Chapter 13.4, Problem 13.178P

(a)

To determine

Find the coefficient of restitution between A and B $\left(e_{AB}\right)$ and between B and C.

Answer to Problem 13.178P

The coefficient of restitution between A and B $\left(e_{AB}\right)$ and between B and C is $\underset{\_}{0.694}$.

Explanation of Solution

Given information:

The weight of the block A $\left(W_A\right)$ is $0.8\text{lb}$.

The weight of the block B $\left(W_B\right)$ is $0.8\text{lb}$.

The weight of the block C $\left(W_C\right)$ is $2.4\text{lb}$.

The coefficient of friction between the block and plane $\left({\mu }_k\right)$ is 0.3.

The initial speed of the block A $\left(v_0\right)$ is $15\text{ft/s}$.

The blocks B and C are at rest.

The distance between the blocks (d) is $12\text{in}\text{.}$.

The width of the each blocks (b) is $3\text{in}\text{.}$.

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

Calculation:

Calculate the mass of the block A $\left(m_A\right)$ using the formula:

$\begin{array}{c}{W}_A=m_{A}g\\ m_A=\frac{W_A}{g}\end{array}$

Substitute $0.8\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$m_A=\frac{0.8}{32.2}$

Calculate the mass of the block B $\left(m_B\right)$ using the formula:

$\begin{array}{c}{W}_B=m_{B}g\\ m_B=\frac{W_B}{g}\end{array}$

Substitute $0.8\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$m_B=\frac{0.8}{32.2}$

Calculate the mass of the block C $\left(m_C\right)$ using the formula:

$\begin{array}{c}{W}_C=m_{C}g\\ m_C=\frac{W_C}{g}\end{array}$

Substitute $2.4\text{lb}$ for $W_C$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$m_C=\frac{2.4}{32.2}$

Show the diagram of the block A just before its impact with block B as in Figure (1).

The expression for the initial kinetic energy of the block A at position ‘1’ $\left(T_1\right)$ as follows:

$T_1=\frac{1}{2}\frac{W_A}{g}{v}_0^2$

Here, $v_0$ is the velocity of the block A at position ‘1’.

The expression for the kinetic energy of the block A at position ‘2’ just before its impact with blocks B $\left(T_2\right)$ as follows:

$T_2=\frac{1}{2}\frac{W_A}{g}{\left(v_A^2\right)}_2$

Here, ${\left(v_A\right)}_2$ is the velocity of the block A at position ‘2’ just before its impact with block B.

The expression for the work done by the block A to overcome frictional force $\left(U_{1-2}\right)$ as follows:

$U_{1-2}=-{\mu }_k{W}_{A}d$

The expression for the principle of work and energy to the block A at position ‘1’ and position ‘2’ just before its impact with block B as follows:

$T_1+U_{1-2}=T_2$ (1)

Substitute $\frac{1}{2}\frac{W_A}{g}{v}_0^2$ for $T_1$, $\frac{1}{2}\frac{W_A}{g}{\left(v_A^2\right)}_2$ for $T_2$, and $-{\mu }_k{W}_{A}d$ for $U_{1-2}$ in Equation (1).

$\begin{array}{l}\frac{1}{2}\frac{W_A}{g}{v}_0^2-{\mu }_k{W}_{A}d=\frac{1}{2}\frac{W_A}{g}{\left(v_A^2\right)}_2\\ \frac{1}{2}\frac{v_0^2}{g}-{\mu }_{k}d=\frac{1}{2}\frac{{\left(v_A^2\right)}_2}{g}\\ {\left(v_A^2\right)}_2=v_0^2-2g{\mu }_{k}d\end{array}$

Substitute $15\text{ft}/\text{s}$ for $v_0$, $32.2\text{ft}/{\text{s}}^2$ for $g$, $0.3$ for ${\mu }_k$, and $12\text{in}\text{.}$ for d.

$\begin{array}{l}{\left(v_A^2\right)}_2={\left(15\text{ft}/\text{s}\right)}^2-2\left(32.2\text{ft}/{\text{s}}^2\right)\left(0.3\right)\left(12\text{in}\right)\\ {\left(v_A^2\right)}_2=225-2\left(32.2\right)\left(0.3\right)\left[12\text{in}\times \left(\frac{0.0833334\text{ft}}{1\text{in}}\right)\right]\\ {\left(v_A^2\right)}_2=225-\left(19.32\right)\left(1\text{ft}\right)\\ {\left(v_A^2\right)}_2=205.68{\text{ft}}^2/{\text{s}}^2\\ {\left(v_A\right)}_2=14.342\text{ft}/\text{s}\end{array}$

Show the diagram of the block A just after its impact with block B as in Figure (2).

