DC Pandey Solutions: Laws of Motion - 2

# DC Pandey Solutions: Laws of Motion - 2 | DC Pandey Solutions for NEET Physics PDF Download

## Introductory Exercise 5.2

Q1. Three blocks of mass 1 kg, 4 kg and 2 kg are placed on a smooth horizontal plane as shown in figure. Find:
(a) the acceleration of the system,
(b) the normal force between 1 kg block and 4 kg block,
(c) the net force on 2 kg block.

Ans: (a) 10 ms-2 (b) 110 N (c) 20 N
Sol: Acceleration of system

= 10 m/s2
Let normal force between 1 kg block and 4 kg block = F1
∴ Net force on 1 kg block = 120 - N

or 10 = 120 - F1
i.e., F1 = 110 N
Net force on 2 kg block = 2*a
= 2*10
=20 N

Q2. Two blocks of mass 2 kg and 4 kg are released from rest over a smooth inclined plane of inclination 30° as shown in figure. What is the normal force between the two blocks?

Ans: zero
Sol: As, 4 g sin 30° > 2 g sin 30°
The normal force between the two blocks will be zero.

Q3. What should be the acceleration ‘a’ of the box shown in figure so that the block of mass m exerts a force mg/4 on the floor of the box?

Ans: 3g/4
Sol:
∴ N = mg/4
As lift is moving downward with acceleration a , the pseudo force on A will be ma acting in the upward direction. For the block to be at rest w.r.t. lift.
N + ma = mg
or

Q4. A plumb bob of mass 1 kg is hung from the ceiling of a train compartment. The train moves on an inclined plane with constant velocity. If the angle of incline is 30°. Find the angle made by the string with the normal to the ceiling. Also, find the tension in the string, (g = 10 m/s2)

Ans: 30°, ION
Sol: Angle made by the string with the normal to the ceiling = θ = 30°
As the train is moving with constant velocity no pseudo force will act on the plumb-bob.
Tension in spring = mg
= 1*10
= 10 N

Q5. Repeat both parts of the above question, if the train moves with an acceleration a = g/2 up the plane.

Ans:
Sol: Pseudo force (= ma) on plumb-bob will be as shown in figure
T cos φ = mg + ma cos (90° - θ)
i.e., T cos φ = mg + ma sin θ …(i)
and       T sin φ = ma cos θ
Squaring and adding Eqs. (i) and (ii),
T2 = m2 g2 + m2 a2 sin2 θ + 2m2 ag sin θ + m2 a2 cos2 θ …(iii)
T2 = m2 g2 + m2 a2 + m2 ag (∴ θ = 30°)

or

Dividing Eq. (i) by Eq. (ii),

i.e.,

Q6. Two blocks of mass 1 kg and 2 kg are connected by a string AS of mass 1 kg. The blocks are placed on a smooth horizontal surface. Block of mass 1 kg is pulled by a horizontal force F of magnitude 8 N. Find the tension in the string at points A and B.

Ans: 4 N, 6 N
Sol:
Net force on 1 kg mass = 8 - T2
∴ 8 - T2 = 1*2
⇒ T2 = 6 N
Net force on 1 kg block = T1
∴ T1 = 2a = 2*2 = 4 N

## Introductory Exercise 5.3

Q1. In the arrangement shown in figure what should be the mass of block A, so that the system remains at rest? Neglect friction and mass of strings.

Ans: 3 kg
Sol: F = 2 g sin 30° = g
For the system to remain at rest
T2 = 2 g …(i)
T2 + F = T1 …(ii)
or          T2 + g = T1 …[ii (a)]
T1 = mg …(iii)
Substituting the values of T1 and T2 from Eqs. (iii) and (i) in Eq. [ii(a)]
2 g + g = mg
i.e., m = 3 kg

Q2. In the arrangement shown in figure, find the ratio of tensions in the strings attached with 4 kg block and that with 1 kg block.

Ans: 4
Sol:
As net downward force on the system is zero, the system will be in equilibrium
∴  T1 = 4 g
and T2 = 1 g

Q3. Two unequal masses of 1 kg and 2 kg are connected by a string going over a clamped light smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped for a moment 1.0 s after the system is set in motion. Find the time elapsed before the string is tight again.

