Question

Consider a system of two toy rail cars (i.e.. frictionless masses)- Suppose that car 1 has mass 6 kg and is traveling at 3 m/s toward the other car- Suppose car 2 has mass 3 kg and is moving toward the other car at 9 m/s. There is a bumper on the second rail car that engages at the moment the cars hit and does not let go (it connects the two cars). The bumper acts as a spring with spring constant 6 N/m. The second car is 10 m from a wall. Let t = 0 be the time that the cars link up. Let Ii be the displacement of the first car from its position at t = 0. and let x2 be the displacement of the second car from its original position. a. Set up a system of second-order differential equations that models this situation. b. Find the solution to this system of differential equations. c. Will the cars be moving toward the wall, away from the wall, or will they be nearly stationary?

Answer 1

(a) Firstly, we apply the initial conditions . First the cars start from the initla positions

$x_1(0)=0, x_2(t)=10$

The first car travels with a speed of 3 m/s and the second car is moving with 9m/s so

$x_1'(0)=3, x_2'(t)=9$

Now applying the following equation

$M x''=-k x$

Writing in the matrix form

$\begin{bmatrix} 6 &0 \\ 0&3 \end{bmatrix} x''=\begin{bmatrix} -6 &6 \\ 6& -6 \end{bmatrix} x$

$x''=\begin{bmatrix} -1 &1 \\ 2& -2 \end{bmatrix} x ....(1)$

We need to find the eigen values that can be worked out as

$\begin{bmatrix} -1-\lambda &1 \\ 2& -2-\lambda \end{bmatrix} =0$

On solving the above determinant we can get the eigen values lambda -0,-3. The corresponding eigen vectors are given as

$\begin{bmatrix} 1\\1 \end{bmatrix}, \begin{bmatrix} 1\\-2 \end{bmatrix}$

and $\omega=\sqrt{3}$

Our general solution becomes

$x=\begin{bmatrix} a_1+b_1t+a_2 \cos \sqrt{3}t+b_2 \sin \sqrt{3}t\\ a_1+b_1t-2a_2 \cos \sqrt{3}t-2b_2 \sin \sqrt{3}t \end{bmatrix}$

Now applying the conditions

$x_1(0)=0, x_2(0)=0,x_1'(0)=3, x_2'(0)=9$

On applying the conditions on the position we get

$x(0)=\begin{bmatrix} a_1+a_2\\a_1-2a_2 \end{bmatrix}=0$

Clearly

$a_1=a_2=0$

Now applying the conditions on the velocity

$x'(t)=\begin{bmatrix} b_1+\sqrt{3}b_2 \cos \sqrt{3}t\\ b_1-2 \sqrt{3}b_2 \cos \sqrt{3}t \end{bmatrix}$

$x'(0)=\begin{bmatrix} b_1+\sqrt{3}b_2 \\ b_1-2 \sqrt{3}b_2\end{bmatrix}=\begin{bmatrix} 3\\9 \end{bmatrix}$

we get

$b_1=5, b_2=-\frac{2}{\sqrt{3}}$

Hence, the solution is

$x^\rightarrow=\begin{bmatrix} 5t-\frac{2}{\sqrt{3}}\sin \sqrt{3}t\\ 5t+\frac{4}{\sqrt{3}}\sin \sqrt{3}t \end{bmatrix}....(2)$

So from the above we conclude

$x''^\rightarrow=\begin{bmatrix} -1 &1 \\ 2& -2 \end{bmatrix} x^\rightarrow$

x(0)=0m

y(0)=0 m

x'(0) =3 m/s

y'(0)=9 m/s

b) From Eq. (2) we conclude

$x(t)= 5t-\frac{2}{\sqrt{3}}\sin \sqrt{3}t$ m

$y(t)= 5t+\frac{4}{\sqrt{3}}\sin \sqrt{3}t$ m

c)After collision they will slightly move away from the wall