Two blocks collide with spring in-between.

In summary, the problem involves two blocks with different masses and velocities colliding on a frictionless surface. A spring with a spring constant of 1120 N/m is attached to one of the blocks. The maximum compression of the spring occurs when both blocks have the same velocity. An attempt to solve the problem using F=ma and the velocity of each block at any position leads to an incorrect answer due to not taking into account the movement of the other end of the spring. The correct solution involves using conservation of energy.
  • #1
Jose Bouza
9
0

Homework Statement


Block 1 (mass 2 kg) is moving rightward at 10 m/s and block 2 (mass 5kg kg) is moving rightward at 3 m/s. The surface is frictionless, and a spring with spring constant of 1120 N/m is fixed on the left side on block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

Homework Equations

The Attempt at a Solution


I realize after the fact that this could probably be solved very easily with conservation of energy and/or momentum but I tried a different route and I'm confused as to why it gave me the wrong answer. What I did was use F=ma to set up a differential equation for the velocity of Block 1 in terms of its compression of the block x.
Since $$F_1 = -kx$$ the velocity at any position x is given by the solution to the equation
[tex]\frac{dv}{dx}m = -kx[/tex]
which is
$$v_1(x) = v_i1 -\frac{kx^2}{2m_1}$$
By Newtons third law the velocity of the second block at any position x should thus be
$$v_2(x) = v_i2 +\frac{kx^2}{2m_2}$$
If we set these equal to each other, plug in the relevant variables and solve for x we get a solution x = 0.1336m. When checked against the solution manual this is wrong though, where did I go wrong? Thanks.
 
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  • #2
Hello Jose,

I think it goes wrong at the very first ##
F_1 = -kx##. This is true for a spring with the other end fixed, but here the other end moves, so you have something with ##
F_1 = -k(x_1 - x_2 - C) ## with C something like ##x_1-x_2## when both blocks touch the end of the spring (without compressing it).
 
  • #3
BvU said:
Hello Jose,

I think it goes wrong at the very first ##
F_1 = -kx##. This is true for a spring with the other end fixed, but here the other end moves, so you have something with ##
F_1 = -k(x_1 - x_2 - C) ## with C something like ##x_1-x_2## when both blocks touch the end of the spring (without compressing it).
Yup you're right. When I add this into my solution I end up with a system of equations that I can't seem to solve for x2-x1 so I guess the energy method is the only one that works here. Thanks anyways.
 

Related to Two blocks collide with spring in-between.

1. What is the concept of "Two blocks collide with spring in-between?"

The concept of "Two blocks collide with spring in-between" refers to a physics problem that involves two blocks colliding with each other while a spring is attached between them. The collision causes the spring to compress or stretch, and the goal is to determine the resulting motion of the blocks and the spring.

2. What factors affect the outcome of the collision between the two blocks?

The outcome of the collision between the two blocks is affected by several factors, including the masses of the blocks, the initial velocities of the blocks, and the stiffness of the spring. The surface on which the blocks are colliding and the coefficient of restitution (a measure of how much energy is lost during the collision) also play a role.

3. How is energy conserved in a collision between two blocks with a spring in-between?

Energy is conserved in a collision between two blocks with a spring in-between due to the principle of conservation of energy. This principle states that the total energy in a closed system remains constant, and it can only be transferred from one form to another. In this scenario, the potential energy stored in the spring is converted into kinetic energy during the collision, and the total energy remains the same.

4. What equations can be used to solve for the motion of the blocks and the spring in this scenario?

The equations of motion, specifically the laws of conservation of energy and momentum, can be used to solve for the motion of the blocks and the spring in this scenario. These equations take into account the masses, velocities, and forces acting on the system to determine the resulting motion.

5. How is the spring constant related to the stiffness of the spring in this problem?

The spring constant, denoted as k, is a measure of the stiffness of the spring and is directly related to the amount of force required to stretch or compress the spring by a certain distance. In this problem, a higher spring constant would indicate a stiffer spring, which would result in a smaller displacement of the spring during the collision between the two blocks.

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