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A man mass 75 kg and a boy mass 25 kg dives off the end of a boat of mass 300 kg so that their relative horizontal velocity with respect to the boat is 3 m/s. if initially the boat is at rest find its final velocity if (1) the two dive off simultaneously (2) the man dives first followed by the boy

  • science - ,

    This is a momentum problem and momentum is conserved:

    M1V+M2V=M3V

    Let

    M1=75 kg
    M2=25 kg
    M3=300 kg
    V1=3 m/s
    V2= 3 m/s
    V3=?
    (3m/s)*(75 kg +25 kg)/300 kg=V3

    v=1 m/s

    Second problem:

    Answer Should be the same; momentum is conserved.


    But I am not 100% sure.

  • science - ,

    Second Problem:
    Momentum is still conserved on each leap (of course) but note that when the boy jumps off at 3m/s _relative_ to the boat, the boat is no longer at rest.

    M1 = 75 kg mass of man
    M2 = 25 kg mass of boy
    M3 = 300 kg mass of boat
    V1 = 3 m/s launch velocity (relative to boat)
    V2 = ? velocity of boat (and boy) after man leaps.
    V3 = ? velocity of boat after boy leaps.

    First jump:
    0 = M1 V1 + (M2+M3) V2
    or
    V2 = -M1 V1 / (M2+M3)

    Second jump:
    (M2+M3) V2 = M2 (V2+V1) + M3 V3
    or
    V3 = V2 - M2 V1 / M3
    Substituting from the first jump:
    V3 = -M1 V1/(M2+M3) - M2 V1/M3
    Thus using the values:
    V3 = -(75)(3)/(75+300) - (25)(3)/(300)
    Finally:
    V3 = -0.85 m/s

    The boat does not move back as fast in this case, because some of the impulse from the man's leap has been imparted to the boy.

  • science - ,

    I do not know how my answer earlier was reposted, but I think I should have went through the math. There is agreement that the velocity will be 1m/s for the first question. However, there is disagreement concerning the velocity for the second question.

    Let

    M1=300 kg
    V1i=0
    V1f=?
    M2=25kg
    V2i=0
    V2f=?
    M3=75kg
    V3i=0
    V3f=-3m/s

    ***I don't know what direction that each is traveling, but I will let the jumpers' velocity be negative when they jump, since that will imply in the opposite direction.

    Momentum Initial=Momentum Final

    M1V1i + M2V2i+M3V3i=M1V1f + M2V2f+M3V3f

    300kg(0) + 25kg(0) + 75kg(0)=300kg(V1f) + 25kg(V2f) + 75kg(-3m/s)

    0=300kg*(V1f) + 25 kg*(V2f) + (75kg)*(-3m/s)

    V1f=V2f, so both equal V.

    0=300kg*(V) + 25 kg*(V)+(75kg)*(-3m/s)

    0=-225J + V(300kg+ 25 kg)

    0=-225J + V(325kg)

    225J/(325kg)=V

    0.692m/s=V

    Now, when the boy jumps off, the boat and the boy have a initial velocity of 0.692m/s, but the final velocity of the boy will be -3m/s in the opposite direction.

    So let

    M1=300 kg
    V1i=0.692 m/s
    V1f=?
    M2=25kg
    V2i=0.692 m/s
    V2f=3 m/s


    M1V1i+M2V2i=M1V1f+M2V2f

    V1f=V2f, so both equal V, which is equal to 0.692 m/s

    M1V+M2V=M1V1f+M2V2f

    Solve for V2f

    (300kg*0.692m/s)+ (25kg*0.692m/s)=(300kg)V1f + (25kg*-3m/s)

    0.692m/s*(300kg+25kg)=(300kg)V1f + (25kg*-3m/s)


    225J=300kg*V1f + (-75J)


    225J+75J=300kg*V1f

    300J=300kg*V1f

    300J/300kg=V1f

    1 m/s=V1f

    It doesn't make a difference if they jump at the same time or right after each other. The only way that the velocity of the boat will change is if there is a long protracted wait after the initial jump.

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