RBSE Solutions Class 11 Physics Chapter 4 Laws of Motion

RBSE Class 11 Physics Chapter 4 Textbook Exercises with Solutions

RBSE Class 11 Physics Chapter 4 Very Short Answer Type Questions

 

Question 1. If the net force on an object is zero, then, what will be its acceleration?
Answer: The acceleration will be zero. This is because, according to the formula \( F = ma \), if the net force \( F \) is zero, then \( 0 = ma \). Since mass \( m \) cannot be zero for an object, the acceleration \( a \) must be zero. This shows that an object with zero net force will not accelerate.
In simple words: If no force acts on something, it won't speed up or slow down; its acceleration will be zero.

🎯 Exam Tip: Remember Newton's First Law: an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

 

Question 2. Write the formula for momentum of a body.
Answer: The formula for momentum of a body is given by \( \vec{p} = m \vec{v} \). Here, \( \vec{p} \) represents the momentum of the body, \( m \) is the mass of the body, and \( \vec{v} \) is the velocity of the body. Momentum is a vector quantity, meaning it has both magnitude and direction, just like velocity.
In simple words: Momentum is found by multiplying a body's mass by its velocity.

🎯 Exam Tip: Always remember that momentum is a vector quantity, so its direction is the same as the velocity's direction.

 

Question 4. What are the directions of action and reaction in Newton's third law of motion?
Answer: In Newton's third law of motion, the action and reaction forces are always opposite to each other in direction. They also always act on different bodies. This means that if object A pushes object B in one direction, object B pushes object A back with the same force but in the opposite direction.
In simple words: Action and reaction forces push in opposite ways and happen to different things.

🎯 Exam Tip: A key point of Newton's third law is that action and reaction forces never cancel each other out because they act on different objects.

 

Question 5. Give an example of a system with variable mass.
Answer: A classic example of a system with variable mass is rocket propulsion. As a rocket burns its fuel, it expels exhaust gases, which causes its total mass to decrease over time. This continuous change in mass is crucial for its movement.
In simple words: A rocket losing fuel as it flies is an example of something with changing mass.

🎯 Exam Tip: Systems with variable mass are often encountered in problems involving rockets, conveyor belts, or leaking containers, where mass is either added or removed from the system.

 

Question 6. Of two contact surfaces, whose value is more: Static or kinetic friction?
Answer: The value of static friction is always more than kinetic friction. Static friction (\( f_s \)) acts when an object is at rest and resists the start of motion, while kinetic friction (\( f_k \)) acts when an object is already moving. It takes more force to get something moving (overcoming static friction) than to keep it moving (overcoming kinetic friction).
In simple words: Static friction is stronger than kinetic friction, meaning it's harder to start moving an object than to keep it moving.

🎯 Exam Tip: Remember that static friction is a self-adjusting force that increases up to a maximum value, which is known as limiting friction, before an object starts to move.

 

Question 7. Whose value is more \( \mu_s \) or \( \mu_k \)?
Answer: The value of \( \mu_s \) (coefficient of static friction) is always more than \( \mu_k \) (coefficient of kinetic friction). This is because static friction (\( f_s \)) is greater than kinetic friction (\( f_k \)). This explains why more force is needed to initiate motion than to maintain it against friction.
In simple words: The number for static friction (\( \mu_s \)) is higher than for kinetic friction (\( \mu_k \)), showing static friction is stronger.

🎯 Exam Tip: The coefficients of friction (\( \mu_s \) and \( \mu_k \)) are dimensionless quantities that depend on the nature of the surfaces in contact.

 

Question 8. Which force acts on a uniform circular motion?
Answer: In uniform circular motion, the force that acts on the object is called centripetal force. This force always points towards the center of the circular path and is responsible for continuously changing the direction of the object's velocity, keeping it in a circle. Without it, the object would fly off in a straight line.
In simple words: Centripetal force keeps an object moving in a circle by pulling it towards the center.

🎯 Exam Tip: Centripetal force is not a new type of force; it is the name given to any force (like tension, gravity, or friction) that causes circular motion.

 

Question 9. How does a vehicle obtain a centripetal force on a levelled circular path?
Answer: A vehicle obtains the necessary centripetal force on a leveled circular path primarily through frictional force. The friction between the vehicle's tires and the road surface pushes the vehicle towards the center of the turn. This frictional force prevents the vehicle from skidding outwards and allows it to follow the curved path safely.
In simple words: On a flat, round turn, friction between the tires and the road gives a vehicle the push it needs to stay in the circle.

🎯 Exam Tip: The maximum speed a vehicle can take a turn on a level road without skidding depends on the coefficient of static friction and the radius of the turn.

 

Question 10. How does a centripetal force other than frictional force is obtained on a banked circular path?
Answer: On a banked circular path, a centripetal force can be obtained from the horizontal component of the normal force, which is given by \( N \sin\theta \). Banking the road means tilting it inwards. This tilt allows a part of the normal force (perpendicular to the road surface) to act horizontally, providing the necessary centripetal force, reducing reliance on friction, especially at higher speeds.
In simple words: On a tilted (banked) road, part of the pushing force from the road itself helps vehicles turn, so they don't only rely on tire grip.

🎯 Exam Tip: Banking of roads is essential for high-speed turns, as it allows vehicles to take curves even with minimal friction, providing safety and stability.

 

Question 2. Why is Newton's first law of motion called law of inertia?
Answer: Newton's first law of motion is called the law of inertia because it describes how objects resist changes in their state of motion. The law states that an object will remain at rest or in uniform motion unless an external force acts on it. Inertia is precisely this property of matter that causes it to resist any change in its state of rest or motion. Therefore, the first law directly defines and explains the concept of inertia.
In simple words: Newton's first law is about inertia because it explains how things naturally resist changing their movement unless a force pushes them.

