Monday, February 16, 2015

Physics 9702 Doubts | Help Page 67

  • Physics 9702 Doubts | Help Page 67


Question 370: [Kinematics > Projectile motion]
Ball is thrown from point P, which is at ground level, as illustrated. 

Initial velocity of the ball is 12.4ms-1 at angle 35o to the horizontal.
Ball just passes over a wall of height h. Ball reaches the wall 0.17s after it has been thrown.
(a) Assuming air resistance to be negligible, calculate
(i) Horizontal distance of point P from wall
(ii) Height h of the wall

(b) A second ball thrown from point P with same velocity as the ball in (a). For this ball, air resistance is not negligible.
Ball hits the wall and rebounds.
Sketch path of this ball between point P and the point where it first hits the ground.

Reference: Past Exam Paper – November 2010 Paper 22 Q2



Solution 370:
(a)
(i)
Horizontal component of velocity, vH = 12.4 cos(36)  (= 10.0ms-1)  
Distance = vHt = 10.0 x 0.17 = 1.7m                         

(ii) Height h of the wall:
Vertical component of velocity, vv = 12.4 sin(36)  (= 7.29ms-1)
Height h = (vvt + ½ at2) = 7.29(0.17) + ½ (-9.81)(0.17)2 = 1.1m

(b)
A smooth curve with the ball hitting the wall below the original
A smooth curve showing rebound to ground with correct reflection at wall










Question 371: [Electric Field]
The electric potentials V are measured at distances x from P along a line PQ. The results are
V / V               13                    15                    18                    21                    23
x / m                0.020               0.030               0.040               0.050               0.060

The component along PQ of the electric field for x = 0.040 m is approximately
A 75 Vm-1 towards P.
B 300 Vm-1 towards Q.
C 300 Vm-1 towards P.
D 450 Vm-1 towards Q.
E 450 Vm-1 towards P.

Reference: Past Exam Paper – J78 / II / 22



Solution 371:
Answer: C.
The electric potential at a point is defined as the work done per unit positive charge in moving a small test charge from infinity to the point. At infinity, the electric potential is defined to be zero.

So, the electric potential at infinity is zero. The potential V increases as we go away from P. Therefore, P is closer to infinity and hence, the component along PQ of the electric field is towards Q.

Since the potential changes with distance, component of electric field = ΔV / Δx

Consider point (x=0.040m, V=18V) along with EITHER (x=0.030m, V=15V) or (x=0.050m, V=21V)

Component of electric field = (18 – 15) / (0.040 – 0.030) = 3 / 0.01 = 300 Vm-1  









Question 372: [Current of Electricity > Internal resistance]
Source of electromotive force (e.m.f.) E has constant internal resistance r and is connected to external variable resistor of resistance R.
As R is increased from value below r to value above r, which statement is correct?
A The terminal potential difference remains constant.
B The current in the circuit increases.
C The e.m.f. of the source increases.
D The largest output power is obtained when R reaches r.

Reference: Past Exam Paper – June 2011 Paper 12 Q34



Solution 372:
Answer: D.
Since the battery has some internal resistance, the terminal potential difference of the battery (p.d. across its terminals – this is the p.d. available to the rest of the circuit) is less than the e.m.f E.

The voltage lost in the batter due to its internal resistance = Ir where I is the current in the circuit.
Current I = E / (R + r)
As R increases, the current I decreases. [B is incorrect]

Terminal pd, V = E – (Ir)
Since current I changes with the resistance R, the terminal p.d. does not remain constant as R is being changed. [A is incorrect]

The e.m.f. of the source is constant, it does not increase. [C is incorrect]

Output power in the load, P = I2R = E2R / (R + r)2
E and r are kept constant while R is being varied. The largest output power is obtained when R reaches r.

This can be proved by differentiating P (in the above equation) with respect to R and then equating to zero.
dP / dR = [E2(R + r)2 – 2E2R(R + r)] / (R + r)4

For maximum power P, dP/dR = 0
E2(R + r)2 – 2E2R(R + r) = 0               {divide by E2(R + r) on both sides,}
(R + r) – 2R = 0
R = r









Question 373: [Electric Field]
In the figure below, the point charge Q1 causes an electric potential of 60 V and electric field strength of 30 Vm-1 at P and the point charge Q2 separately causes a potential of 120 V and a field strength of 40 Vm-1 at P.

Which one of the following gives possible values of potential and field strength at P due to the joint action of Q1 and Q2.
            Potential / V                Field / Vm-1
A         180                              70
B         180                              50
C         135                              50
D         60                                10
E          135                              10

Reference: Past Exam Paper – J84 / II / 1



Solution 373:
Answer: B.
The electric potential is a scalar while the electric field strength is a vector (it has a direction).

So, the electric potential is obtained by simple addition.
Electric potential = 60 + 120 = 180 V            [C, D and E are incorrect]

The electric field strength is obtained by vector addition. Since the directions of the electric field strengths due to Q1 and Q2 are different, the field at P should be less than (30 + 40 =) 70 Vm-1. [A is incorrect]




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