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Some Physic question on kinematics, mechanics that i couldn't solve

  1. Aug 7, 2005 #1

    Wen

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    Here are some physic question i encounter in a book for physic olympiad. I couldn't work them out.


    Questions:
    1)
    A particle moves along a straight line ABC with B
    being the midpoint between A and C. The acceleration
    of the particle from A to B is constant at a1, while
    its acceleration from B to C is constant at a2. Given
    that

    vB = (vA + vC) / 2

    where vA, vB and vC are the velocities of the particle
    at points A, B and C respectively, compare the
    magnitudes of a1 and a2.

    2)
    Two points A and B are a distance s apart along a
    straight line. Distance s is divided into n equal
    sections, such that each section is a distance s/n
    long. A stationary particle starts travelling from A
    such that its inital acceleration is a constant a
    along the first section. At the end of each section,
    the acceleration of the particle increases by a/n,
    until the particle moves past B. Find the velocity vB
    of the particle after it has moved past point B.

    3)

    Problem 1

    A particle travelled half of a certain distance with a
    velocity v0. The remaining part of the distance was
    covered with velocity v1 for half the time, and with
    velocity v2 for the other half of the time. Find the
    mean velocity of the particle over the entire distance
    travelled?

    Problem 2

    A uniform ladder of mass m1 and length 2l rests
    against a smooth vertical wall with the foot of the
    ladder placed on a rough horizontal floor some
    distance away from the bottom of the wall. The
    coefficient of static friction between the foot of the
    ladder and the floor is ?, and a man of mass m2 climbs
    a distance l1 up the ladder from the bottom.

    (a) If the ladder makes an acute angle ? with the
    floor, how far can the man climb without the ladder
    slipping?

    (b) If the man is to climb a distance l2 from the foot
    of the ladder, what angle should the ladder make with
    the horizontal floor?

    4)
    boat is rowed from A across the river. If the boat is
    always pointing in the direction perpendicular to the
    river, it will eventually reach C on the opposite bank
    in 10 min. If the rower wants to reach point B
    directly opposite A instead, he has to point his boat
    in the direction towards D when he set off from A, and
    maintain that direction as he rows across. In this
    case he will take 12.5 min to cross the river.



    Given that BC is 120 m, find

    (a) the boat's speed v relative to the river;
    (b) the river's width L;
    (c) the speed u of river flow; and
    (d) the angle ? between AD and AB.

    5)
    Problem 1

    A boat can travel at a speed of 3 m s-1 on still
    water. A boatman wants to cross a river whilst
    covering the shortest possible distance. In what
    direction should he row with respect to the bank if
    the speed of the water is

    (i) 2 m s-1,
    (ii) 4 m s-1?

    Assume that the speed of the water is the same
    everywhere.

    Problem 2

    The suspension springs of all four wheels of a car are
    identical. By how much does the body of a car
    (considered rigid) rise above each of the wheels when
    its right front wheel is parked on an 8-cm-high
    pavement? Does the result change when the car is
    parked with both right wheels on the pavement? Does
    the result depend on the number and positions of the
    people sitting in the car? :grumpy:
     
  2. jcsd
  3. Aug 8, 2005 #2

    Fermat

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    Use the suvat eqn v² = u² + 2as along each section, AB and BC.

    Compare results.
     
    Last edited: Aug 8, 2005
  4. Aug 8, 2005 #3

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    This is just a case of developing a recurrence relation.

    You have n sections. Use the suvat eqn v² = u² + 2as on each section.

    first develop a recurrence relation for the accln, [tex]a_r[/tex], on the rth section.
    second develop a recurrence relation for the velocity, [tex]v_r[/tex], (using suvat) on each section incorporating [tex]a_r[/tex].

    You should end up with,

    [tex]v_n^2 = v_0^2 + \frac{as}{n}(3n-1)[/tex]

    where [tex]v_0 = v_A[/tex] and [tex]v_n = v_B[/tex]
     
  5. Aug 8, 2005 #4
    Are you Fermat from TSR?
     
  6. Aug 8, 2005 #5

    Wen

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    For Question 1, i've tried the similar method. The answer i obtained was

    a1/a2=( Vc^2 +2VaVc- Va^2)/(Vc^2-2VaVc-Va^2)
    Is this correct because i find my working too simple to be true.

    For Question 3 problem 1,
    my answer is : {2(V1+V2) Vo}/{2Vo+V1+V2}
    Is it correct?

    Thanks for the help.
     
  7. Aug 8, 2005 #6

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    yes. --------
     
    Last edited: Aug 8, 2005
  8. Aug 8, 2005 #7

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    The top line (numerator) should be (Vb² - Va²) Also check signs in the denominator.

    Now simplify it with "vB = (vA + vC) / 2" given earlier in the question.

    it's fairly straightforward.
     
  9. Aug 8, 2005 #8

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    Have'nt started Q3 yet. Starting t'others now.
     
  10. Aug 8, 2005 #9

    Fermat

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    yes, it's correct.
     
  11. Aug 8, 2005 #10

    Wen

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    Thanks alot
     
  12. Aug 8, 2005 #11

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    I only did part (a) for this one. I got,

    [tex]l_1 = l\left(1 - \frac{(m_1+m_2)}{m_2}\{1 - 2\mu tan\alpha\}\right)[/tex]

    not sure if it's right - what did you get?
     
    Last edited: Aug 8, 2005
  13. Aug 8, 2005 #12

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    This one seemed a bit difficult to answer they way they asked, but I managed to get it solved by working backwards! I got the river speed first, then the angle, then the relative speeds, then the river's width.

