Download Newton`s Law of motion 2

Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Newton’s Law of motion 2
- Mechanics -

Newton’s 3rd Law – Action = Reaction
~ Inertia, Friction

Conservation of Momentum
一粒砂裏有一個世界，一朵花裏有一個天堂；
把無窮無盡握於手掌，永恆寧非剎那之間。
詩人 – 布雷克
- Mechanics – Newton’s Law of motion 2: page 1 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Newton’s Third Law of motion – action-and-reaction pair
A) Newton’s 3rd Law of motion – Action = reaction
Case 1: Roller-skater pushes against a wall
He is exerting a force on the wall,

meanwhile the wall exerts a force on
him.
Case 2: A sprinter at the start of a race
He pushed backwards on the starting

block, meanwhile the block pushing
him forward.
Case 3: (demo experiment)

When A alone pulls, both A and B move towards each other.

A exerts a force on B, while

B exerts a force on A in the opposite direction
Newton’s 3rd Law of motion:
For every action force there is an equal magnitude and opposite direction
reaction.
Principle of Water Rocket
1.
The rocket is partially filled with water and air is
pumped into it.
2.
When the trigger is pulled, the compressed air forces
the water downwards and the rocket lift-off.
3.
The compress air exerts a force on the water, and
4.
the water exerts an equal but opposite force on the
air, and hence the rocket
- Mechanics – Newton’s Law of motion 2: page 2 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
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B) Action and reaction pair: Common false example
A person sits on a chair inside a stationary lift. What force acting on the person and
which of these forces are ‘action and reaction’ pair.
Action & Reaction pair
Weight of a person
(a force acting on person due to the Earth)
Normal reaction act on the person by chair
Weight of chair
Normal reaction act on the chair by ground
Remark:

