ratio of their masses?

An object at rest is suddenly broken apart into two fragments by an explosion. The first fragment (with mass m_1) acquires twice the kinetic energy of the second one (with mass m_2).

What is the ratio of their masses? (m_1/m_2)

4 Answers

  • by conservation of momentum m1 v1 = m2 v2 , where v1 and v2 are the velocities of the fragments after explosion. then, v1 / v2 = m2 / m1

    now, as per question, K1 = 2 K2, where K1 andK2 are their kinetic energies.hence, 1/2 m1 v1^2 = 2 x 1/2 m2 v2^2

    ie., 2x m2 / m1 = (v1 /v2)^2= (m2 / m1)^2

    then, m1 / m2 = 1 / 2 ie., m1 : m2 = 1 : 2 is the required ratio.

  • Using the relation KE = P² / 2m where P is the momentum and KE the kinetic energy.

    m_1 has momentum P_1 and kinetic energy P²_1 / 2m_1

    m_2 has momentum P_2 and kinetic energy P²_2 / 2m_2

    By conservation of momentum, |P_1| = |P_2| = P (say)

    We are told KE of m_1 is twice KE of m_2

    so P² / 2m_1 = 2 P² / 2m_2

    Multiplying through by 2 / P²

    1 / m_1 = 2 / m_2

    From which 2m_1 = m_2

    2m_1 / m_2 = 1

    m_1 / m_2 = 1/2

    Ratio of mass 1 to mass 2 is 1 : 2

  • nicely the respond is A yet that's not the style you clean up it. Q=3 A=2 Q/A ratio is .574 so then enable us to divide Q by using A that's 3/2=a million.5 Now multiply your answer by using the ratio. a million.5*.574=.861

  • The momentums have to be equal, so we have:

    m₁V₁ = m₂V₂

    From the KE statement

    2*½m₁V₁² = ½m₂V₂²

    2m₁V₁² = m₂V₂²

    now we have 2 equations

    m₁V₁ = m₂V₂

    2m₁V₁² = m₂V₂²

    m₁/m₂ = V₂/V₁

    from the other equation

    2m₁V₁² = m₂V₂²

    V₁² / V₂² = m₂/2m₁

    V₁ / V₂ = √(m₂/2m₁)

    substituting

    m₁/m₂ = √(2m₁/m₂)

    squaring

    m₁²/m₂² = 2m₁/m₂

    multiplying both sides by m₂/m₁

    m₁/m₂ = 2

    .

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