Chapter 13.4, Problem 13.188P

When the rope is at an angle of $a=30°$ , the 1-Ib sphere A has a speed $v_0=4$ 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 $\theta =20°$ . 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

(a)

The velocities of A and B just after the impact.

Answer to Problem 13.188P

$\to v_B{}^1=3.228ft/s$

$v{\prime }_A=2.3636ft/s$ At $\beta =63.772°+20°=83.772°$

Explanation of Solution

Given information:

$\begin{array}{l}\alpha =30°\\ v_0=4ft/s\\ l=2.6ft\\ \theta =20°\\ e=0.7\\ k=2lb/in\end{array}$

Mass of sphere is $1lb$.

Mass of wedge is $2lb$.

Concept used:

The total linear momentum of two particles is conserved. Therefore:

$m_A{v}_A+m_B{v}_B=m_A{v}_A{}^1+m_B{v}_B{}^1$

The co-efficient of restitution is defined as.

$v_B{}^1-v_A{}^1=e\left(v_A-v_B\right)$

The principle of conservation of energy is defined as.

“When a particle moves under the action of conservation of forces. the sum of kinetic energy and potential energy of that particle remains constant.”

$T_1+V_1=T_2+V_2$

Calculation:

At initial stage:

$h_0=l(1-cos\alpha )=\left(2.6\right)(1-cos30°)=0.34833ft$

Find the Kinetic and potential energies.

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

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

Just before the impact:

$\begin{array}{l}{T}_1=\frac{1}{2}{m}_A{v}_A{}^2\\ V_1=0\end{array}$

Therefore.

$\begin{array}{l}{m}_{A}g{h}_0+\frac{1}{2}{m}_A{v}_0{}^2=\frac{1}{2}{m}_A{v}_A{}^2\\ v_A{}^2=2g{h}_0+v_0{}^2\end{array}$

Substitute and solve:

$\begin{array}{l}{v}_A{}^2=2\left(32.2ft/s^2\right)\left(0.34833f\right)+{\left(4ft/s\right)}^2\\ v_A=6.1994ft/s\end{array}$

Draw impulse momentum diagram.

Apply conservation of momentum in t direction.

$m_A{v}_{A}sin\theta +0=m_A{(v{\prime }_A)}_t$

Therefore:

${(v{\prime }_A)}_t=v_{A}sin\theta =6.1994sin20°=2.1203ft/s$

For both A and B :

Apply conservation of momentum in x direction.

$m_A{v}_A+0=m_A{(v{\prime }_A)}_{t}sin\theta +m_A{(v{\prime }_A)}_{n}cos\theta +m_B{v}_B{}^1$

Substitute:

$\left(\frac{1lb}{g}\right)\left(6.1994\right)=\left(\frac{1lb}{g}\right)\left(2.1203\right)\sin 20°+\left(\frac{1lb}{g}\right)\cos 20°{(v{\prime }_A)}_n+\left(\frac{2lb}{g}\right)v_B{}^1$

Therefore:

$\cos 20°{(v{\prime }_A)}_n+2v_B{}^1=5.4742\to \left(1\right)$

Apply co-efficient of restitution equation.

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

Substitute:

$v_B{}^1\cos 20°-{(v{\prime }_A)}_n=e{v}_A\cos 20°$

Solve.

$v_B{}^1\cos 20°-{(v{\prime }_A)}_n=0.7\left(6.1994\right)\cos 20°\to \left(2\right)$

Solve equation 1 and 2.

$\begin{array}{l}{(}_v=-1.0446ft/s\\ v_B{}^1=3.228ft/s\end{array}$

Find the magnitude of $v{\prime }_A$ :

$v{\prime }_A=\sqrt{{(v{\prime }_A)}_n{}^2+{(v{\prime }_A)}_t{}^2}=\sqrt{{\left(-1.0446ft/s\right)}^2+{\left(2.1203ft/s\right)}^2}=2.3636ft/s$

Find the angle.

$\begin{array}{l}\tan \beta =\frac{{(v{\prime }_A)}_t}{-{(v{\prime }_A)}_n}=\frac{2.1203}{1.0446}\\ \beta =63.772°\end{array}$

Conclusion:

The velocities of A and B just after the impact is equal to:

$\to v_B{}^1=3.228ft/s$. $v{\prime }_A=2.3636ft/s$ At $\beta =63.772°+20°=83.772°$

To determine

(b)

The maximum deflection of the spring.

Answer to Problem 13.188P

Maximum deflection of the spring is $x=1.97in$ .

Explanation of Solution

Given information:

$\begin{array}{l}\alpha =30°\\ v_0=4ft/s\\ l=2.6ft\\ \theta =20°\\ e=0.7\\ k=2lb/in\end{array}$

Mass of sphere is $1lb.$

Mass of wedge is $2lb.$

Concept used:

The principle of conservation of energy is defined as:

“When a particle moves under the action of conservation of forces. the sum of kinetic energy and potential energy of that particle remains constant”

$T_1+V_1=T_2+V_2$

Calculation:

According to the conservation of energy:

Just after the impact.

$\frac{1}{2}{m}_B{\left(v_B{}^1\right)}^2=\frac{1}{2}k{x}^2$

Rearrange:

$x=\sqrt{\frac{m_B}{k}}\left(v_B{}^1\right)$

According to sub part a.:

$\to v_B{}^1=3.228ft/s$

Substitute and solve:

$x=\sqrt{\frac{m_B}{k}}\left(v_B{}^1\right)=\sqrt{\frac{\left(\frac{2lb}{32.2ft/s^2}\right)}{\left(2lb/in\right)\left(12in/ft\right)}}\left(3.228ft/s\right)$

Therefore.

$\begin{array}{l}x=0.1642ft\\ x=1.97in\end{array}$

Conclusion:

The maximum deflection is equal to $x=1.97in$.