🎯 Exam Tip: Inertia is an inherent property of mass; the more massive an object, the greater its inertia and the more force needed to change its motion.

 

Question 3. Explain why a cricketer moves his hand backwards while holding a catch?
Answer: A cricketer moves their hand backwards while catching a ball to increase the time interval over which the ball's momentum changes. According to the impulse-momentum theorem (\( Impulse = F \cdot \Delta t = \Delta p \)), if the change in momentum (\( \Delta p \)) is constant (the ball's momentum goes from a high value to zero), increasing the time of impact (\( \Delta t \)) will reduce the average force (\( F \)) applied to the hands. This lessens the impact force, preventing injury to the cricketer's hands.
In simple words: Pulling hands back when catching a ball spreads out the stopping time, which makes the force on the hands much smaller and safer.

🎯 Exam Tip: This principle of increasing interaction time to reduce force is used in many safety designs, such as airbags in cars or padded sports equipment.

 

Question 4. Define force.
Answer: Force is a physical quantity that attempts to change a body's state of motion, meaning it can cause an object at rest to move, or an object in motion to speed up, slow down, or change direction. It is a vector quantity, possessing both magnitude and direction, and its SI unit is the Newton (N). Force is essentially a push or a pull.
In simple words: Force is a push or pull that can make an object move, stop, or change direction.

🎯 Exam Tip: Remember that force is a vector, so both its strength and direction are important when describing it.

 

Question 5. When the acceleration of a particle is measured in an inertial system as zero. Can we say that no force acts on the particle? Explain.
Answer: Yes, if a particle's acceleration is measured as zero in an inertial reference frame, we can definitively say that no *net* force acts on it. This is a direct consequence of Newton's first and second laws of motion, which are valid in inertial frames. An inertial frame is one where Newton's laws hold true. According to Newton's second law, \( F_{net} = ma \). If \( a = 0 \), then \( F_{net} = 0 \). Even if multiple forces are acting, they must all cancel each other out, resulting in zero net force. For example, a book resting on a table has gravitational force pulling it down and normal force pushing it up, but these cancel out, so its acceleration is zero.
In simple words: If an object isn't speeding up or slowing down in an inertial system, it means all the pushes and pulls on it are perfectly balanced.

🎯 Exam Tip: An inertial frame of reference is one that is either at rest or moving with a constant velocity, not accelerating. Earth is often approximated as an inertial frame for many everyday phenomena.

 

Question 6. According to Newton's third law of motion, in a game of tug, each team pulls the opposing team with equal force. How can one team win and the other one loses?
Answer: In a tug-of-war, Newton's third law applies to the forces between the two teams. Team A pulls Team B with the same force that Team B pulls Team A. However, a team wins not by pulling harder on the other team, but by applying a greater force on the ground. When a team pulls the rope, they push their feet against the ground. The ground then exerts an equal and opposite force on the team's feet, pushing them forward. The team that manages to generate a larger forward friction force from the ground, by pushing harder and more effectively, will be able to overcome the other team's grip on the ground and pull them across the line. So, winning depends on the force a team exerts on the ground, not directly on the opposing team.
In simple words: In tug-of-war, both teams pull the rope equally, but the winning team pushes the ground harder, getting a bigger push-back from the ground to move forward.

🎯 Exam Tip: Winning a tug-of-war is about maximizing the static friction force with the ground, not about overpowering the opponent's pull on the rope itself.

 

Question 8. Define impulse acting on a body.
Answer: Impulse is the total effect of a force on a body's motion. It is defined as the product of the average force applied to a body and the time interval over which that force acts. Mathematically, impulse \( \vec{I} \) is given by \( \vec{I} = \vec{F}_{avg} \cdot \Delta t \). Impulse is a vector quantity and represents the change in momentum of the body.
In simple words: Impulse is how much a force changes an object's motion, calculated by multiplying force by the time it acts.

🎯 Exam Tip: Remember that impulse has the same units as momentum (Newton-seconds or kg m/s) and is directly related to the change in an object's momentum.

 

Question 9. What is an impulsive force?
Answer: An impulsive force is a very large force that acts for a very short duration of time. Examples include the force during a collision, the impact force when catching a ball, or the force of a hammer hitting a nail. Although the time of contact is brief, the magnitude of the force is significant, leading to a substantial change in momentum.
In simple words: An impulsive force is a strong push or pull that happens very quickly, like during a crash.

🎯 Exam Tip: While an impulsive force is large, its total effect, or impulse, depends on both its magnitude and the short time duration for which it acts.

 

Question 10. Write the impulse-momentum theorem?
Answer: The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its linear momentum. In simpler terms, when a force acts on an object for a certain period, it changes the object's momentum. This change is precisely what the impulse represents. Mathematically, it is expressed as \( I = \Delta \vec{p} = \vec{p}_2 - \vec{p}_1 \), where \( I \) is the impulse and \( \Delta \vec{p} \) is the change in momentum.
In simple words: The impulse-momentum theorem says that a push or pull over time (impulse) directly causes a change in an object's movement (momentum).

🎯 Exam Tip: This theorem is very useful for analyzing collisions and impacts where forces are not constant over time.

 

Question 11. Write the law of conservation of momentum.
Answer: The law of conservation of momentum states that if no net external force acts on a system of bodies, then the total linear momentum of the system remains constant. This means that within an isolated system, momentum cannot be created or destroyed, only transferred between objects. This principle is derived from Newton's second law, \( F = \frac{dp}{dt} \), where if \( F = 0 \), then \( \frac{dp}{dt} = 0 \), implying that momentum \( p \) is constant.
In simple words: If nothing from outside pushes or pulls a group of objects, their total movement always stays the same.

🎯 Exam Tip: This law is fundamental for understanding collisions and explosions in isolated systems, where the total momentum before and after the event remains the same.

