## geerky42 Group Title Need help on #11, please. (Momentum) one year ago one year ago

1. geerky42 Group Title

2. geerky42 Group Title

I solved #10 and the answer is 109.217 m/s if this help you all.

3. geerky42 Group Title

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

4. wio Group Title

By the way, how did you find the velocity?

5. geerky42 Group Title

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

6. VeritasVosLiberabit Group Title

$\frac{ 1 }{ 2 }m _{b}v _{b1}^{2}=\frac{ 1 }{ 2 }m _{b}v _{b2}^{2}+\frac{ 1 }{ 2 }kx$ 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. VeritasVosLiberabit Group Title

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

8. VeritasVosLiberabit Group Title

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

9. geerky42 Group Title

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

10. VeritasVosLiberabit Group Title

ok, I'm going to retry it then.

11. geerky42 Group Title

Any idea for #11, @wio ?

12. wio Group Title

oh sorry I was idle...

13. wio Group Title

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

14. wio Group Title

Did you try putting the elastic energy as energy lost?

15. geerky42 Group Title

I did and the answer is incorrect...

16. wio Group Title

Give me a second...

17. wio Group Title

What about the conservation of momentum?

18. wio Group Title

Was momentum give to the block on the spring?

19. wio Group Title

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. wio Group Title

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. wio Group Title

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. wio Group Title

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

23. wio Group Title

It's a really unfair question in my opinion.

24. wio Group Title

@geerky42 Have you been looking at the momentum?

25. geerky42 Group Title

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

26. wio Group Title

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

27. geerky42 Group Title

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. wio Group Title

How did you find the speed of the block then?

29. geerky42 Group Title

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

30. wio Group Title

Yeah well the block started with no energy right?

31. geerky42 Group Title

Right.

32. wio Group Title

It got elastic energy

33. wio Group Title

you converted that into kinetic energy?

34. geerky42 Group Title

Well, it moves so it should have kinetic energy.

35. wio Group Title

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

36. geerky42 Group Title

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 Group Title

Kinetic energy instantly after collision.

38. geerky42 Group Title

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

39. geerky42 Group Title

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

40. wio Group Title

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

41. geerky42 Group Title

No, how can this help me?

42. wio Group Title

Because it would be the amount of energy the bullet lost

43. geerky42 Group Title

None, still don't work...

44. wio Group Title

What are you getting for it though?

45. geerky42 Group Title

17.143 J

46. wio Group Title

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

47. wio Group Title

Meaning the first KE minus the new KE

48. geerky42 Group Title

I did this, I got this answer.

49. wio Group Title

your original KE wasn't 375?

50. wio Group Title

and new KE isn't 17?

51. geerky42 Group Title

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

52. wio Group Title

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

53. geerky42 Group Title

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

54. geerky42 Group Title

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

55. wio Group Title

yeah sorry

56. geerky42 Group Title

Thanks for your time and patience, though.