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Few Problems

  1. Sep 14, 2006 #1

    danago

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    Hey. Ive got a physics test coming up soon, and i had a few questions that im wanting confirmed. All help is greatly appreciated.

    _______________________________________________________​

    1. IN a horrific car crash, a car skids 85m before striking the rear of a parked car. Just before the moment of impact, the car is moving with a velocity of 15 [itex]ms^{-1}[/itex]. The cars become locked together and skid for another 5.2m before coming to rest. The mass of the first car is 1350kg, and the mass of the second is 1520kg.

    a) What is the velocity of the two cars just after impact?

    b) What is the impulse on each car during the collision?

    c) What is the average size of the frictional forces between the cars and the road that finally bring the cars to rest?


    a) For part a, i just used the fact that momentum is conserved.

    [tex]
    \begin{array}{l}
    \sum {p_i } = 1350(15) = 20250 \\
    \sum {p_f } = (1350 + 1520)v = 2870v \\
    \sum {p_i } = \sum {p_f } \\
    \therefore 2870v = 20250 \\
    \therefore v = 7.0557ms^{ - 1} \\
    \end{array}
    [/tex]

    Therefore, after impact, the velocity of the combined masses is ~7[itex]ms^{-1}[/itex].

    b) Part b, i just found the change in momentum.

    [tex]
    \begin{array}{c}
    I = \Delta p \\
    = m\Delta v \\
    = m(v - u) \\
    = 1350({\rm{7}}{\rm{.0557 - 15}}) \\
    = - 10724.8{\rm{ }}Ns \\
    \end{array}
    [/tex]

    Therefore, the impulse experienced by each car is 10724.8 Ns.

    c) I had the most trouble with part c. For this question, i found the acceleration of the cars, from ~7[itex]ms^{-1}[/itex] to rest in 5.2m.

    [tex]
    \begin{array}{l}
    v^2 = u^2 + 2as \\
    \therefore 0 = 7.0557^2 + 2(5.2)a \\
    \therefore a = - 4.79ms^{ - 2} \\
    \end{array}
    [/tex]

    Therefore, the masses were deccelerating at a rate of 4.79[itex]ms^{-2}[/itex]. For this decceleration to occur in a mass of 2870kg, i used newtons formula to calculate the force required:

    [tex]
    \begin{array}{c}
    \sum F = ma \\
    = 2870( - 4.79) \\
    = - 13738N \\
    \end{array}
    [/tex]

    Therefore, an average force of ~13738N is required to bring the two cars to rest within 5.2m of collision.

    _______________________________________________________​

    2. The following image depicts a roller coaster:

    [​IMG]

    a) Assuming that the speed of the cart at point A is 4[itex]ms^{-1}[/itex], find the speed of the cart at points B, C and D. Also assume that no energy is lost due to friction/drag.

    b) It is found that at point C, the cart stops. What is the energy loss due to friction between points A and C?


    I was a little confused with this one at first. I started by finding the total mechanical energy at point A:

    [tex]
    \begin{array}{c}
    E_M = E_K + E_P \\
    = 8m + 294m \\
    \end{array}
    [/tex]

    Assuming that this is a perfectly elastic energy transformation, the mechanical energy at point B should be the same as the kinetic energy, since potential energy is zero, because of a zero height relative to the ground. So:

    [tex]
    \begin{array}{l}
    \frac{{mv^2 }}{2} = 8m + 294m \\
    \therefore v^2 = 604 \\
    \therefore v = \sqrt {604} = 24.6ms^{ - 1} \\
    \end{array}
    [/tex]

    At point C, since energy is still being conserved, and this is an elastic situation, the mechanical energy should equal the mechanical energy used in the previous calculation. So that:

    [tex]
    \begin{array}{l}
    245m + \frac{{mv^2 }}{2} = 8m + 294m \\
    \therefore v = 10.7ms^{ - 1} \\
    \end{array}
    [/tex]

    And i used the same process for point D to get a velocity of 19.2[itex]ms^{-1}[/itex].

    For question b, im quite stuck. Im not sure where to start from. Any help please? Would the answer just be the kinetic energy the cart has at point C, which would be ~57m J? Also, how can i calculate the efficiency of the track from A to C?

    _______________________________________________________​

    3. A ball of mass 0.1kg strikes a raquet with velocity of 10[itex]ms^{-1}[/itex], and rebounds with a velocity of 9.5[itex]ms^{-1}[/itex]. Using the force/displacement graph, find the maximum displacement of the strings.

    [​IMG]


    This question im completely stuck. Everything ive tried leads nowhere. My physics teacher couldnt even do it.


    _______________________________________________________​

    Thanks very much for the help. For the questions ive answered, id just like some confirmation as to weather ive done them correctly.

    Thanks again,
    Dan.
     