    I'd post the solution/working, but I don't think I'm allowed to do that am I?
    The answers I got were,

    1) river speed u = 12 m/min
    2) angle between AD and AB is 36.87 deg
    3) relative speeds boat:river = v:u = 20:12 = 5:3
    4) river's width L = 200m
     
  14. Aug 8, 2005 #13

    Wen

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    My answer is

    l1= l{( 2m1/m2 +2) tan alpha- m1/m2}
     
  15. Aug 8, 2005 #14

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    I found this one a bit awkward to visualise at first.
    I've got a solution but i'm not sure if it's right - what do you think?

    Here's my case for how I solved it.

    See attachment

    Argument
    If car is raised at spring 3, then the two springs 2 and 4 are also raised/extended.
    Since 2 and 4 are extended then the compressive load from these springs, supporting the car, is reduced.
    .: Additional load is required from springs 1 and 3 to support the car.
    By symmetry, springs 1 and 3 are equally compressed, by an amount x, say. And the springs 2 and 4 are equally extended, also by an amount x.

    I was able to get x = 2.

    i.e. springs 1 and 3 are compressed by 2 cm - car body lowers by 2 cm above the wheels
    springs 2 and 4 are extended by 2 cm - car body rises by 2 cm above the wheels.

    If both right wheels are on the pavement, then all springs ae giving the same load/support, so the car body won't rise/lower wrt the wheels.

    All results are indepndent of the numbers of passengers or their positions provided they represent a UDL. (uniformly distribiuted load).
     

    Attached Files:

  16. Aug 8, 2005 #15

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    I corrected my own expression.

    It rearranges into your expression - which looks promising.

    I'm assuming you just missed out a factor of μ in your own post.
     
  17. Aug 9, 2005 #16

    Wen

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    Argument
    If car is raised at spring 3, then the two springs 2 and 4 are also raised/extended.
    Since 2 and 4 are extended then the compressive load from these springs, supporting the car, is reduced.
    .: Additional load is required from springs 1 and 3 to support the car.
    By symmetry, springs 1 and 3 are equally compressed, by an amount x, say. And the springs 2 and 4 are equally extended, also by an amount x.

    I agree with this part

    I tried to find X : elastic potential energy EPE of the 2 compressed spring = EPE of the 2 extented ones + the GPE gained by the centre of mass of the car

    But if the springs are compressed and extended by X equally, they woulf cancel out.

    I am stuck at this part. :frown:
     
  18. Aug 9, 2005 #17

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    I just drew a diagram! Of the springs and car (modelled as a UDL board) see attachment in post #14.
    If you think about it ...
    Suppose springs 1 and 3 are rigid. Now lift up spring 3 by 8 cm. Springs 2 and 4 are half-way across the car, so will lift up by 4 cm - i.e. 2 and 4 get an extension of 4 cm. Spring 1, being rigid, does not change.
    Now let springs 1 and 3 be normal springs again. They will both compress by 2 cm. The top of springs 2 and 4, being attached to the car/board will also lower by 2 cm. But they had been extended by 4 cm. So, in effect, they get an extension of 4 - 2 = 2 cm.
    This agrees with the Argument. That springs 1 and 3 compress by an amount x = 2, and springs 2 and 4 extend by an amount x = 2.
     
  19. Aug 9, 2005 #18

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    I'm not sure how you would do it this way. We don't know what the spring constants are. I don't think you can use the mass of the car for GPE since we don't know the position of the com of the car. And the lift given to it would vary depending on the geometry of the car - which we don't know.

    As I said, I just drew a few diagrams - see following post.
     
  20. Aug 9, 2005 #19

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    In Fig 1, I have modelled the car as a board, representiong a UDL, un-equally supported at the four corners.

    Take a view-point along the diagonal from spring 2 to spring 4.
    This gives View 'A', Fig 2, where springs 2 and 4 overlap and are equi-distant from springs 1 and 3.

    The line PQR is the top of spring 1. Assume springs 1 and 3 are rigid.

    Spring 3 is 8 cm above the road - this will give an extension of 4 cm in springs 2 and 4.

    Now let springs 1 and 3 be normal springs. The whole car/board will lower by the same amount, x cm say. See Fig 3.

    Spring 1 is compressed and goes x cm below the base line/level PQR.

    Springs 2 and 4 are lowered by x cm, such that they are now (4-x) cm above the base line, i.e. they have an extension of (4-x) cm.


    But by the Argument, this extension is x cm.


    .: 4 - x cm = x cm
    4 = 2x
    x = 2 cm
    =======
     

    Attached Files:

  21. Aug 10, 2005 #20

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    Q5 part 1

    i) this part is easy. Draw the velocity vector diagram such that the resultant vcelocity of the boat is directly across the river, perpindicular to the flow of water.

    ii) Look at Fig 1. The boat travels, at velocity u, at an angle α to the river bank and the resultant velocity, w, is at an angle β to the river bank.
    The cross-river velocity is wsinβ. Time to cross the river is T = L/wsinβ, where L is the width of the river.
    Distance travelled by boat is S = wT.
    S = w*L/wsinβ
    S = L/sinβ
    S is a minimum when sinβ, or β, is a maximum.
    β is a max. when the angle between v and w is 90º. You would have to prove this. There are a couple of ways of doing that.
    The vector triangle is now a right-angled triangle with u = 4 and v = 3 and w = √7.
    sinα = w/u = √7/4
    α = 41.41º
    ========
     

    Attached Files:

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