 

Question 13. Why a gun recoils backward when a bullet is fired?
Answer: A gun recoils backward when a bullet is fired due to the law of conservation of momentum. Before firing, both the gun and the bullet are at rest, so the total momentum of the system is zero. When the gun is fired, it exerts a forward force on the bullet, causing the bullet to move forward with a certain momentum. To conserve the total momentum of the system (keeping it at zero), the gun must move backward with an equal and opposite momentum. This backward movement of the gun is called recoil.
In simple words: A gun kicks back when fired because for every action (bullet flying forward), there's an equal and opposite reaction (gun moving backward) to keep the total movement balanced.

🎯 Exam Tip: The gun's recoil velocity is much smaller than the bullet's velocity because the gun's mass is significantly larger than the bullet's mass.

 

Question 14. How many types of friction are there?
Answer: There are primarily two types of friction:
(i) Static friction: This type of friction acts on a body when it is at rest and an external force tries to move it. It opposes the initiation of motion. The maximum value of static friction, just before the body starts to move, is known as "Limiting friction."
(ii) Dynamic or kinetic friction: This type of friction acts on a body when it is already in motion. It opposes the relative sliding motion between the surfaces in contact. The value of kinetic friction is generally less than the maximum static (limiting) friction.
Static friction keeps objects from moving, while kinetic friction slows down moving objects.
In simple words: There are two main types of friction: static (when things are still) and kinetic (when things are moving).

🎯 Exam Tip: Static friction is a variable force that adjusts itself to be equal and opposite to the applied force, up to its limiting value.

 

Question 15. Define centripetal acceleration.
Answer: Centripetal acceleration is the acceleration experienced by an object moving in a circular path. This acceleration always points towards the center of the circle, perpendicular to the object's velocity. It is responsible for continuously changing the direction of the object's velocity, thereby keeping it on the circular path, without changing its speed. The continuous change in direction of velocity is what causes acceleration.
In simple words: Centripetal acceleration is the pull towards the center that keeps an object moving in a circle, constantly changing its direction.

🎯 Exam Tip: Centripetal acceleration is given by \( a_c = \frac{v^2}{r} \) or \( a_c = \omega^2 r \), where \( v \) is tangential speed, \( \omega \) is angular speed, and \( r \) is the radius of the circle.

 

Question 16. Why is a road banked on a circular turn?
Answer: A road is banked on a circular turn to safely provide the necessary centripetal force for vehicles. When a road is banked, it is tilted inwards. This tilt allows a component of the normal force (the force exerted by the road perpendicular to its surface) to act horizontally, towards the center of the circular path. This horizontal component provides part, or all, of the centripetal force required to make the turn, reducing the reliance on friction. Banking helps prevent vehicles from skidding outwards, especially at higher speeds, and keeps all wheels in contact with the road, enhancing safety and stability.
In simple words: Roads are tilted on curves (banked) so that part of the road's push helps cars turn, making it safer and reducing the need for tire grip.

🎯 Exam Tip: Banking helps maintain vehicle stability at curves, as it shifts the responsibility of centripetal force from friction to the normal force, making it safer in adverse weather conditions too.

 

Question 18. Define non-inertial frame of reference.
Answer: A non-inertial frame of reference is one in which Newton's first and second laws of motion do not appear to be valid without the introduction of fictitious (or pseudo) forces. Unlike inertial frames, non-inertial frames are accelerating with respect to an inertial frame. Examples include an accelerating car, a rotating merry-go-round, or a falling elevator. In these frames, objects may appear to accelerate even when no real force is acting on them.
In simple words: A non-inertial frame is a moving viewpoint where Newton's laws don't seem to work without adding "fake" forces.

🎯 Exam Tip: The key characteristic of a non-inertial frame is that it is accelerating, which leads to the apparent need for fictitious forces like the centrifugal force or Coriolis force.

 

Question 19. Is Earth an inertial frame of reference?
Answer: Strictly speaking, Earth is not an inertial frame of reference because it is constantly revolving around the Sun and rotating on its axis, both of which involve centripetal acceleration. Therefore, it possesses acceleration and should be treated as a non-inertial frame. However, the centripetal acceleration of Earth due to its rotation and revolution is negligibly small compared to the acceleration due to gravity (g = \( 9.8m/s^2 \)). For most everyday terrestrial experiments and calculations, this small acceleration can be ignored, and the Earth is often considered an *approximately* inertial frame of reference. For instance, the centripetal acceleration on Earth is about \( 0.006 m/s^2 \).
Answer: Earth undergoes centripetal acceleration because it revolves around the Sun. Therefore, it is technically a non-inertial frame of reference. We can calculate this acceleration:
\( a_c = \frac{v^2}{R} = \omega^2 R = \left(\frac{2\pi}{T}\right)^2 R = \frac{4\pi^2 R}{T^2} \)
Putting \( R = 1 \text{ AU} = 1.496 \times 10^{11} \text{ m} \) and \( T = 365\frac{1}{4} \text{ days} \)
\( T = 365.25 \times 24 \times 60 \times 60 \text{ sec} = 3.15576 \times 10^7 \text{ sec} \)
\( a_c = \frac{4\pi^2 (1.496 \times 10^{11})}{(3.15576 \times 10^7)^2} \approx 0.006 \text{ m/s}^2 \)
This value is very small compared to the acceleration due to gravity (\( g = 9.8 \text{ m/s}^2 \)). Hence, for most practical experiments on Earth, it can be considered an inertial frame.
In simple words: Earth is always moving, so it's not truly an inertial frame. But because its acceleration is very tiny, we usually treat it as one for most experiments.

🎯 Exam Tip: Always specify whether Earth is being treated as "approximately inertial" or "strictly non-inertial" based on the context and precision required by the problem.

