Conservation of momentum with a spring

In summary, the conversation discusses the use of a compressed spring with spring constant K to shoot a ball of mass m into a stationary block of the same mass. The ball sticks to the block after the collision. The correct equations for finding the final velocity of the ball + block are discussed, with an emphasis on the use of conservation of energy and momentum.
  • #1
magnifik
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Homework Statement


A spring with spring constant K is compressed by x and used to shoot a ball of mass m into an initially stationary block also of mass m. The ball sticks to the block after the collision. Calculate the final velocity of the ball + block.


Homework Equations





The Attempt at a Solution


F=kx
m1v1 + m2v2 = (m1+m2)v2
1/2mv^2 + 1/2kx^2 = (m1+m2)v2
v2 = 1/4v1^2 + kx^2/4m

i think my value for the spring's energy is wrong and that messed everything up??
 
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  • #2
magnifik said:
F=kx
m1v1 + m2v2 = (m1+m2)v2

So far so good. o:)

1/2mv^2 + 1/2kx^2 = (m1+m2)v2
v2 = 1/4v1^2 + kx^2/4m

That's not quite right. The spring's potential energy is completely converted into the ball's kinetic energy (ignoring friction and the mass of the spring itself).

You can use conservation of energy to find the speed of the ball before it hits the block, but you need to stop there.

When the ball hits the block, kinetic energy is not conserved, because some of the energy is lost to heat.

On the other hand, momentum is conserved before and after the collision between the ball and the block. So use conservation of momentum to find the velocity of the ball+block.
 

Related to Conservation of momentum with a spring

1. How is the conservation of momentum applied to a spring system?

The conservation of momentum in a spring system is based on the principle that the total momentum of the system remains constant. This means that the initial momentum of the spring, including any objects attached to it, will be equal to the final momentum after the spring has been compressed and released. This is because the forces acting on the system are internal (between the objects and the spring) and therefore do not change the total momentum of the system.

2. What factors affect the conservation of momentum in a spring system?

The conservation of momentum in a spring system is affected by the mass and velocity of the objects attached to the spring, as well as the stiffness of the spring itself. The more massive the objects and the higher their velocities, the greater the momentum of the system. Additionally, a stiffer spring will result in a greater change in momentum during compression and release.

3. Can the conservation of momentum be violated in a spring system?

No, the conservation of momentum is a fundamental law of physics and cannot be violated. In a spring system, any changes in momentum are due to internal forces and will not affect the total momentum of the system. This is true even if there is energy loss due to friction or other factors.

4. How is the conservation of momentum related to the conservation of energy in a spring system?

The conservation of momentum and the conservation of energy are closely related in a spring system. Both principles state that the total amount of the respective quantity remains constant. In a spring system, the potential energy stored in the spring is converted to kinetic energy as the spring is compressed and released, while the total momentum of the system remains constant.

5. Are there real-world applications of the conservation of momentum in spring systems?

Yes, the conservation of momentum in spring systems has many real-world applications. This principle is used in engineering and design, such as in shock absorbers for vehicles, where the compression and release of a spring helps to absorb and dissipate energy. It is also important in sports, such as in the mechanics of a pogo stick or a trampoline, where the conservation of momentum allows for the repeated bouncing motion.

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