Anonymous
Anonymous asked in Science & MathematicsPhysics · 8 years ago

Kinetic energy and momentum relation?

If kinetic energy of a body is increased by 300% then percentage change in momentum will be

1) 100%

2) 150%

3) 265%

4) 73.2%

5 Answers

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  • knr
    Lv 6
    8 years ago
    Best Answer

    the percentage of increase in momentum = 10[ ( square root of (300+100 ) -10 ] = 100 %

    so the answer is (1)

  • 8 years ago

    Pretend the mass is 2kg and the object is travelling at 1m/s.

    KE = ½ x 2 x 1² = 1J

    Momentum = 2 x 1 = 2kgm/s

    If the kinetic energy is increased by 300%, that means it is increased by 3J, and the new kinetic energy is 1+3 = 4J.

    The new speed is given by ½ x 2 x v² = 4 giving v= 2m/s.

    New momentum = 2x2 = 4kgm/s

    The momentum has doubled (from 2 to 4), which is an increase of 100%. So the correct answer is 1).

  • 8 years ago

    let the initial KE be x.

    as the mass of the body is fixed, let it be m.

    now new KE is 3x. the relation between KE and momentum is p=root(2mKE)

    chang in momentum is {root(6mKE) -root(2mKE)}. so the percentage will be root(3)-1 X 100 = 73 % proximate . . hit maximum points cuz i did lot of hard work for it! have fun

  • Ivan
    Lv 6
    8 years ago

    KE E=p^2/2m p is momentum m=mass

    taking log and differentiating

    dE/E=2Dp/p

    or DE/E %=2Dp/p%

    300=2x(%Dp)

    or %Dp=300/2=150

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  • cavitt
    Lv 4
    3 years ago

    ok.E preliminary = 0.5 * m * v1* v1 very final ok.E = 0.5 * m * v2* v2 ok.E preliminary/ok.E very final = a million/2 so, v1^2/v2^2 = 0.5 v2/v1 = a million/0.seventy one = a million.4 Momentum preliminary = m*v1 Momentum very final = m*v2 = m * a million.4 * v1 So momentum is better by way of a million.4 cases

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