Inelastic collision and Hook’s law

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Homework Help Overview

The discussion revolves around an inelastic collision problem involving a bullet and a block, where the bullet compresses a spring upon impact. The participants are tasked with determining the mass of the block using principles of momentum and energy conservation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between kinetic energy and elastic potential energy, questioning the correct application of formulas. There is discussion about whether the bullet sticks to the block and how that affects the equations used.

Discussion Status

Some participants have attempted to derive expressions for velocity in terms of the masses involved and have shared their calculations. There is an ongoing exploration of the correct setup for the equations, with some expressing confusion about the cancellation of terms and the implications of the bullet-block interaction.

Contextual Notes

Participants are working under the constraints of a homework assignment, which limits guidance from instructors. There is uncertainty regarding the correct interpretation of the energy conservation principles in the context of the problem.

yellowmax31
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Homework Statement


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

Homework Equations


m1v1=(m1+m2)v`

Ee=Ek

Et=Et`

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
 
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Does the bullet stick to the block? If so, ½kx² should be equal to ½(m1 + m2)v² (instead of ½m1v²).
 
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:
Solve for v in terms of m1 and m2 and substitute it back in the momentum equation.
 
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:
yellowmax31 said:
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

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

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