#g481

Quantum physics The photon: introduction

Light has wave properties, as shown by Young's double slit experiment.

Light is simultaneously produced in packets (or quantums) of electromagnetic energy called photons.

The energy of a photon depends on the frequency of the light wave:

E = hf

Mechanics Questions on kinetic energy and potential energy Cont'd from "Energy"

1. A loaded truck with a mass of 30000kg leaves the motorway with a speed of 20ms-1. The upward slope of the exit road along with the friction provides a constant stopping force of 84000N. The truck stops at the top of the exit road.

Find the work done as a result of this loss of kinetic energy.

F = ma a = F ÷ m   = 84000N ÷ 30000kg   = -2.8ms-2

v2 = u2 + 2as s = (v2 - u2) ÷ 2a   = (0 - 202ms-1) ÷ (2 × 2.8ms-2)

  = 400ms-1 ÷ 5.6ms-2   = 71.4m

W = Fx    = 84000N × 71.4m    = 6×109J = 6MJ

2. A ball bearing of mass 5.83 grams is dropped from an electromagnet at time 0. It passes through two light gates, separated by a distance of 40.7cm, at times 0.016s and 0.289.

(i) Show that g is roughly 9.81ms-2.

s = ut + ½at2 asfih

s = ½g(0.016s)2for the first gate s + 0.407m = ½g(0.289s)2 for the second gate

so 0.407m = ½g(0.2892  - 0.0162) f

g = 2 × 0.407             0.0832    = 9.78ms-2.

(ii) Find the loss in potential energy of the ball between the two gates.

Ep = mgh     = 5.83×10-3kg × 9.81ms-2 × 0.407m     = 0.0233J

(iii) Find the kinetic energy of the ball as it pasts the second gate.

Ek = ½mv2     = ½ × m × (a × t)2     = ½ × 5.83×10-3kg × (9.81ms-2 × 0.289s)2     = 0.0234J

3. An object of mass 20kg, is stopped by a constant force of 250N in the time of 5s.

Find the work done on the object.

F = ma a = F ÷ m    = 250N ÷ 20kg    = 12.5ms-2 ×

s = ut + ½at2   = (12.5 × 5) × 5 + ½ × (12.5 × 52)   = (62.5) × 5 + ½ × (12.5 × 25)   = 468.5m

Mechanics Energy

Definition: Energy The stored ability to do work.

At the basic level, there are two main energies. These are:

Kinetic energy Ek Energy where movement is taking place.

Potential energy Ep Energy due to to a body's position in a field of energy. Can be either a gravitational, magnetic, electric or nuclear field.

There are other types of energy, such as chemical energy, nuclear energy, sound energy etc., but these are not relevant in the course until later units.

Conservation of energy

The principle of conservation of energy states that: In any closed system, energy may be converted from one form into another but it can not be created or destroyed, so it is conserved.

A Sankey diagram is used to illustrate this principle.

Below is a simplified Sankey diagram for a car travelling at night at a constant speed:

Mechanics Questions on work Cont'd from "Work done"

1. A caravan of mass 400kg is pulled at constant speed along a horizontal road against a drag force of 240N.

(i) Calculate the work done to pull the caravan 500m.

work = Fx        = 240N × 500m        = 1.2×105J

(ii) Explain why the weight of the caravan does not enter the calculation.

Mechanics Work done

Work in physics is the effect of a force in a certain distance. It is a quantity of energy.

Definition: Work done work = force × distance moved in direction of the force

W = Fx cosθ

Where W is work, F is force and x is distance. cosθ is for if the work force is acting at angle θ to the force.

This equation can be found on the Physics A data sheet⇗, and is explained on my G481 formulae page.

The unit for pressure is J (joules).

Mechanics General G481 revision questions See "search/G481"

1. A shot putter's throw is shown below. The shot is stationary at position A. The shot leaves the putter's hand at point B.

θ (angle from horizontal at which the shot leaves hand) = 34.5° u (velocity at which shot leaves hand) = 12.9ms-1

(i) Show that the initial vertical component of u is about 7ms-1.

12.9ms-1 cos 55.5° = 7.3ms-1

(ii) Calculate the time between the athlete releasing the shot at B and it reaching C (the highest point of path).

v = u + at 0 = (12.9ms-1 cos 55.5°) + -9.81ms-2 × t 9.81ms-2 × t = 7.3ms-1 t =  7.3ms-1        9.81ms-2    = 0.7s

(iii) Calculate the horizontal distance travelled by the shot between B and D.

v = 12.9ms-1 cos 34.5° = 10.8ms-1 ½ s = ½ (u + v) t       = ½ (10.8ms-1) × 1.71s       = 9.1m s = 9.1m × 2 = 18.2m

(iv) The shot has a mass of 5.00kg. Show that the shot's kinetic energy Ek as it leaves the putter's hand at B is about 420J.