The Speed of a Bullet Fired into a Ballistic Spring System

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SUMMARY

The discussion focuses on deriving the speed of a bullet fired into a ballistic spring system, where a spring with spring constant k supports a block of mass M. The bullet of mass m impacts the block, causing maximum spring compression d. The correct expression for the bullet's initial speed, derived using conservation of momentum and energy principles, is Vb = ((kd^2 - Mgd)(m+M))/m^2. The initial attempt at the solution was incorrect, but the final expression accurately reflects the relationship between the variables involved.

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[SOLVED] Spring System

1. Homework Statement

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A spring whose spring constant is k is suspended from the ceiling. A block of mass M hangs from the spring. A bullet of mass m is fired vertically upward into the bottom of the block. The spring's maximum compression d is measured.

Find an expression for the bullet's speed.
Express your answer in terms of the variables m, M, k, d, and constant g.


2. Homework Equations

Ki + Ug + Usp = Kf + Ug + Usp

3. The Attempt at a Solution

I used conservation of momentum to find the final velocity of the bullet+block

(m+M)vf=mvi + Mvi

so...

vf=(m/m+M)*vB

where vB is the initial speed of the bullet.

Next, I used:
Ki + Ug + Usp = Kf + Ug + Usp

(1/2)mvi^2 + (1/2)kd^2 + Mgd = (1/2)mvf^2 + (1/2)kd^2 + Mgd
0 + (1/2)kd^2 + 0 = (1/2)(m+M)[(m+M)^2*vB^2] + 0 + Mgd

I found the answer to be:

((kd^2 - Mgd)(m+M))/m^2 = Vb

but this was not correct...please help me.
 
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Nevermind, I got it right...finally
 


wait can you show how you did this because I have the same problem thanks!
 

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