Laws Of Motion Test

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Laws Of Motion Test - 18

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Question 1
1 / -0

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. The speed is then increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks.

A
The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

B
The stone moves radially inwards.

C
The stone flies off tangentially from the instant the string breaks.

D
The stone moves radially outwards.
Solution
The stone is moving in a circular path with constant speed, this is uniform circular motion.
In uniform circular motion, at any instant velocity (magnitude speed) acts tangentially to the circular path.
Hence, when string is cut the stone will fly off tangentially to circular path with velocity of magnitude speed.
Question 2
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A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s−1, what would be the reading on the scale?

A
105 kg

B
75 kg

C
70 kg

D
35 kg

Solution

Mass of the man, m = 70 kg
Acceleration, a = 0
Using Newton’s second law of motion, We can write the equation of motion as,
R – mg = ma
∴ R = mg = 70 × 10 = 700 N
∴ the weighing scale = 700 / g = 700 / 10 = 70 kg
Question 3
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Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 ms⁻², the stone being at rest relative to the train. Neglect air resistance.

A
0.1 N along the direction of motion

B
0.2 opposite to the direction of motion

C
0.1 N opposite to the direction of motion

D
1.0 N in the direction of motion
Solution
As the stone is lying on the floor of the train, its acceleration is same as that of the train.
∴ Force acting on stone,
F = ma = 0.1 x 1 = 0.1 N.
This force is along the horizontal direction of motion of the train.
Note that in each case, the weight of the stone is being balanced by the normal reaction.
Pseudo force is considered when we are not in the same frame as that of the object.
Question 4
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Newton’s third law states that when two bodies interact.

A
they exert forces on each other that at each instant are equal in magnitude and opposite in direction.

B
they exert forces on each other that at each instant are equal in magnitude and same in direction.

C
they exert forces on each other that at some instants are equal in magnitude and opposite in direction.

D
they exert forces on each other that at some instants are equal in magnitude and same in direction.
Solution
The third law states that all forces between two objects exist in equal magnitude and opposite direction.
If one object A exerts a force F_A on a second object B, then B simultaneously exerts a force F_B on A, and the two forces are equal in magnitude and opposite in direction, F_A = −F_B
Newton's third Law:
Question 5
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Force is required

A
to start a stationary object and to stop a moving object.

B
only to keep an object moving.

C
only to stop a moving object.

D
only to start a stationary object moving.
Solution
Force is required to start a stationary object and to stop a moving object due to inertia.
Inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
This is also called law of inertia or newton's first law of motion.
Newton's first Law:
Question 6
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According to Newton's third law of motion, the action and reaction forces are

A
not equal in magnitude and opposite in direction

B
not equal in magnitude and same in direction

C
equal in magnitude and same in direction

D
equal in magnitude and opposite in direction
Solution
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
To every action there is always opposed an equal reaction or the mutual actions of two bodies upon each other are always equal and opposite in direction.
Question 7
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According to first law of motion

A
body acted on by no net force moves with increasing velocity and negative acceleration

B
body acted on by no net force moves with constant velocity (which may be zero) and zero acceleration

C
body acted on by net force moves with constant velocity (which may be zero)

D
body acted on by zero net force moves with non zero acceleration

Solution

If net force act on a body then acceleration of the body will also be zero. Hence velocity will not be changed i.e. it continues in its existing state of rest or uniform motion in a straight line.
Question 8
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Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg lying on the floor of a train which is accelerating with 1 ms⁻², the stone being at rest relative to the train. Neglect air resistance.

A
1.0 N in the direction of motion

B
0.1 N opposite to the direction of motion

C
0.1 N along the direction of motion

D
0.2 opposite to the direction of motion

Solution

Weight of the stone is balanced by the reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
a = 1ms^-2
Force in horizontal direction
F = ma = 0.1 x 1 = 0.1N
Question 9
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A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 ms⁻² what would be the reading on the scale?

A
150 kg

B
120 kg

C
130 kg

D
140 kg

Solution

According to Newton's second Law of motion,
Apparent weight, R = m(a+g)
= 70(10+10)
= 1400N
mass = R/g
= 1400/10
= 140Kg
Question 10
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A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms⁻². The crew and the passengers weigh 300 kg. Give the magnitude and direction of the force on the floor by the crew and passengers,

A
3000 N upward

B
4500 N upward

C
7500 N upward

D
1500 N upward
Solution
The moving helicopter is a non-inertial frame and hence we do need to apply pseudo force onto the body so as to solve equations with Newton's laws.
The pseudo force needed to be applied, f = ma = 300 x 15 N = 4500N
Thus the net downward force due to the passengers is 7500N.
Question 11
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A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What is the angle of banking required to prevent wearing out of the rail?

