A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s . It has a head-on collision with

Question

A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s . It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.27 m/s . Suppose the collision is elastic.1. Find the magnitude of the final velocity of the 0.157kg glider.
2. Find the magnitude of the final velocity of the 0.306kg glider.

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Mít Mít 5 months 2021-08-11T00:17:00+00:00 1 Answers 10 views 0

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    2021-08-11T00:18:10+00:00

    Answer:

    v1 = −2.201946 m/s ( to the left)

    v2 = 0.7780534 m/s ( to the right)

    Explanation:

    Given the following :

    Mass of first glider (m1) = 0.149kg

    Initial Speed of first glider (u1) = 0.710 m/s

    Mass of second glider (m2) = 0.308kg

    Initial Speed of second glider (u2) = 2.27m/s

    For elastic collision:

    m1u1 + mu2u2 = m1v1 + m2v2

    Where V1 and v2 = final velocities if the body after collision.

    Taking right as positive ; left as negative

    u1 = 0.710m/s ; u2 = – 2.27m/s

    u1 – u2 = – (v1 – v2)

    0.710 – – 2.27 = – v1 + v2

    v2 – v1 = 2.98 – – – – (1)

    From:

    m1u1 + mu2u2 = m1v1 + m2v2

    (0.149 * 0.710) + ( 0.308 * – 2.27) = (0.149 * v1) + (0.308 * v2)

    0.10579 + (-0.69916) = 0.149 v1 + 0.308v2

    −0.59337 = 0.149 v1 + 0.308v2

    Dividing both sides by 0.149

    v1 + 2.067v2 = −0.59337 – – – – – (2)

    From (1)

    v2 = 2.98 + v1

    v1 + 2.067(2.98 + v1) = −0.59337

    v1 + 6.16 + 2.067v1 = −0.59337

    3.067v1 = −0.59337 – 6.16

    3.067v1 = −6.75337

    v1 = −6.75337 / 3.067

    v1 = −2.201946 m/s ( to the left)

    From v2 = 2.98 + v1

    v2 = 2.98 + (-2.201946)

    v2 = 0.7780534 m/s ( to the right)

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