Search results

  1. B

    Steel Ball Collision problem

    I tried using 1/2(m1+m2)v2=(m1+m2)gy. I tried solving for y, but I am not getting the right answer. Can you explain where I am going wrong?
  2. B

    Steel Ball Collision problem

    alright so I did do it right. The initial height of ball 1 equals the final height of ball 2 because the collision is elastic. So how do I approach part B?
  3. B

    Steel Ball Collision problem

    so I set 1/2mv^2 equal to mgh, and solved for the velocity. I got 1.95 m/s. Am I doing this right?
  4. B

    Steel Ball Collision problem

    How do I find the height of the first ball?
  5. B

    Steel Ball Collision problem

    can you explain part A more. How do I apply conservation of energy and conservation of momentum?
  6. B

    Steel Ball Collision problem

    [SOLVED] collision problem Homework Statement Two identical steel balls, each of mass 3.4 kg, are suspended from strings of length 27 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle theta = 74o with the vertical and let it...
  7. B

    Initial momentum of the system

    [SOLVED] Conservation of Momentum Homework Statement Two blocks with masses m1 = 2.6 kg and m2 = 3.6 kg are at rest on a frictionless surface with a compressed spring between them. The spring is initially compressed by 60.0 cm and has negligible mass. When both blocks are released...
  8. B

    Horizontal Speed of bullet fragment

    never mind I got it, thanks
  9. B

    Horizontal Speed of bullet fragment

    What do I put in for m1 and m2?
  10. B

    Horizontal Speed of bullet fragment

    so then I just take 125 and divide by -9.8?
  11. B

    Horizontal Speed of bullet fragment

    can you explain to me how to do that?
  12. B

    Horizontal Speed of bullet fragment

    So the initial velocity is 125 m/s. The accel is -9.8 m/s^2. I still need the distance right? And then I solve for the final velocity?
  13. B

    Horizontal Speed of bullet fragment

    I am still confused, I don't understand how to figure out the height.
  14. B

    Horizontal Speed of bullet fragment

    I'm guessing that you are to assume that all of the KE goes into the second half. Can you help me out, I'm lost?
  15. B

    Horizontal Speed of bullet fragment

    [SOLVED] Horizontal Speed of bullet fragment Homework Statement A 14-kg shell is fired from a gun with a muzzle velocity 125 m/s at 33o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the...
  16. B

    Force due to wind and rolling friction

    alright, I got it, thanks!
  17. B

    Force due to wind and rolling friction

    Then once I get the acceleration, I just multiply by 2450 kg?
  18. B

    Force due to wind and rolling friction

    The acceleration is negative, right. So doesn't that make the force negative? The van reaches 49.27 mph at 11 sec. So what other point should I use to find the slope?
  19. B

    Molecular mass

    molecular mass of oxygen as a molecule is 32 grams because it is O_2. The molar mass is 16 grams for oxygen atom.
  20. B

    Force due to wind and rolling friction

    Homework Statement The graph shows the speed as a function of time for a minivan, coasting on neutral along a straight, level road, loaded with windsurfing equipment and towing a boat.The total mass is 2450kg (weight = 5402lbs). Find the size of the force due to wind and rolling friction when...
  21. B

    Block on loop-the-loop track

    or do I just set mgh=204 and solve for h?
  22. B

    Block on loop-the-loop track

    So I do 1/2mv_1^2+mgy_1=1/2mv_2^2+mgy_2? And then I am solving for y_1 right?
  23. B

    Block on loop-the-loop track

    Can you show me how to set up the equations that I need?
  24. B

    Block on loop-the-loop track

    So then can the lowest the height at Point P only be equal to the height of Point A which is 2r?
  25. B

    Block on loop-the-loop track

    I haven't done this before so I am really confused right now. I am not sure what it tells about the mechanical energy at point P.
  26. B

    Block on loop-the-loop track

    I understand that it is a constant, so does that mean KE+PE equals the answer from part A?
  27. B

    Block on loop-the-loop track

    I am not sure what to do from here. That's where I am stuck at
  28. B

    Block on loop-the-loop track

    So when I take the velocity from the centripetal acceleration, and plug it into KE=1/2mv^2, I get 114.63 J. Is this right so far?
  29. B

    Block on loop-the-loop track

    I am not sure how to do that. Could you please explain it more to me?
  30. B

    Block on loop-the-loop track

    For part C I don't even know where to start. mv^2/r must be greater than or equal to mg, right?
  31. B

    Block on loop-the-loop track

    For part B, I just take the KE that I solved for and set it equal to 1/2mv^2 and solve for v. Then v^2/r=a, right?
  32. B

    Block on loop-the-loop track

    Alright so I got part A. I found the PE at point P and subtracted the PE at point A. For part B, wouldn't the acceleration just be equal to g?
  33. B

    Block on loop-the-loop track

    I don't know v at P...to get that do I set KE=2PE
  34. B

    Block on loop-the-loop track

    I am confused on point A. How would I solve for KE if I don't have a velocity? How do I solve for that quantity?
  35. B

    Block on loop-the-loop track

    How do I figure out a velocity to use in the equation?
  36. B

    Block on loop-the-loop track

    is that where you use KE_1+PE_grav?
  37. B

    Block on loop-the-loop track

    They are both constant? right?
  38. B

    Block on loop-the-loop track

    Well what I tried to do was set 1/2mv^2=2mgh, and solve for a velocity. Then I tried to plug that into KE=1/2mv^2, but I don't think that is right. Is the total mechanical energy conserved?
  39. B

    Block on loop-the-loop track

    Homework Statement A small block of mass m = 1.3 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point P at rest a distance h = 52.0 m above the bottom of the loop of radius R = 18.0 m. A)What is the kinetic energy of the mass at the point A on the...
  40. B

    Speed of Block against spring

    W_f=umgd KE=1/2mv^2 W_spring=1/2kx^2 How do you relate these three equations to get a speed? That is what I am struggling with.
  41. B

    Speed of Block against spring

    Does the kinetic energy go against the friction force. So would it be F_fric=U_k*mgd?
  42. B

    Speed of Block against spring

    Homework Statement A moving 1.3 kg block collides with a horizontal spring whose spring constant is 491 N/m.The block compresses the spring a maximum distance of 5.0 cm from its rest postion. The coefficient of kinetic friction between the block and the horizontal surface is 0.49.What is the...
  43. B

    Spring Problem, help!

    Oh never mind I got part B. Can you help me with part C though?
  44. B

    Spring Problem, help!

    Is it F_k=U_k *N?
  45. B

    Spring Problem, help!

    Is it -U_k*mgd?
  46. B

    Spring Problem, help!

    Alright, it accepted the answer. I am lost when it comes to Part B
  47. B

    Spring Problem, help!

    So then the answer should be -.61375 J ? So then how do I do the same thing with friction for part B?
  48. B

    Spring Problem, help!

    I tried it again and it said that it was wrong. We must have gone wrong somewhere...
  49. B

    Spring Problem, help!

    It says that the answer for part A is wrong though. Doesn't the coefficient of the kinetic friction play into the problem somewhere?
  50. B

    Spring Problem, help!

    So is that all you have to do for part A?
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