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A2 Physics Unit 4 June 2014 Multichoice A2 Physics Unit 4 June 2014 Multichoice Mr D Powell

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Raw Marks.... A = B = C = D = E = U =

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D Ft = p Ft = mv-mu

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A (mv-mu) /F = t mv/F = t 584.6 x 10 -6

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B p = momentum = mv = m/V V = m p = ( V) * v v = 2 x 10 -4 m 3 / 7.2 x 10 -4 ms -1 = 0.277ms -1 1000kgm -3 x 2.0 x 10 -4 m 3 x 0.277ms -1 = 0.05555 kgms -1 = 0.056 kgms -1

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C /2 means it travels 90 per second. Hence, in 6 seconds it must have travelled 6 /2 = 3 = 1.5x the circle Hence displacement must be opposite to P so 1.6m

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D F = mv 2 /r acts towards centre

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B T = mv 2 /l - mg As force of mass on centre is mv 2 /l due to F = mv 2 /r However, F = ma = mg acts downwards against the mass

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C x = Acos(2 ft) x = Pcos(Qt) So A = P so answer must be B or C As Q = 2 f f = Q/2 C!

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D v = 2 f (A 2 -x 2 ) v = 2 fA (as x = 0) v = 2 x 320 x 0.5 = 320 mms -1

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A Damping reduces all activity!

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C F = Gm 1 m 2 /r 2 So we need to use d = r Hence B or C as 50/50. But as at midpoint effect of masses is due to the difference between M 1 and M 2 so you must subtract! M2M2 M1M1 d Midpoint m

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C Cannot be A (see graph all negative) B not true see graph R E D (V units are... Jkg -1 ) Must be C from grad

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B A ca be repulsive +/- or -/- or +/+ C - no as r 2 D if Q changes so does force B see formula...

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B d= 20mm, F = 0.5 d= Xmm, F = 1

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C See formula, the force depends in m, v, r and q, v, B So must be C

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C 12.5V 25V 8.33V Each field decreases as you move away from it. Two fields added together as 2m mark give 25V so each contributes 12.V each at this point. If you move away from P by 1m you go down to 8.33V, towards up to 25V. So if you add the 25V field from P 1m away to the 8.33V contribution from Q you get answer C

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D A true- crosses equi potentials B true not crossing equi potentials C work overall cancels out, some in and some out! Hence,D

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A Electric field is lower in Y so direction is X to Y, and direction a positive charge would move. Hence, electron will slow and decelerate in this field. (as it is negative) (F towards X) It gains +30eV of PE (as it is a negative charge) as it moves from X->Y field

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A Considering E-field... E = V/ d so.... E d = V Hence must be A or C Considering G field.... g = V/ r g r = V (Must be A)

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B Q=CV & E = 0.5CV 2 & E = 0.5QV Maths Method Quadratic.... E = 0.5CV 2 1600 x 10 -6 = 0.5CV 2 - eq 1 400 x 10 -6 = 0.5C(V-2) 2 - eq 2 Divide 1 eq by 2 to give quadratic and roots of 3V 2 16V + 16 = 0 Solve for V = 4 or V = 4/3 = 1.33V. Implied that V = 4V originally. Hence... Sub into eq.... E = 0.5CV 2 => 2E/V 2 = C = 1600 x 10 -6 x 2 / 4 2 = 200 x 10 -6 Quick Method... E V 2 So E results in (v/2) 2. So V must have halved from 4V to 2V. Sub into E = 0.5CV 2 => 1600 x 10 -6 x 2 / 4 2 = 200 x 10 -6 http://www.mathsisfun.com/quadratic-equation-solver.html

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B The half value period is 36ms. Hence, in this time the energy would fall to of this as E/4 = 0.5C(V/2) 2 In a second time period of 36ms the voltage will halve again so energy become 1/16 of original.

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B If Ber = p If e doubles and p doubles r is the same!

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D Capacitor of course as this is an electric field!

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D a faster moving magnet would induce a greater EMF & would pass through a vertical coil more rapidly Hence, D

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A = BAN sin( t) / = BANsin( t) When sin( t) =1 at maximum when coil within field at 90 / = BANsin( t) / 2 f = BAN = magnetic flux linkage

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C Two stage calc... 1 work out output V Work out % eff from P = VI