Ans: 1/3 s
Sol: 2 g - T = 2a
T - 1 g = 1a
1 g = 3 a
∴ a = g/3
Velocity of 1kg block 1 section after the system is set in motion
v = 0 + at

On stopping 2 kg, the block of 1kg will go upwards with retardation g. Time ( t' ) taken by the 1 k g block to attain zero velocity will be given by the equation.

⇒
If the 2 kg block is stopped just for a moment (time being much-much less than 1/3 s), it will also start falling down when the stopping time ends.
In  time upward displacement of 1 kg block

Downward displacement of 2 kg block

As the two are just equal, the string will again become taut after time 1/3 s.

Q4. Two unequal masses of 1 kg and 2 kg are connected by an inextensible light string passing over a smooth pulley as shown in figure. A force F = 20 N is applied on 1 kg block. Find the acceleration of either block. (g = 10 m/s2).

Ans:
Sol: F + 1g - T = 1a      … (i)
and T - 2g = 2a …(ii)
F - 1g = 3a

## Introductory Exercise 5.4

Q1. Consider the situation shown in figure. Both the pulleys and the string are light and all the surfaces are smooth.
(a) Find the acceleration of 1 kg block.
(b) Find the tension in the string.
(g = 10 m/s 2).

Ans: (a) 2g/3, (b) 10/3 N
Sol: 2T = 2*a            … (i)
and     1g - T = 2a …(ii)
Solving Eqs. (i) and (ii),
α = g/3
∴  Acceleration of 1 kg block

Tension in the string

Q2. Calculate the acceleration of either blocks and tension in the string shown in figure. The pulley and the string are light and all surfaces are smooth.

Ans:
Sol: Mg - T = Ma             …(i)
T = Ma …(ii)
Solving Eqs. (i) and (ii)
α = g/2
and T = Mg

Q3. Find the mass M so that it remains at rest in the adjoining figure. Both the pulley and string are light and friction is absent everywhere, (g = 10 m/s 2).

Ans: 4 . 8 kg
Sol: Block of mass M will be at rest if
T = Mg …(i)
For the motion of block of mass 3 kg
....(ii)
For the motion of block of mass 2 kg
....(iii)
g = 5a
i.e.,  α = g/5
Substituting above value of a in Eq. (iii),
T/2 = 2(g+a)
or
= 24g/5
Substituting value of above value of T in Eq. (i),
M = 24/5
= 4.8 kg

Q4. In figure assume that there is negligible friction between the blocks and table. Compute the tension in the cord connecting m2 and the pulley and acceleration of m2 if m1 = 300 g, m2 = 200g and F = 0.401V.

Ans:
Sol: T/2 = m1.2a
i.e., T = 4 m1a …(i)
F - T = m2a …(ii)
or      F - 4m1a = m2a
or

∴ T = 4 m1a
= 4*0.3* 2/7
= 2.4/7
= 12/35 N

Introductory Exercise 5.5

Ques 1: In figure m1 = 1 kg and m2 = 4 kg. Find the mass M o f the hanging block which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction less and the strings and the pulleys are light

Note In exercises 2 to 4 the situations described take place in a box car which has initial velocity v = 0 but acceleration

Ans: 6 .83 kg
Sol: Block on triangular block will not slip if

m1a cos θ= m1g sin θ
i.e., a = g tanθ …(i)
N = m1g cosθ + m1asinθ …(ii)
For the movement of triangular block
T - N sinθ = m2α …(iii)
For the movement of the block of mass M
Mg - T = Ma …(iv)
Mg - N sinθ = ( m2 + M )α
Substituting the value of N from Eq. (ii) in the above equation
Mg - (m1g cosθ + m1α sinθ) sinθ
=( m2α + Mα)
i.e., M (g - α) = m1g cosθ sinθ + (m2 + m1 sin2θ)α
Substituting value of a from Eq. (i) in the above equation,
M(1 - tanθ) = m1 cosθ sinθ + ( m2 + m1 sin2 θ) tanθ

