The diagram shows an arrangement to stop trains that are travelling too fast. Trains coming from the left travel at a speed of 50 m s-1.

Question 36

The diagram shows an arrangement to stop trains that are travelling too fast.

Trains coming from the left travel at a speed of 50 m s^-1. At marker 1, the driver must apply the brakes so that the train decelerates uniformly in order to pass marker 2 at no more than 10 m s^-1.

The train carries a detector that notes the times when the train passes each marker and will apply an emergency brake if the time between passing marker 1 and marker 2 is less than 20 s.

How far from marker 2 should marker 1 be placed?

A 200 m                      B 400 m                      C 500 m                      D 600 m

Reference: Past Exam Paper – June 2013 Paper 12 Q7

Solution:

Answer: D.

Initial speed u = 50 m s^-1

Final speed (max) v = 10 m s^-1

Minimum time for passing between marker 1 and 2: t = 20 s

{If time is less than 20 s, it means that the train is moving too fast. So an emergency brake is then applied.}

Since the train undergoes uniform deceleration, we can use the equations of uniformly accelerated motion.

Distance between markers = s = ???

s = ut + ½ at²

To obtain the acceleration a:

a = (v-u) / t = (10 – 50) / 20 = - 2 m s^-2  

Replacing a in the equation,

s = ut + ½at²

s = (50×20) + (½×-2×20)²

s = 600 m

A battery of e.m.f. 6.0 V and negligible internal resistance is connected to three resistors, each of resistance 2.0 kΩ, and a thermistor, as shown in Fig. 9.1.

Question 13

A battery of e.m.f. 6.0 V and negligible internal resistance is connected to three resistors, each of resistance 2.0 kΩ, and a thermistor, as shown in Fig. 9.1.

Fig. 9.1

The thermistor has resistance 2.8 kΩ at 10 °C and resistance 1.8 kΩ at 20 °C.

(a) Calculate the potential

(i) at point A, [1]

(ii) at point B for the thermistor at 10 °C, [2]

(iii) at point B for the thermistor at 20 °C. [1]

(b) The points A and B in Fig. 9.1 are connected to the inputs of an ideal operational amplifier (op-amp), as shown in Fig. 9.2.

Fig. 9.2

The thermistor is warmed from 10 °C to 20 °C.

State and explain the change in the output potential VOUT of the op-amp as the thermistor is warmed. [4]

Reference: Past Exam Paper – November 2015 Paper 41 & 42 Q9

Solution:

(a)

(i) (+) 3.0 V

{Since the resistors have the same resistance, the e.m.f. is equally divided.}

(ii)

{Potential divider equation: V1 = e.m.f. × (R1 / (R1+R2))

The lower resistors are connected to the negative terminal (0 V) of the battery. To find the potential at B, we need to consider the lower resistor.}

potential = 6.0 × {2.0 / (2.0 + 2.8)}

potential = 2.5 V                                 

(iii)

potential = 6.0 × {2.0 / (2.0 + 1.8)}

potential = 3.2 V

(b)

{Recall:

The potential at A (V_A) is fixed at 3 V (as calculated in (a)(i)). It is connected to V^-.

The potential at B (V_B) depends on the temperature of the thermistor. It is connected to V⁺.}

At 10 °C, VA > VB, (since V_B = 2.5 V as calculate in (a)(ii)).

The output voltage V_OUT is -9.0 V (since V^- > V⁺).

At 20 °C, the output voltage VOUT is +9.0 V (as V⁺ is now greater than V^-).

There is a sudden switch (from –9 V to +9 V) when VA = VB.

{As the temperature is increased, V_B would become equal to V_A at some point. It is at this point that the polarity changes.}