Tuesday, October 6, 2015

Physics 9702 Doubts | Help Page 207

  • Physics 9702 Doubts | Help Page 207



Question 996: [Dynamics > Momentum]
A wooden block is freely supported on brackets at a height of 4.0 m above the ground, as shown. 

A bullet of mass 5.0 g is shot vertically upwards into the wooden block of mass 95 g. It embeds itself in the block. The impact causes the block to rise above its supporting brackets.
The bullet hits the block with a velocity of 200 m s–1. How far above the ground will the block be at the maximum height of its path?
A 5.1 m                       B 5.6 m                       C 9.1 m                       D 9.6 m

Reference: Past Exam Paper – June 2015 Paper 13 Q13



Solution 996:
Answer: C.
From the conservation of momentum,
Sum of momentum before collision = Sum of momentum after collision

Momentum of block before collision = 0        [since it is not moving]
Momentum of bullet before collision = 5.0g × 200ms-1

The bullet embeds itself into the block upon the collision. Let their velocity be u.
Momentum after collision = (5.0g + 95g) × u = 100u

100u = 5 × 200
Velocity v = 1000 / 100 = 10ms-1

At the maximum height, the velocity of the block would be zero. The acceleration of free fall is downwards, opposite the upward motion of the block.

Consider the upward motion of the block:
Distance travelled = s. Initial velocity, u = 10ms-1. Final velocity, v = 0. Acceleration, a = -9.81ms-2.
Equation for uniformly accelerated motion: v2 = u2 + 2as
0 = 102 + 2×(-9.81)×s
Distance travelled, s = 100 / (2×9.81) = 5.1m

Height of block above the ground at maximum height = 4 + 5.1 = 9.1m












Question 997: [Applications > Communicating Information > Modulation]
(a) (i) Describe what is meant by frequency modulation.

(ii) A sinusoidal carrier wave has frequency 500 kHz and amplitude 6.0 V. It is to be frequency modulated by a sinusoidal wave of frequency 8 kHz and amplitude 1.5 V.
The frequency deviation of the carrier wave is 20 kHz V–1.
Describe, for the carrier wave, the variation (if any) of
1. the amplitude,
2. the frequency.

(b) State two reasons why the cost of FM broadcasting to a particular area is greater than that of AM broadcasting.

Reference: Past Exam Paper – June 2008 Paper 4 Q11



Solution 997:
(a) (i) For frequency modulation, the frequency of the carrier wave varies in synchrony with the displacement of the information signal.

(ii)
1. Zero (accept constant)
{The amplitude of the carrier wave does not vary.}

2.
{The carrier wave is to be frequency modulated by a sinusoidal wave of frequency 8 kHz and amplitude 1.5 V. Frequency deviation of carrier wave = 20 kHz V–1.
The amplitude of 1.5V could correspond to a frequency deviation of 1.5 × 20 = 30 kHz.}
Upper limit = 500 + 30 = 530 kHz
Lower limit = 500 – 30 = 470 kHz
The frequency changes from upper limit → lower limit → upper limit at 8000times s–1 {since the frequency of the sinusoidal wave is 8 kHz = 8000 Hz.}

(b) Choose any 2: e.g.
More radio stations required / shorter range
More complex electronics
Larger bandwidth required











Question 998: [Matter > Kinetic theory]
In a sample of gas at room temperature, five atoms have the following speeds:
1.32 × 103 m s–1
1.50 × 103 m s–1
1.46 × 103 m s–1
1.28 × 103 m s–1
1.64 × 103 m s–1.
For these five atoms, calculate, to three significant figures,
(a) the mean speed,

(b) the mean-square speed,

(c) the root-mean-square speed.

Reference: Past Exam Paper – June 2015 Paper 42 Q2



Solution 998:
(a) Mean speed = 1.44 × 103 m s–1
{Mean speed = (1.32 + 1.50 + 1.46 + 1.28 + 1.64) × 103 / 5 = 1.44 × 103 m s–1}

(b) We need to sum the individual squared speeds
Mean square speed = 2.09 × 106 m2 s–2
{Mean square speed = (1.322 + 1.502 + 1.462 + 1.282 + 1.642) × (103)2 / 5 = 2.09 × 106 m2 s–2 }

(c) Root-mean-square speed = 1.45 × 103 m s–1
{This is the square root of the mean square speed.}










Question 999: [Electric field]
A potential difference is applied between two metal plates that are not parallel.
Which diagram shows the electric field between the plates?


Reference: Past Exam Paper – June 2011 Paper 11 Q28 & Paper 13 Q29



Solution 999:
Answer: A.
Electric field strength, E = V / d
The closer the plates (the smaller the value of d), the greater is the electric field strength.

The electric field strength is represented by the spacing between the field lines. The close the field lines together, the greater is the electric field strength. Thus, the field lines should be closer at the bottom of the plates. [B and C are incorrect]

But the field lines should always be drawn perpendicular to the plates. Since the plates are inclined, joining the lines from both plates would result into curved field lines, instead of straight lines. [D is incorrect]



1 comment:

  1. God bless you, Admin. This website is such great help.

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