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8 | F2. Force and Newton’s Laws
Solution Guide (3rd Ed.)
F2. Force and Newton’s Laws
Multiple-choice questions
1.
Susan is holding…
A When the balloon is released, from stationary
to moving ⇒ acceleration ⇒ subjected to an
unbalanced force. So, (2) and (3) are incorrect.
2.
On a horizontal snow surface…
Uniform speed ⇒ net force = zero. With F
and at uniform speed ⇒ a resistant force. So,
without F , the sled will come to a stop.
B
6.
A
A block is suspended…
At rest ⇒ mg is balanced by the pulling force.
Only the net force on the block is zero among the
three options. So, only (3) is correct.
B
4.
A wheel…
Although the whole wheel is moving forwards, the contact point on the wheel tends to
slide backwards. To prevent the sliding, friction is
acting forwards.
A
Const. v ⇒ net force on Susan is zero.
In fact, Susan is only acted by the gravitational force and
the normal reaction from the passenger conveyor.
Note the term uniform speed.
3.
Susan is standing…
7.
A horizontal force…
Before the block moves, the frictional force
is equal to F in magnitude. The net force on the
block is zero.
C
8.
A shuttlecock…
D Consider the free body diagrams below,
where f is the air resistance.
Friction opposes the relative motion of the two contacting
surfaces. Note the contact points when dealing with friction.
5.
Train yourself with appendix exercise: Drawing free body
diagrams, if you find it difficult to draw them.
A note is stuck…
The gravitational force is balanced by the
frictional forces on the two sides of the note. All
these two forces act in the vertical direction.
A
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9.
A steel bead…
B The spacing between successive images increases and then becomes steady. This implies
that the speed of the bead is getting higher, but at
a decreasing rate.
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F2. Force and Newton’s Laws | 9
Solution Guide (3rd Ed.)
10.
For the previous question…
18.
11.
Bob bends his knees…
A
19.
C
0 − 22 = 2 × a × 10
When tackling action-reaction pairs, identify the two objects
involved first.
Note that action and reaction must act on different objects.
So, m = F /a = (−1)/(−0.2) = 5 kg.
20.
A train…
Applying s = ut + 12 at 2 , the acceleration is
given by
B
A block of mass…
B
14.
∴ a = −0.2 m s−2
A hockey puck…
B
13.
A sled initially…
Applying v 2 − u 2 = 2as , the acceleration is
given by
The force acting on Bob by the ground and the
force acting on the ground by Bob form an actionreaction pair.
A
12.
A Porsche…
The car reaches the inal speed v =
100 km h−1 = 100/3.6 m s−1 in 3.70 s. So, average
net force F = ma = 1200 × (v/3.70) ≈ 9010 N.
Force of gravity is const., while luid friction
increases with speed. The steel bead reaches terminal v when the two forces are equal.
C
810 = 0 +
Acceleration a = F /m = 2/2 = 1 m s−2 .
∴ a = 0.4986 m s−2
A toy compartment…
C
1
× a × 572
2
Consider the free body diagram:
So, F = 320 000 × 0.4986 = 1.60 × 105 N.
21.
A small disc…
C
Applying v 2 − u 2 = 2as ,
(
)
0.01mg
0−u = 2× −
× 15
m
2
Note that T − f = ma . So, T = ma + f = 0.2 × 0.5 +
0.1 = 0.2 N.
15.
∴ u = 1.716 ≈ 1.72 m s−1
22.
C
On a smooth horizontal ground…
Before S 1 breaks, net force F = 20 N. So,
m = F /a = 20 kg.
23.
After S 1 breaks, net force becomes 80 N. So,
a ′ = 80/20 = 4 m s−2 .
C Before S 1 breaks, net force F = 10 × 1 = 10 N.
So, there is friction f = 10 N (to the left).
After S 1 breaks, net force = 80 − f = 70 N. So,
a = 70/10 = 7 m s−2 .
1
3
:
1
6
= 2 : 1.
Michael applies…
C The force exerted on the clock by Michael and
the force exerted on Michael by the clock forms
an action-reaction pair. So, F = 300 N.
See the next question.
On a horizontal ground…
From a = (v − u)/t and F = ma ,
F1 : F2 = a1 : a2 =
A
16.
A block…
24.
In the above question…
B
25.
f = 300 − 50 × 0.2 = 290 N.
At the instant shown…
D
Consider the following free body diagram.
See the previous question.
17.
A force…
C
F is const. ⇒ a is const. From s = ut + 12 at 2 , if
u = 0, s ∝ t 2 .
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10 | F2. Force and Newton’s Laws
Solution Guide (3rd Ed.)
The force on the gymnast is given by
31.
