Sample solutions: IB

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Exam level: HI | Edition: 2017 | Season: S | Timezone: TZ2 | Number: 26 |
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E maximum potential = 1 /2 C x
If x = 1 /2 x E pot = 1/ 4 E maximum potential
Thus, for x = 1/ 2 x
Answer C is correct. | ||||

The question | ||||

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Exam level: HI | Edition: 2017 | Season: S | Timezone: TZ2 | Number: 7 |
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An alpha decay can be considered as an explosion. The momentum before the explosion is zero and due to the law of conservation of momentum the momentum after the explosion is zero as well.
Let u be the velocity of the nucleus after the decay.
0 = m nucleus x u + m alpha x v An alpha particle is built up by a helium nucleus and its mass is 4 atomic units. The nucleus after the decay has a mass (210 – 4) = 206 Thus, 0 = 206 x u + 4 x v
u = (-) 4 v ÷ 206
Answer D is correct. | ||||

The question | ||||

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Exam level: SL | Edition: 2017 | Season: S | Timezone: TZ1 | Number: 16 |
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If the intensity of the unporalized light = I The polarizing sheets are initially parallel and the intensity after sheet 2 is 1 /2 I However, by turning sheet 2 by an angle θ the intensity after sheet 2 becomes according to Malus’s Law 1/ 2 I To get an intensity after sheet 2 equal to 1 /4 I And cos θ = √1/ 2 = 1/2 √ 2 . And thus, the angle θ = cos It does not matter whether you turn sheet 2 and keep sheet 1 fixed or you turn sheet 1 and keep sheet 2 fixed.
Answer C is correct. | ||||

The question | ||||

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Exam level: SL | Edition: 2017 | Season: S | Timezone: TZ1 | Number: 3 |
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Apply the formula s = 1/ 2 g t The fourth second lasts from t = 3 s to t = 4 s. t = 3 seconds s = 1 /2 x 10 x (3) t = 4 seconds s = 1 /2 x 10 x (4) Distance travelled in the fourth second = 80 m - 45 m = 35 m
Answer C is correct. | ||||

The question | ||||

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