RBSE Class 11 Physics Chapter 4 Long Answer Type Questions

 

Question 1. What is Newton's second law of motion? Define it. Obtain Newton's first law from it.
Answer:
**Momentum and Newton's Second Law of Motion**
Linear momentum (\( \vec{p} \)) is a physical quantity that represents the motion possessed by a body. It is mathematically defined as the product of the body's mass (\( m \)) and its velocity (\( \vec{v} \)). The unit of momentum is Kilogram meter per second (Kg m/s).
Momentum is a vector quantity, meaning it has both magnitude and direction, which is the same as the direction of velocity.
For example, if a ball of mass \( m_1 \) and a car of mass \( m_2 \) (\( m_2 > m_1 \)) move with the same velocity \( \vec{v} \), their momenta are \( \vec{p_1} = m_1 \vec{v} \) and \( \vec{p_2} = m_2 \vec{v} \). Since \( m_2 > m_1 \), it follows that \( p_2 > p_1 \), meaning the car has greater momentum.
Similarly, if two objects of the same mass are thrown at different velocities, the one with greater velocity will have greater momentum. If two objects of masses \( m_1 \) and \( m_2 \) have equal momentum (\( m_1 \vec{v_1} = m_2 \vec{v_2} \)), then if \( m_2 > m_1 \), it must be that \( v_2 < v_1 \), meaning the lighter body possesses greater velocity.
Momentum can also be related to kinetic energy (\( E_k \)). We know that \( E_k = \frac{1}{2} mv^2 \) and \( p = mv \). From \( p = mv \), we can write \( v = p/m \). Substituting this into the kinetic energy equation gives:
\( E_k = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{1}{2} m \frac{p^2}{m^2} = \frac{p^2}{2m} \)
**Newton's Second Law of Motion:**
Newton's second law states that the rate of change of an object's linear momentum is directly proportional to the net external force applied to it. This change in momentum occurs in the direction of the applied force. Mathematically, it is expressed as:
\( \vec{F} = k \frac{d\vec{p}}{dt} \)
Where \( \vec{F} \) is the net external force, \( \frac{d\vec{p}}{dt} \) is the rate of change of momentum, and \( k \) is a proportionality constant. In the SI system, \( k=1 \), so:
\( \vec{F} = \frac{d\vec{p}}{dt} \)
Since \( \vec{p} = m\vec{v} \), we can write:
\( \vec{F} = \frac{d(m\vec{v})}{dt} \)
If the mass \( m \) is constant (which is often the case for most everyday objects):
\( \vec{F} = m \frac{d\vec{v}}{dt} \)
Since \( \frac{d\vec{v}}{dt} = \vec{a} \) (acceleration), the law becomes:
\( \vec{F} = m\vec{a} \)
This equation shows that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This fundamental law explains how forces cause objects to accelerate, providing a precise measure of force.
**Derivation of Newton's First Law from the Second Law:**
Newton's first law (law of inertia) can be obtained directly from the second law.
According to Newton's second law:
\( \vec{F} = \frac{d\vec{p}}{dt} \)
If the net external force \( \vec{F} \) acting on the body is zero, then:
\( 0 = \frac{d\vec{p}}{dt} \)
This implies that the momentum \( \vec{p} \) is constant.
If \( \vec{p} = m\vec{v} \) is constant, and the mass \( m \) is constant, then the velocity \( \vec{v} \) must also be constant.
If \( \vec{v} \) is constant, it means the object's speed and direction are unchanging. This covers two scenarios:
1. If the object was initially at rest (\( \vec{v} = 0 \)), it will remain at rest.
2. If the object was initially in motion (\( \vec{v} \neq 0 \)), it will continue to move with the same constant velocity in a straight line.
This is precisely the statement of Newton's first law of motion, which describes the behavior of objects when no net external force acts on them.
In simple words: Newton's second law says that force makes things accelerate. If there's no force, then there's no acceleration, which means the object stays still or keeps moving at a steady speed, explaining Newton's first law.

🎯 Exam Tip: When deriving the first law from the second, clearly state the assumption that external force is zero and explain how constant momentum leads to constant velocity.

 

Question 2. Explain Newton's third law of motion with the help of two examples.
Answer:
**Newton's Third Law of Motion**
Newton's third law of motion states that for every action, there is always an equal and opposite reaction. This means that if an object 'A' exerts a force on an object 'B', then object 'B' simultaneously exerts a force of equal magnitude and opposite direction back on object 'A'. Forces always occur in pairs; one body cannot exert a force on another without experiencing a force itself. This law highlights a fundamental symmetry in nature.
**Example 1: A Book on a Table**
Consider a book placed on a table. The book exerts a downward force on the table due to its weight (\( W = mg \)), which is the "action" force. According to Newton's third law, the table simultaneously exerts an equal and opposite force on the book in the upward direction. This upward force is called the normal force (\( N \)). So, \( N = mg \) in magnitude, and its direction is opposite to the weight. Both forces act on two different objects (book on table, table on book), fulfilling the conditions of the third law.

**Example 2: Recoil of a Gun**
When a bullet is fired from a gun, the gun exerts a force on the bullet, accelerating it forward (action). In response, the bullet exerts an equal and opposite force on the gun, causing the gun to move backward (recoil). This is the "reaction" force. Because the gun has a much larger mass than the bullet, its backward velocity (recoil) is much smaller. The backward movement is felt as a "kick" at the shoulder of the person firing the gun.

The backward movement of the gun is called recoil.
In simple words: Newton's third law says forces come in pairs: if you push something, it pushes you back. Like a book on a table, the book pushes down, and the table pushes up. Or when a gun fires, the bullet goes forward, and the gun kicks backward.

🎯 Exam Tip: Always specify that action and reaction forces act on *different* bodies and are simultaneous, not sequential.

