phy1 2003

6731 Unit Test PHY1

1. Magnitude of resultant force

4 cm line S / 1.7 cm line N 1

8 cm line NE / 8N resolved into two perp. components (5.7E & 1.7N or 5.7N) 1

Correct construction for vector sum 1

5.7 – 6.1 N 1

Name of physical quantities

Vectors 1

Two other examples

Any two named vectors other than force (if >2, must all be vectors) 1

2. Calculation of average velocity

Use of v = s/t 1

v = 1.86 m s -1 / 1.9 m s -1 1

Acceleration of trolley

Selecting u = u + 2as 1

Correct substitutions 1

2.87 m s /2.9 m s /3.0 m s-1 1

Tension in string

Use of F = ma 1

2.73 N / 2.76 N / 2.85 N 1

Assuming no action other horizontal force/table smooth/light string/inextensible string 1

Explanation

Suspended mass/system is accelerating 1

Idea of resultant force on the 0.4 kg mass 1

4. Addition to diagram

Downwards arrow Y through middle third of left leg 1

Downward arrow Z with correct line of action 1

[Ignore lengths of arrows and point of action] [Must have at least one correct label to get 2 marks; no labels gets max 1 out of 2] [One correct label can get 2 marks]

Explanation

Quality of written communication

Clockwise moments = Anticlockwise when balanced

Y is smaller than X but acts further from P

Moment of XP /Moment of Y = F x YP

Z has little or no moment about P/Z acts through P

Gravitational potential energy

Use of mgh 1

Vertical drop per second = (8.4 m) sin (3) 1

-.9 x 10 J/Js -1/W 1

What happens to this lost gpe

Becomes internal energy/used to do work against friction and or heat energy. 1[mention of K.E. loses the mark]

Estimate of rate at which cyclist does work

Rate of working = 2 x 3.9 x 10 W 1

=7.8 to 10 W 1

[3.9 x 10 W earns 1 out of 2]

6. Momentum and its unit

Momentum = mass x velocity 1

Kgms-1 or N s 1

Momentum of thorium nucleus before the decay

Zero 1

Speed of alpha particle/radium nucleus and directions of travel

Alpha particle because its mass is smaller/lighter 1

So higher speed for the same (magnitude of) momentum OR Newton 3rd Law argument 1

Opposite directions/along a line 1

Nuclear equation

Correct symbol and numbers for tin OR beta 1

Correct symbols and numbers for the other two 1

Decay constant

Use of lambda = 0.69/ half life 1~

1.57 x 10^-15 y-1 OR 4.99 x 10 s-1 1

Activity of source and comparison with normal background count

rate

Use of A = lambda N 1

0.11/0.12 (Bq) 1

Lower (than background) [Allow ecf – assume background = 0.3 to 0.5] 1

8. Radiation tests

Alpha:

Test 2 or 2 and 1 1

Count drops when alphas have been stopped by the air / alphas have a definite range / (only) alpha have a short range (in air) 1

Beta:

Test 3/3 and 1, because 1 mm aluminium stops (some) beta/does not stop any gamma rays 1

Gamma:

Test 4 or 4 and 1, because 5 mm aluminium will stop all the betas, (so there must be gamma too)/gamma can penetrate 5 mm of aluminium

Table

Target for Alpha scattering

Gold atoms/gold foil gold leaf/gold film/very thin sheets of gold/metal foil etc. [NOT thin gold sheet] 1

Target for Deep inelastic scattering Protons/neutrons/nucleons /liquid hydrogen/nuclei 1

Conclusions

(i) Atom mainly empty space/nucleus is very small 1

Nucleus dense/massive 1

(ii) Nucleons have a substructure 1

Made of quarks 1