An object hangs by means of two cords around two rods, as shown. The object is held in equilibrium by the forces F1 and F2.

Question 1

An object hangs by means of two cords around two rods, as shown.

The object is held in equilibrium by the forces F1 and F2. The object weighs 10 N. There is negligible friction between the rods and cords. Angle θ is 90°.

Which row of the table gives an angle θ of 90°?

F1 / N               F2 / N

A          4.0                   6.0

B          6.0                   4.0

C         6.0                   8.0

D         8.0                   6.0

Reference: Past Exam Paper – March 2017 Paper 12 Q14

Solution:

Answer: C.

The 2 forces F₁ and F₂ will also be along the cords.

Since the object is held in equilibrium, the resultant of the forces F₁ and F₂ should be equal to the weight of the object (= 10N).

The resultant of 3 vectors can be obtained by the triangle method.

From the question, we know that the angle between F₁ and F₂ is 90°. So, the triangle formed is actually a right-angled triangle. Also, the weight of the object is taken as the hypotenuse (since the other 2 forces formed the right angle).

From Pythagoras’ theorem,

F₁² + F₂² = 10²

It can be seen that both choices C and D apply to this equation.

But, now, let’s say that both forces were of the same magnitude, the object would have been midway between the 2 rods. But since the object is closer to the rod on the right, the force should be greater there.

So, F₂ > F₁.

The power P required to move an object through a medium at constant speed depends on the speed v and the resistive force F acting on the object.

Question 2

The power P required to move an object through a medium at constant speed depends on the speed v and the resistive force F acting on the object.

The resistive force F also depends on the speed v.

Which row shows a possible relationship between speed v, resistive force F and power P?

resistive force F                      power P

A          proportional to v                      constant

B          proportional to v                      proportional to v

C         proportional to v²                           proportional to v²

D         proportional to v²                           proportional to v³

Reference: Past Exam Paper – June 2016 Paper 11 Q17

Solution:

Answer: D.

We know that

Power = Work done / time = F × d / t = F × (d/t) = F × v

P = Fv

Now, the resistive force also depends on speed v. (But we do not know to which ‘power’; e.g. v or v² or v³ or v⁴ or … ???)

But, whatever the relationship between resistive force F and speed v, power P = Fv. So, if

Resistive force F ∝ vⁿ

Power P = Fv and is proportional to (vⁿ × v =) vⁿ⁺¹

That is, the exponent of v for power will be more than that for the force by 1.

Radon 222 86 Rn is the start of a decay chain that forms bismuth 214 83 Bi by alpha and beta emission.

Question 2

Radon ²²²₈₆Rn is the start of a decay chain that forms bismuth ²¹⁴₈₃Bi by alpha and beta emission.

For the decay of each nucleus of radon, how many α−particles and β−particles are emitted?

α−particles     β−particles

A                      1                      1

B                      2                      1

C                     1                      2

D                     2                      2

Reference: Past Exam Paper – June 2012 Paper 12 Q40

Solution:

Answer: B.

α-particle = ⁴₂He and β-particle = ⁰_-1e

Let ‘m’ be the number of α-particles emitted and ‘n’ be the number of β-particles emitted.

²²²₈₆Rn        -->       ²¹⁴₈₃Bi +          m ⁴₂He            +          n  ⁰_-1e

The mass number and proton number should be conserved.

Mass number: 222 = 214 + 4m + 0n

4m = 8 giving m = 2    (so, 2 α-particles emitted)

Proton number: 86 = 83 + 2m – 1n

2m – n = 3

n = 4 – 3 = 1                since m = 2 (from above)

So, 1 β-particle is emitted.