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:
Go toThe 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]
For Q371 answer option is C but your explanation says its B. Can you please confirm this?
ReplyDeletethanks. the explanation has been updated.
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