 

Question 3. Write impulse-momentum theorem and prove it. How will you find out the impulse from a graph?
Answer:
**Impulse-Momentum Theorem**
The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its linear momentum. In simpler terms, the total effect of a force acting for a certain duration (impulse) results in an equivalent change in the object's motion (momentum).
An impulsive force is a force that acts for a very short interval of time, such as hitting a ball, jumping, diving, or catching. Such forces don't remain constant; they often start from zero, reach a maximum, and then return to zero.
**Proof of Impulse-Momentum Theorem**
According to Newton's second law of motion, the net external force \( \vec{F} \) acting on a body is equal to the rate of change of its linear momentum \( \vec{p} \):
\( \vec{F} = \frac{d\vec{p}}{dt} \)
Rearranging this equation, we get:
\( d\vec{p} = \vec{F} dt \)
Now, we integrate both sides over a time interval from \( t_1 \) to \( t_2 \). Let the momentum change from \( \vec{p_1} \) to \( \vec{p_2} \) during this interval:
\( \int_{t_1}^{t_2} d\vec{p} = \int_{t_1}^{t_2} \vec{F} dt \)
\( [\vec{p}]_{p_1}^{p_2} = \int_{t_1}^{t_2} \vec{F} dt \)
\( \vec{p_2} - \vec{p_1} = \int_{t_1}^{t_2} \vec{F} dt \)
The term \( \int_{t_1}^{t_2} \vec{F} dt \) is defined as the impulse (\( \vec{I} \)). Therefore,
\( \vec{I} = \vec{p_2} - \vec{p_1} = \Delta \vec{p} \)
This proves the impulse-momentum theorem: the impulse applied is equal to the change in momentum.
The dimensional formula of impulse is \( [M^1 L^1 T^{-1}] \), which is the same as that of momentum. Its SI unit is Newton-second (N-s) or kilogram meter per second (kg m/s).
**Finding Impulse from a Force-Time Graph**
Impulse can be determined from a force-time graph. Since impulse is given by \( \vec{I} = \int \vec{F} dt \), it represents the area under the force-time curve.

If you have a graph of force versus time, the impulse delivered over a certain time interval (\( t_1 \) to \( t_2 \)) is simply the area under that curve between \( t_1 \) and \( t_2 \). For constant force, this area is a simple rectangle. For variable forces, you might need to use integration or graphical methods (like counting squares or approximating shapes) to find the area.
In simple words: The impulse-momentum theorem says that a push's effect equals the change in an object's movement. You can find this "effect" by measuring the area under a force-time graph.

🎯 Exam Tip: When dealing with force-time graphs, remember that the area under the curve always represents the impulse, regardless of whether the force is constant or variable.

 

Question 4. Write the law of conservation of momentum for a system of N particles. Obtain it from Newton's second law of motion. Explain the law of conservation of momentum with the help of an example.
Answer:
**Principle of Conservation of Linear Momentum**
The law of conservation of linear momentum states that if the net external force acting on a system of bodies (or particles) is zero, then the total linear momentum of the system remains constant. This means that the initial total momentum of the system equals the final total momentum, even if individual momenta within the system change due to internal forces (like collisions or explosions).
For a system of N particles, if the vector sum of all external forces is zero (\( \vec{F}_{ext} = \sum \vec{F}_i = 0 \)), then the total linear momentum \( \vec{P}_{total} = \sum \vec{p}_i = m_1\vec{v_1} + m_2\vec{v_2} + ... + m_N\vec{v_N} \) remains constant.
**Obtaining the Law from Newton's Second Law**
Newton's second law for a system of particles states that the net external force acting on the system is equal to the rate of change of its total linear momentum:
\( \vec{F}_{ext} = \frac{d\vec{P}_{total}}{dt} \)
If the net external force acting on the system is zero, i.e., \( \vec{F}_{ext} = 0 \), then:
\( 0 = \frac{d\vec{P}_{total}}{dt} \)
This implies that \( \vec{P}_{total} \) must be a constant vector.
Thus, if no external force acts on a system, its total linear momentum is conserved.
**Example: Recoil of a Gun**
Consider a system consisting of a gun and a bullet. Before firing, both the gun and the bullet are at rest. Therefore, their initial velocities are zero (\( \vec{u}_{bullet} = 0 \), \( \vec{u}_{gun} = 0 \)). The total initial momentum of the system is:
\( \vec{P}_{initial} = m_{bullet}\vec{u}_{bullet} + m_{gun}\vec{u}_{gun} = m_{bullet}(0) + m_{gun}(0) = 0 \)
When the gun is fired, internal forces cause the bullet to move forward with a velocity \( \vec{v}_{bullet} \) and the gun to recoil backward with a velocity \( \vec{v}_{gun} \). The total final momentum of the system is:
\( \vec{P}_{final} = m_{bullet}\vec{v}_{bullet} + m_{gun}\vec{v}_{gun} \)
Since the firing process involves only internal forces (the force between the gun and the bullet), and assuming no significant external forces act on the system during the brief firing interval, the total momentum must be conserved:
\( \vec{P}_{initial} = \vec{P}_{final} \)
\( 0 = m_{bullet}\vec{v}_{bullet} + m_{gun}\vec{v}_{gun} \)
This equation shows that \( m_{bullet}\vec{v}_{bullet} = -m_{gun}\vec{v}_{gun} \). The negative sign indicates that the gun's recoil velocity (\( \vec{v}_{gun} \)) is in the opposite direction to the bullet's velocity (\( \vec{v}_{bullet} \)). This illustrates that the forward momentum of the bullet is balanced by the backward momentum of the gun, keeping the total momentum of the system constant at zero. The larger mass of the gun ensures its recoil velocity is smaller than the bullet's velocity (\( |\vec{v}_{gun}| = \frac{m_{bullet}}{m_{gun}} |\vec{v}_{bullet}| \)).
This principle is crucial for understanding how rockets work, as they expel gas in one direction to gain momentum in the opposite direction, and for analyzing collisions.
In simple words: The law of momentum conservation says that if no outside forces push on a group of objects, their total movement stays the same. For example, when a gun fires, the bullet goes forward, and the gun kicks back, but their total movement remains balanced.

