Collision between two moving blocks with a spring

In summary, the two masses slid along a frictionless table with velocities of +10 m/s and +3.0 m/s, respectively. A massless spring with a spring constant of 1120 N/m was attached to the second block. After the collision, the spring was compressed a distance of X.
  • #1
Tadams92
2
0

Homework Statement


A block of mass m1 = 4.0 kg slides along a frictionless table with a velocity of +10 m/s. Directly in front of it, and moving with a velocity of +3.0 m/s, is a block of mass m2 = 9.0 kg. A massless spring with spring constant k = 1120 N/m is attached to the second block as in the figure below. After the collision, the spring is compressed a distance X, find X.

(Attempt at a drawing)
block1 Spring...Block 2
[m1] |--/\/\/\/[m2]

Everything is moving in the +x direction, or to the right.

Homework Equations


1/2*m*v^2
1/2*k*x^2
Pi=Pf
P=m*v

The Attempt at a Solution


I thought that you would use the difference between their KE's to find the PE stored in the spring. However, when I run the numbers into the equation PE = KE1-KE2. I end up getting slightly over. So I'm doing something wrong and I cannot find where it is at.

Edit: I accidentally hit backspace and reloaded the page, so it freaked out and copied everything.
 
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  • #2
Hello Tadams92,

Welcome to Physics Forums!
Tadams92 said:

The Attempt at a Solution


I thought that you would use the difference between their KE's to find the PE stored in the spring. However, when I run the numbers into the equation PE = KE1-KE2. I end up getting slightly over.
If you want help, you're going to have to run the numbers and show us your work here. It's part of the forum rules. :smile:
So I'm doing something wrong and I cannot find where it is at.
Again I'm not going to be able to help you until you show your work.

But just as a general hint, which might possibly relate to what's going wrong, when you are comparing the different kinetic energies (PE = KE1-KE2), you shouldn't be comparing the kinetic energies of Block 1 vs. Block 2 (before the collision). Instead you need to compare the kinetic energy of the entire system before the collision to the kinetic energy of the entire system after the collision at the point in time that the spring is at maximum compression. Conservation of momentum is a necessary part of the process. :wink:
 
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  • #3
Sorry for the late reply here, but I figured it out today when i asked my physics teacher about it. She basically said the same thing you did collins. compare 2 blocks before and after collision. Thank you for the response though. Good to know I can come here for some help.

Thanks for letting me know how the forums works a little more. Don't want to be trying to make you guys do my homework for me. I would rather lose a point or two and understand the concept than get the few points. Test points >>> homework points. I'll let you know as much as possible when I ask questions down the road. Thanks again.
 

FAQ: Collision between two moving blocks with a spring

How does the speed of the blocks affect the collision with the spring?

The speed of the blocks will determine the amount of energy transferred during the collision. The faster the blocks are moving, the more energy will be transferred to the spring. This will result in a larger displacement of the spring and a greater force exerted on the blocks.

What factors influence the amount of energy transferred during the collision?

The amount of energy transferred during the collision is influenced by the mass and velocity of the blocks, as well as the stiffness of the spring. A heavier block or a faster moving block will result in a greater amount of energy being transferred.

How can the collision between the blocks and the spring be modeled mathematically?

The collision between the blocks and the spring can be modeled using the principles of conservation of energy and momentum. This involves calculating the initial and final velocities of the blocks and the spring, as well as the force exerted by the spring on the blocks.

What happens to the energy of the system after the collision?

The energy of the system will remain constant, as energy cannot be created or destroyed. However, the energy will be transferred from the blocks to the spring during the collision, resulting in a change in the kinetic and potential energy of the system.

How does the stiffness of the spring affect the collision between the blocks?

The stiffness of the spring will determine the amount of force exerted on the blocks during the collision. A stiffer spring will exert a greater force and result in a larger displacement of the spring. This will also affect the amount of energy transferred during the collision.

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