Collision with Spring: Solving for Distance and Final Velocities

In summary, the problem involves two blocks, one moving at 8m/s and the other at rest with a spring in front of it. The spring has a spring constant of 6 N/m. The goal is to determine the maximum distance the spring is compressed and the final velocities of the blocks after the collision. To solve for the distance, the equation 1/2mv^2' = 1/2kx^s + 1/2mv^2 can be used. As for the final velocities, the conservation of momentum law can be applied, with the assumption that the two blocks stick together after the collision and that the initial velocity of the first block remains at 8m/s. The center of mass frame
  • #1
kle89
4
0
Problem

One block (3kg) is moving at 8m/s on a frictionless plane to collide with another block at rest with the same mass, but has a spring in front of it. Spring constant = 6 N/m

a. Determine the maximum distance the spring is compressed.
b. Determine the final velocities of the blocks after the collision. You should assume that the 1 kg and 2 kg blocks remain stuck together.

Honestly i have no idea how to set the equation up? Would i use 1/2mv^2' = 1/2kx^s + 1/2mv^2 to find the distance of the spring? The problem is i wouldn't know what the final velocity would be after hitting the spring so i can't solve exactly for x. Also the other block with the spring could move after the collision so wouldn't its velocity be equal to that of the original block after hitting the spring but in the opposite direction? Any hints to get me on the right track would be appreciated. Thanks in advance.
 
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  • #2
maybe you can assume that at the point where the 1st block contacts the spring of the 2nd block, the velocity of the 1st block is still 8 m/s... Then you can just look focus on the time frame in which the whole spring compression occurs, knowing that the initial velocity of the 1st block when it contacted the spring was 8m/s.

Part b would most probably involved conservation of momentum law. But you need to solve a.
 
Last edited:
  • #3
use center of mass frame!
 

1. What is a "collision with spring" in physics?

A collision with spring is a situation in physics where an object collides with a spring, causing the spring to be compressed or stretched. This results in a transfer of energy between the object and the spring.

2. How is the energy conserved in a collision with spring?

In a collision with spring, the total energy of the system (object + spring) remains constant. This means that the initial kinetic energy of the object is converted into potential energy of the spring when it is compressed or stretched, and vice versa.

3. What is the difference between an elastic and inelastic collision with spring?

An elastic collision with spring is one in which the total kinetic energy of the system remains constant before and after the collision. In an inelastic collision with spring, some of the kinetic energy is lost as heat or sound during the collision, resulting in a decrease in the total kinetic energy of the system.

4. How does the mass of the object affect the collision with spring?

The mass of the object affects the collision with spring by determining the amount of kinetic energy that is transferred to the spring. A heavier object will transfer more energy to the spring compared to a lighter object, resulting in a greater compression or stretching of the spring.

5. What is the equation for calculating the spring constant in a collision with spring?

The spring constant (k) in a collision with spring can be calculated using the equation k = (F - mg)/x, where F is the force applied to the spring, m is the mass of the object, and x is the displacement of the spring. This equation is derived from Hooke's Law, which states that the force applied to a spring is directly proportional to the amount it is stretched or compressed.

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