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Sunday, January 26, 2020

Two cells with electromotive forces E1 and E2 and internal resistances r1 and r2 are connected to a resistor R as shown.


Question 38
Two cells with electromotive forces E1 and E2 and internal resistances r1 and r2 are connected to a resistor R as shown.



The terminal potential difference across cell 1 is zero.

Which expression gives the resistance of resistor R?






Reference: Past Exam Paper – November 2019 Paper 12 Q34





Solution:
Answer: A.


Let the current in the circuit be I.

From Kirchhoff’s law,
Sum of e.m.f. in the circuit = Sum of p.d. in the circuit
E1 + E2 = I (r1 + r2 + R)
R + r1 + r2 = (E1 + E2) / I                     eqn (1)

However, none of the options available contains the current I. So, we need to substitute I by other quantities.


Consider the cell 1.
e.m.f. of cell 1 = terminal p.d. + lost volts

It is given that the terminal p.d. is zero. So,
e.m.f. of cell 1 = lost volts
E1 = I × r1
Current I = E1 / r1


Replace the current I in equation (1),
R + r1 + r2 = (E1 + E2) × r1 / E1
R + r1 + r2 = E1r1/E1 + E2r1/E1
R + r1 + r2 = r1 + E2r1/E1
R = r1 + E2r1/E1 – r1 – r2
R = E2r1/E1 – r2

Note that r2 can be written as E1r2/E1 so as to obtain the same denominator.
R = E2r1/E1 – E1r2/E1
R = (E2r1 –1r2) / E1

Friday, January 24, 2020

Fig. 2.1 shows an object M on a slope. M moves up the slope, comes to rest at point Q and then moves back down the slope to point R.


Question 35
Fig. 2.1 shows an object M on a slope.


Fig. 2.1

M moves up the slope, comes to rest at point Q and then moves back down the slope to point R.

M has a constant acceleration of 3.0 m s-2 down the slope at all times.

At time t = 0, M is at point P and has a velocity of 3.6 m s-1 up the slope.

The total distance from P to Q and then to R is 6.0 m.

(a) Calculate, for the motion of M from P to Q,
(i) the time taken, [2]
(ii) the distance travelled. [1]


(b) Show that the speed of M at R is 4.8 m s-1. [2]


(c) On Fig. 2.2, draw the variation with time t of the velocity v of M for the motion P to Q to R.


Fig. 2.2
[3]


(d) The mass of M is 450 g.
Calculate the difference in the kinetic energy of M at P and at R. [2]





Reference: Past Exam Paper – November 2015 Paper 22 Q2





Solution:
(a)
(i)
v = u + at

{At Q, the object comes to rest, so v = 0.
The motion is up the slope while the acceleration is down the slope. So, the acceleration is negative as it opposes the motion.}
0 = 3.6 – 3.0t
t (= 3.6 / 3.0) = 1.2 s

(ii)
{Consider the motion from Q to P,
Average speed = distance / time
(3.6+0)/2 = s / 1.2
s = 3.6 × 1.2 / 2}
(distance to rest from P = (3.6 × 1.2) / 2 =) 2.2 (2.16) m

{Note that the simple formula
speed = distance / time
can only be used when the speed is constant (there is no acceleration).

The formula
Average speed = total distance (being considered) / time
can be used when there is a constant acceleration.}


(b)
{Total distance from P to Q and then to R = 6.0 m
Distance from Q to P = 2.16 m}
Distance (from P to R) = 6.0 – 2.16 (= 3.84)

{From P directly to R,}
v2 = u2 + 2as = 2 × 3.0 × 3.84 (= 23.04)
v2 = 3.62 + 2×1.68×3.0        (= 23.04)        
v = 4.8 m s-1                                                                    


(c)
straight line from v = 3.6 m s–1 to v = 0 at t = 1.2 s
straight line continues with the same gradient as v changes sign
straight line from v = 0 intercept to v = –4.8 m s-1

{The graph is a straight line passing through (0, 3.6) {at P} to (1.2, 0) {at Q}. The straight line continues up to v = - 4.8 m s-1 (at R).
It is a straight line as the acceleration remains the same throughout the motion. SO, the gradient is also constant.}


(d)
difference in KE = ½ m(v2 u2)

{Mass = 450 g = 0.45 kg}

difference in KE = 0.5 × 0.45 (4.82 – 3.62) [= 5.184 – 2.916]
difference in KE = 2.3 (2.27) J
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