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Simple Harmonic Motion (Bullet fired into a block attached to spring)

by fluffymastr
Tags: attached, block, bullet, fired, harmonic, motion, simple, spring
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fluffymastr
#1
May4-11, 01:41 PM
P: 1
1. The problem statement, all variables and given/known data

A 5g bullet is fired horizontally into a 0.50kg block of wood resting on a frictionless table. The block, which is attached to a horizontal spring, retains the bullet and moves forward, compressing the spring. The block-bullet system goes into SHM with a frequency of 9Hz and amplitude of 15cm.

A) Determine the speed of the bullet


2. Relevant equations

Fs=-kx
a=-(k/m)x
PEs= .5kx2
v=[(k/m)(A2-x2)].5
T=2π(m/k).5

3. The attempt at a solution

I wasn't sure how to even approach this but I think I need to determine the spring constant k using a form of T=2π(m/k).5.

So:
T=2π(m/k).5
T2=4π2(m/k)
T2=4π2(m/k)
k=(4π2m)/(T2)
k=(4π2(.5kg+.005kg))/(92)
k=.24613kg/s2

Then plug that into:
v=[(k/m)(A2-x2)].5
v=[(.24613/.505)(.152-02)].5
v=.1047m/s

I have NO IDEA if I am right or not. Anyone want to confirm or deny?
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ideasrule
#2
May4-11, 04:07 PM
HW Helper
ideasrule's Avatar
P: 2,322
T=2π(m/k).5
T2=4π2(m/k)
T2=4π2(m/k)
k=(4π2m)/(T2)
k=(4π2(.5kg+.005kg))/(92)
k=.24613kg/s2
The frequency is 9 Hz, not the period.
Then plug that into:
v=[(k/m)(A2-x2)].5
v=[(.24613/.505)(.152-02)].5
v=.1047m/s

I have NO IDEA if I am right or not. Anyone want to confirm or deny?
That would be right if you used the right value of k. The equation comes from the conservation of energy, which says that 1/2*k(A^2-x^2)=1/2mv^2.


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