🎯 Exam Tip: Always clearly define the "system" for which momentum is conserved; this helps identify whether forces are internal or external.

 

Question 5. Explain the motion of a rocket and obtain the important formula for its velocity?
Answer:
**Motion of a Rocket**
Rocket motion is an excellent example of a variable mass system and is governed by the principle of conservation of momentum. A rocket works by expelling high-velocity exhaust gases from its rear downwards. According to Newton's third law, the force exerted by the rocket on the exhaust gases is equal and opposite to the force exerted by the exhaust gases on the rocket. This upward force on the rocket, known as thrust, propels the rocket forward. As the fuel burns and gases are expelled, the mass of the rocket continuously decreases.

**Formula for Rocket Velocity (Rocket Equation)**
To obtain the formula for the rocket's velocity, we apply the conservation of momentum to the rocket-fuel system.
Let \( M \) be the mass of the rocket at time \( t \), and \( v \) be its velocity.
At time \( t + dt \), a small mass \( dm \) of gas is ejected with a relative velocity \( v_g \) (velocity of exhaust gas relative to the rocket) downwards, and the rocket's mass becomes \( M - dm \), and its velocity becomes \( v + dv \).
The total momentum of the system at time \( t \) is \( P_t = Mv \).
The total momentum of the system at time \( t+dt \) is \( P_{t+dt} = (M - dm)(v + dv) + dm(v - v_g) \). (Assuming \( v_g \) is positive downwards, and rocket velocity \( v \) is positive upwards, so the absolute velocity of exhaust gas is \( v - v_g \)).
Using the principle of conservation of momentum, if no external forces act on the system during this small time interval:
\( P_t = P_{t+dt} \)
\( Mv = (M - dm)(v + dv) + dm(v - v_g) \)
\( Mv = Mv + Mdv - vdm - dmdv + vdm - v_gdm \)
Canceling \( Mv \) from both sides and neglecting the term \( dmdv \) (which is a product of two small differentials and hence very small):
\( 0 = Mdv - v_gdm \)
\( Mdv = v_gdm \)
\( \frac{dv}{dt} = \frac{v_g}{M} \frac{dm}{dt} \) (This is the thrust equation, where \( v_g \frac{dm}{dt} \) is the thrust force).
Now, considering an external force, like gravity (\( -Mg \)), acting downwards (assuming upwards as positive):
\( M\frac{dv}{dt} = v_g \frac{dm}{dt} - Mg \)
\( dv = v_g \frac{dm}{M} - g dt \)
Integrating this equation from an initial state (\( M_0, v_0 \)) to a final state (\( M, v \)):
\( \int_{v_0}^{v} dv = \int_{M_0}^{M} v_g \frac{dm}{M} - \int_{0}^{t} g dt \)
Assuming \( v_g \) and \( g \) are constant:
\( v - v_0 = v_g [\ln M]_{M_0}^{M} - g[t]_{0}^{t} \)
\( v - v_0 = v_g (\ln M - \ln M_0) - gt \)
\( v - v_0 = v_g \ln \left(\frac{M}{M_0}\right) - gt \)
Since \( \ln \left(\frac{M}{M_0}\right) = -\ln \left(\frac{M_0}{M}\right) \):
\( v - v_0 = -v_g \ln \left(\frac{M_0}{M}\right) - gt \)
Or, \( v = v_0 - v_g \ln \left(\frac{M_0}{M}\right) - gt \)
This is the fundamental rocket equation, also known as the Tsiolkovsky rocket equation, which gives the velocity of the rocket at any time \( t \). If the rocket starts from rest (\( v_0 = 0 \)) and we use base 10 logarithm:
\( v = -2.303 v_g \log_{10} \left(\frac{M_0}{M}\right) - gt \)
The negative sign indicates that the exhaust velocity is opposite to the rocket's motion. This formula shows that the final velocity depends on the exhaust speed, the ratio of the initial mass (rocket + fuel) to the final mass (rocket only), and the gravitational acceleration. The term \( -gt \) accounts for the reduction in velocity due to gravity. The speed acquired when all fuel is burnt is called the burn-out speed.
The rocket's velocity increases significantly as the mass of the rocket decreases (when fuel is expelled), demonstrating how the conservation of momentum leads to propulsion.
In simple words: A rocket moves by shooting gas out its back. This push from the gas makes the rocket go forward, and as it loses mass (burns fuel), it speeds up. The main formula for its speed depends on how fast the gas leaves and how much mass the rocket loses.

🎯 Exam Tip: Remember to clearly define all variables in the rocket equation (initial mass, final mass, exhaust velocity, initial velocity, gravitational acceleration) and explain the significance of each term.

 

Question 6. How many types of friction are there? Write their laws.
Answer: Newton's second law of motion states that the rate at which an object's linear momentum changes is directly proportional to the external force applied to it. This change always happens in the direction of the applied force. The law also helps define force quantitatively.
In simple words: Newton's second law says that how fast an object's movement changes depends on how much force is pushed on it. The object will speed up or slow down in the same direction as the push.

🎯 Exam Tip: When defining Newton's second law, always mention both the direct proportionality to force and the alignment of the change in momentum with the force's direction to score full marks.

 

Question 7. Explain how will you find out the direction of static friction?
Answer: Static frictional force is the friction that acts between a resting object and the surface it is on. This force keeps the object from moving and must be overcome for the object to start sliding. The direction of static friction is always opposite to the direction in which the body *tends* to move. For instance, if you try to push an object forward, static friction will act backward. Static friction depends on the nature of the contact surfaces and only appears when an external force attempts to move the body. It's a self-adjusting force that increases up to a maximum (limiting friction) before motion begins.
In simple words: Static friction acts against how an object tries to move when it's still. If you push an object forward, the static friction will push backward to stop it from moving.

