Q: Q2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion; (b) during its downward motion

Let the mass of the pebble be m = 0.05 kg.\nThe force due to gravity acting on the pebble is F_g = mg, directed vertically downwards. (g ≈ 9.8 m/s²)\nF_g = 0.05 kg × 9.8 m/s² = 0.49 N, vertically downwards.\n\nAir resistance (F_air) always opposes the direction of motion.\n\n(a) **During its upward motion:**\nWhen the pebble is moving upwards, both the force of gravity (F_g) and the air resistance (F_air) act in the downward direction (opposite to the direction of motion). Therefore, the net force is the sum of these two forces.\n\n* **Direction:** Vertically downwards.\n* **Magnitude:** F_net = F_g + F_air = mg + F_air. (The magnitude will be greater than 0.49 N, as F_air is a positive value).\n\n(b) **During its downward motion:**\nWhen the pebble is moving downwards, the force of gravity (F_g) acts downwards, but the air resistance (F_air) acts upwards (opposite to the direction of motion). The net force is the difference between these two forces.\n\n* **Direction:** Vertically downwards (since F_g is generally greater than F_air during free fall until terminal velocity is reached).\n* **Magnitude:** F_net = F_g - F_air = mg - F_air. (The magnitude will be less than 0.49 N).\n\n(c) **At the highest point where it is momentarily at rest:**\nAt the highest point, the pebble's instantaneous velocity is zero. Since its velocity is momentarily zero, there is no air resistance acting on it at that instant (as air resistance depends on velocity). The only force acting on the pebble at this precise moment is the force due to gravity.\n\n* **Direction:** Vertically downwards.\n* **Magnitude:** F_net = F_g = mg = 0.49 N.

Question 1

Q1. Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed; (b) a cork of mass 10 g floating on water; (c) a kite skillfully

Accepted Answer

According to Newton's First Law of Motion, if an object is moving with a constant velocity (which includes being at rest), its acceleration is zero. Consequently, the net external force acting on the object must also be zero.

(a) **A drop of rain falling down with a constant speed:**
Since the rain drop is falling with a constant speed, its acceleration is zero. Therefore, the net force acting on it is zero.

(b) **A cork of mass 10 g floating on water:**
Since the cork is floating on water, it is at rest relative to the water surface (assuming still water). Its velocity is zero, and thus its acceleration is zero. Therefore, the net force acting on it is zero.

(c) **A kite skillfully held stationary in the sky:**
As the kite is held stationary, its velocity is zero, and consequently, its acceleration is zero. Therefore, the net force acting on it is zero.

(d) **A car moving with a constant velocity of 30 km/h on a rough road:**
Since the car is moving with a constant velocity, its acceleration is zero. Therefore, the net force acting on it is zero. This implies that the forward force generated by the engine is exactly balanced by the retarding forces like friction and air resistance.

(e) **A high-speed electron in space far from all material objects, and free of electric and magnetic fields:**
In this scenario, the electron is far from all material objects (negligible gravitational influence) and free of electric and magnetic fields (no electromagnetic forces). Since there are no external forces acting on it, according to Newton's First Law, its acceleration is zero. Therefore, the net force acting on it is zero. It will continue to move with its constant high speed in a straight line.

Question 2

Q2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion; (b) during its downward motion

Accepted Answer

Let the mass of the pebble be m = 0.05 kg.
The force due to gravity acting on the pebble is F_g = mg, directed vertically downwards. (g ≈ 9.8 m/s²)
F_g = 0.05 kg × 9.8 m/s² = 0.49 N, vertically downwards.

Air resistance (F_air) always opposes the direction of motion.

(a) **During its upward motion:**
When the pebble is moving upwards, both the force of gravity (F_g) and the air resistance (F_air) act in the downward direction (opposite to the direction of motion). Therefore, the net force is the sum of these two forces.

*   **Direction:** Vertically downwards.
*   **Magnitude:** F_net = F_g + F_air = mg + F_air. (The magnitude will be greater than 0.49 N, as F_air is a positive value).

(b) **During its downward motion:**
When the pebble is moving downwards, the force of gravity (F_g) acts downwards, but the air resistance (F_air) acts upwards (opposite to the direction of motion). The net force is the difference between these two forces.

*   **Direction:** Vertically downwards (since F_g is generally greater than F_air during free fall until terminal velocity is reached).
*   **Magnitude:** F_net = F_g - F_air = mg - F_air. (The magnitude will be less than 0.49 N).

(c) **At the highest point where it is momentarily at rest:**
At the highest point, the pebble's instantaneous velocity is zero. Since its velocity is momentarily zero, there is no air resistance acting on it at that instant (as air resistance depends on velocity). The only force acting on the pebble at this precise moment is the force due to gravity.

*   **Direction:** Vertically downwards.
*   **Magnitude:** F_net = F_g = mg = 0.49 N.