A
27^∘

B
37^∘

C
42^∘

D
32^∘

Solution

Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/s
Mass of the train, m = 106 kg
The centripetal force is provided by the lateral thrust of the rail on the wheel.
As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail.
The angle of banking θ, is related to the radius (r) and speed (v) by the relation:
tan θ = v² / rg
= 152 / (30 x 10)
= 225 / 300
⇒ θ = tan^-1 (0.75) = 36.870
Therefore, the angle of banking is about 36.87degree = 37degree.
Question 12
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According to second law of motion

A
The momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts

B
The change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts

C
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts

D
The momentum of a body is directly proportional to the applied force and takes place in the direction opposite to the force
Solution
According to Newton s Second Law of Motion, also known as the Law of Force and Acceleration, a force upon an object causes it to accelerate according to the formula net force = mass x acceleration.
It states that the force applied on a body is proportional to the rate of change of momentum of the body.
So the acceleration of the object is directly proportional to the force and inversely proportional to the mass.
Question 13
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According to the law of conservation of momentum,

A
the momentum of an isolated system of interacting particles is conserved.

B
the total momentum of an interacting system of particles and fields is conserved.

C
the total momentum of an isolated system of interacting particles is conserved.

D
the momentum of an interacting system of particles and fields is conserved.

Solution

When no external forces act on a system of several interacting particles, then the total momentum of system is conserved.
Law of conservation of Momentum:
Question 14
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A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s⁻¹How long does the body take to stop?

A
1.7 s

B
6.0 s

C
5.0 s

D
none

Solution

According to Newton's second Law
Force = mass x accleration
F = m.a
= m.v/t
t = m.v/F
= 20x15/50
= 6 sec
Question 15
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A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 ms⁻² what would be the reading on the scale?

A
125 kg

B
105 kg

C
95 kg Hz

D
115 kg

Solution

Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 upward
Using Newton’s second law of motion, we can write the equation of motion as:
R – mg = ma
R = m(g + a)
= 70 (10 + 5) = 70 x 15
= 1050 N
∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg
Question 16
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A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs up with an acceleration of 6 ms⁻²

A
640 N

B
600 N

C
760 N

D
740 N

Solution

Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 6 m/s2 upward
Using Newtons second law of motion, we can write the equation of motion as:
T mg = ma
T = m(g + a)
= 40 (10 + 6)
= 640 N
Question 17
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Force is a

A
quantitative measure of the momentum between two bodies.

B
qualitative measure of the momentum of two bodies.

C
qualitative measure of the interaction between two bodies.

D
quantitative measure of the interaction between two bodies.

Solution

Newton's second law of motion gives the quantitative definition of force. The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Question 18
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When bodies are in contact with each other then

A
contact forces are equal and in same direction.

B
contact forces are not equal and in same direction.

C
contact forces are equal and opposite in direction.

D
contact forces are not equal and opposite in direction.
Solution
According to newton's third law of motion for every action there is always equal and opposite reaction or the mutual actions of two bodies upon each other.
Whenever two objects interact with each other, then the pair of forces acting on them are equal in magnitude and opposite in direction. Contact forces are mutual forces (action and reaction).
Question 19
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A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.

A
365000 N

B
378000 N

C
300000 N

D
350000 N

Solution

Initial thrust = upthrust required to impart acceleration + uthrust to overcome gravity
⇒ ma + mg = m(a+g)
⇒ 20000(5+10) = 300000 N
Question 20
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Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. a horizontal force F = 600 N is applied to the 20 kg mass so as to pull it. What is the tension in the string?

A
250 N

B
300 N

C
150 N

D
200 N

Solution

For some instance assume both masses as one system, thus we get that
600 = 30a
Where a is the common acceleration of the system.
Now if we consider the 10kg block we get that
T = 10a
And a = 20m/s²
Thus we get T = 200N
Question 21
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A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs down with an acceleration of 4 ms⁻²

A
200 N

B
300 N

C
240 N

D
280 N

Solution

Mass of the monkey, m = 40 kg
Acceleration due to gravity, g = 10 m/s
Maximum tension that the rope can bear, Tmax = 600 N
Acceleration of the monkey, a = 4 m/s2 downward
Using Newton’s second law of motion, we can write the equation of motion as:
mg T = ma
T = m (g- a)
= 40(10-4)
= 240 N
Question 22
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When several forces act on a body

A
the effect on its motion is the same as when a single force acts on the body

B
the effect on its motion is the same as when a single force, equal to the vector sum (resultant) of the forces, acts on the body

C
the effect on its motion is the same as when a single force, not equal to the vector sum (resultant) of the forces, acts on the body

D
none

Solution

When an object has several forces acting on it, the effect of force is the same as one force acting on the object in a certain direction and this overall force is called the ‘resultant force’. The resultant force is essential to change the velocity of an object.
Question 23
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Friction is

A
perpendicular to contact surface and aids motion

B
parallel to contact surface and aids motion

C
perpendicular to contact surface and opposes i motion

D
parallel to contact surface and opposes the relative motion
Solution
Friction opposes the relative motion between the surfaces in contact with each other. Friction is the force that opposes relative motion or tendency of relative motion between two surfaces in contact.
When two surfaces move relative to each other or they have a tendency to move relative to each other, at the point (or surface) of contact, there appears a force which opposes this relative motion or tendency of relative motion between two surfaces in contact.
It acts on both the surfaces in contact with equal magnitude and opposite directions (Newton's 3rd law). also Friction force tries to stop relative motion between two surfaces in contact.
Question 24
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A truck starts from rest and accelerates uniformly at 2.0 ms⁻². At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms⁻² ) of the stone at t = 11s? (Neglect air resistance.)