Between 2 bodies only (A act on B, and B act on A)
- Mechanics – Newton’s Law of motion 2: page 3 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
「小心思考：亞里士多德與牛頓」
牧羊童亞里士多德養了一頭驢子，亞里士多德稱呼驢子為牛頓。每天早上，
亞里士多德都帶牛頓上市場作買賣；牛頓也乖乖的拖普車子走，每天如是。唯一
有趣的，牛頓是一頭「好學」的驢子，特別時偏好物理學；每天晚上，牛頓總是
要念一會物理，做一會習題，才可以好好的睡一覺。
這天晚上，驢子牛頓正讀著力學的經典，當他念到第 3 定律時，竟然被他誤
出一個道理，就是這個想法，直叫他整夜也睡不著，一直開心到日出東方。亞里
士多德這一早也如常的把牛頓帶出來，將車子套在牛頓的脖子上。可惜的是，牛
頓竟再不願意拖著車子走了…
牛頓竟對亞里士多德這樣說：
「根據物理學的定律，我是不能把車子拉動的：
我用多大的力拉車子，車子就用多大的力拉我；無論如何，我也不能把車子拉動
了。」
這回慘了，請你幫一幫亞里士多德，將牛頓錯誤的思想糾正過來…
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- Mechanics – Newton’s Law of motion 2: page 4 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
97/09
Which of the following pairs of forces is/are action and reaction pair(s) according to Newton’s
third law of motion ?
(1) The weight of a man standing on a chair.
The gravitational force acting on the earth by
and
The force acting on the man by the chair.
and
The gravitational force acting on the moon by
(2)
the moon.
the earth.
The force exerted by a swimmer on the water
and
The force exerted by the water to push the
(3)
to push the water backward.
swimmer forward.
A. (1) only
B. (2) only
C. (1) and (3) only
D. (2) and (3) only
E. (1), (2) and (3)
01 / 09
A man is pulling a suitcase along the horizontal ground as shown below.
Which of the following pairs of forces is/are action and reaction pair(s) according to Newton’s third
law of motion ?
The gravitational force exerted by the earth and The gravitational force exerted by the man on
(1)
on the man
the earth
The pulling force exerted by the man on the and The friction exerted by the ground on the
(2)
suitcase
suitcase
The gravitational force exerted by the earth and The normal reaction exerted by the ground on
(3)
on the suitcase
the suitcase
A. (1) only
B. (3) only
D. (2) and (3) only
E. (1), (2) and (3)
C. (1) and (2) only
- Mechanics – Newton’s Law of motion 2: page 5 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
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Remark:
95/11
The above diagram shows a block resting on the ground.
Let
W be the weight of the block.
F
be the gravitational force acting on the earth by the block, and
R
be the force acting on the block by the ground.
Which of the following pairs of forces is/are action and reaction pair(s) according to Newton’s
third of motion ?
(1) R and W
(2) W and F
A. (1) only
B. (2) only
D. (1) and (2) only
E. (2) and (3) only
(3) F and R
C. (3) only
96/10
The above diagram shows a man lifting a ball vertically upwards with uniform acceleration.
Let
F1 be the force acting on the ball by the man.
F2 be the force acting on the man by the ball.
F3 be the gravitational force acting on the ball.
Which of the following correctly describes the relation between the magnitudes of the forces ?
A. F1 = F2 > F3
B. F1 = F3 > F2
D. F1 > F2 > F3
E. F1 > (F2 + F3)
C. F1 > F2 = F3
- Mechanics – Newton’s Law of motion 2: page 6 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
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Remark:
92/10
Which of the following pairs of forces F1 and F2 is/are action and reaction pair(s) according to
Newton’s third law of motion ?
A. (2) only
B. (3) only
D. (1) and (3) only
E. (1), (2) and (3)
C. (1) and (2) only
80 / 03
Two block A and B of masses m1 and m2 respectively are connected by a light spring on a horizontal
frictionless table.
The spring is stretched by moving the blocks apart.
What is the ratio of the
acceleration of A to that B at the moment when they are released ?
A. 1 : 1
B. m1 : m2
D. m12 : m22
E. m22 : m12
C. m2 : m1
00 / 02
It is said that Galileo Galilei (1564 – 1642), an Italian scientist, dropped a small iron ball and a large
cannon ball from the top of the Leaning Tower of Pisa.
He found that the two balls reached the
ground at almost the same time.
Which of the following is/are correct deduction(s) from this experiment ?
(1) The two balls fell with the same acceleration.
(2) A body will maintain uniform motion if there is no external force acting on it.
(3) The gravitational forces acting on the two balls were identical.
A. (1) only
B. (3) only
D. (2) and (3) only
E. (1), (2) and (3)
C. (1) and (2) only
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B)
Teacher notes
Remark:
Free body diagram (concern force act-on it only)
In the case with more than one body, use “free-body diagram” method to fix it.

