# 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 35^{o}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, v

_{H}= 12.4 cos(36) (= 10.0ms^{-1})
Distance = v

_{H}t = 10.0 x 0.17 = 1.7m
(ii) Height h of the wall:

Vertical component of velocity, v

_{v}= 12.4 sin(36) (= 7.29ms^{-1})
Height h = (v

_{v}t + ½ at^{2}) = 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 = Î”V / Î”x

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 = I

^{2}R = E

^{2}R / (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 = [E

^{2}(R + r)

^{2}– 2E

^{2}R(R + r)] / (R + r)

^{4}

For maximum power P, dP/dR = 0

E

^{2}(R + r)

^{2}– 2E

^{2}R(R + r) = 0 {divide by E

^{2}(R + r) on both sides,}

(R + r) – 2R = 0

R = r

__Question 373: [Electric Field]__In the figure below, the point charge Q

_{1}causes an electric potential of 60 V and electric field strength of 30 Vm

^{-1}at P and the point charge Q

_{2}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 Q

_{1}and Q

_{2}.

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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