🎯 Exam Tip: Remember that static friction always opposes the *tendency* of motion, not necessarily the actual motion itself, which is a key distinction.

 

Question 8. Define the coefficients of static and kinetic friction. How will you find out their value?
Answer:
Static Frictional Force: This is the friction that acts between a stationary object and the surface it rests on. It keeps the object at rest and must be overcome for the object to start moving. The static friction \( f_s \) depends on the nature of the surfaces in contact, only appears when a force tries to move the body, and is a self-adjusting force. For an object to move, the applied force must overcome this static friction.
Coefficient of Static Friction (\( \mu_s \)): This is the ratio of the maximum static friction (\( (f_s)_{\max} \)) to the normal reaction force (N) between the surfaces. It is found using the formula: \( \mu_s = \frac{(f_s)_{\max}}{N} \). It has no units and depends on the surface properties (roughness, wetness), not the apparent contact area.

Kinetic Frictional Force: When an applied force is greater than the limiting static friction, the body starts moving. The friction acting between the moving surfaces, which opposes their motion, is called kinetic frictional force (\( f_k \)). Its value is usually less than the maximum static friction.

Coefficient of Kinetic Friction (\( \mu_k \)): Similar to static friction, kinetic friction is directly proportional to the normal reaction force (N): \( f_k \propto N \), so \( f_k = \mu_k N \). Thus, \( \mu_k = \frac{f_k}{N} \). It also has no units and depends on the nature of the contact surfaces. The value of \( \mu_k \) is always less than \( \mu_s \).
In simple words: Static friction stops things from moving, while kinetic friction slows them down when they are already moving. Their "coefficients" are numbers that tell us how sticky or slippery two surfaces are, helping us calculate how much friction there will be.

🎯 Exam Tip: Clearly differentiate between static and kinetic friction, noting that static friction prevents initial motion while kinetic friction acts during motion, and that \( \mu_s \) is generally greater than \( \mu_k \).

 

Question 9. Explain the circular motion of an object in horizontal plane and determine the formula for its time period.
Answer: When an object moves in a circular path on a horizontal plane, like a mass \( m \) tied to a string of length \( L \) and whirled around, it experiences centripetal force. The string sweeps out a cone, with the angle \( \theta \) between the string and the vertical normal. The forces acting on the mass are tension \( T \) from the string and its weight \( mg \). The vertical component of tension, \( T \cos \theta \), balances the weight \( mg \). The horizontal component, \( T \sin \theta \), provides the necessary centripetal force for circular motion.
To find the time period \( \tau \):

From the geometry, the radius of the circular path is \( r = L \sin \theta \).
The velocity of the mass is \( v = r\omega \), where \( \omega = \frac{2\pi}{\tau} \).
So, \( v = r \frac{2\pi}{\tau} = \frac{2\pi L \sin \theta}{\tau} \).

The centripetal force \( F_c = \frac{mv^2}{r} \) is provided by \( T \sin \theta \).
From vertical equilibrium, \( T \cos \theta = mg \).

Dividing the two equations:
\( \frac{T \sin \theta}{T \cos \theta} = \frac{mv^2/r}{mg} \)
\( \tan \theta = \frac{v^2}{rg} \)

Substitute \( v = \frac{2\pi r}{\tau} \):
\( \tan \theta = \frac{(2\pi r/\tau)^2}{rg} = \frac{4\pi^2 r^2}{rg\tau^2} = \frac{4\pi^2 r}{g\tau^2} \)

From \( r = L \sin \theta \), we get \( \tau^2 = \frac{4\pi^2 r}{g \tan \theta} = \frac{4\pi^2 (L \sin \theta)}{g (\sin \theta / \cos \theta)} = \frac{4\pi^2 L \cos \theta}{g} \)

Therefore, the time period is:
\( \tau = 2\pi \sqrt{\frac{L \cos \theta}{g}} \)

This formula shows that the time period depends on the length of the string, the angle it makes with the vertical, and the acceleration due to gravity. The speed of the object must be just right to maintain the conical path.
In simple words: When an object swings in a circle on a flat surface, the string holding it makes a cone shape. The forces pulling it keep it in this circle. The time it takes to complete one full circle can be found using a specific formula that depends on the string's length and how wide the cone is.

🎯 Exam Tip: When deriving the time period, ensure you correctly resolve the tension force into its horizontal and vertical components, and equate the horizontal component to the centripetal force.

 

Question 10. Explain the circular motion of an object in vertical plane. Find out the formula for the tension produced at the highest and lowest points in the string.
Answer: When an object of mass \( m \) is whirled in a vertical circle with a string of radius \( r \), its motion is more complex than horizontal circular motion because gravity continuously affects its speed and the tension in the string. The velocity of the body and the tension in the string change at different points in the circle. The tension in the string is highest at the bottom and lowest at the top.

At the lowest point (A):
The tension \( T_1 \) acts upwards, and the weight \( mg \) acts downwards. The net upward force provides the centripetal force.
\( T_1 - mg = \frac{m v_{1}^{2}}{r} \)
Here, \( v_1 \) is the velocity at the lowest point. This means the tension at the bottom must support the weight and also provide the force needed to keep the object moving in a circle.

At the highest point (B):
Both the tension \( T_2 \) and the weight \( mg \) act downwards, contributing to the centripetal force.
\( T_2 + mg = \frac{m v_{2}^{2}}{r} \)
Here, \( v_2 \) is the velocity at the highest point.

Minimum velocity to complete a vertical circle without the string slackening:
At the highest point, for the string not to slacken, the tension \( T_2 \) must be at least zero.
If \( T_2 = 0 \), then \( mg = \frac{m v_{2}^{2}}{r} \)
\( v_{2}^{2} = gr \)
\( v_2 = \sqrt{gr} \) (This is the minimum velocity at the highest point.)