Apply net F = ma to the whole system. So,
a = 5/(2 + 3) = 1 m s−2 . Apply net F = ma to block
A . So, T = 2 × 1 = 2 N.
A
50 × 6 = F − 50 × 9.81
∴ F = 791 N
26.
Blocks A and B are connected…
Formally:
A girl is hugging…
The girl and her teddy bear have the same
speed and acceleration. But being more massive,
the girl is exerted by a larger net force.
{
T = mAa
B
27.
Solving, we get a = F /(m A + m B ) and thus T .
32.
Ken of mass…
B
F − T = mB a
Consider the free body diagram:
Two blocks A and B…
Note that the net force F on the whole system is m B g . Apply net F = ma , we get a =
C
/
F /(m B + m A ) = m B g /(m B + m A ) = g (1 + m B /m A ).
33.
Some buckets…
B Actually, T A = 2mg , TB = mg and TC = mg ,
where m = 1 kg in these cases.
34.
Two blocks A and B…
D
T2 = 0.6g + 10 = 15.9 N.
(1) is incorrect, since the grav. force is 70 × 9.81 =
686.7 N, not 600 N.
A further question is, which string breaks first if you apply
a great force? The answer depends. Pulling slowly, string 1
(∵ T1 > T2 ). With a jerk, string 2 (∵ inertia).
(2) is incorrect, since the balance reads the supporting force by the balance, i.e. 600 N.
(3) is correct, since the acceleration is (686.7 −
600)/70 ≈ 1.24 m s−2 .
28.
35.
T − mAg = mAa
mB g − T = mB a
Put m A = 1 kg and m B = 3 kg.
{
Don’t mix up velocity and force! Drawing a free body diagram helps.
T −g = a
3g − T = 3a
John is taking…
D
Solving, we get T = 1.5 × 9.81 N.
Falling freely ⇒ acted by the grav. force only.
36.
Two blocks A and B are placed…
A
−2
For (1), a = 20/(3 + 5) = 2.5 m s .
For (2), B is more massive. With the same acceleration, the net force on B is greater.
For (3), the force on B by A and that on A by B
form an action-reaction pair. They should be the
same in magnitude.
i
Consider blocks A and B .
{
A woman is going…
Neglecting air resistance, the ball is only acted
by the grav. force (i.e. its weight) when it is in the
air. So, the net force is a constant.
30.
Two blocks A and B are connected…
D
B
29.
T1 = (0.4 + 0.6)g + 10 = 19.8 N.
Two blocks A and B are connected…
C Const. speed ⇒ no friction. Note that the
spring balance reads the tension T . Consider the
free body diagram of A or B . Obviously, T = 10 N.
See the next question.
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F2. Force and Newton’s Laws | 11
Solution Guide (3rd Ed.)
37.
(b) The net force needed should be doubled.
Blocks A and B…
The acceleration of the system is a = (10 −
5)/(m + 4m) = 1/m . Apply Newton’s 2nd law to B .
So, 10 − T = m × (1/m), and thus T = 9 N.
C
4.
49.1 N
.
(1M+1A)
(b) The net force on the bucket cannot be
greater than 80 − 49.05 = 30.95 N.
Two blocks A and B are stacked…
(1M)
That means, max. acceleration is 30.95/5 =
6.19 m s−2 .
The two blocks have the same acceleration.
Applying net F = ma , we get F A : FB = m A : m B
= 1 : 2.
A
39.
Fred wants to get…
(a) W = mg = 5 × 9.81 = 49.05 ≈
See the previous question.
38.
(1A)
Applying s = ut + 12 at 2 , the min. time is given
by
Block B with block A…
10 = 0 +
Suppose F = (m A + m B )a , the net force on A
is m A a , and thus smaller than F in magnitude. So,
(2) is correct.
C
∴t ≈
1
× 6.19 × t 2
2
(1M)
1.80 s or −1.80 s (rejected)
(1A)
Given s , find t ⇒ use s = ut + 12 at 2 . The question is, a =?
When F is removed, friction between A and B
becomes zero. So, A and B move together afterwards at const. velocity, and (3) is incorrect.
5.
A man of mass…
(a) The balance reading is equal to the weight of
the man.
(1A)
Short questions
(b) (i) Consider the free body diagram:
1.
A man lifts a rook…
(a) The ingers exert the same magnitude of
force such that the horizontal net force is
zero.
(b) f min = 0.05 × 9.81 + 0.05 × 1 = 0.5405
≈ 0.541 N .
2.
(2A)
(1M+1A)
A block is initially…
(a) The train slows down.
(b) amax =
f max
m
1.96 m s−2
=
0.2×m×9.81
m
(1A)
.