Draw each object separately

Draw ALL forces act on each object

Use Newton’s 2nd Law of motion on each object

Concern “the forces” act on it ONLY.
Example 1
(a)
Two blocks of mass 1 kg and 3 kg are tied together on a smooth horizontal table.
The system is pulled by a horizontal force P of 16 N as shown in Figure.
(b)
(i)
Find the acceleration of the system.
(ii)
Find the tension in the string connecting the blocks.
The horizontal force P is applied to the 3 kg mass instead of the 1 kg mass as
shown in Figure.
(i)
Find the acceleration of the system.
(ii)
Find the tension in the string
connecting the blocks.
(c)
Explain, in terms of inertia, the difference in the tensions of the string in (a)
and (b).
- Mechanics – Newton’s Law of motion 2: page 8 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Example 2
Two identical blocks, A and B, each of mass 2 kg are connected by a light string which
passes over a smooth pulley as shown in Figure.
Initially, block B rests on the floor
and block A is 4 m above the floor.
(a)
A mass of 1 kg is then placed on block A. Find
(i) the acceleration of the system.
(ii) the time taken for block A to reach the floor.
(iii) the velocity of the system just before block A hits
the floor.
(b)
After block A has reached the floor,
(i) what is the resultant force acting on block A ?
(ii) what is the resultant force acting on block B ?
(iii) what is the maximum height reached by block B ?
- Mechanics – Newton’s Law of motion 2: page 9 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Example 3
Block A and B are connected by string as shown in figure.
Mass of A and B are 50 kg and 70 kg respectively.
(a)
If
all
surfaces
are
smooth,
what
is
the
acceleration of the blocks A and B ?
(b)
If there is friction of 30 N between the
surfaces and the blocks, what is the acceleration
of A and B now ?
Example 4
A block A of mass 50 kg is placed as shown in figure and
connected by string to another block B.
The friction
between block A and the surface is 20 N.
When A
descends down the slope with acceleration of 3 ms-2, what
is the mass of B ?
- Mechanics – Newton’s Law of motion 2: page 10 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Example 5
Two scale pans each of mass 0.1 kg are attached to a thread which passes over a
smooth pulley.
A mass of 0.5 kg is placed on one pan and a mass of 0.2 kg is placed
on the other.
(a)
Find the acceleration of the system.
(b)
Find the tension in the thread.
(c)
Find the force each mass exerted on the scale pan.
- Mechanics – Newton’s Law of motion 2: page 11 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Example 6
A man of mass 60 kg stands on a wooden platform of mass 20 kg which is suspended
by a string as shown in Figure.
The string passes over a fixed smooth pulley.
One
end of the string is pulled by the man and the platform rises with an acceleration of 2
ms –2.
(a)
Find the tension in the string.
(b)
Find the pull of the man on the string.
(c)
Suppose the man stands on a weighing machine
which is placed on the platform, what is the
reading on the machine ?
- Mechanics – Newton’s Law of motion 2: page 12 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
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Remark:
91 / 02
Two blocks of equal mass are placed on a smooth horizontal surface as shown above.
A constant
force of 12 N is applied to block A so that the two blocks move towards the right together.
The
force acting on A by B is
A. zero.
B. 6 N to the left.
D. 12 N to the left.
E. 12 N to the right.
C. 6 N to the right.
94 / 04
Two blocks X and Y of weight 2 N and 8 N respectively are suspended by two light strings as
shown above.
A downward force of A applied to X.
Find the tensions T 1 and T2 in the two
strings.
T1
T2
A.
2N
8N
B.
4N
10 N
C.
4N
14 N
D.
6N
12 N
E.
6N
14 N
- Mechanics – Newton’s Law of motion 2: page 13 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
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Remark:
95 / 02
In the above diagram, the two blocks are connected by a light string S.
-2
acceleration of 2 m s under the action of a constant force F.
They move with uniform
What will the accelerations of the
blocks become if S suddenly breaks ?
2 kg block
4 kg block
A.
6 m s-2
0 m s-2
B.
6 m s-2
2 m s-2
C.
2 m s-2
0 m s-2
D.
2 m s-2
2 m s-2
E.
0 m s-2
3 m s-2
92 / 09
In the above diagram, block A and B are connected by a light inextensible string and rest on a
smooth horizontal table.
The masses of A and B are 2 kg and 3 kg respectively.
pulled by a force of 2 N.
Find the tension in the string S.
A. 0.4 N
B. 0.8 N
C. 1.0 N
D. 1.2 N
Block A is
E. 2.0 N
89 / 03
Three blocks of equal mass are placed on a smooth horizontal surface as shown.
A constant force
F is applied to block A so that the three blocks move towards the right with the same acceleration.
The net force acting on block B is
A. 0
B. F/3
C. F/2
D. 2F/3
E. F
- Mechanics – Newton’s Law of motion 2: page 14 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
80 / 08
In the figure above, X and Y are blocks of mass 1 kg and 2 kg respectively.
S is a spring balance
of negligible mass and P is a smooth pulley fixed at the top of two smooth inclined planes.
What
is the reading of S when X is held stationary ?
A. 5 N
B. 10 N
C. 15 N
D. 20 N
E. 30 N
- Mechanics – Newton’s Law of motion 2: page 15 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Conservation of Momentum
< Group Magic Experiment >
Object

to determine the “momentum change” in the INELASTIC collision
Apparatus & Set-up

A motion sensor and 2 PASCar were used
Procedure

Motion sensor is set to survey the velocity of 1st PASCar, m1 = _______ kg

2nd PASCar is set stationary in the middle of the 1.2 m track

push the 1st PASCar collides the 2nd PASCar with inelastic

Repeat the experiment in difference masses of the 1st PASCars
Result & Analysis

[Def.] momentum (p) = mass (m)  velocity (v)

momentum is a vector, and Unit is ___________ (kg ms-1)

