Spring Compression: 2.5kg Block + 4.5kg Block

In summary: At this point, they will move together.) And what does conservation of energy say about the kinetic energy of the two blocks at this point?In summary, the problem involves a collision between a 2.5 kg block moving at 6.0 m/s and a 4.5 kg block at rest, with a coil spring attached to the second block. The spring has a spring constant of 860 N/m, and the question asks for the maximum compression of the spring. Using conservation of momentum and energy, the correct approach is to equate the kinetic energy of the first block to the combined potential energy of the second block and the compressed spring. By setting the relative velocity of the two blocks to zero at maximum compression
  • #1
Oomair
36
0

Homework Statement



A 2.5 kg block slides along a frictionless tabletop at 6.0 m/s toward a second block (at rest) of mass 4.5 kg. A coil spring, which obeys Hooke's law and has spring constant k = 860 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block

Homework Equations





The Attempt at a Solution



i first used M1V1 = M2V2, M2 is the block at rest, i calculated the speed of block at rest and i got 3.33 m/s, and then i use the energy equations to find the spring compression

.5 m1(6)^2 = .5k(x^2) + .5m2(3.33)^2, i solve for x, but it comes out wrong, anything I am missing here?
 
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  • #2
You forgoet to mention what the question was asking, but...You should be able to forget the first blcok after the collision. I would just use momentum conservation to find the velocity of the second block after the collision. Then the kinteic energy associated with this velocity will be converted completely to the potential energy to be stored in the spring once its compressed.
 
  • #3
in other words, 1/2mv^2 = 1/2kx^2 (all for the second block), where v is the velocity of the second block after the collision, or the start point of the new system.

Edit: you might also want to see if the collision is completely elastic or not, because this would change your answer
 
Last edited:
  • #4
Oomair said:
i first used M1V1 = M2V2, M2 is the block at rest, i calculated the speed of block at rest and i got 3.33 m/s,
This assumes that all the momentum of the first block is transferred to the second. That's not true.

I assume you are trying to find the maximum spring compression. Hint: How will the velocities of the two blocks be related when the spring is at maximum compression?

(But you are correct to be using conservation of momentum and energy. So you are close.)
 
  • #5
The book is saying that the collision is perfectly elastic, but I am confused on how i would relate the speeds in terms of energy
 
  • #6
Yes, the collision is perfectly elastic. You'll need that fact later.

When the spring is at maximum compression, what must be the relative speed of the two blocks?
 

1. What is spring compression?

Spring compression is the change in length or displacement of a spring when a force is applied to it. It is a result of Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of compression or extension of the spring.

2. What does the 2.5kg and 4.5kg block represent in this scenario?

The 2.5kg and 4.5kg blocks represent the masses attached to the ends of the spring. The 2.5kg block is the lighter object and the 4.5kg block is the heavier object. These masses will affect the amount of compression of the spring when a force is applied.

3. How is the spring compression calculated?

The spring compression can be calculated using the formula: F = kx, where F is the force applied to the spring, k is the spring constant, and x is the amount of compression or extension of the spring. The spring constant is a measure of the stiffness of the spring and is determined by the material and design of the spring.

4. What factors can affect the spring compression?

The spring compression can be affected by several factors such as the mass of the blocks attached to the spring, the spring constant, the force applied to the spring, and the initial length of the spring. The surface on which the blocks are placed can also affect the spring compression.

5. Why is spring compression important?

Spring compression is important because it allows us to study the relationship between force and displacement of a spring, which is a fundamental concept in physics. It is also used in many practical applications, such as in shock absorbers, car suspensions, and mechanical watches. Understanding spring compression can also help in designing and improving various mechanical systems.

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