What is the spring constant of the spring?

  • Thread starter Thread starter JumpinJohny
  • Start date Start date
  • Tags Tags
    Spring
AI Thread Summary
The discussion centers on calculating the spring constant of a spring attached to a block of wood that absorbs a bullet, leading to simple harmonic motion. The bullet, weighing 5g and traveling at 650 m/s, embeds in a 2.5kg block, resulting in a combined mass of 2.505kg. Participants emphasize using conservation of momentum to determine the velocity of the block and bullet immediately after the collision, which is calculated to be approximately 1.297 m/s. This velocity can then be applied to further solve for the spring constant and total energy of the system. The conversation highlights the importance of understanding momentum conservation in collision scenarios.
JumpinJohny
Messages
15
Reaction score
0

Homework Statement


A pellet gun fires a bullet into a stationary block of wood that is attached to a spring on a frictionless surface. When the bullet enters the wood, it reamins inside, and the bullet and block enter into simple harmonic motion with amplitude = 11cm. The bullet ( m = 5g) was initially traveling at 650 m/s before hitting the block of wood ( m = 2.5kg).

A: What is the spring constant of the spring?
B: What is the total energy of the system after the collision?(I know how to solve this one, I just need A to solve it.)
C: What is the maximum acceleration of the bullet/block system once it begins its oscillation?
D:Where will the bullet and block reach 0 velocity?(I think this is at the equilibrium position, but I'm not sure how to solve for it)

We know:
the amplitude of the block-and-bullet is .11m
the mass of the bullet is .005kg
the mass of the block is 2.5kg
the mass of the bullet-and-block is 2.505kg
the velocity of the bullet before it hits the block is 650 m/s.

Homework Equations


I know for B that you use the equation for total energy, kinetic energy + potential energy.

The Attempt at a Solution


My attempt at A; it wasn't so good. I tried using F = -kx, but that of course didn't work since we don't know the acceleration to solve for F. Then I considered kx=mg, but that doesn't work either because that only works for a spring that is suspended. So I'm not really sure where to go with this :eek:
 
Physics news on Phys.org
Hint: What's the speed of "block+bullet" immediately after the collision?
 
I really don't know Al.
 
Hint: What's conserved in any collision?
 
Errrrr it's either energy or momentum. I think momentum.
 
Good. Use a conservation law to figure out the speed of "block+bullet" after they collide.
 
Alright. p=mv, so then .005*650=3.25; so 3.25/2.505= 1.297.
So 1.297 would be the new velocity?
 
Is that correct?
 
JumpinJohny said:
Is that correct?
Perfect!

Now see if you can make use of that speed to solve part A (and part B, for that matter).
 
Back
Top