# Inelastic collision and Hook’s law

1. May 11, 2009

### yellowmax31

1. The problem statement, all variables and given/known data
a bullet of mass 0.005kg traveling at 120m/s hits a block at rest that falls and compresses a spring (k=99n/m) to a maximum compression of 0.121m what is the mass of the block? (solve using momentum)
m1=0.005kg
m2=?
v1=120m/s
v=?
k=99n/m
x=0.121m

2. Relevant equations
m1v1=(m1+m2)v

Ee=Ek

Et=Et

3. The attempt at a solution
attempt one
Ee=Ek
1/2kx^2=1/2m1v^2
v=squareroot((kx^2)/m1)
v=17m/s
m1v1=(m1+m2)v
0.005(120)= (0.005+ m2)17
0.005(120)/17=(0.005+ m2)
0.03529=(0.005+ m2)
0.03529-0.005=m2
0.030=m2

I got it wrong and my teacher will not tell me the answer I can't see were i went wrong

2. May 11, 2009

### dx

Does the bullet stick to the block? If so, ½kx² should be equal to ½(m1 + m2)v² (instead of ½m1v²).

3. May 11, 2009

### yellowmax31

yes the bullet sticks to the block and i am trying to figure out what m2 is so I am not sure how to incorporate ½(m1 + m2)v² (instead of ½m1v²).

Last edited: May 11, 2009
4. May 11, 2009

### dx

Solve for v in terms of m1 and m2 and substitute it back in the momentum equation.

5. May 11, 2009

### yellowmax31

v=squareroot((kx^2)/(m1+m2)

m1v1=(m1+m2)v

m1v1=(m1+m2)squareroot((kx^2)/(m1+m2)

m1^2v1^2=m1+m2((kx^2)/(m1+m2))

then the m1+m2 cancel iam sorry i not getting this i just can't see where iam going wrong

Or did you mean
m1v1=(m1+m2)v

v=m1v1/(m1+m2) & v=squareroot((kx^2)/(m1+m2)
so
m1v1/(m1+m2)=squareroot((kx^2)/(m1+m2)

(m1v1)^2/(m1+m2)^2 = (kx^2)/(m1+m2)

(m1v1)^2/(m1+m2) = (kx^2) (brought m1+m2 over)

(0.005*120)^2/(m1+m2) = 1.449459

0.36=1.4 (m1+m2) (mutiplied both sides by (m1=+m2))

0.36= 0.007247295 + 1.4m2

0.352752705= 1.4m2
0.2519=m2
???? is that it or did i mess up again

Last edited: May 11, 2009
6. May 11, 2009

### dx

They don't cancel. One of the (m1 + m2)'s is under a square root.