## geerky42 3 years ago Need help on #11, please. (Momentum)

1. geerky42

2. geerky42

3. geerky42

I'm not sure exactly what #11 mean; I don't see how this system lost energy during collision.

4. anonymous

By the way, how did you find the velocity?

5. geerky42

Are you telling me that the block lost energy to the spring?

6. anonymous

$\frac{ 1 }{ 2 }m _{b}v _{b1}^{2}=\frac{ 1 }{ 2 }m _{b}v _{b2}^{2}+\frac{ 1 }{ 2 }kx$ was the formula I got for 10. $\sqrt{2[\frac{ 1 }{ 2 }(.003kg)(500\frac{ m }{ s })^{2}-\frac{ 1 }{ 2 }(859\frac{ N }{ m })(.02m)]/.003kg}=v _{b2}$

7. anonymous

my final answer for #10 was 494.24 m/s which makes more sense to me than 109.217 m/s

8. anonymous

but I'm not sure. are you sure your answer for #10 is correct?

9. geerky42

Well, I submitted my answer on online homework and it said it's correct.

10. anonymous

ok, I'm going to retry it then.

11. geerky42

Any idea for #11, @wio ?

12. anonymous

oh sorry I was idle...

13. anonymous

So you figured out how much energy was transferred to the block, and the remaining energy was used to calculate velocity, right?

14. anonymous

Did you try putting the elastic energy as energy lost?

15. geerky42

I did and the answer is incorrect...

16. anonymous

Give me a second...

17. anonymous

What about the conservation of momentum?

18. anonymous

Was momentum give to the block on the spring?

19. anonymous

Figure out the momentum before hand and figure out the momentum later. That would tell you how much momentum the block received. Compare that to the elastic energy.

20. anonymous

I mean, use the momentum of the block to find its velocity and figure out the kinetic energy of the block. Then compare that to the elastic energy.

21. anonymous

That might be what they're asking. It's true that I find it hard to tell. Obviously this is a closed system an no energy is being lost. So maybe what they mean is how much kinetic energy do we expect the block to have, which it won't have.

22. anonymous

If that doesn't work then I'm really stumped here.

23. anonymous

It's a really unfair question in my opinion.

24. anonymous

@geerky42 Have you been looking at the momentum?

25. geerky42

None, It doesn't work... I guess I need to ask my physics teacher a question. Thanks.

26. anonymous

@geerky42 I'm not getting your 109 velocity through energy methods alone

27. geerky42

Sorry I just use conservation of energy to find the speed of block after collision, I then plug the speed of block into conservation of momentum to find the speed of bullet after collision. My brain is dead, sorry about that.

28. anonymous

How did you find the speed of the block then?

29. geerky42

"Sorry I just use conservation of energy to find the speed of block"

30. anonymous

Yeah well the block started with no energy right?

31. geerky42

Right.

32. anonymous

It got elastic energy

33. anonymous

you converted that into kinetic energy?

34. geerky42

Well, it moves so it should have kinetic energy.

35. anonymous

How did you find out how much bullet energy went into kinetic energy then?

36. geerky42

I don't find it from bullet, I did work "backward." We know that the spring has been compressed by 2 cm and from that we can find elastic energy and then from that, we can convert it into kinetic energy. That's how I find the speed of block.

37. geerky42

Kinetic energy instantly after collision.

38. geerky42

I don't see how system lost the energy, spring absorb energy from block that has been hit by bullet.

39. geerky42

The surface is frictionless. I don't know what else cause it to lost energy...

40. anonymous

Oh, did you try entering the change in kinetic energy of the bullet?

41. geerky42

No, how can this help me?

42. anonymous

Because it would be the amount of energy the bullet lost

43. geerky42

None, still don't work...

44. anonymous

What are you getting for it though?

45. geerky42

17.143 J

46. anonymous

No, that's the new KE of the bullet. I said the change.

47. anonymous

Meaning the first KE minus the new KE

48. geerky42

I did this, I got this answer.

49. anonymous

50. anonymous

and new KE isn't 17?

51. geerky42

Forgot to square one of the velocity, I got 357.107 J

52. anonymous

Yes, that's what I'm getting as the change in bullet eneryg.

53. geerky42

Submitted this answer; it's still wrong...

54. geerky42

I guess I have to see my teacher and talk about it.

55. anonymous

yeah sorry

56. geerky42

Thanks for your time and patience, though.