Substituting θ = 30° , m1 = 1 kg and m2 = 4 kg

= 6.82 kg

Ques 2: A 2 kg object is slid along the friction less floor with initial velocity (10 m/s)î (a) Describe the motion of the object relative to car (b) when does the object reach its original position relative to the box car.
Ans:
(a) x = x0 + 10t - 2.5t2 , v = 10 - 5t (b) t = 4s
Sol: (a) Using
Displacement of block at time t relative to car would be

Velocity of block at time t (relative to car) will be

(b) Time (t) for the block to arrive at the original position (i.e., x = x0) relative to car
x0 = x+ 10 t - 2.5 t2
⇒ t = 4s

Ques 3: A 2 kg object is slid along the friction less floor with initial transverse velocity (10 m/s). Describe the motion (a) in car’s frame (b) in ground frame.
Ans:
(a) x = x0 - 2.5t2, z = z0 + 10t, vx = - 5t, vz = 10 ms-1
(b) x = x0, z = z0 + 10t, vx = 0, vz = 10 ms-1
Sol: (a) In car’s frame position of object at time t would be given by
In car’s frame

x = x0 + 0*t + 1/2(-5)t2
i.e., x = x0 - 2.5 t2 …(i)
and z = z + 10t …(ii)
Velocity of the object at time t would be

and

(b) In ground frame the position of the object at time t would be given by
In ground frame

x = x0
and z = z0 + 10t
Velocity of the object at time t would be

ans

Ques 4: A 2 kg object is slid along a rough floor (coefficient of sliding friction = 0.3) with initial velocity (10 m/s)î. Describe the motion of the object relative to car assuming that the coefficient of static friction is greater than 0.5.
Ans:
x = x0 + 10f - 4t2 , K = 10 - 8 t for 0<f<1.25 s object stops at t = 1.25 s and remains at rest relative to car.
Sol: m = 2 kg

Normal force on object = mg
Maximum sliding friction = μsmg
= 0.3*2*10 = 6 N
Deceleration due to friction = 6/2 =3 m/s2
Deceleration due to pseudo force  = 5 m/s2
∴ Net deceleration = (3 + 5) m/s2
= 8 m/s2
∴ Displacement of object at any time t (relative to car)

Thus, velocity of object at any time t (relative to car)

The object will stop moving relative to car when
10 - 8t = 0 i.e., t = 1.25s
∴ vx = 10 - 8t for  0<t<1.25 s

Ques 5: A block is placed on an inclined plane as shown in figure. What must be the frictional force between block and incline if the block is not to slide along the incline when the incline is accelerating to the right at

Ans:
9/25 mg
Sol: For block not to slide the frictional force ( f ) would be given by

f + ma cosθ = mg sinθ
or
f = mg sinθ - ma cosθ

The document DC Pandey Solutions: Laws of Motion - 2 | DC Pandey Solutions for NEET Physics is a part of the NEET Course DC Pandey Solutions for NEET Physics.
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## FAQs on DC Pandey Solutions: Laws of Motion - 2 - DC Pandey Solutions for NEET Physics

