SHM problem with Momentum and Energy

AI Thread Summary
The discussion revolves around a physics problem involving a block attached to a spring and a bullet that embeds itself in the block, leading to oscillations. The user calculated the frequency, maximum velocity, and energy of the system, ultimately determining that the bullet had 12.47J of kinetic energy. However, discrepancies arose when comparing results from energy conservation and momentum conservation methods, with the user finding different speeds for the bullet. The confusion stems from the inelastic nature of the collision, as energy is not conserved in such cases, which is a crucial point in solving the problem. The user seeks clarification on why the energy approach yielded an incorrect result, emphasizing the importance of understanding energy conservation in inelastic collisions.
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Homework Statement



A 1.00kg block is attached to a horizontal spring with spring constant 2500N/m. The block is at rest on a frictionless surface. A 10g bullet is fired into the block, in the face opposite the spring, and sticks.

a. What was the bullet's speed if the subsequent oscillations have an amplitude of 10.0cm


Homework Equations



frequency = 1/(2*pi)*sqrt(k/m) (k is spring constant, and m is mass)

Vmax = 2*pi*f*A (f is frequency, A is amplitude, V is velocity)

M1*V1 = M2*V2 (momentum) (M is mass, V is velocity)

E = 1/2m(Vmax)^2 (energy of a spring system)


The Attempt at a Solution



1.) find frequency

The total mass in the spring system would be the mass + the bullet (1.01kg)

using the frequency equation listed above:

7.918Hz = 1/(2*pi)*sqrt(2500/1.01)

2.) find Vmax

I know A is 10cm, or 0.1m

using the Vmax equation listed above:

4.97 = 2*pi*7.918*0.1

3.) Find system energy

Now that i have all the information I need, I may use the Energy equation listed above and find that

12.47 = 1/2(1.01)*(4.97^2)

4.) 12.47J entered the system when bullet hit M.

  • This means that the bullet had 12.47J of kinetic energy

the equation for kinetic energy is K=1/2*m*V^2

using that equation I find that:

12.47 = 1/2(0.01)*(V^2)

V = 49.93m/s

3b.) Use Momentum

using the momentum equation above:

1.01*4.97 = 0.01*V2

V2 = 505m/s (the correct answer)



Now this is where I need help. I've received two different answers! The original way I did this problem was with energy because I wanted to give it a try. For some reason I've been running into trouble with all of these oscillation equations (I've got a couple more I'm planning on posting)

Why is it that the energy approach isn't working? The answer seems to be off by about a factor of 10, but I can't find a problem with my units! Is this a problem with my math, or is this not a scenario of which using the energy equations is appropriate?
 
Physics news on Phys.org
The collision is inelastic, as the bullet sticks to the mass. Can you conserve energy in inelastic collisions? :wink:
 
Oh sheesh! Ok that makes sense. I'm scared that I didn't realize that! :bugeye:

Thanks for helping me
 
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