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.

To determine

The final velocities of the car A and C.

Answer to Problem 14.105RP

The value of final velocities of the car A is V_A = 0.641ft/s and for car C is V_c=4.41ft/s

Explanation of Solution

Given information:

Based on the principle of impulse momentum system before impact and after impact is same and equal

First consider the momentum of the car A hits car B at first time:

$\begin{array}{l}m{v}_A+m{v}_B=m{v}_A^{\text{'}}+m{v}_B^{\text{'}}\\ m\left(5.76ft/s\right)+m\left(0\right)=m{v}_A^{\text{'}}+m\left(5.04ft/s\right)\\ 5.76=v_A^{\text{'}}+5.04\\ v_A^{\text{'}}=0.72ft/s\end{array}$

And then consider the momentum when the car B hits car C we get

$\begin{array}{l}m{v}_B^{\text{'}}+m{v}_c=m{v}_B^{\text{'}\text{'}}+m{v}_c^{\text{'}}\\ m\left(5.04ft/s\right)+m\left(0\right)=m\left(0.630ft/s\right)+m{v}_c^{\text{'}}\\ 5.04=0.630+v_c^{\text{'}}\\ v_c^{\text{'}}=4.41ft/s\end{array}$

Formulate the momentum when car A hits car B for second consideration

$\begin{array}{l}m{v}_A^{\text{'}}+m{v}_B^{\text{'}\text{'}}=m{v}_A^{\text{'}\text{'}}+m{v}_B^{\text{'}\text{'}\text{'}}\\ m\left(0072ft/s\right)+m\left(0.630ft/s\right)=m{v}_A^{\text{'}\text{'}}+m\left(0.709ft/s\right)\\ 0.72+0.630=v_A^{\text{'}\text{'}}+0.709\\ v_A^{\text{'}\text{'}}=1.35-0.709\\ =0.641ft/s\end{array}$

To determine

The coefficient of restitution for each of the collision.

Answer to Problem 14.105RP

The value of coefficient of restitution is e₁ = 0.75, e₂=0.75, e₃=0.75

Explanation of Solution

Here the coefficient of restitution is a difference between relative velocities to the after and before impact.

First, we must formulate the coefficient of restitution when the condition is A hits B for first consideration.

$\begin{array}{l}{e}_1=\left|\frac{v_A^{\text{'}}-v_B^{\text{'}}}{v_A-v_B}\right|\\ e_1=\left|\frac{0.72ft/s-5.04ft/s}{5.76ft/s-0}\right|\\ =\left|\frac{-4.32}{5.76}\right|\\ =0.75\end{array}$

Similarly formulate coefficient of resolution the condition is B hits C, we get:

$\begin{array}{l}{e}_2=\left|\frac{v_B^{\text{'}\text{'}}-v_C^{\text{'}}}{v_B^{\text{'}}-v_C}\right|\\ e_2=\left|\frac{0.630ft/s-4.41ft/s}{5.04ft/s-0}\right|\\ =\left|\frac{-3.78}{5.04}\right|\\ =0.75\end{array}$

Formulate the coefficient of resolution then consider A hits B for the second consideration.

$\begin{array}{l}{e}_3=\left|\frac{v_A^{\text{'}\text{'}}-v_B^{\text{'}\text{'}\text{'}}}{v_A^{\text{'}}-v_B^{\text{'}\text{'}}}\right|\\ e_3=\left|\frac{0.641ft/s-0.709ft/s}{0.72ft/s-0.630ft/s}\right|\\ =\left|\frac{-0.068}{0.09}\right|\\ =0.756\end{array}$