 1. What are the laws of motion?
Ans. The laws of motion are three fundamental principles proposed by Sir Isaac Newton in his work "Mathematical Principles of Natural Philosophy". These laws describe the relationship between the motion of an object and the forces acting upon it. The laws of motion are: 1. Newton's First Law (Law of Inertia): An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity, unless acted upon by an external force. 2. Newton's Second Law (Law of Acceleration): The acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. It can be mathematically represented as F = ma, where F is the net force, m is the mass of the object, and a is its acceleration. 3. Newton's Third Law (Law of Action-Reaction): For every action, there is an equal and opposite reaction. This means that whenever an object exerts a force on another object, the second object exerts an equal and opposite force on the first object.
 2. What is the significance of the laws of motion?
Ans. The laws of motion are significant because they provide a framework for understanding and predicting the motion of objects in the presence of forces. They are essential in various fields of science, engineering, and everyday life. Some of the key significance of the laws of motion are: 1. Predicting and analyzing motion: The laws of motion allow us to predict and analyze the motion of objects by considering the forces acting upon them. This is crucial in fields like physics, engineering, and sports. 2. Designing effective transportation systems: Understanding the laws of motion helps in designing efficient transportation systems like cars, airplanes, and rockets. It enables engineers to calculate the forces required to achieve desired accelerations and velocities. 3. Explaining celestial phenomena: The laws of motion provide the basis for understanding celestial phenomena, such as the motion of planets, moons, and satellites. They help explain phenomena like gravitational attraction, orbits, and tides. 4. Safety and stability: Applying the laws of motion helps ensure the safety and stability of structures and objects. Engineers use these laws to design structures that can withstand forces like wind, earthquakes, and collisions. 5. Improving sports performance: Athletes and coaches can optimize performance by understanding and applying the laws of motion. They can analyze factors like force, acceleration, and momentum to improve techniques and achieve better results.
 3. How do the laws of motion relate to everyday life?
Ans. The laws of motion have numerous applications in everyday life. Some examples of how these laws relate to our daily experiences are: 1. Driving a car: When a car accelerates, the passengers feel a backward force pushing them into the seats. This is due to Newton's second law, which states that the acceleration of an object is directly proportional to the force applied to it. Similarly, when the car suddenly stops, the passengers lurch forward due to their inertia. 2. Playing sports: Sports activities like running, jumping, and throwing involve the application of the laws of motion. Athletes utilize the principles of force, acceleration, and momentum to achieve better performance. 3. Balancing on a bicycle: Riding a bicycle requires a balance between the gravitational force pulling downwards and the force exerted through pedaling. This demonstrates Newton's third law, as the force exerted on the pedals results in an equal and opposite reaction force propelling the bicycle forward. 4. Projectile motion: Throwing a ball or shooting a projectile follows the laws of motion. The trajectory of the projectile depends on the initial velocity, angle of projection, and the force applied. Understanding these laws helps in activities like playing baseball, basketball, or even throwing a frisbee. 5. Earth's rotation and tides: The laws of motion help explain phenomena like the rotation of the Earth and the occurrence of tides. The gravitational force between the Earth, Moon, and Sun influences the motion of the Earth and causes the periodic rise and fall of tides.
 4. How are the laws of motion different from each other?
Ans. The laws of motion differ in terms of the concepts they address and the relationships they describe. Here's how they are different from each other: 1. Newton's First Law: This law, also known as the Law of Inertia, describes the behavior of objects at rest or in constant motion. It states that an object at rest tends to stay at rest, and an object in motion tends to stay in motion with a constant velocity unless acted upon by an external force. This law focuses on the concept of inertia. 2. Newton's Second Law: This law, also known as the Law of Acceleration, relates the net force acting on an object to its mass and acceleration. It states that the acceleration of an object is directly proportional to the net force applied and inversely proportional to its mass. This law focuses on the concept of force and its effect on acceleration. 3. Newton's Third Law: This law, also known as the Law of Action-Reaction, describes the interaction between two objects. It states that for every action, there is an equal and opposite reaction. This law focuses on the concept of forces acting in pairs. While the first law deals with the concept of inertia, the second law relates force, mass, and acceleration, and the third law explains the action-reaction relationship between two objects.
 5. How do the laws of motion apply to space travel?
Ans. The laws of motion are crucial in understanding and enabling space travel. Here's how they apply to space travel: 1. Newton's First Law: In the absence of external forces, an object in space will continue moving with a constant velocity. This law is essential for spacecraft to remain in their orbits around celestial bodies or to travel in a straight line through space. 2. Newton's Second Law: Spacecraft utilize this law to achieve acceleration and change their velocity. By firing rocket engines, a spacecraft can exert a force in one direction, resulting in acceleration in the opposite direction. This law helps calculate the force required to achieve desired accelerations and velocities. 3. Newton's Third Law: When a spacecraft fires its engines, the expelled propellant generates a backward force. As per Newton's third law, an equal and opposite reaction force propels the spacecraft forward. This action-reaction principle is vital for the propulsion of spacecraft. In summary, the laws of motion enable space travel by providing a foundation for calculating trajectories, understanding propulsion systems, and ensuring the stability and maneuverability of spacecraft.

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