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Homework Help: Direct collision with a barrier at an angle

  1. Jun 27, 2013 #1
    1. The problem statement, all variables and given/known data
    A ball of mass m moves horizontally with speed 20 m/s towards a smooth barrier xy which makes an angle tan-inverse(4/3) with the horizontal.The coefficient of restitution is (3/4)

    (i) The speed of the ball after the impact
    (ii) The magnitude of the impulse due to the impact
    (iii) the loss in kinetic energy
    (iv) the tan of the angle between the ball's new path and the horizontal

    2. Relevant equations

    3. The attempt at a solution
    let barrier xy = x axis
    i = horizontal component vector
    j = vertical component vector
    resolving 20 m/s into component vectors 20(3/5) = 12i 20(4/5) = -16j
    12i - 16j

    i component remains the same

    16(3/4) 12j

    12i +12 j (changes direction) using Pythagoras theorem to find magnitude = 12√2 m/s

    since only the j vector changes

    12m -(-16m ) = 28m Ns

    (iii) (1/2)m(20^2) - (1/2)m(12√2)^2 = 56m j

    (12i - 16j) + (12i + 12j) = 24i - 4j j/i = tanβˆ… = 4/24 = 1/6

    I got the right answers for part (i), (ii) and (iii)
    (iv) is the part im stuck on, is it not as simple as adding the two vectors and finding the angle? the answer at the back of my book is 7
  2. jcsd
  3. Jun 27, 2013 #2
    No it is not that simple. You have got to add angles not Vectors.
  4. Jun 27, 2013 #3
    Thanks I forgot that the ratio and the angle don't increase proportionally. Is this correct?
    tan-inverse(4/3) +tan-inverse(12/12) = 98.1301

    tan98.1301= -7
    does it matter that its - 7?
    can I just take it as positive?
  5. Jun 27, 2013 #4
    They didn't specify whether the positive x axis or the negative x axis. So answer you got is correct!
  6. Jun 27, 2013 #5
    Alright thanks.
  7. Jun 27, 2013 #6
    You are welcome!
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