(1M+1A)
A car of mass…
90 × 1000
60 × 60
)2
(1A)
(ii) The normal reaction R acting on the
man is given by
The balance reading is thus
∴ a = −9.301 m s−2
i
.
∴ R = 377.65 N
= 2 × a × 33.6
Net force is 1800 × (−9.301) ≈
768 N
65 × (−4) = R − 65 × 9.81
(a) Applying v 2 − u 2 = 2as , the acceleration is
0−
(1M)
∴ R = 767.65 N
The balance reading is thus
(1A)
(
ma = R − mg
65 × 2 = R − 65 × 9.81
= 1.962 ≈
(c) The train changes its direction of motion.
3.
The normal reaction R acting on the
man is given by
−16 700 N
(1M)
.
(1M+1A)
378 N
.
(1A)
(iii) The case is the same as (i). The balance
reading is thus 768 N .
(1A)
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12 | F2. Force and Newton’s Laws
6.
Solution Guide (3rd Ed.)
8.
A block of mass…
(a) T = 2 × 9.81 = 19.62 ≈
19.6 N
.
A bungee jumper…
(a) From A to B , the tension is zero. So, the
jumper is falling freely.
(1M+1A)
(b) (i) For T = 10 N, consider the free body
diagram below:
2
(1A)
2
Applying v − u = 2as , his speed at B is
given by
v 2 − 0 = 2 × 9.81 × 20
∴v ≈
(1M)
2a = 2 × 9.81 − 10
∴ a = 4.81 m s−2
So, the acceleration is
downwards.
(1A)
(b) From B to C , tension T < weight, so net
force points downwards, and his speed is
increasing.
The acceleration is given by
ma = mg − T
19.8 m s−1
4.81 m s−2
(1A)
9.
(ii) For T = 30 N,
From C to D , tension T > weight, so the
net force points upwards. His speed is
decreasing.
(1A)
The jumper’s speed is highest at C .
(1A)
In Hong Kong, candidates…
(a) Free body diagrams:
(i)
(ii)
ma = mg − T
(1A)
(1A+1A)
2a = 2 × 9.81 − 30
∴ a = −5.19 m s−2
So, the acceleration is
upwards.
7.
5.19 m s−2
(1A)
(b) Applying v 2 − u 2 = 2as , the speed is given by
Two identical blocks…
(a) T = 2.5 × 9.81 = 24.525 ≈
24.5 N
0 − u 2 = 2 × (−9.81) × 0.533
(1M+1A)
∴u≈
(b) The tension T is given by
{
T − mAg = mAa
mB g − T = mB a
(2M)
49.05 − T = 5a
32.7 N
.
(1A)
For the same a , F ∝ m .
T − 24.525 = 2.5a
Solving, we get T =
(1A)
A horizontal force…
(a) The 3 kg box is exerted by the greatest net
force.
Put m A = 2.5 kg, mB = 5 kg, and
g = 9.81 m s−2 .
{
10.
3.23 m s−1
(1M)
(1A)
(b) (i) Applying F = ma , the common accelera12
= 2 m s−2 .
tion is a = 1+2+3
(1M)
The contact force between the 1 kg and
the 2 kg box is thus (2+3)×2 = 10 N . (1A)
(ii) The contact force between the 2 kg and
the 3 kg box is thus 3 × 2 = 6 N .
(1A)
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F2. Force and Newton’s Laws | 13
Solution Guide (3rd Ed.)
11.
The average net force F is 48 × (−41.49) =
(1M+1A)
−1990 N, i.e. 1990 N upwards.
A baseball player…
(a) Both are exerted the same magnitude of
force.
(b) Applying v 2 − u 2 = 2as , the acceleration of
the ball is given by
0 − 302 = 2 × a × 0.15
Note that the net force F is not equal to the force f
exerted by the trampoline, where f = F − mg .
(1A)
2.
(a) The max. speed is
(1M)
∴ a = −3000 m s−2
F
a
=
−800
−3000
≈
0.267 kg
.
1.28 m s−1
(b) The can leaves the hand at
.
0.29 s
(1A)
.
(1A)
The can stops speeding up when it leaves
the hand of the student.
Applying F = ma , the mass is
m=
A student…
(1A)
(1A)
(c) The distance travelled is 21 × (1.02 − 0.12) ×
1.28 = 0.58 m .
(1M+1A)
The −ve sign suggests that the direction of F and a
is opposite to the travelling direction of the ball.
(d) The average frictional force is
F = ma = 0.316 × 0−(−1.28)
1.03−0.29 ≈
Long questions
1.
3.
0.547 N
.