The expression for the kinetic energy of the block A immediately after the impact $\left({T^{\prime }}_2\right)$ as follows:

${T^{\prime }}_2=\frac{1}{2}\frac{W_A}{g}{\left({v^{\prime }}_A\right)}_2^2$

Here, ${\left({v^{\prime }}_A\right)}_2$ is the velocity of block A immediately after the impact.

The block finally comes to stop after the impact. Thus, $T_3=0$.

The expression for the work done by the block A after the collision to overcome the frictional force $\left(U_{2-3}\right)$ as follows:

$U_{2-3}=-{\mu }_k{W}_{A}b$

The expression for the principle of work and energy to the block A after it collides with block B to find the velocity of the block A after its impact with B as follows:

${T^{\prime }}_2+U_{2-3}=T_3$ (2)

Substitute $\frac{1}{2}\frac{W_A}{g}{\left({v^{\prime }}_A\right)}_2^2$ for ${T^{\prime }}_2$, 0 for $T_3$, and $-{\mu }_k{W}_{A}b$ for $U_{2-3}$ in the equation (2).

$\begin{array}{l}{T^{\prime }}_2+U_{2-3}=T_3\\ \frac{1}{2}\frac{W_A}{g}{\left({v^{\prime }}_A\right)}_2^2-{\mu }_k{W}_{A}b=0\\ \frac{1}{2}\frac{W_A}{g}{\left({v^{\prime }}_A\right)}_2^2={\mu }_k{W}_{A}b\\ {\left({v^{\prime }}_A\right)}_2^2=2g{\mu }_{k}b\end{array}$

Substitute $32.2\text{ft}/{\text{s}}^2$ for g, $0.3$ for ${\mu }_k$, and $3\text{in}\text{.}$ for b.

$\begin{array}{l}{\left({v^{\prime }}_A\right)}_2^2=2\left(32.2\text{ft}/{\text{s}}^2\right)\left(0.3\right)\left(3\text{in}\right)\\ {\left({v^{\prime }}_A\right)}_2^2=\left(19.32\right)\left[3\times \left(\frac{0.0833334\text{ft}}{1\text{in}}\right)\right]\\ {\left({v^{\prime }}_A\right)}_2^2=\left(19.32\right)\left(0.25\right)\\ {\left({v^{\prime }}_A\right)}_2^2=4.83{\text{ft}}^2/{\text{s}}^2\\ {\left({v^{\prime }}_A\right)}_2=2.198\text{ft}/\text{s}\end{array}$

Show the momentum impact diagram of the blocks A and B as in Figure (3).

The expression for the principle of conservation of momentum to the collision between the block A and block B as follows:

$m_A{\left(v_A\right)}_2+m_B{v}_B=m_A{\left({v^{\prime }}_A\right)}_2+m_B{v^{\prime }}_B$

Here, $v_B$ is the initial velocity of the block B before impact and ${v^{\prime }}_B$ is the velocity of the block B after its impact with block A.

Substitute $\frac{0.8\text{lb}}{32.2}$ for both $m_A$, $\frac{0.8\text{lb}}{32.2}$ for $m_B$, 0 for $v_B$, $14.342\text{ft}/\text{s}$ for ${\left(v_A\right)}_2$, and $2.198\text{ft}/\text{s}$ for ${\left({v^{\prime }}_A\right)}_2$.

$\begin{array}{l}\left(\frac{0.8\text{lb}}{32.2}\right)\left(14.342\text{ft}/\text{s}\right)+\left(\frac{0.8\text{lb}}{32.2}\right)\left(0\right)=\left(\frac{0.8\text{lb}}{32.2}\right)\left(2.198\text{ft}/\text{s}\right)+\left(\frac{0.8\text{lb}}{32.2}\right){v^{\prime }}_B\\ {v^{\prime }}_B+2.198=14.342\\ {v^{\prime }}_B=14.342-2.198\\ {v^{\prime }}_B=12.144\text{ft}/\text{s}\end{array}$

Calculate the coefficient of restitution for the impact between the block A and block B$\left(e_{AB}\right)$ using the formula:

$\begin{array}{l}\left[{\left(v_A\right)}_2-v_B\right]e_{AB}={v^{\prime }}_B-{\left({v^{\prime }}_A\right)}_2\\ e_{AB}=\frac{{v^{\prime }}_B-{\left({v^{\prime }}_A\right)}_2}{{\left(v_A\right)}_2-v_B}\end{array}$

Substitute 0 for $v_B$, $14.342\text{ft}/\text{s}$ for ${\left(v_A\right)}_2$, $2.198\text{ft}/\text{s}$ for ${\left({v^{\prime }}_A\right)}_2$, and $12.144\text{ft}/\text{s}$ for ${v^{\prime }}_B$.

