Two thin and identical steel rods move with equal speeds v=1 m/s in opposite directions and collide longitudinally.The collision is perfectly elastic and the rods swap their velocities. How long does the collision last in seconds if the length of the rods is l=30 cm. Assume that the speed of sound in the steel rods is c=5000 m/s?
This problem is fundamentally flawed, because you cannot treat the collision as elastic (in the sense that all the energy stays in the center of mass motion). You can't on the one hand consider the effect of the finite elasticity coefficient making the collision last a finite time and the same time pretend that the rods won't oscillate.
This is how you have to solve the problem. The relation between the deformation tensor e_{i,j} and the stress tensor t_{i,j} is
e_{i,j} = (1+nu)/E t_{i,j} - nu/E t_{k,k} delta_{i,j}
where
e_{i,j} = 1/2 [du_i/dx_j + du_j/dx_i]
with u the displacement vector and
t_{i,j} is the ith component of the force per unit area with normal in the jth direction that the part the normal is pointing in exerts on the other part.
Taking the 1-direction to be the direction the rods move in, we can assume that the only nonzero component of t is t_{1,1}. At the point where the two rods meet this will be positive at first. If you then pretend that the two rods form one big rod, the collision time is the time during which t_{1,1} at that points is larger than zero.
The equation of motion is (Newton's second law in local form):
rho d^u_i/dt^2 = d/dx_j t_{i,j}
Express t_{i,j} in terms of u_{i,j} from:
e_{i,j} = (1+nu)/E t_{i,j} - nu/E t_{k,k} delta_{i,j}
Take the trace:
e_{i,i} = (1-2nu)/E t_{k,k} ----->
t_{k,k} = E/(1-2nu) e_{k,k}
so, you can write:
t_{i,j} =
E/(1+nu) e_{i,j} + E nu/[(1-2nu)(1+nu)] e_{k,k} delta_{i,j}
The equation of motion
rho d^2u_i/dt^2 = d/dx_j t_{i,j}
can then be written in terms of the displacement u alone as:
rho d^2u_i/dt^2 =
1/2 E/(1+nu) d^2u_i/dx_jdx_j +
E/[2(1+nu)(1-2nu)] d^2u_j/dx_jdx_i
You need to solve these partial differential equation with the appropriate boundary and initial conditions to solve the problem.
0.00012
To find out how long the collision lasts, we need to determine the time it takes for the sound wave to travel across the rods.
Given:
Speed of sound in the steel rods, c = 5000 m/s
Length of the rods, l = 30 cm = 0.3 m
We can calculate the time it takes for the sound wave to cross one rod using the formula:
t = l / c
Substituting the values:
t = 0.3 / 5000
t ≈ 0.00006 seconds (approximately)
Since the rods collide longitudinally, both rods will produce sound waves that need to traverse the length of the other rod, so we need to consider the total time for the collision:
Total time = 2t = 2 * 0.00006
Total time ≈ 0.00012 seconds
Therefore, the collision lasts approximately 0.00012 seconds.