^2\\ 0.3239\text{lb}\cdot \text{ft}=\frac{1}{2}k{\left(\Delta x\right)}^2\end{array}$

Substitute $2\text{lb}/\text{in}\text{.}$ for k.

$\begin{array}{l}0.3239\text{lb}\cdot \text{ft}=\frac{1}{2}\left(2\text{lb}/\text{in}\text{.}\right)\left(\frac{12\text{in}\text{.}}{1\text{ft}}\right){\left(\Delta x\right)}^2\\ {\left(\Delta x\right)}^2=0.0269{\text{lb}}^2\\ \Delta x=\left(0.1642\text{ft}\right)\left(\frac{12\text{in}\text{.}}{1\text{ft}}\right)\\ \Delta x=1.97\text{in}\text{.}\end{array}$

Therefore, the maximum deflection of the spring $\left(\Delta x\right)$ is $\underset{\_}{1.97\text{in}\text{.}}$.