Question 3

Q3. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Accepted Answer

The problem asks for the magnitude and direction of the acceleration of a body subjected to two perpendicular forces. We will first find the resultant force using vector addition and then apply Newton's Second Law to find the acceleration.

Question 4

Q4. 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. How long does the body take to stop?

Accepted Answer

This problem involves finding the time taken for a body to stop under the influence of a constant retarding force. We will use Newton's Second Law to find the acceleration and then a kinematic equation to find the time.

Question 5

Q5. A constant force acts on a body of mass 3.0 kg initially at rest. If the body moves a distance of 2.0 m in 3.0 s in the same direction as the force, find the magnitude of the force

Accepted Answer

This problem requires us to find the magnitude of a constant force. We will first use kinematic equations to determine the acceleration of the body and then apply Newton's Second Law to calculate the force.

Question 6

Q6. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s–1 to 3.5 m s–1 in 25 s. The direction of the motion of the body remains unchanged. What is the magni

Accepted Answer

The magnitude of the force is 0.18 N. Since the speed increases and the direction of motion remains unchanged, the force acts in the same direction as the motion of the body.

Question 7

Q7. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Accepted Answer

The magnitude of the acceleration of the body is 2 m s⁻². The direction of the acceleration is 36.87° with respect to the 8 N force (or 53.13° with respect to the 6 N force).

Question 8

Q8. The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child.

Accepted Answer

The average retarding force on the vehicle is 1162.5 N.

Question 9

Q9. A rocket with a lift-off mass 20000 kg is blasted upwards with an initial acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast.

Accepted Answer

The initial thrust of the blast is 2.96 x 10⁵ N.

Question 10

Q10. A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻¹ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant

Accepted Answer

The position of the body at different times is calculated by considering the phases of motion: constant velocity before the force is applied, constant acceleration while the force is applied, and constant velocity again after the force stops acting. We define the North direction as positive (+x) and South direction as negative (-x).

Question 11

Q11. A truck starts from rest and accelerates uniformly at 2.0 m s⁻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 are

Accepted Answer

When the stone is dropped from the moving truck, it initially possesses the horizontal velocity of the truck at that instant. Once dropped, it undergoes projectile motion under gravity, meaning its horizontal velocity remains constant (neglecting air resistance) and its vertical motion is governed by acceleration due to gravity.

Question 12

Q12. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Accepted Answer

To find the acceleration of the body, we first need to determine the net force acting on it. Since the two forces are perpendicular, we can find their resultant using vector addition. Once the net force is known, Newton's Second Law (F_net = ma) can be used to find the acceleration.

Question 13

Q13. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s–2. Calculate the initial thrust (force) of the blast.

Accepted Answer

The initial thrust (force) of the blast is 2.96 x 10^5 N.

Question 14

Q14. An astronaut accidentally gets separated from his small spaceship accelerating in intersteller space at a constant rate of 100 m s–2. What is the acceleration of the astronaut the

Accepted Answer

According to Newton's First Law of Motion, an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. When the astronaut gets separated from the spaceship, the force that was accelerating him (due to the spaceship's propulsion) no longer acts on him. Since there are no nearby stars to exert gravitational force, and no other external forces (like air resistance in space) are mentioned, the net force on the astronaut becomes zero.

Therefore, the astronaut will continue to move with the velocity he had at the instant of separation. His acceleration will be zero.

Question 15

Q15. 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. What is the tension in the strin

Accepted Answer

The tension in the string provides the necessary centripetal force for the stone to move in a circle. We will first calculate the tension for the given speed and then determine the maximum speed for the given maximum tension.

Question 16

Q16. If in Exercise 5.15, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajec

Accepted Answer

The correct option is (a).

**Explanation:**
When an object is moving in a circular path, its velocity vector at any instant is tangential to the circle at that point. The string provides the centripetal force, which continuously changes the direction of the velocity vector, keeping the object in a circular path. If the string breaks, the centripetal force suddenly ceases to act. According to Newton's First Law of Motion (Law of Inertia), the stone will continue to move in the direction it was moving at the instant the string broke. This direction is tangential to the circular path at that point. Gravity will, of course, act on the stone, causing it to follow a parabolic trajectory in the vertical plane, but its initial motion *at the instant of breaking* will be tangential to the circle.

Question 17

Q17. Explain why:
(a) a horse cannot pull a cart and run in empty space.
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly.
(c) it is easier to pu

Accepted Answer

Let's explain each phenomenon based on the fundamental laws of physics:

### (a) A horse cannot pull a cart and run in empty space.

**Principle:** Newton's Third Law of Motion and the necessity of external reaction forces.

**Explanation:**
For a horse to pull a cart and move forward, it needs to exert a force on the ground. According to Newton's Third Law, the ground then exerts an equal and opposite reaction force on the horse. It is this reaction force from the ground that propels the horse (and thus the cart) forward. In empty space, there is no ground or any other medium for the horse to push against. Therefore, there is no external reaction force available to accelerate the horse and cart. Without this external force, the horse cannot move itself or the cart forward, regardless of how much internal effort it exerts.

