# 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 72 months agoFavorite 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 1m.it 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.