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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Monday, March 30, 2020

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

Saturday, March 28, 2020

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 (VA) is fixed at 3 V (as calculated in (a)(i)). It is connected to V-.
The potential at B (VB) depends on the temperature of the thermistor. It is connected to V+.}

At 10 °C, VA > VB, (since VB = 2.5 V as calculate in (a)(ii)).
The output voltage VOUT 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, VB would become equal to VA at some point. It is at this point that the polarity changes.}
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