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Collision of cballs of clay help

  1. Mar 18, 2005 #1
    All of these problems are the last of a series of ten, which i cannot solve. Ive given extensive efforts but am now nearly out of time (12pm est). if anone could be of help please feel free.


    A 20 g ball of clay traveling east at 2m/s collides with a 30 g ball of clay traveling 30 degrees south of west at 1m/s. what is the speed of the resulting blob, m/s?

    Dan is gliding on his skateboard at 2.00m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 7.00m/s(as measured by an observer on the ground). Dan's mass is 60.0 kg and the skateboard's mass is 6.00kg. how fast is dan going as his feet hit the ground

    A 10.0g ball of clay traveling east at 7.00m/s collides and sticks together with a 10.0g ball of clay traveling north at 5.50m/s. what is the speed of the resulting clay

    A spaceship of mass 2.3×10^6kg is cruising at a speed of 6×10^6m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.8×10^5kg, is blown straight backward with a speed of 2.3×10^6m/s. A second piece, with mass 7.6×10^5kg, continues forward at 1.3×10^6m/s.
     
    Last edited: Mar 18, 2005
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  3. Mar 18, 2005 #2

    arildno

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    Do you agree that 1&3 is basically the same problem?
    What do you know of inelastic collisions?
     
  4. Mar 18, 2005 #3
    i just solved the spaceship problem.

    they are all momentum problems, and i feel as though i understand them all but something isnt quite right. and i know of ineleastic collisions.

    for the skateboard question, i did 60kg(Vf)=6kg*2m/s+60kg*2m/s

    for blob of clay question (without degrees) i did (.01kg+.03kg)Vf=.02kg*3m/s+.03kg*2m/s
     
    Last edited: Mar 18, 2005
  5. Mar 18, 2005 #4

    arildno

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    The "blob of clay" questions need you to regard velocities as VECTORS, not scalars as you have done
     
  6. Mar 18, 2005 #5
    could you be more specific, do i find x and y components and then add? im almost positive im doing it correctly. for the first clay question... (.01kg +.01kg)Vf=.07kgm/s+.055kgm/s final answer of 6.25 is incorrect
     
    Last edited: Mar 18, 2005
  7. Mar 18, 2005 #6

    arildno

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    Right, in so far as you mean to say you add componentwise.
    For example, the first one:
    Let the unit vector "east" be [tex]\vec{i}[/tex] and north [tex]\vec{j}[/tex]
    Then, the unit vector which makes 30 degrees angle to the unit "west" vector in the southern direction is: [tex]\vec{t}=-\cos(30)\vec{i}-\sin(30)\vec{j}[/tex]
    Hence, the final velocity, [tex]\vec{v}_{f}[/tex] fulfills:
    [tex](m_{1}+m_{2})\vec{v}_{f}=m_{1}v_{1}\vec{i}+m_{2}v_{2}(-\cos(30)\vec{i}-\sin(30)\vec{j})[/tex]
    [tex]m_{1}=20g, v_{1}=2\frac{m}{s}, m_{2}=30g, v_{2}=1\frac{m}{s}[/tex]
     
  8. Mar 18, 2005 #7
    thank you so much, but what about the skateboarder, i cant figure it out.

    edit: that equation does not work. but im out of time. thanks for the help
     
    Last edited: Mar 18, 2005
  9. Mar 18, 2005 #8

    arildno

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    Do you know what speed is?
    It is not the same as velocity, I thought you knew the difference.
     
  10. Mar 18, 2005 #9
    their desired answer is in the form of m/s, the correct answer is .410. how did they come to this?
     
  11. Mar 18, 2005 #10

    arildno

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    Well, you must calculate the speed from the velocity and plug in the numbers.
    To continue with the one I started with, the final speed, [tex]||\vec{v}_{f}||[/tex]
    satisfies:
    [tex]||\vec{v}_{f}||=\frac{1}{m_{1}+m_{2}}\sqrt{(m_{1}v_{1}-m_{2}v_{2}\cos(30))^{2}+m_{2}^{2}v_{2}^{2}\sin^{2}(30)}=\frac{1}{20+30}\sqrt{(20*2-30*1*\frac{\sqrt{3}}{2})^{2}+30^{2}*1^{2}*\frac{1}{4}}[/tex]
     
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