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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_Ht = 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_vt + 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:

Go to
The electric potentials V are measured at distances x from P along a line PQ. The results are








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]