Bullet embedding in block on spring with k=14.0N/m

[SnowOwl18]

Ok I've been trying this problem :

---A 21.0kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k=14.0N/m. A 2.10×10-2kg bullet traveling with a speed of 530m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion? Recall that the amplitude is the maximum displacement from equilibrium.----


And I thought I could use the equation (1/2)mv^2 = (1/2)kA^2 ...but I got the answer wrong...and the hint says to first use conservation of momentum, and then conservation of energy. It seems like i couldn't do that, since it seems like I don't have enough information. Any help? Thanks :)

 

[quantumdude]

Ok I've been trying this problem :
And I thought I could use the equation (1/2)mv^2 = (1/2)kA^2 ...but I got the answer wrong...and the hint says to first use conservation of momentum, and then conservation of energy.

That's right. In general, energy is not conserved in collisions, but momentum is.

It seems like i couldn't do that, since it seems like I don't have enough information. Any help? Thanks :)

You do have enough information. You know the mass of the bullet (m) and the mass of the block (M). You also know the initial velocity of the bullet (v_i), and you want to know the final velocity of the bullet+block system (v_f).

Can you write the law of conservation of momentum in terms of those symbols?

 

[SnowOwl18]

Oh my bad...I wasn't thinking of the bullet block system. Thank you..I'll try that and see if it works.

 

[Duarh]

well, since the collision is not elastic (the bullet sticks), PE+KE energy is not conserved (dissipated into sound, block heating up/breaking and so forth). so you should calculate the velocity of the system after impact using conservation of momentum and then use this new velocity to determine the energy that must be equal to kA^2/2

 

[SnowOwl18]

oh i got it, thanks!