A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely with no frict

Question

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely with no friction. Determine: a) the final velocity of the cart, b) the impulse exerted by the cart on the package, c) the fraction of the initial energy lost in the impact.

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Vân Khánh 2 months 2021-07-31T23:51:38+00:00 1 Answers 1 views 0

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    2021-07-31T23:52:57+00:00

    Answer:

    (a) the final velocity of the cart is 0.857 m/s

    (b) the impulse experienced by the package is 21.43 kg.m/s

    (c) the fraction of the initial energy lost is 0.71

    Explanation:

    Given;

    mass of the package, m₁ = 10 kg

    mass of the cart, m₂ = 25 kg

    initial velocity of the package, u₁ = 3 m/s

    initial velocity of the cart, u₂ = 0

    let the final velocity of the cart = v

    (a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

    m₁u₁  + m₂u₂ = v(m₁  +  m₂)

    10 x 3   + 25 x 0   = v(10  +  25)

    30  = 35v

    v = 30 / 35

    v = 0.857 m/s

    (b) the impulse experienced by the package;

    The impulse = change in momentum of the package

    J = ΔP = m₁v – m₁u₁

    J = m₁(v – u₁)

    J = 10(0.857 – 3)

    J = -21.43 kg.m/s

    the magnitude of the impulse experienced by the package = 21.43 kg.m/s

    (c)

    the initial kinetic energy of the package is calculated as;

    K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

    the final kinetic energy of the package;

    K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

    the fraction of the initial energy lost;

    = \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

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