At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/s)ˆx. One of

Question

At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/s)ˆx. One of the particles is at the origin. The other particle has a mass of 0.10 kg and is at rest on the x-axis at x = 8.0 m. a) What is the mass of the particle at the origin? b) Calculate the total momentum of this system. c) What is the velocity of the particle at the origin?

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Latifah 3 years 2021-08-06T16:39:43+00:00 1 Answers 198 views 0

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  1. lethu
    0
    2021-08-06T16:40:46+00:00
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    Answer:

    Explanation:

    Given that,

    At one instant,

    Center of mass is at 2m

    Xcm = 2m

    And velocity =5•i m/s

    One of the particle is at the origin

    M1=? X1 =0

    The other has a mass M2=0.1kg

    And it is at rest at position X2= 8m

    a. Center of mass is given as

    Xcm = (M1•X1 + M2•X2) / (M1+M2)

    2 = (M1×0 + 0.1×8) /(M1 + 0.1)

    2 = (0+ 0.8) /(M1 + 0.1)

    Cross multiply

    2(M1+0.1) = 0.8

    2M1 + 0.2 =0.8

    2M1 = 0.8-0.2

    2M1 = 0.6

    M1 = 0.6/2

    M1 = 0.3kg

    b. Total momentum, this is an inelastic collision and it momentum after collision is given as

    P= (M1+M2)V

    P = (0.3+0.1)×5•i

    P = 0.4 × 5•i

    P = 2 •i kgm/s

    c. Velocity of particle at origin

    Using conversation of momentum

    Momentum before collision is equal to momentum after collision

    P(before) = M1 • V1 + M2 • V2

    We are told that M2 is initially at rest, then, V2=0

    So, P(before) = 0.3V1

    We already got P(after) = 2 •i kgm/s in part b of the question

    Then,

    P(before) = P(after)

    0.3V1 = 2 •i

    V1 = 2/0.3 •i

    V1 = 6 ⅔ •i m/s

    V1 = 6.667 •i m/s

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