Before collision: m2 = ______ kg, and p2 = m2u2 = ( ____ ) (0) = ______ kg ms-1
Before collision
m1
u1
After collision
p1 = m1u1
mt = m1+m2
v
pf= mtv
Case 1
Case 2
Case 3
Case 4
Discussion & Improvement

Total momentum before (pi = p1 + p2) & total momentum after (pf)

Compare total momentum before and after collision: pi ___ pf
Summary

Total momentum before and after collision is conserved
- Mechanics – Newton’s Law of motion 2: page 16 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Collision
Vehicles exert force on one another on impact. The force of impact causes damage
to the vehicles and changes their velocities (both magnitude and direction).
A) Linear Momentum (p)
Greater Damage by:

greater mass (m), and

faster / greater velocity (v)
[Def.]
momentum (p) =
mass (m)  velocity (v)
Unit: kg ms-1
Example 1
A table-tennis ball of mass 0.01 kg travelling at 20 m s -1 hits a wall and bounces off
with the same velocity. Find
(a) the initial momentum of the ball
(b) the final momentum of the ball, and
(c) the change of momentum of the ball
What kind of collision it is? Elastic or Inelastic?
Elastic collision: separate after the collision (difference velocity)
Inelastic collision: stick together after collision (same velocity)
RReem
maarrkk:: PPhhyyssiiccaall m
meeaanniinngg ooff ““m
moom
meennttuum
m””
In our textbook (Physics Today) said: “the amount of MOTION of an object
depends on both its mass and velocity”. I am not total agree with him. In my opinion,
Momentum is a physical quantity by mathematical operation: mass  velocity. There
is no direct or indirect physical meaning. However, it is a useful quantity, especial in
the case of collision.
- Mechanics – Newton’s Law of motion 2: page 17 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Example 2
Remark:
(momentum)
Find the momentum of :
(a) a car with 1000 kg travelling at 70 km h-1
(b) a man about 60 kg running at 30 km h-1
(c) an oil tanker with mass 107 kg sailing at 10 kmh-1
(d) a 0.1 kg golf ball travelling 100 km h-1
87 / 02
89 / 07
Which of the following is / are the correct unit(s) for momentum?
(1) kg m s-1
(2) kg m s-2
(3) N s
A. (1) only
B. (2) only
D. (1) and (3) only
E. (2) and (3) only
C. (3) only
- Mechanics – Newton’s Law of motion 2: page 18 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
< Optional Demo Magic Experiment >
Object

to study the force during the period collision
Apparatus & Set-up

A motion sensor, PASCar and a force sensor were used
Procedure

Motion sensor is set to survey the velocity of PASCar, with a force sensor

Push the PASCar to crash the wall

Plot the force & velocity against time
Result & Analysis

Mass of the PASCar with force sensor (m) : __________ kg

velocity before collision (u) = ______ m s-1, velocity after collision (v) = ______ m s-1

Momentum change =  p = m v = m ( v - u ) = ( ___ ) ( ____ - ____ )= ______ kg ms-1

Print out the graph of “FORCE against TIME”, and draw a copy below
Discussion & Improvement

