physics help please?

A bundle of positrons (the anti-matter equivalent of electrons) is travelling East along a beamline in a linear accelerator. The bundle contains 2.329E16 positrons. Each positron weighs 9.62174e-30kg. The bundle accelerates at 0.0313 m/s^2 East. How much work is the force of gravity doing on the bundle for every 1 m travelled?

1 Answer

  • ?
    Lv 7
    2 months ago
    Favorite Answer

    Question is incorrectly formulated.  Try this...

    Rest mass of positron is same as electron, 9.10938e-31kg, so we can find positron’s speed:

    m = m₀/√(1 - v²/c²)

    9.62174e-31 = 9.10938 e-31/√(1 - v²/c²)

    I get v = 0.322c = 0.322*3.00e8 = 9.66e7 m/s but check my arithmetic.

    The bundle has a horizontal acceleration aₓ = 0.0313m/s² and a vertical acceleration aᵧ = 9.81m/s².   aₓ is very small compared to aᵧ so we can neglect it

    We treat the bundle as ta projectile with horizontal velocity component v = 9.66e7 m/s.

    The path of the bundle will be almost horizontal. So we can take the distance travelled as the horizontal distance.

    Time to cover 1m  = 1/v = 1/9.66e7 = 1.035e-8 s

    Assume vertical acceleration is from rest so when the bundle falls bundle travels falls a distance Δh in time 1.035e-8 s

    After 1 metre, height change: Δh = ½aᵧt² = ½*9.81*( 1.035e-8)² = 5.25e-16 m

    Work done by the force of gravity per metre travelled =mgΔh

    = 2.329e16 * 9.62174e-30 *  5.25e-16

    = 1.18e-38 J

    The time to travel 2 metres is twice the time to travel 1m.

    After  2metre, height change: Δh = ½aᵧt² = ½*9.81*( 2*1.035e-8)² = 2.10e-15 m

    Note this is *four times* the height change for 1m

    So the height change per metre is now 2*1.18e-38 J  = 2.39e-38J

    In other words the work done per metre by gravity is not a constant quantiity so the question can’t be answered.

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