$\begin{array}{c}{e}_{AB}=\frac{12.144\text{ft}/\text{s}-2.198\text{ft}/\text{s}}{14.342\text{ft}/\text{s}-0}\\ =\frac{9.946}{14.342}\\ =0.694\end{array}$

Therefore, the coefficient of restitution between A and B $\left(e_{AB}\right)$ and between B and C is $\underset{\_}{0.694}$.

(b)

To determine

Find the displacement (x) of block C.

Answer to Problem 13.178P

The displacement (x) of block C is $\underset{\_}{8.84\text{in}\text{.}}$.

Explanation of Solution

Given information:

The weight of the block A $\left(W_A\right)$ is $0.8\text{lb}$.

The weight of the block B $\left(W_B\right)$ is $0.8\text{lb}$.

The weight of the block C $\left(W_C\right)$ is $2.4\text{lb}$.

The coefficient of friction between the block and plane $\left({\mu }_k\right)$ is 0.3.

The initial speed of the block A $\left(v_0\right)$ is $15\text{ft/s}$.

The blocks B and C are at rest.

The distance between the blocks (d) is $12\text{in}\text{.}$.

The width of the each blocks (b) is $3\text{in}\text{.}$

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

Calculation:

Show the diagram of the block B just before its impact with block C as in Figure (4).

The expression for the kinetic energy of the block B at position ‘2’ just after the impact with block A $\left(T_2\right)$ as follows:

$T_2=\frac{1}{2}\frac{W_B}{g}{\left({v^{\prime }}_B\right)}^2$

The expression for the kinetic energy of the block B just before its impact with blocks C at the position ‘4’ $T_4$ as follows:

$T_4=\frac{1}{2}\frac{W_B}{g}{\left({v^{″}}_B\right)}^2$

Here, ${v^{″}}_B$ is the velocity of the block B just before its impact with block C.

The expression for the work done by the block B to overcome the frictional force in reaching position ‘4’ from position ‘2’ as follows:

$U_{2-4}=-{\mu }_k{W}_{B}d$

The expression for the principle of work and energy to the block B just before its impact with block C at the position ‘2’ and position ‘4’ as follows:

$T_2+U_{2-4}=T_4$ (3)

Substitute $\frac{1}{2}\frac{W_B}{g}{\left({v^{\prime }}_B\right)}^2$ for $T_2$, $\frac{1}{2}\frac{W_B}{g}{\left({v^{″}}_B\right)}^2$ for $T_4$, and $-{\mu }_k{W}_{B}d$ for $U_{2-4}$.

$\begin{array}{l}{T}_2+U_{2-4}=T_4\\ \frac{1}{2}\frac{W_B}{g}{\left({v^{\prime }}_B\right)}^2-{\mu }_k{W}_{B}d=\frac{1}{2}\frac{W_B}{g}{\left({v^{″}}_B\right)}^2\\ \frac{{\left({v^{\prime }}_B\right)}^2}{2g}-{\mu }_{k}d=\frac{{\left({v^{″}}_B\right)}^2}{2g}\\ {\left({v^{″}}_B\right)}^2={\left({v^{\prime }}_B\right)}^2-2g{\mu }_{k}d\end{array}$

Substitute $12.144\text{ft}/\text{s}$ for ${v^{\prime }}_B$, $32.2\text{ft}/{\text{s}}^2$ for g, $0.3$ for ${\mu }_k$, and $12\text{in}\text{.}$ for d.

$\begin{array}{l}{\left({v^{″}}_B\right)}^2={\left(12.144\text{ft}/\text{s}\right)}^2-2\left(32.2\text{ft}/{\text{s}}^2\right)\left(0.3\right)\left(12\text{in}\right)\\ {\left({v^{″}}_B\right)}^2=147.477-\left(19.32\right)\left[12\text{in}\times \left(\frac{0.0833334\text{ft}}{1\text{in}}\right)\right]\\ {\left({v^{″}}_B\right)}^2=147.477-\left(19.32\right)\left(1\right)\\ {\left({v^{″}}_B\right)}^2=128.157\\ {v^{″}}_B=11.321\text{ft}/\text{s}\end{array}$

Show the momentum impact diagram of the blocks B and C as in Figure (5).

