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.

The graph shows the vertical velocity of a parachutist during the first 20 s of her jump.

Question 1

The graph shows the vertical velocity of a parachutist during the first 20 s of her jump.

Approximately how far does she fall before opening the parachute?

A 390 m                      B 570 m                      C 710 m                      D 770 m

Reference: Past Exam Paper – November 2016 Paper 12 Q8

Solution:

Answer: B.

Opening the parachute causes a large decrease in velocity. From the graph, the approximate time at which the parachute is opened is t = 13s (the moment BEFORE the large deceleration).

Distance travelled = area under v-t graph. So, we need to find the area from t = 0s to t = 13s. This region can be approximated as follows:

Area of triangle from t=0 to t=5: distance = ½ × 5 × 44 = 110 m

Area of trapezium from t=5 to t=10: distance = ½ × (44+60) × 5 = 260 m

Area of rectangle from t=10 to t=13: distance = 60 × 3 = 180 m

Total distance travelled = 110 + 260 + 180 = 550 m