Sunday, January 17, 2016

Physics 9702 Doubts | Help Page 232

  • Physics 9702 Doubts | Help Page 232

Question 1091: [Forces > Turning effect]
(a) Define the torque of a couple.

(b) A torque wrench is a type of spanner for tightening a nut and bolt to a particular torque, as illustrated in Fig. 3.1.

The wrench is put on the nut and a force is applied to the handle. A scale indicates the torque applied.
The wheel nuts on a particular car must be tightened to a torque of 130 N m. This is achieved by applying a force F to the wrench at a distance of 45 cm from its centre of rotation C. This force F may be applied at any angle θ to the axis of the handle, as shown in Fig. 3.1.

For the minimum value of F to achieve this torque,
(i) state the magnitude of the angle θ that should be used,
(ii) calculate the magnitude of F.

Reference: Past Exam Paper – June 2009 Paper 21 & 22 Q3

Solution 1091:
(a) The torque of a couple is defined as the product of (the magnitude of one) force and the perpendicular distance between the forces

(i) Angle θ = 90°
{From the definition of torque, we force being considered and the distance should be perpendicular to each other. When considering the distance of 45cm, the force that should be taken is the vertical component of force F, i.e. F sin θ.
For a minimum value of force F, sin θ should have a maximum value. sin θ changes from -1 to +1. sin θ is maximum when θ = 90°. That is, the direction of the force F is itself vertical in this case.}

130 = F × 0.45
Force, F = 290N

Question 1092: [Kinematics > Linear motion – Reaction time]
A car is travelling along a straight road at speed v. A hazard suddenly appears in front of the car. In the time interval between the hazard appearing and the brakes on the car coming into operation, the car moves forward a distance of 29.3 m. With the brakes applied, the front wheels of the car leave skid marks on the road that are 12.8 m long, as illustrated in Fig. 2.1.
 Fig. 2.1

It is estimated that, during the skid, the magnitude of the deceleration of the car is 0.85 g, where g is the acceleration of free fall.
(a) Determine
(i) the speed v of the car before the brakes are applied,
(ii) the time interval between the hazard appearing and the brakes being applied.

(b) The legal speed limit on the road is 60 km per hour.
Use both of your answers in (a) to comment on the standard of the driving of the car.

Reference: Past Exam Paper – November 2008 Paper 2 Q2

Solution 1092:
{We cannot use velocity = distance / time since there is a deceleration (speed is not constant) and we do not have data about the time the car takes to travel the distance of 29.3m.
Final speed, u = 0.  u2 = v2 +2as. 0 = v2 + 2(-0.85g)(12.8)}
v2 = 2as = 2 × (0.85×9.8) × 12.8
v = 14.6ms-1               

(ii) Time = (distance / speed =) 29.3 / 14.6 = 2.0s

EITHER 60kmh-1 = 16.7ms-1 OR 14.6ms-1 = 53kmh-1 OR 22.1ms-1 = 79.6kmh-1
So, the driving is within the speed limit but the reaction time is too long / too slow

Question 1093: [Current of Electricity > Kirchhoff’s laws]
(a) A student has been asked to make an electric heater. The heater is to be rated as 12 V 60W, and is to be constructed of wire of diameter 0.54 mm. The material of the wire has resistivity 4.9 × 10–7 Ωm.
(i) Show that the resistance of the heater will be 2.4Ω.
(ii) Calculate the length of wire required for the heater.

(b) Two cells of e.m.f. E1 and E2 are connected to resistors of resistance R1, R2 and R3 as shown in Fig. 7.1.

Fig. 7.1
The currents I1, I2 and I3 in the various parts of the circuit are as shown.
(i) Write down an expression relating I1, I2 and I3.
(ii) Use Kirchhoff’s second law to write down a relation between
1. E1, R1, R2, I1 and I3 for loop ABEFA,
2. E1, E2, R1, R3, I1 and I2 for loop ABCDEFA.

Reference: Past Exam Paper – November 2001 Paper 2 Q7

Solution 1093:
P = V2 / R                    OR P = VI and V = IR
R = V2 / P = 122 / 60
R = 2.4 Ω

Resistance R = ρL / A
2.4 = (4.9×10-7 × L) / {π × (0.27×10-3)2}
Length L = 1.12 m

(i) I1 + I2 = I3
{Consider junction B. From Kirchhoff’s law, the sum of current entering a junction is equal to the sum of current leaving the junction.}

1. E1 = I1R1 + I3R2
{Consider loop ABEFA. The sum of p.d. in a loop is equal to the e.m.f. in the loop.}

2. E1 – E2 = I1R1 – I2R3
{Consider loop ABCDEFA. The sum of p.d. in a loop is equal to the overall e.m.f. in the loop. But current flows from the positive terminal of a battery. Considering the loop, current I1 and I2 opposes each other.
Take the anticlockwise direction to be positive. Thus, E1 and I1 and positive and E2 and I2 are negative. Now the current leaving a battery is the same as the current entering the battery. So, the current flowing through R3 is I2.}

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