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Momentum and Energy

  1. May 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A 64kg student is standing in the middle of a frozen pond with a frictionless surface while carrying 2 books, a physics book with a mass of 2.5kg and a biology book with a mass of 3.5kg. The physics book is thrown with a speed of 6m/s relative to the student to the right and then the biology book is thrown in the opposite direction with a speed of 4m/s relative to the student. Calculate the velocity of the student after the biology textbook is thrown.

    2. Relevant equations
    E = 1/2 mv(squared), initial moment = final momentum, therefore mv (initial) = mv (final), maybe more, im not sure

    3. The attempt at a solution

    Since the student is at rest, the initial momentum is obvisiously 0. But once the first book is thrown and then the second book, i don't know how to calculate the velocity of the student after each time a book is thrown.
     
  2. jcsd
  3. May 3, 2008 #2

    olgranpappy

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    use conservation of momentum. write down the final momentum in terms of the sum of: the students momentum (which you want to solve for), the physics book's momentum, and the biology book's momentum. and then set that sum equal to the initial momentum (zero).
     
  4. May 4, 2008 #3
    ok i got that part, but it says that the biology textbook is thrown with a speed of 4m/s "relative to the student." Wouldn't the student already be moving after throwing the first textbook?? therefore how would i calculate the velocity in terms of the speed of the textbook relative to the ground. this question gives the speed of the textbook relative to the 'already moving person.' how would i figure that out?
     
  5. May 4, 2008 #4

    olgranpappy

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    so, do the problem in two steps. first find the velocity of the student+biobook after the physics book is thrown. then find the velocity of the student after the bio book is thrown. in the first
    step the initial momentum is zero. in the second step the initial momentum is (M_student+M_bio)*v_1. where v_1 is the velocity of the student + biobook you found in the first step. since you know v_1 the velocity of the bio book to use in the second step is v_1-4 (taking positive to be to the right...n.b., v_1 is also negative)
     
  6. May 5, 2008 #5
    so to solve for v_1 would it be:
    0 = (64 + 3.5)v_1 + 2.5(6)

    but once i solve for v_1, how would i set up the equation to solve for v_1-4?
     
  7. May 6, 2008 #6

    olgranpappy

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    you don't want to solve for v_1 - 4... you already know what v_1 is, thus you already know what v_1 - 4 is.

    Okay, so here's what you can do now. You know v_1. You know the total momentum of all the objects much equal zero. You know that the phys book is moving to the right at 6 m/s. The student throws the bio book now so that it is moving to the left at (|v_1|+4) m/s or rather to the right at v_1 - 4 m/s (v_1 is negative). Then let the student's velocity be called v_2.

    So, you have, at the very end, the velocity of the physics book is +6, the velocity of the bio book is (v_1-4), the velocity of the student is v_2. Write down the total momentum at the very end and set it equal to zero.
     
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