(1M+1A)
A helicopter…
(a) W = mg = 12 000 × 9.81 = 117 720 ≈
A gymnast…
118 000 N
(a) Free body diagram:
(1A)
(1M+1A)
(b) (i) Free body diagrams:
(1A)
(b) Applying s = ut + 12 at 2 , taking downward
direction as positive,
2 = −1.5t +
1
× 9.81 × t 2
2
(1M)
=
0 = 4.905t 2 − 1.5t − 2
∴ t = 0.8095 s
It takes
0.810 s
or −0.5037 s (rejected)
.
(1A)
(c) Consider the speed of the gymnast when
she just reaches the trampoline. Applying
v 2 − u 2 = 2as , we get
0.79 m s−2
(1M+1A)
Note that m tot = 14 000 kg, and F −
m tot g = m tot a , where F is the lifting
force. So, F is m tot (g + a) = 148 000 N . (1A)
Alternatively, consider the free body diagram of
the helicopter. Note that F − mg − T = ma . So,
F = (12000)(9.81 + 0.79) + T = 148 000 N.
(iii) Applying s = ut + 12 at 2 , the time is given
by
v 2 − (−1.5)2 = 2 × 9.81 × 2
∴ v 2 = 41.49 m2 s−2
(ii) aheli = aveh = 21 200−2000×9.81
2000
(1M)
Applying v 2 − u 2 = 2as , the average acceleration is given by
100 = 0 +
∴t ≈
1
× 0.79 × t 2
2
15.9 s
(1M)
(1A)
0 − 41.49 = 2 × a × 0.5
∴ a = −41.49 m s−2
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(1M)
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14 | F2. Force and Newton’s Laws
Solution Guide (3rd Ed.)
Figure 2: for LQ 4b
4.
(b) The force F exerted by spittlebug is given by
Jane and her friends…
(a) The distance is 12 × (3 + 6.5) × 4 =
19 m
.
F − mg = ma
(1M+1A)
(b) See Fig. 2.
2A for correct shape of the graph
1A for correct axes with labels
F = m(a + g )
(3A)
F = 12 × 10−6 × (400 + 1) × 9.81
∴F =
(c) Applying v 2 − u 2 = 2as , the acceleration of
the lift is given by
the order of 10−2 N.
So, the force is 1.44 × 10 N.
(1A)
Note that the net force F is not equal to the
braking force f , where f = F − mg . To find f ,
we can also consider the conservation of energy:
f × 7.5 = 21 m(12)2 + mg (7.5), and solve for f .
1.
The net force on the tank is zero from t1 to t2 ,
implying that the tank can be at rest or in uniform
motion. The water tends to move forward from
t 3 to t 4 , implying that the tank is slowing down.
Thus (1) is incorrect while (2) and (3) are correct.
0 − u 2 = 2 × 9.81 × 0.7
∴ u = 3.706 m s−1
(1M)
.
A tank of water…
D
(a) Applying v 2 − u 2 = 2as , the initial velocity is
given by
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Shoot-the-stars questions
A spittlebug…
9.44×10−4 s
(1A)
(1M)
4
≈
(1A)
(1M)
F = 1500 × (−9.6) = −1.44 × 104 N.
3.706−0
400×9.81
(1A)
F man is in the order of 1000 N while F bug is in
∴ a = −9.6 m s−2
The time is
0.0472 N
(c) A man exerts a greater force.
0 − 122 = 2 × a × 7.5
5.
(1M)
2.
A sealed box…
A Terminal v ⇒ const. v ⇒ no net force is
acting on the weight. So, the balance reading (i.e.
the supporting force from the balance) is equal to
2 × 9.81 N.
In fact, this case is just the same as a lift descending at a
const. speed.
(1M+1A)
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F2. Force and Newton’s Laws | 15
Solution Guide (3rd Ed.)
3.
Block B is hanged…
Both blocks accelerate at the same rate from
rest. So their speeds are always the same. The
tension in the string becomes zero immediately
after A is released.
C
6.
A pulley…
(a) Consider the free body diagram:
See the next question.
4.
Block B is hanged…
B Note that, at the instant when the system is
released, net force F A = mg + T , but net force
mg −T
F B = mg − T . So, a B = m < g .
See the previous question.
5.
A basket…
B
Note that mg − 2T = ma . ∴ T = m(g − a)/2 =
10 × (9.81 − 0.5)/2 = 46.55 ≈ 46.6 N.
F = 2T = 2 × (5 × 9.81) =
98.1 N
(1M+1A)
(b) Tension T is given by
{
mAg − T = mAa
T − mB g = mB a
(2M)
Put m A = 5 kg, m B = 2 kg, and
g = 9.81 m s−2 .
{
49.05 − T = 5a
T − 19.62 = 2a
Solving, we get T = 28.03 N. So, F = 2T =
2 × 28.03 ≈ 56.1 N .
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(1A)
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