Calculate Amplitude of SHM for Block Connected to Spring and Embedded Bullet

[PhysicsPhun]

A 35.0 kg block at rest on a horizontal frictionless table is connected to the wall via a spring with a spring constant k=25.0 N/m. A 3.60×10-2 kg bullet traveling with a speed of 490 m/s embeds itself in the block. What is the amplitude of the resulting simple harmonic motion?


Okay.
So If I'm not mistaken, we're looking for Amplitude or Max displacement.

What I did first was try to find the intial velocity of the block with :

V= ((m_1)/(m_1+m_2))*V_1_i

V_1_i = 490 m/s
m_1 = .036 kg
m_2 = 35 kg

I got apprx. .5 m/s for the final object which weighs 35.036 kg.

i then found angular frequency with :

w^2= (k/m)

Then if found the period:

w= (2*pi/T)

With the period i found the frequency:

T = (1/f)

.. this is kinda where i get stuck ..
I then assumed the time it takes to get to the max displacement is half of the period. I don't know if i can do that, but i did : ).

And even after assuming that, I'm stuck.

I think my knowns are, Velocity, Mass, Angular frequency, Spring constant, Period, frequency, and time.

Did i do something wrong, If not, where can i go from here to find the amplitude?

 

[jamesrc]

You started off correctly, but your subsequent steps are not necessary to solve this problem. Try using the idea that energy is conserved (remember, no friction). So, you've got a mass (the block+bullet) with some initial velocity (that you got from conservation of momentum). At maximum displacement (which will be equal to the amplitude of the SHM), the kinetic energy will be zero (the block has momentarily stopped moving) and there will be some potential energy (given by .5kx²). That ought to be all the info you need to find the answer. Post back if you need more help.

 

[cookiemonster]

Edit: What james said.

Don't worry about all that stuff.

You correctly got the velocity of the block, so that's good. But once you get that, use an energy argument. All of the kinetic energy is converted to elastic potential energy at the point of maximum displacement.

cookiemonster

 

[PhysicsPhun]

Thanks, a lot I got it.. We're learning new stuff fast, so I just assumed we were supposed to use maybe the new Energy equations involving all those little variables. Totally forgot about the simpler equations.

Thanks a lot both of you.