n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of kinetic energ

Question

n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of kinetic energy were released in the explosion, how much kinetic energy did the heavier piece acquire

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Nem 4 years 2021-08-29T05:49:25+00:00 2 Answers 9 views 0

Answers ( )

  1. danthu
    0
    2021-08-29T05:50:48+00:00
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    Answer:

    The heavier piece acquired 2800 J  kinetic energy

    Explanation:

    From the principle of conservation of linear momentum:

    0 = M₁v₁ – M₂v₂

    M₁v₁ = M₂v₂

    let the second piece be the heavier mass, then

    M₁v₁ = (2M₁)v₂

    v₁  = 2v₂ and v₂ = ¹/₂ v₁

    From the principle of conservation of kinetic energy:

    ¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J

    ¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400

    ¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400

    K.E₁ + ¹/₂K.E₁ = 8400

    Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁

    1.5 K.E₁ = 8400

    K.E₁ = 8400/1.5

    K.E₁ = 5600 J

    K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J

    Therefore, the heavier piece acquired 2800 J  kinetic energy

  2. Sapo1
    0
    2021-08-29T05:50:51+00:00
    Reply

    Answer:

    The heavier piece acquired 2800J of energy.

    Explanation:

    Let the velocity of the first piece be v1 and that of the second piece be v1 and also let the masses be m1 and m2 respectively.

    m2 = 2m1

    From the principles of conservation of momentum

    m1v1 = m2v2

    m1v1 = 2m1v2

    Dividing through by m1

    That is v1 = 2v2

    v2 = 1/2v1

    Total kinetic energy = 8400J

    1/2m1v1² + 1/2m2v2² = 8400

    1/2m1v1² + 1/2(2m1) × (1/2v1)² = 8400

    1/2m1v1² + 1/4m1v1² = 8400

    Let 1/2m1v1 = K1

    K1 + 1/2K1 = 8400

    3/2K1 = 8400

    K1 = 8400× 2/3 = 5600J

    K2 = 8400 – 5600 = 2800J

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