A
8

B
14

C
10.0

D
12

Solution

Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s2
Time, t = 10 s
As per the first equation of motion, final velocity is given as:
v = u + at
⇒ 0 + 2 X 10 = 20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontal component (v_x) of velocity, in the absence of air resistance, remains unchanged, i.e.,
v_x = 20 m/s
The vertical component (v_y) of velocity of the stone is given by the first equation of motion as:
v_y = u + a_yδt
Where, δt = 11 – 10 = 1 s and a_y = g = 10 m/s2
∴v_y = 0 + 10 X 1 = 10 m/s
Question 25
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A monkey of mass 40 kg climbs on a rope which can stand a maximum tension of 600 N. What is the tension in the rope if the monkey climbs up with a uniform speed of 5 ms⁻¹

A
400 N

B
315 N

C
206 N

D
275 N

Solution

When monkey climbs with uniform speed, the acceleration is zero, so
T = mg = 40x10 = 400 N
Question 26
1 / -0

A stationary ball weighing 0.25kg acquires a speed of 10m/s when hit by a hockey stick. The impulse imparted to the ball is

A
1.5 Ns

B
0.5 Ns

C
2.5 Ns

D
2.0 Ns

Solution

Initial momentum of the ball,
P_i = 0
Final momentum of the ball,
P_f = mv
⇒ 0.25×10 = 2.5 kg m/s
Impulse imparted on the ball,
I = P_f − P_i
⇒ I = 2.5 − 0 = 2.5 N s
Question 27
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Static friction force

A
increases until the body starts to move.

B
decreases before moving and increases later.

C
remains constant.

D
is a strong interaction force.
Solution
It is the maximum friction, where a body just starts to move over the surface.
For example a large block of mass m is placed on a horizontal table. Apply a small force the block does not move due to the friction force which balances the applied force.
Now Gradually increase the applied force until it start moving. This maximum applied force acting in opposite direction is called as Static Friction.
Question 28
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A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms⁻¹. What is the trajectory of the bob if the string is cut when the bob is at one of its extreme positions

A
bob will go upwards

B
bob will fall vertically downwards

C
bob will go downwards at an angle

D
bob will go down in a parabolic path

Solution

At the extreme position of the oscillation, the speed of the bob is zero. So the bob is momentarily at rest. if the string is cut, the bob will fall vertically downwards.
Question 29
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If second law is applied to a rigid body

A
the acceleration is that of the centre of mass

B
the acceleration is the average of all particles in the body

C
the acceleration is that of any particle in the body

D
none of the above

Solution

The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates).
Question 30
1 / -0

Rolling friction is

A
greater than or equal to static

B
greater than static friction

C
equal to static friction

D
less than static friction

Solution

Rolling friction is always less than static friction because in order for an object to roll the force of friction between it and surface must be large enough to keep the object from sliding. Hence rolling friction is always greater than static force.

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Laws Of Motion Test - 18

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Laws Of Motion Test - 18

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Question 1

The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.

The stone moves radially inwards.

The stone flies off tangentially from the instant the string breaks.

The stone moves radially outwards.

Solution

Question 2

105 kg

75 kg

70 kg

35 kg

Solution

Question 3

0.1 N along the direction of motion

0.2 opposite to the direction of motion

0.1 N opposite to the direction of motion

1.0 N in the direction of motion

Solution

Question 4

they exert forces on each other that at each instant are equal in magnitude and opposite in direction.

they exert forces on each other that at each instant are equal in magnitude and same in direction.

they exert forces on each other that at some instants are equal in magnitude and opposite in direction.

they exert forces on each other that at some instants are equal in magnitude and same in direction.

Solution

Question 5

to start a stationary object and to stop a moving object.

only to keep an object moving.

only to stop a moving object.

only to start a stationary object moving.

Solution

Question 6

not equal in magnitude and opposite in direction

not equal in magnitude and same in direction

equal in magnitude and same in direction

equal in magnitude and opposite in direction

Solution

Question 7

body acted on by no net force moves with increasing velocity and negative acceleration

body acted on by no net force moves with constant velocity (which may be zero) and zero acceleration

body acted on by net force moves with constant velocity (which may be zero)

body acted on by zero net force moves with non zero acceleration

Solution

Question 8

1.0 N in the direction of motion

0.1 N opposite to the direction of motion

0.1 N along the direction of motion

0.2 opposite to the direction of motion

Solution

Question 9

150 kg

120 kg

130 kg

140 kg

Solution

Question 10

3000 N upward

4500 N upward

7500 N upward

1500 N upward

Solution

Question 11

27∘

37∘

42∘

32∘

Solution

Question 12

The momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts

The change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts

The momentum of a body is directly proportional to the applied force and takes place in the direction opposite to the force

Solution

27^∘

37^∘

42^∘

32^∘