Question 18

Q18. A body of mass 2 kg is pulled by a uniform string. The string passes over a frictionless pulley and is attached to a block of mass 4 kg. The 2 kg mass is on a smooth horizontal tab

Accepted Answer

The acceleration of the system is 6.67 m/s² and the tension in the string is 13.34 N.

Question 19

Q19. A body of mass 0.4 kg moving initially with a constant speed of 10 m/s on a rough horizontal table is subjected to a constant retarding force of 10 N. Calculate the distance covere

Accepted Answer

The distance covered by the body before it stops is 2 meters.

Question 20

Q20. A body of mass 5 kg is suspended by a string. Calculate the tension in the string when:
(a) the body is at rest.
(b) the body moves up with an acceleration of 2.0 m/s².
(c) the

Accepted Answer

The tension in the string for the different scenarios is:
(a) 50 N
(b) 60 N
(c) 40 N
(d) 50 N
(e) 50 N

Question 21

Q21. A body of mass 10 kg is placed on a horizontal table. The coefficient of kinetic friction between the body and the table is 0.2. A horizontal force of 30 N is applied to the body.

Accepted Answer

The acceleration of the body is 1.0 m/s².

Question 22

Q22. A body of mass 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle of 30° with the horizontal, the body just begins to slide. Calculate the coeffici

Accepted Answer

The coefficient of static friction between the body and the plane is approximately 0.577.

Question 23

Q23. A nucleus is at rest in the laboratory frame of reference. If it disintegrates into two smaller nuclei, the products must move in opposite directions.

Accepted Answer

Yes, this statement is correct. This phenomenon can be explained by the principle of conservation of linear momentum.

**Explanation:**
1.  **Initial State:** The nucleus is initially at rest in the laboratory frame of reference. This means its initial velocity is zero. Therefore, the total initial linear momentum of the system (the nucleus) is zero.
    `P_initial = M * V_initial = M * 0 = 0`

2.  **Disintegration:** When the nucleus disintegrates into two smaller nuclei (let's call them fragment 1 and fragment 2), the total linear momentum of the system must remain conserved, according to the law of conservation of linear momentum, in the absence of any external forces. Since nuclear forces are internal to the system, there are no external forces acting during disintegration.

3.  **Final State:** Let `m1` and `m2` be the masses of the two fragments, and `v1` and `v2` be their respective velocities after disintegration. The total final linear momentum of the system is `P_final = m1 * v1 + m2 * v2`.

4.  **Conservation of Momentum:** Applying the conservation of linear momentum:
    `P_initial = P_final`
    `0 = m1 * v1 + m2 * v2`
    `m1 * v1 = - m2 * v2`

5.  **Conclusion:** This equation shows that the momentum of fragment 1 (`m1 * v1`) must be equal in magnitude and opposite in direction to the momentum of fragment 2 (`m2 * v2`). Since mass is a scalar and always positive, for their momenta to be opposite in direction, their velocities (`v1` and `v2`) must also be in opposite directions. Thus, the products must move in opposite directions.

Question 24

Q24. An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV (10^4 eV) and the second with 100 keV. Which is faster, the electron or the p

Accepted Answer

To determine which particle is faster and to find the ratio of their speeds, we will use the formula for kinetic energy, K.E. = (1/2)mv², and the given values for masses and kinetic energies.

Question 25

Q25. A rocket is in a steady vertical ascent with a constant thrust of 2 × 10^5 N. The mass of the rocket is 10^4 kg. The exhaust speed is 1000 m/s. Calculate the rate of consumption of

Accepted Answer

This problem involves the principle of rocket propulsion, where thrust is generated by the expulsion of exhaust gases. The thrust force is directly related to the exhaust speed and the rate of fuel consumption.

Question 26

Q26. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Accepted Answer

To find the acceleration of the body, we first need to determine the net force acting on it. Since the two forces are perpendicular, we can use vector addition to find the resultant force. Then, we will apply Newton's second law (F_net = ma) to find the acceleration.

Question 27

Q27. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Accepted Answer

The magnitude of the acceleration of the body is 2 m/s² and its direction is 36.87° with respect to the 8 N force.

Question 28

Q28. A body of mass 10 kg is acted upon by a force of 30 N. If the coefficient of kinetic friction between the body and the surface is 0.2, calculate the acceleration of the body.

Accepted Answer

The acceleration of the body is 1.04 m/s².

Question 29

Q29. A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Accepted Answer

The magnitude of the acceleration of the body is 2 m/s². The direction of the acceleration is approximately 36.87° with respect to the 8 N force (or 53.13° with respect to the 6 N force).

Question 30

Q30. A body of mass 10 kg is acted upon by a force of 50 N. If the initial velocity of the body is 10 m/s, what is its velocity after 4 s, assuming the force acts along the direction of

Accepted Answer

The velocity of the body after 4 s is 30 m/s.

Chapter 5: Laws of Motion