    Last edited: Sep 14, 2006
  2. jcsd
  3. Sep 14, 2006 #2

    danago

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    Ok. Sorry about accidently posing before finishing my post. Its all done now though :)
     
  4. Sep 14, 2006 #3

    J77

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    May help

    For part 3 - the slope of the force-displacement curve gives you the stiffness coefficient.
     
  5. Sep 14, 2006 #4

    danago

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    Stiffness coefficient? Thats something that ive never come across or heard of. Where can i find more information on this?
     
  6. Sep 14, 2006 #5

    Kurdt

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    Well presented. I don't have tim to check them all immediately but i've checked the first. All the methods are correct and the numbers seem fine nice work.
     
  7. Sep 14, 2006 #6

    andrevdh

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    #3. Many assumptions need to be made in order to get to an answer for this one. I think the basic idea is that the raquet acts like a spring storing the kinetic energy of the ball temporarily and then returning it back to the ball (like an object being dropped onto a spring, but in this case only kinetic energy needs to be considered). It seems there is some energy lost in this interaction, but I would assume that the strings would deflect according to the initial kinetic energy of the ball. Which means that one need to find the displacement at which the area under the graph (the work done by the strings) is the same as the initial kinetic energy of the ball.
     
    Last edited: Sep 14, 2006
  8. Sep 14, 2006 #7
    wheres point A on the rollercoaster?
     
  9. Sep 14, 2006 #8

    danago

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    Oops accidently cut it off. Its the point 30m from the ground :)

    And thanks for the replies everyone. Im still not quite sure about question 3 though
     
  10. Sep 14, 2006 #9

    J77

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    The work done in part 3 is [tex]\frac12 K x^2[/tex] where K's the stiffness coefficient.

    Can this be equated to anything?
     
  11. Sep 14, 2006 #10

    danago

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    ok. So the stiffness coefficient is 200. Do i then just find the maximum value of x in:

    [tex]
    W = 100x^2
    [/tex]

    ?
     
  12. Sep 14, 2006 #11

    danago

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    Nvm. Just realised the quadratic has no maximum.
     
  13. Sep 14, 2006 #12

    Astronuc

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    Stiffness is the propensity to resist displacement when subject to a force. The greater the stiffness, the higher the spring constant, i.e. the more force is takes for a given displacement. This is the meaning of Hooke's law.

    http://en.wikipedia.org/wiki/Hooke's_law - relates force and displacement
    http://en.wikipedia.org/wiki/Young's_modulus - relates stress and strain (which are related to stress and strain)

    http://phoenix.phys.clemson.edu/labs/124/shm/index.html
     
  14. Sep 14, 2006 #13

    danago

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    Thanks for that astro.
     
  15. Sep 14, 2006 #14

    Astronuc

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    Along the lines of what J77 is alluding, what is the origin of the energy in the spring (tennis racket springs) when it deflects and what is the velocity of the ball at maximum spring deflection?
     
  16. Sep 14, 2006 #15

    danago

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    The energy to displace the strings would be the kinetic energy from the ball, and at maximum displacement, the balls velocity is zero?
     
  17. Sep 14, 2006 #16

    Astronuc

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    At maximum displacement, the spring has stopped moving (deflecting) - then it starts moving in the opposite direction. That's more or less the definition of an extremum.
     
  18. Sep 14, 2006 #17

    danago

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    Im still not really seeing how to solve the problem :(
     
  19. Sep 14, 2006 #18

    Astronuc

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    Looking back at post #6 (andrevdh) and post #9 (J77), one could do an equivalence between the spring energy 1/2 kx2 and the initial kinetic energy of the ball 1/2 m v2. Now, as andrevdh implied, requires the assumption that the raquet plane does not deflect itself, i.e. all the deflection goes into the springs.

    The above relationship simplies to kx2 = m v2, which is where J77 was leading you.

    Now the other part is the fact that the ball rebounds with a velocity of 9.5 m/s, so there is a loss of energy somewhere, which could be internal friction due to inelastic collision (deformation of the ball). One could assume half the energy loss during the extension of the spring and the other half during restoration, so then the effective kinetic energy going into the strings may be based on the mean velocity (speed) of 9.75 m/s.

    Conservation of energy applies to elastic collision, and not inelastic collision.
     
  20. Sep 14, 2006 #19

    J77

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    I was thinking of equating the work done by the 'spring' with the change in kinetic energy of the ball...

    Would this not work?
     
  21. Sep 14, 2006 #20

    Kurdt

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    For #2 part b. you know the speed it should be going if there is no friction, so due to the friction all that speed has diminished, therefore the energy loss is as you thought. Efficiency is normally defined as the real performance/ideal performance. I suggest using energy in such a manner

    efficiency= energy lost/initial energy

    seems most logical to me. I could be wrong because I've never had to apply it to this situation though so if somebody could check me.
     
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