Bullet-in-Block with Spring

  • #1

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 travelling 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:
 

Answers and Replies

  • #2
Doc Al
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Hint: What's the speed of "block+bullet" immediately after the collision?
 
  • #3
:yuck:I really don't know Al.
 
  • #4
Doc Al
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Hint: What's conserved in any collision?
 
  • #5
Errrrr it's either energy or momentum. I think momentum.
 
  • #6
Doc Al
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Good. Use a conservation law to figure out the speed of "block+bullet" after they collide.
 
  • #7
Alright. p=mv, so then .005*650=3.25; so 3.25/2.505= 1.297.
So 1.297 would be the new velocity?
 
  • #9
Doc Al
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Is that correct?
Perfect!

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

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