Assessment Schedule 2017 Scholarship Physics (93103)

Transcription

1 Scholarhip Phyic (93103) 201 page 1 of 5 Aement Schedule 201 Scholarhip Phyic (93103) Evidence Statement Q Evidence 1-4 mark 5-6 mark -8 mark ONE (a)(i) Due to the motion of the ource, there are compreion and tretching of wavelength while the velocity remain contant. Thi lead to a change in frequency. The intantaneou change in frequency mean that the plane ha been aumed to pa directly through the oberver. fvw f a = v - v w = - v fvw f b = v + v w = + v ly (ly) Subtituting give = m 1 Component of the plane velocity at emiion point toward 113 oberver = vcof = 113 = 3.8 m 1 f = f v w = v w v ( 3.8) = Hz Vertical force on plane (after the drop) = 4000 g 3000g = 1000 g Vertical acceleration of plane = F m = 1000g 3000 = 1 3 g Vertical ditance moved by plane in 1.5 = ½ at 2 = = 3.68 m Vertical ditance fallen by pod = = m Vertical eparation = 14.2 m.

2 Scholarhip Phyic (93103) 201 page 2 of 5 Q Evidence 1-4 mark 5-6 mark -8 mark TWO (a) GPE lot = KE gained mgh = ½ mv 2 v 2 = 2gh Speed of 1 kg ma at colliion = 2gh = m 1 Speed of platform and ball after colliion uing conervation of momentum mv = 3mV (V = peed after colliion) V = v 3 = KE after = 1 2 mv 2 = = g J Total initial energy wa mgh = = 2g J Miing energy = Heat radiated = 2g ( 2 3 )g = 4 3 g =13.1 J x = initial compreion due to platform = mg k = = m 100 y = extra compreion due to addition of the falling 1kg ma. KE + Initial EPE + lo of GPE = Final EPE ½ mv 2 + ½kx 2 + mgy = ½k(x + y) y ly (ly) of phyic = 50( y + y 2 ) y = y + 50y 2 50y y 6.54 = 0 y = 9.81± ( ) ± = = 4.28 (ignore negative root) Extra compreion = y = m New equilibrium point i compreion of cm. Max diplacement from untretched length = cm Amplitude to 3.4 cm (d) By comparing the total energy at the bottom of the motion with the total energy at the top of the motion. Energy tored at bae = ½k(x + y) 2 = ( ) 2 = J Energy tored at top = mg h + ½ k = (2 0.34) = = J

3 Scholarhip Phyic (93103) 201 page 3 of 5 Q Evidence 1-4 mark 5-6 mark -8 mark THREE (a) hf = work function + E k E k = J = J hf = = J work function = J (ly) Diameter of the beam = m Area illuminated = πr 2 = m 2 Energy of a ingle photon = hf = hc λ = J The number of photon per econd = = The number of photon 1 m 2 = = Conervation of momentum tate that: h = h λ 2 + m e v ly Conervation of energy tate that: hf 1 = hf m e v2 and ince c = fλ h = h λ c m e v2 Finally 2hc = 1 2 m e v2 + m e vc For 4.00 kev electron the kinetic energy = J Uing E k = 1 2 mv 2 give v = m 1 2hc = = The final wavelength = m (d) Each electron come with it own proton. The 1 gram of hydrogen i almot entirely ingle proton, o there are about a many electron a there are proton. In all the other light element, half the ma i compoed of neutron o half the ma i proton, meaning half the number of electron in the 1 gram ma. The number of electron i only approximately half becaue both hydrogen and the light element have iotope with differing number of neutron which uually erve to reduce the number of proton (and hence the number of electron) in any ma. (But not in the cae of He3).

4 Scholarhip Phyic (93103) 201 page 4 of 5 Quetion Evidence 1-4 mark 5-6 mark -8 mark FOUR (a)(i) The phae angle i the ame for phaor diagram involving reactance and reitance a below. (ly) (i) The triangle formed with tanφ = X L X C R C increae; thi reduce the capacitive reactance a X i proportional to 1/C, but increae the overall reactance becaue X L i contant o X L -X c i larger, o from the above diagram the angle mut increae. V = IX = 1.5 X with V = 185 V So net X = ohm R = = ohm. tan 30 ly Capacitance ha doubled, a they are in parallel. The reactance (capacitive) ha halved. Since it lead the current X L > X C. The overall value i tan = 5.2 (iii) 1 ωc ω L = ω L 1 2ωC = 5.2 Solving thee imultaneouly give: C = F L = 0.69 H

5 Scholarhip Phyic (93103) 201 page 5 of 5 Quetion Evidence 1-4 mark 5-6 mark -8 mark FIVE (a)i) By conidering the force due to gravity acting down the board: of phyic. (ly) So the vertical acceleration component i ginj. At top the vertical velocity i zero. Setting a = ginj And howing that the initial vertical velocity i v 0 inq. Then uing the kinematic equation v f 2 = v i 2 + 2ad Rearranging provide the reult. The horizontal component of velocity i v 0 coq. The ditance, d will be travelled in time D t no force ly of phyic. of phyic. (iii) are acting thi i a contant velocity ituation Δt = d v v + at = 0 at top o Δt = v 0 inθ ginϕ of phyic. Uing the reult to the two previou quetion it can be hown that d v 0 coθ = 2v inθ 0 ginϕ Uing the trig identity provided give v 0 2 in2θ = gd inϕ Simple rearrangement give θ = 1 gd inϕ 2 in 1 2 v 0 Initial energy 1 2 Iω mv02 = 10 mv 2 0 Final energy at the top where v f i the horizontal velocity (ince it i unchanged) i 10 mv 2 f + mgδyinϕ v f = v 0 coθ Equate energie and rearrange give 10 mv 2 0 (1 co 2 θ) = mgδyinϕ uing trig identity give 10 mv 2 0 (in 2 θ) = mgδyinϕ ( ) 2 Therefore Δy = v 0 inθ 10 ginϕ (d)(i) Thi will reult in D y tending to infinity a imple conequence of Newton firt law. (d) More energy input due to additional rotational component.