The expression for the principle of conservation of momentum to the collision between the block B and block C as follows:

$m_B{v^{″}}_B+m_C{v}_C=m_B{v^{‴}}_B+m_C{v^{\prime }}_C$

Substitute $\frac{W_B}{g}$ for $m_B$ and $\frac{W_C}{g}$ for $m_C$.

$\left(\frac{W_B}{g}\right){v^{″}}_B+\left(\frac{W_C}{g}\right)v_C=\left(\frac{W_B}{g}\right){v^{‴}}_B+\left(\frac{W_C}{g}\right){v^{\prime }}_C$

Here, ${v^{‴}}_B$ is the velocity of the block B just after the impact with block C which is zero, as the Block B comes to rest immediately after the impact, $v_C$ is the initial velocity of the block C before impact and ${v^{\prime }}_C$ is the velocity of the block C just after the impact with block B.

Substitute $0.8\text{lb}$ for $W_B$, $2.4\text{lb}$ for $W_C$, 0 for $v_C$, $11.321\text{ft}/\text{s}$ for ${v^{″}}_B$, and 0 for ${v^{‴}}_B$.

$\begin{array}{l}\left(\frac{0.8\text{lb}}{g}\right)\left(11.321\text{ft}/\text{s}\right)+0=0+\left(\frac{2.4\text{lb}}{g}\right){v^{\prime }}_C\\ 2.4{v^{\prime }}_C=9.0568\\ {v^{\prime }}_C=3.774\text{ft}/\text{s}\end{array}$

Calculate the coefficient of restitution for the impact between the block B and block C$\left(e_{BC}\right)$ as follows:

$e_{BC}=\frac{{v^{\prime }}_C-{v^{‴}}_B}{{v^{″}}_B{}_2-v_C}$

Substitute 0 for $v_C$, $11.321\text{ft}/\text{s}$ for ${v^{″}}_B$, $3.774\text{ft}/\text{s}$ for ${v^{\prime }}_C$, and 0 for ${v^{‴}}_B$.

$\begin{array}{c}{e}_{BC}=\frac{3.774\text{ft}/\text{s}-0}{11.321\text{ft}/\text{s}-0}\\ =0.333\end{array}$

Show the diagram of the block C after its impact with Block B as in Figure (6).

The expression for the kinetic energy of the block C immediately after its impact with blocks B at position ‘4’ $\left(T_4\right)$ as follows:

$T_4=\frac{1}{2}\frac{W_C}{g}{\left({v^{\prime }}_C\right)}^2$

Finally, at the position ‘5’, the block C comes to rest. Thus, $T_5=0$

The expression for the work done by the block C to overcome the frictional force in reaching the position ‘5’$\left(U_{4-5}\right)$ as follows:

$U_{4-5}=-{\mu }_k{W}_{C}x$

Here, x is the distance travelled by the block C before coming to rest.

The expression for the principle of work and energy to the block C after its impact with block B as follows:

$T_4+U_{4-5}=T_5$ (4)

Substitute $\frac{1}{2}\frac{W_C}{g}{\left({v^{\prime }}_C\right)}^2$ for $T_4$, 0 for $T_5$, and $-{\mu }_k{W}_{C}x$ for $U_{4-5}$ in the Equation (4).

$\begin{array}{l}\frac{1}{2}\frac{W_C}{g}{\left({v^{\prime }}_C\right)}^2-{\mu }_k{W}_{C}x=0\\ {\mu }_k{W}_{C}x=\frac{1}{2}\frac{W_C}{g}{\left({v^{\prime }}_C\right)}^2\\ x=\frac{{\left({v^{\prime }}_C\right)}^2}{2{\mu }_{k}g}\end{array}$

Substitute $3.774\text{ft}/\text{s}$ for ${v^{\prime }}_C$, $32.2\text{ft}/{\text{s}}^2$ for g, and $0.3$ for ${\mu }_k$.

$\begin{array}{c}x=\frac{{\left(3.774\text{ft}/\text{s}\right)}^2}{2\left(0.3\right)\left(32.2\text{ft}/{\text{s}}^2\right)}\\ =\frac{14.24}{19.32}\\ =0.737\text{ft}\times \left(\frac{12\text{in}}{1\text{ft}}\right)\\ =8.84\text{in}\text{.}\end{array}$

Therefore, displacement (x) of block C is $\underset{\_}{8.84\text{in}\text{.}}$