Test: Laws of Motion- 2


30 Questions MCQ Test Physics For JEE | Test: Laws of Motion- 2


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QUESTION: 1

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.

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

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?

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

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.

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

Newton’s third law states that when two bodies interact.

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 FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal in magnitude and opposite in direction, FA = −FB
  • Newton's third Law:
    Engineering Car Crash Safety with Newton's Third Law | Lesson Plan
QUESTION: 5

Force is required

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

According to Newton's third law of motion, the action and reaction forces are

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

According to first law of motion

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

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.

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

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?

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

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,

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

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?

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

According to second law of motion

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

According to the law of conservation of momentum,

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

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?

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

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?

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

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

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

Force is a

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

When bodies are in contact with each other then 

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

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.

Solution:

Initial thrust = upthrust required to impart acceleration + uthrust to overcome gravity

⇒ ma + mg = m(a+g)
⇒ 20000(5+10) = 300000 N

QUESTION: 20

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?

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/s
Thus we get T = 200N

QUESTION: 21

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

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

When several forces act on a body

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

Friction is

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

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.)

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 (vx) of velocity, in the absence of air resistance, remains unchanged, i.e.,

vx = 20 m/s

The vertical component (vy) of velocity of the stone is given by the first equation of motion as:

vy = u + ayδt

Where, δt = 11 – 10 = 1 s and ay = g = 10 m/s2

∴vy = 0 + 10 X 1 = 10 m/s

QUESTION: 25

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

Solution:

When monkey climbs with uniform speed, the acceleration is zero, so

T = mg = 40x10 = 400 N

QUESTION: 26

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

Solution:

Initial momentum of the ball,
Pi​ = 0
Final momentum of the ball,
Pf  = mv
⇒ 0.25×10 = 2.5 kg m/s
Impulse imparted on the ball,
I = Pf − Pi 
⇒ I = 2.5 − 0 = 2.5 N s

QUESTION: 27

Static friction 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

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

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

If second law is applied to a rigid body

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

Rolling friction is

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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