Compare the unit of momentum & are of force-time graph

Unit of momentum:________ , unit of area of force-time graph: ________
Summary

Area of Force-time graph = change of momentum
- Mechanics – Newton’s Law of motion 2: page 19 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
B) Force and Impulse
Impulse is a continue varies force during the time of collision. F(t)t.
[Def.]
Impulse of the force =
Area of the shallow of the “force-time graph” during the collision
Unit = _________
Average force, Fave
= The average value of force
during collision
= (Area of shallow) / t
where: t is contact time
Mathematical device:
Start from: Newton’s 2nd Law of motion
Fave = m a = m (v – u) / t
<< Impulse (Area of shallow) = change of momentum (p) >>
Newton’s 3rd Law of motion is valid during the collision, that mean:
A continue force acts on A by B during the collision (in t), meanwhile
a NEGATIVE continue force acts on B by A at the same time.
- Mechanics – Newton’s Law of motion 2: page 20 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Example 3
Remark:
(contact time)
A 0.2 kg tennis ball travelling at 10 m s-1 It shit by a racket and returns at 20 m s-1.
The ball is in contact with the racket for a time of 0.1 s. Find
(a) the change in momentum of the ball ;
(b) the average force exerted on the ball by the racket ; and
(c) the average force exerted on the racket by the ball.
89 / 08
Bullets, each of mass m, are fired at the rate of n bullets per second. They hit a vertical wall with
horizontal speed v and rebound from the wall with the same horizontal speed v. Which of the
following statements is / are correct?
(1) The total change in momentum of the bullets is zero.
(2) The total change in momentum of the bullets in one second is 2 mnv.
(3) The average force exerted on wall is 2 mnv.
A. (1) only
B. (2) only
D. (1) and (3) only
E. (2) and (3) only
C. (3) only
- Mechanics – Newton’s Law of motion 2: page 21 All rights reserved. No part of this publication could be reproduced,
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Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
C) Conservation of momentum
Consider a point mass collides another point mass.
Case 1: Inelastic
For m1 : FN = Fave =
For m2 : FN = -Fave =
Case 2: Elastic
In conclusion:
total momentum before collision = total momentum after collision
m1u1 + m2u2
or
= (m1 + m2) v
inelastic case,
= m1v1 + m2v2
elastic case
This is called: “Conservation of momentum” (動量守恆)
- Mechanics – Newton’s Law of motion 2: page 22 All rights reserved. No part of this publication could be reproduced,
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- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Example 4
Remark:
(conservation of momentum)
A 2000 kg car travelling to the right at 15 m s-1 collides and couples with a 1000 kg
car initially to left at 20 m s-1. What is the velocity of the pair just after the
collision ?
Example 5
(inelastic collision)
A 1.5 kg trolley carrying a lump of plasticine rests on a level frictionless track. A
pellet of mass 0.002 kg is fired from an air rifle into the plasticine along the track
and becomes embedded in it.
After impact, the trolley moves forwards at a velocity of 0.4 m s-1.
(a) Suggest a method to measure the velocity of the trolley.
(b) Use the conservation of momentum to calculate the velocity of the pellet.
- Mechanics – Newton’s Law of motion 2: page 23 All rights reserved. No part of this publication could be reproduced,
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- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Example 6
Remark:
(Elastic collision)
Two bodies of mass m1 and m2 undergo a head-on collision, their velocities before
and after collision are shown.
(a) Determine the ratio m1 and m2.
(b) If the collision is completely inelastic, find the common velocity
Example 7
(explosion)
Find the recoil velocity of a rifle of mass 5 kg after it shoots a 50 g bullet at a
speed of 300 m s-1.
- Mechanics – Newton’s Law of motion 2: page 24 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Example 8
Water leaves the fireman’s hose at a rate of 7.2 kg s-1 and with a velocity of 21 m s-1.
Calculate the force exerted by this water jet on the wall of a building. What
assumption have you made in the calculation?
Example 9
A rocket consumes 100 kg of fuel per second and the hot gas is ejected with a
velocity of 2800 m s-1, relative to the rocket. Calculate the forward thrust
experienced by the rocket at that moment.
- Mechanics – Newton’s Law of motion 2: page 25 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Example 9
Remark:
(Advanced)
Two blocks A and B of masses 1 kg an d3 kg respectively are connected by a loose,
long string. B is at rest and A is moving with a velocity of 8 m s-1 just before the
string becomes taut.
(a) What is the common velocity of the blocks after the string becomes taut ?
