Usual disclaimers. These are in no sense official and my contain typos and errors

OCR A G484 Newtonian World 29 June 2010

I thought this was a pretty easy paper - and easy to predict if you looked at what wasn't on the Jan 2010 paper

Q1 a) 1 and 3. Impulse is same / Total energy is conserved [2]

b) i) Loss in KE = Final KE - Initial KE = 1/2 x 0.06 x (12^2 - 9^2) = 1.89J [2]

ii) Force = change in momentum / time = 0.06 x (12 - (-9) ) / 0.15s = 8.4 N (watch for signs) [2]

iii) Newton 3rd law says same as ii) = 8,4N [1]

c) i) particles are in constant random motion with (MB distribution of velocities)

no intermolecular forces

vol of molecules is negligible compared to vol of gas

duration of collision is negligible compared to time between collisions

can apply newtonian mechanics

etc [3]

ii) Molecules collide elastically with wall and bounce off

Wall exerts force on molcules changing their momentum

so molcules exert force on wall (=change in momentum per unit time)

Pressure = total force / area

(pretty much a repeat of steps in b)

(p = 1/3 rho csquared bar) [3]

Total 13

Q2 a) i) Vertical component of L = weight

Horizontal component of L = centripetal force [2]

ii) F = mv^2/r 1.8E6 = 1.2E5 x V^2 / 2E3 so v = 1.73E2 ms-1 [2]

b) F = ma

so GMm/r^2 = mv^2/r cancel and sub v = 2 pi r / T then rearrange [3]

c) i) Period is 24h + orbit is over equator [2]

ii) r^3 = GMT^2/4pi^2 sub T = (24 x 60 x 60) gives r = 4.2E7m [3]

Total 12

Q3 a)1) acc is proportional to displacment

2) acc is directed towards a fixed point / is in opposite direction to displacment [2]

b) i) KE graph is zero at end s and max at x = 0 / reflection of PE graoph in line E = 9mJ [2]

ii) Total Energy is horizontal line through E=18mJ [1]

c) i) Amplitude = max displacment = 0.04m [1]

ii) KEmax = 18E-3J so use 1/2 mv^2 to get v=0.548ms-1 [2]

iii) vmax = 2 pi f A so f = 2.18Hz [2]

d) i) Useful eg MRI - RF pulse at Larmor frequency makes presessing proton move to higher energy state [2]

ii) problem : bridge oscillating with large amplitude when freq of

gusting wind / steps of walkers = natural freq of bridge [2]

Total 14

Q4 a) i) Brownian motion [1]

ii) Air molecules move randomly in all directions at high speeds [2]

b) i) PV = nRT so n = 0.602 mol so m= 1.68E-2 kg [4]

ii) 1) Thermal equilibrium means gains heat at same rate as loses it / constant temp [1]

2) m and V are constant so p/T is constant p = 2.6E5 * 273/290 = 2.45E5 Pa [2]

Total 10

Q5 a) i) KE = 1/2 mv^2 (again!) = 3.54E5 J [1]

ii) Work done = change in energy = Fd

so F = 3.54E5/40 = 8.84E3 N (or use uvast and F=ma)

assume : braking force is only force acting ie ignore other resistive forces like drag [3]

b) i) 1/4 energy loss at each brake = 8.85E4J

= mcdT so dT = 141K = 141 C (trap a change of temp is same in Kelvin or Celsius) [2]

ii) Front brakes do more work so will be higher temp at front and lower at back

Discs will lose heat due to forced convection so wont reach as high a temp

Drag also act so less force needed so lower temps. [4}

iii) Use larger disc / bigger mass

Use material with higher shc

Use more discs / twin discs at each wheel. [1]

Total 11

So there we have it. Most of my students doing resits seem to have found it pretty straightforward

(apart from one who didnt spot the last question!

)

Col

(edited typos)