A perfectly inelastic collision?

  • Thread starter 00PS
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Homework Statement



A 5.00 g bullet moving with an initial speed of 410 m/s is fired into and passes through a 1.00 kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 965 N/m.

1. If the block moves 5.20 cm to the right after impact, find the speed at which the bullet emerges from the block.
wrong check mark


Homework Equations



p=mv (momentum)
(1/2)mv^2 (kinetic energy)
(1/2kx^2 (elastic potential energy of a spring, where x is the distance the spring is compressed)
-Energy is always conserved in a isolated system
-Momentum is always conserved in a closed isolated system

The Attempt at a Solution



I am having a hard time grappeling what type of problem this is. A perfectly inelastic collision? But the bullet then exits the block...so a little tricky. Here is my attempt:

(1/2)mv^2(initial)=(1/2)mv^2 (final) - (1/2)kx^2

solving for v (final) i got 28.4 m/s

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
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you are on the right track; this can definitely be solved with Energy considerations. First, you should check the sign on your 'spring energy' expression. What you are implying with that statement is that as the bullet compresses the spring, your bullet obtains more energy, when it is the opposite.
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

A 5.00 g bullet moving with an initial speed of 410 m/s is fired into and passes through a 1.00 kg block. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 965 N/m.

1. If the block moves 5.20 cm to the right after impact, find the speed at which the bullet emerges from the block.
wrong check mark

Homework Equations



p=mv (momentum)
(1/2)mv^2 (kinetic energy)
(1/2kx^2 (elastic potential energy of a spring, where x is the distance the spring is compressed)
-Energy is always conserved in a isolated system
-Momentum is always conserved in a closed isolated system

The Attempt at a Solution



I am having a hard time grappeling what type of problem this is. A perfectly inelastic collision? But the bullet then exits the block...so a little tricky.

Hi 00PS! Welcome to PF! :smile:

No, energy is not always conserved in a isolated system … in fact energy is never conserved unless the question tells you it is! :smile:

There is no way that a bullet tearing a hole in something conserves energy! :rolleyes:

In this case, you have to work backwards …

you know how far the spring compresses, so you know what energy it got.

That energy came from the block, not the bullet, so you can work out the KE of the block after the bullet left it.

Then use conservation of momentum to find the speed of the bullet. :smile:
 
  • #4
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oops...already making use of my user name. :smile:

Hi 00PS! Welcome to PF! :smile:

No, energy is not always conserved in a isolated system … in fact energy is never conserved unless the question tells you it is! :smile:


I simply forgot the word 'mechanichal'. According to my textbook, mechanical energy of a isolated system is conserved. Nonetheless, I asked my professor about the problem this morning in class, and he explained it exactly as you did, thanks! I knew conservation of momentum was involved somehow, but could not put my finger on how to apply it.

btw, thanks for the welcome! I found this site sort of as a joke using google, and to my surpise a phyics homework help does exist!

EDIT: Do you know how you could find how much mechanichal energy was transferred to internal energy?
 
Last edited:

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