Chapter 14, Problem 14.105RP

Three identical cars are being unloaded from an automobile carrier. Cars B and C have just been unloaded and are at rest with their brakes off when car A leaves the unloading ramp with a velocity of 5.76 ft/s and hits car B, which hits car C. Car A then again hits car B. Knowing that the velocity of car B is 5.04 ft/s after the first collision, 0.630 ft/s after the second collision, and 0.709 ft/s after the third collision, determine (a) the final velocities of cars A and C, (b) the coefficient of restitution for each of the collisions.

Fig. P14.105

To determine

Find the final velocity of the car A and C.

Answer to Problem 14.105RP

The final velocity of the car A and C are $\underset{\_}{0.641\text{ft/s}}$ and $\underset{\_}{4.41\text{ft/s}}$.

Explanation of Solution

Given information:

Consider the mass of car A, car B, and C is denoted by m.

The initial velocity of the car is ${\left(v_A\right)}_1=5.76\text{ft/s}$.

The initial velocity of the car B and C is zero.

The velocity of the car B after the first, second and third collisions are ${\left(v_B\right)}_2=5.04\text{ft/s}$, ${\left(v_B\right)}_3=0.630\text{ft/s}$, and ${\left(v_B\right)}_4=0.709\text{ft/s}$

Calculation:

The Horizontal momentum of the system is conserved as no horizontal force is acting on the system.

Show the Event $1\to 2$ of car A hitting the car B as shown in Figure 1.

Refer to Figure 1.

Show the conservation of horizontal momentum as follows:

$m{\left(v_A\right)}_1+m{\left(v_B\right)}_1=m{\left(v_A\right)}_2+m{\left(v_B\right)}_2$ (1)

Substitute $5.76\text{ft/s}$ for ${\left(v_A\right)}_1$, 0 for ${\left(v_B\right)}_1$, and $5.04\text{ft/s}$ for ${\left(v_B\right)}_2$ in Equation (1).

$\begin{array}{l}m\left(5.76\right)+m\times 0=m{\left(v_A\right)}_2+m\left(5.04\right)\\ 5.76m=m{\left(v_A\right)}_2+5.04m\\ {\left(v_A\right)}_2=0.72\text{ft/s}\end{array}$

Show the Event $2\to 3$ of car B hitting the car C as shown in Figure 2.

Refer to Figure 2.

Show the conservation of horizontal momentum as follows:

$m{\left(v_B\right)}_2+m{\left(v_C\right)}_1=m{\left(v_B\right)}_3+m{\left(v_C\right)}_3$ (2)

Substitute $0.630\text{ft/s}$ for ${\left(v_B\right)}_3$, 0 for ${\left(v_C\right)}_1$, and $5.04\text{ft/s}$ for ${\left(v_B\right)}_2$ in Equation (2).

$\begin{array}{l}m\left(5.04\right)+m\times 0=m\left(0.630\right)+m{\left(v_C\right)}_3\\ 5.04m=0.63m+m{\left(v_C\right)}_3\\ {\left(v_C\right)}_3=4.41\text{ft/s}\end{array}$

Thus, the final velocity of the car C is $\underset{\_}{4.41\text{ft/s}}$.

Show the Event $3\to 4$ of car A hitting the car B as shown in Figure 3.

Refer to Figure 3.

Show the conservation of horizontal momentum as follows:

$m{\left(v_A\right)}_2+m{\left(v_B\right)}_3=m{\left(v_A\right)}_4+m{\left(v_B\right)}_4$ (3)

Substitute $0.630\text{ft/s}$ for ${\left(v_B\right)}_3$, $0.72\text{ft/s}$ for ${\left(v_A\right)}_2$, and $0.709\text{ft/s}$ for ${\left(v_B\right)}_4$ in Equation (3).

$\begin{array}{l}m\left(0.72\right)+m\left(0.63\right)=m{\left(v_A\right)}_4+0.709m\\ 1.35m=m{\left(v_A\right)}_4+0.709m\\ {\left(v_A\right)}_4=0.641\text{ft/s}\end{array}$

Thus, the final velocity of the car A is $\underset{\_}{0.641\text{ft/s}}$.

To determine

Find the coefficients of restitution of each collision.

Answer to Problem 14.105RP

The coefficients of restitution of first, second and third collision are $\underset{\_}{0.75}$, $\underset{\_}{0.75}$, and $\underset{\_}{0.756}$.

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

Consider the coefficient of restitution corresponding to collision Event $1\to 2$, $2\to 3$, $3\to 4$ are denoted by $e_{1\to 2}$, $e_{2\to 3}$, and $e_{3\to 4}$.

Calculate the coefficient of restitution corresponding to collision Event $1\to 2$ using the relation:

$e_{1\to 2}=\left|\frac{{\left(v_A\right)}_2-{\left(v_B\right)}_2}{{\left(v_A\right)}_1-{\left(v_B\right)}_1}\right|$ (4)

Substitute $0.72\text{ft/s}$ for ${\left(v_A\right)}_2$, $5.04\text{ft/s}$ for ${\left(v_B\right)}_2$, $5.76\text{ft/s}$ for ${\left(v_A\right)}_1$, and $0$ for ${\left(v_B\right)}_1$ in Equation (4).

$\begin{array}{c}{e}_{1\to 2}=\left|\frac{0.72-5.04}{5.76-0}\right|\\ =0.75\end{array}$

Calculate the coefficient of restitution corresponding to collision Event $2\to 3$ using the relation:

$e_{2\to 3}=\left|\frac{{\left(v_B\right)}_3-{\left(v_C\right)}_3}{{\left(v_B\right)}_2-{\left(v_C\right)}_2}\right|$ (5)

Substitute $0.63\text{ft/s}$ for ${\left(v_B\right)}_3$, $4.41\text{ft/s}$ for ${\left(v_C\right)}_3$, $5.04\text{ft/s}$ for ${\left(v_B\right)}_2$, and $0$ for ${\left(v_C\right)}_2$ in Equation (4).

$\begin{array}{c}{e}_{2\to 3}=\left|\frac{0.63-4.41}{5.04-0}\right|\\ =0.75\end{array}$

Calculate the coefficient of restitution corresponding to collision Event $3\to 4$ using the relation:

$e_{3\to 4}=\left|\frac{{\left(v_A\right)}_4-{\left(v_B\right)}_4}{{\left(v_A\right)}_3-{\left(v_B\right)}_3}\right|$ (6)

Substitute $0.641\text{ft/s}$ for ${\left(v_A\right)}_4$, $0.709\text{ft/s}$ for ${\left(v_B\right)}_4$, $0.72\text{ft/s}$ for ${\left(v_A\right)}_3$, and $0.63$ for ${\left(v_B\right)}_3$ in Equation (6).

$\begin{array}{c}{e}_{3\to 4}=\left|\frac{0.641-0.709}{0.72-0.63}\right|\\ =0.756\end{array}$

Thus, the coefficients of restitution of first, second and third collision are $\underset{\_}{0.75}$, $\underset{\_}{0.75}$, and $\underset{\_}{0.756}$.