(b) The blocks then subjected to the same resisting force 5 N and are gradually
brought to rest.
(i) What is the time taken for each block to come to rest?
(ii) What is the distance travelled by each block before coming to rest?
(iii) If the length of the string is shortened to 0.7 m, would block B collide with
block A before it comes to rest?
- Mechanics – Newton’s Law of motion 2: page 26 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
91/11
Two particles A and B of masses 2 kg and 1 kg respectively move in opposite directions.
The
initial velocity of A is 4 ms-1 towards the right, while that of B is 2 ms-1 towards the left.
They
collide head on.
-1
After the collision, the velocity of A becomes 1 ms towards the RIGHT.
What would be the velocity of particle B ?
A. 2 ms-1 towards the right
B. 3 ms-1 towards the right
C. 4 ms-1 towards the right
C. 6 ms-1 towards the right
E. 7 ms-1 towards the right
90 / 03
A trolley moves with constant speed along a horizontal surface. A lump of plasticine having the
same mass as the trolley is dropped onto the trolley and sticks to it. Which one of the following
ticket-tapes best represents the motion of the trolley?
80 / 06
Two objects P and Q of mass 2 kg and 3 kg respectively have the same momentum. They are then
subjected to the same constant resisting force and gradually brought to rest. What is the ratio of the
stopping distance of P to that of Q.
A. 4 : 9
B. 2 : 3
C. 1 : 1
D. 3 : 2
E. 9 : 4
- Mechanics – Newton’s Law of motion 2: page 27 All rights reserved. No part of this publication could be reproduced,
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- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Structure Type Question
Question 1
Two trolleys of mass 1 kg and 2 kg separate explosively. The 1 kg trolley moves with
a constant velocity of 2.4 m s-1 and collides with a wall 3 m away as shown.
(a) What is the velocity of the 2 kg trolley after the explosion?
(b) What is the time taken for the 1 kg trolley to reach the wall?
(c) If the average force acting on the 1 kg trolley by the wall is 80 N and it lasts
for 0.05 s, calculate the velocity of the trolley on the rebound.
(d) Will the 1 kg trolley collide with the 2 kg block for a second time?
- Mechanics – Newton’s Law of motion 2: page 28 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Question 2
A boy of mass 60 kg is stepping onto the bank
from a boat of mass 200 kg.
(a) Initially both the boy and the boat are
stationary, what are their total momentum?
(b) The boy moves with a velocity of 3.3 m s-1 onto the bank.
(i)
Calculate the momentum of the boy.
(ii)
Find the recoil velocity of the boat.
(c) (i)
(ii)
A rope is always used to tie the boat to the bank. Explain briefly.
Suppose the boy leaves the boat in 0.5 second, what is the tension in the
rope?
- Mechanics – Newton’s Law of motion 2: page 29 All rights reserved. No part of this publication could be reproduced,
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- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
(00/04)
A car of mass 1000 kg moves along a straight road with a speed 10 m s -1.
of mass 3000 kg, which is initially at rest.
-1
forward with a speed 4.5 m s .
It collides with a lorry
Immediately after the collision, the lorry moves
The time of contact of the car and the lorry is 0.5 s.
(a)
the speed of the car immediately after the collision,
(b)
the average force acting on the lorry during the collision,
(c)
the average force acting on the car during the collision.
Find
(5 marks)
- Mechanics – Newton’s Law of motion 2: page 30 All rights reserved. No part of this publication could be reproduced,
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- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
Question 3
A fire breaks out in the middle part of a ball building. Some residents go to the top
of the building and jump down to a safety net opened by the firemen on the ground
(a) The mass of a certain resident is 60 kg and the
height of the building is 100 m.
(i)
Find the speed of the resident at the
moment when he reaches the safety net.
(ii)
Suppose the resident comes to rest after
reaching the safety net, calculate the
change in momentum of him.
(iii) Owing to the elasticity of the safety net,
the time of contact between the residents
and the net is 2 seconds. Find the force
acting on the resident.
(b) If the resident lands on the ground, the time of contact is only 0.04 second.
Find the force acting on the resident.
(c) Explain why a safety net is used.
- Mechanics – Newton’s Law of motion 2: page 31 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.
Chinese YMCA College
Teacher: Joey Fung, chi-lam
Date:
Teacher notes
Remark:
- Mechanics – Newton’s Law of motion 2: page 32 All rights reserved. No part of this publication could be reproduced,
or transmitted, in any form without prior permission of the author.
- © 17/05/06, Joey’s Library (Publishing) Co. Ltd.