eh asked in Science & MathematicsPhysics · 2 months ago

Help me with physics question pls!!!?

A ball with mass mA= 0.500 kg, moving to +x direction with a speed of 4.50 m/s has a head-on collision with a 1.000 kg ball, which is initially at rest. Considering that the collision is perfectly elastic, find the speed and direction of each ball after the collision.

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  • 2 months ago
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    Answer:

    Velocity first ball vA2 = -1.5 m/s

    Velocity second ball vB2 = 3 m/s

    "collision is perfectly elastic" means that both conservation of energy and conservation of impulse applies.

    Energy before and after

    mA vA1^2 / 2 = mA vA2^2 / 2 + mB vB2^2 / 2

    or

    mA vA1^2 = mA vA2^2 + mB vB2^2

    here

    mA and mB = masses

    vA1 = velocity ball A before

    vA2 = velocity ball A after

    vB1 = velocity ball B before (vB1=0)

    vB2 = velocity ball B after

    Impulse before and after

    mA vA1 = mA vA2 + mB vB2

    From equation of impulse

    mA vA2 = mA vA1 - mB vB2

    vA2 = (mA vA1 - mB vB2) / mA

    square and substitute into equation for energy

    mA vA1^2 = (mA vA1 - mB vB2)^2 / mA + mB vB2^2

    mA^2 vA1^2 = (mA vA1 - mB vB2)^2 + mAmB vB2^2

    0 = - 2 mA mB vA1 vB2 + mB^2 vB2^2 + mAmB vB2^2

    (mB^2 + mAmB) vB2^2 - 2 mA mB vA1 vB2 = 0

    ((mB^2 + mAmB) vB2 - 2 mA mB vA1) vB2 = 0

    discard solution vB2=0

    (mB^2 + mAmB) vB2 - 2 mA mB vA1 = 0

    vB2 = 2 mA mB vA1 / (mB^2 + mAmB)

    vB2 = 2 * 0.5 * 1 * 4.5 / (1^2 + 0.5*1)

    vB2 = 3m/s

    vA2 = (0.5 * 4.5 - 1 * 3) / 0.5 = -1.5 m/s

    Check

    Energy before/after

    0.5 * 4.5^2 = 0.5 * (-1.5)^2 + 1 * 3^2

    10.125 = 10.125 ....... OK

    Impulse

    0.5 * 4.5 = 0.5 (-1.5) + 1 * 3

    2.25 = 2.25

    Edit:

    "prettier" formula layout added

    Attachment image
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