Linear momentum and collision

  • Thread starter ch010308
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  • #1
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Homework Statement



A 5.00-g bullet moving with an initial speed of 400 m/s is fired into and passes through a 1.00-kg block, as in Figure P9.67. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring of force constant 900 N/m. If the block moves 5.00 cm to the right after impact, find
(a) the speed at which the bullet emerges from the block and
(b) the mechanical energy converted into internal energy in the collision.

Homework Equations



conservation of momentum
KE=1/2 mv2

The Attempt at a Solution



I solved part a, getting an answer of 100m/s for the speed at which the bullet emerges from the block. But I'm stuck at part b.
I interpreted the question as: finding the mechanical energy tat was converted = change in mechanical energy = change in kinetic energy of the bullet = 1/2 m(vf -vi) = (1/2)(5/1000)(100-400)2 = 225J

The correct answer should be 374J.

Pls help, thanks! :)
 

Attachments

  • Fig P9.67.jpg
    Fig P9.67.jpg
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Answers and Replies

  • #2
rl.bhat
Homework Helper
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What is the role played by the spring in this problem?
 
  • #3
lanedance
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hi

havenl't been through it all, but one thing that stands out is your last calc of cahnge in KE

should be
1/2 m(vf^2 -vi^2)

NOT
1/2 m(vf -vi)^@2as you've done in your calc
 
  • #4
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should be
1/2 m(vf^2 -vi^2)

NOT
1/2 m(vf -vi)^@2as you've done in your calc

oh shoot! I cant believe I made tt mistake again n again. :tongue2: Thanks for pointing it out!

What is the role played by the spring in this problem?

ok, i think i get it.

change in KE of bullet = 1/2 m(vf2 -vi2)
= (1/2)(5/1000)(1002-4002) = -375J

change in KE of spring = (1/2)(1)(1.52 - 02) = 1.125J

total change in KE = -375 + 1.125 = -373.875 = -374J (round up to 3sf)

Is this correct? :)
 

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