Bullet fired into a block attached to a spring

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Homework Help Overview

The problem involves a bullet fired into a block attached to a spring on a frictionless surface. The scenario includes a completely inelastic collision where the bullet embeds into the block, leading to questions about the block's velocity after impact and the bullet's initial speed.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the collision and the relationship between kinetic energy and potential energy in the context of the spring's compression. There are questions about how to relate the kinetic energy of the block to the potential energy stored in the spring.

Discussion Status

Participants are exploring the relationship between kinetic energy and potential energy, with some suggesting that the kinetic energy of the block just after impact corresponds to the potential energy of the compressed spring. There is an ongoing examination of the implications of this relationship.

Contextual Notes

There are multiple unknowns in the problem, and participants are questioning the necessity of using work done by the spring versus its potential energy. The discussion reflects uncertainty regarding the application of relevant equations.

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Homework Statement



A bullet mass(b) .008 kg is fired (and embedded) into a block mass(B) 0.992 kg on a horizontal frictionless surface. The spring is compressed 0.15 m on impact. The spring's constant is 300 N/m.
Q: What is the block's velocity after impact
Q: What is the initial speed of the bullet?

Homework Equations



I want to say that this is a completely inelastic collision since the two masses have the same final velocity (2)? so in that case... (k2/k1) = m(b)/[(m(b)+m(B)]
also,
m(b)v(b) + m(B)v(B) = [(m(b)+m(B)]v2
and U(el) ...potential energy of an elastic... is 1/2kx^2
W(el) ...work done by spring... is -(change in potential energy).

The Attempt at a Solution



Im not sure what to do with so many unknowns. Also not sure if I need to use the work done by a spring or the potential energy of a spring. Any help to get me in the right direction is much appreciated.
 
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You know after impact, the block must have enough initial kinetic energy to depress the spring by 0.15 m. How much kinetic energy does that much elastic potential energy correspond to?
 
so the elastic potential energy of the spring when it is compressed is 0.5(300)(0.15)^2 = 3.375 N*m. So are you saying that the initial kinetic energy of the block 0.5mv^2 = 3.375 ?
 
Yeah, where "initial" in this context means, just after impact.
 

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