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So I've been working on this problem for a while and can't seem to come up with anything and I don't really understand why I'm wrong in what I did. I could just get the answer online, but I actually want to understand what's happening.

A rifle bullet with mass 8.12g and initial horizontal velocity 250m/s strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0cm. After the impact, the block moves in SHM. Calculate the period of this motion.

(Period) T = 2pi*sqrt(m/k)

(Kinetic Energy) K = .5*m*v^2

(Potential Energy) U(spring) = .5*k*x^2

(Momentum) P = mv

I'll use m for the mass of the bullet and M for the mass of the block.

So first I used conservation of momentum to assume that m*v(bullet) = (M+m)(v(final)).

Plugging in I got (.00812kg)(250m/s) = (.922kg+.00812kg)*v(final)

v(final) = 2.182514m/s

So then I assumed that the energy that the system has right after the bullet makes contact is fully kinetic, since in SHM all of the energy is kinetic when at the origin. I think I'm right in assuming this, but maybe that's where I went wrong. But anyways, continuing with what I took as the assumption, I then used the formula for kinetic energy.

K = .5*(M+m)(v(final))^2 So,

K = .5*(.922kg+.00812kg)*(2.182514m/s)^2 = 2.214492J

So then I went and assumed that all energy would be potential when the spring is fully compressed.

U = .5*k*x^2

2.214492J = .5*k*(.18m)^2

k = 136.697066N/m

So then I simply plugged the k into the formula for period and got:

T = 2*pi*sqrt((M+m)/k)

T = 2*pi*sqrt((.922kg+.00812kg)/136.697066N/m)

T = .518286s

Apparently this solution is wrong though, so I was wondering if anyone could help me out so that I could better understand this stuff. : /

## Homework Statement

A rifle bullet with mass 8.12g and initial horizontal velocity 250m/s strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0cm. After the impact, the block moves in SHM. Calculate the period of this motion.

## Homework Equations

(Period) T = 2pi*sqrt(m/k)

(Kinetic Energy) K = .5*m*v^2

(Potential Energy) U(spring) = .5*k*x^2

(Momentum) P = mv

## The Attempt at a Solution

I'll use m for the mass of the bullet and M for the mass of the block.

So first I used conservation of momentum to assume that m*v(bullet) = (M+m)(v(final)).

Plugging in I got (.00812kg)(250m/s) = (.922kg+.00812kg)*v(final)

v(final) = 2.182514m/s

So then I assumed that the energy that the system has right after the bullet makes contact is fully kinetic, since in SHM all of the energy is kinetic when at the origin. I think I'm right in assuming this, but maybe that's where I went wrong. But anyways, continuing with what I took as the assumption, I then used the formula for kinetic energy.

K = .5*(M+m)(v(final))^2 So,

K = .5*(.922kg+.00812kg)*(2.182514m/s)^2 = 2.214492J

So then I went and assumed that all energy would be potential when the spring is fully compressed.

U = .5*k*x^2

2.214492J = .5*k*(.18m)^2

k = 136.697066N/m

So then I simply plugged the k into the formula for period and got:

T = 2*pi*sqrt((M+m)/k)

T = 2*pi*sqrt((.922kg+.00812kg)/136.697066N/m)

T = .518286s

Apparently this solution is wrong though, so I was wondering if anyone could help me out so that I could better understand this stuff. : /

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