Question 1

(a)An impactor mass of 45 kg is used to represent the weight of child reasonably regarded to be involved in an accident with glass or plastics.

(b)The BS standards gives the manufacturers a clear set of standards that their products need to achieve to be safe and fit for the purpose that they have been designed for. It also gives the purchaser the expectation that the item has reached the standards set down by the BSI and will be a safe for the expected life of the item.

(c) (i)The energy on impact is calculated by using the following equations:

Potential Energy (PE) = mass x gravity x height

This gives the potential energy at the height when the ball is held at the start of the test. This can be used as a check to for the Kinetic energy (KE) equation as the law of energy conversation states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant. So the PE when the ball is held at height will be the same as the KE just before the impact with the glass.

To calculate the KE use the equation

KE = ½mv2

(ii)

Using the KE equation from question (i)

KE = ½mv2

|u = initial velocity |

|v = final velocity |

|a = acceleration |

|s = distance |

Insert the constants of

Mass = 45kg

To calculate the v2 using the equation

v2 = u2 + 2 x a x s

For test 1

KE = ½mv2

For the v2

v2 = 02 + 2 x 9.8 x 305

Gives 5978

Insert in to KE = ½mv2 to give

KE = ½ x 45 x 5978

To give 134505 = 135J to 3 sf

To check use PE = mgh

45 x 9.8 x 305 = 134505

Round up to 3 sf to give 135 J

So PE =KE

135J is as given in BS 6206:1981.

For test 2

KE = ½mv2

For the v2

v2 = 02 + 2 x 9.8 x 457

Gives 8957.2

Insert in to KE = ½mv2 to give

KE = ½ x 45 x 8957.2

To give 201537 = 202J to 3 sf

To check use PE = mgh

45 x 9.8 x 457 = 201537

Round up to 3 sf to give 202 J

So PE =KE

202J is as given in BS 6206:1981.

For Test 3

KE = ½mv2

For the v2

v2 = 02 + 2 x 9.8 x 1219

Gives 23892.4

Insert in to KE = ½mv2 to give

KE = ½ x 45 x 23892.4

To give 537579 = 538J to 3 sf

To check use PE = mgh

45 x 9.8 x 1219 = 537579

Round up to 3 sf to give 538 J

So PE =KE

538J is as given in BS 6206:1981

(iii) The velocity that the impactor strikes the glass when it is dropped from a height of 1219 mm is calculated as above using:

v2 = u2 + 2 x a x s

v2 = 02 + 2 x 9.8 x 1219 = 23892.4

v = (23892.4= 154.6 m s-1

Question 2

(a)There are 3 main features of an invention to make it patentable. It has to have something new about it; this could be an improvement on an existing item. Also it must have a purpose (useful) and be able to be manufactured but this is not as important with today’s technologies as a software program can be patented.

(b) (i) The advantages of using a hollow shape for lintels are that a solid lintel is both heavy and cumbersome when it is in transit and when being manoeuvred into position at the build site. The added weight of the lintel will also require the supporting wall to be at a required strength to support the lintel and the load above it. They can be considered that they can be over designed for the job that they are...