Using the principle of conservation of energy between the highest (B) and lowest (A) points:
Total Energy at A = Total Energy at B
\( \frac{1}{2} m v_{1}^{2} = mg(2r) + \frac{1}{2} m v_{2}^{2} \)
(The height difference is \( 2r \))
Multiply by \( \frac{2}{m} \):
\( v_{1}^{2} = 4gr + v_{2}^{2} \)

Substitute \( v_2 = \sqrt{gr} \) into the equation for \( v_1 \):
\( v_{1}^{2} = 4gr + (\sqrt{gr})^2 \)
\( v_{1}^{2} = 4gr + gr \)
\( v_{1}^{2} = 5gr \)
\( V_1 = \sqrt{5gr} \) (This is the minimum velocity at the lowest point.)

Now, we can find the maximum tension at the lowest point, when \( v_1 = \sqrt{5gr} \):
\( T_1 = mg + \frac{m v_{1}^{2}}{r} = mg + \frac{m (\sqrt{5gr})^2}{r} = mg + \frac{m(5gr)}{r} = mg + 5mg = 6mg \)

This means the tension at the lowest point must be six times the object's weight to ensure it completes the loop.
In simple words: When you swing something in a full circle up and down, gravity makes it tricky. The rope is tightest at the bottom and loosest at the top. To make sure it doesn't fall, the object needs a certain minimum speed at the top, and this speed requires a much stronger pull on the rope at the bottom.

🎯 Exam Tip: Remember that tension at the highest point can be zero for minimum velocity, while tension at the lowest point must be \( 6mg \) for the loop to be completed.

 

Question 11. Explain the motion of a vehicle on a circular path and give the formula for maximum velocity.
Answer: When a vehicle moves on a circular path, it needs a centripetal force to keep it from skidding outwards.

Motion on a Level Circular Path:
On a flat, circular road, the weight of the car (\( mg \)) acts downwards, and the normal reaction (\( N \)) from the road acts upwards. These two forces balance each other, so \( N = mg \). The necessary centripetal force \( \left( \frac{mv^2}{r} \right) \) is provided entirely by the frictional force (\( F \)) between the tires and the road.
For safe driving without skidding, the centripetal force must be less than or equal to the maximum static friction:
\( \frac{mv^2}{r} \leq F \)
Since \( F \leq \mu_s N \) and \( N = mg \), where \( \mu_s \) is the coefficient of static friction:
\( \frac{mv^2}{r} \leq \mu_s mg \)
\( v^2 \leq \mu_s rg \)
The maximum safe velocity \( v_{\max} \) for a level circular road is:
\( v_{\max} = \sqrt{\mu_s rg} \)
This maximum velocity depends on the road's radius, the friction coefficient, and gravity, but not on the vehicle's mass.

Motion on a Banked Circular Path:
To allow vehicles to travel at higher speeds on curved roads without relying solely on friction, roads are often "banked." This means the outer edge of the road is raised higher than the inner edge, creating an angle \( \theta \) with the horizontal. This banking helps to manage the centripetal force.
On a banked road, three main forces act on the vehicle:
1. Weight (\( mg \)) acting vertically downwards.
2. Normal reaction (\( N \)) from the road, perpendicular to the surface.
3. Frictional force (\( F \)) acting along the surface, either up or down the slope depending on the speed.

By resolving these forces, the horizontal component of the normal force, and potentially the friction, contribute to the centripetal force, allowing for higher safe speeds. If friction is also considered up to its maximum value, the maximum velocity on a banked road is given by:
\( v = \sqrt{\frac{rg(\mu_{s} + \tan \theta)}{1-\mu_{s} \tan \theta}} \)
This formula ensures that the vehicle can maintain contact with the road and receive enough centripetal force without skidding. The banking angle helps reduce the reliance on friction, making turns safer at higher speeds.
In simple words: When a car turns in a circle, it needs a side force to keep it from sliding off. On a flat road, friction from the tires provides this force, but there's a speed limit. On a "banked" road (tilted road), the road's tilt helps provide this turning force, letting cars go faster around curves more safely.

🎯 Exam Tip: For level roads, friction alone provides centripetal force; for banked roads, both the normal force component and friction contribute. Remember that banking is essential for higher speeds and safety.

 

Question 13. Explain the motion of a body on an inclined plane.
Answer: When an object is placed on an inclined plane (a sloped surface at an angle \( \theta \) to the horizontal), several forces act on it. These forces determine whether the object will slide down, stay at rest, or move up the plane.

The primary forces are:
1. The object's weight (\( mg \)), which acts vertically downwards.
2. The normal reaction force (\( N \)), which acts perpendicular to the inclined surface.
3. The frictional force (\( f_s \) or \( f_k \)), which acts parallel to the surface, opposing the tendency of motion.

We resolve the weight \( mg \) into two components:
- \( mg \sin \theta \), acting parallel to the plane, pulling the object downwards along the slope.
- \( mg \cos \theta \), acting perpendicular to the plane, pressing the object against the surface.

The normal reaction force \( N \) balances the perpendicular component of weight: \( N = mg \cos \theta \).

Static Condition:
If the object is at rest, the static frictional force \( f_s \) balances the component of weight pulling it down the slope: \( f_s = mg \sin \theta \).
The object will remain at rest as long as \( mg \sin \theta \leq (f_s)_{\max} \), where \( (f_s)_{\max} = \mu_s N = \mu_s mg \cos \theta \).
So, \( mg \sin \theta \leq \mu_s mg \cos \theta \)
\( \tan \theta \leq \mu_s \).
The maximum angle at which the object can stay at rest is called the angle of repose (\( \theta_s \)), where \( \tan \theta_s = \mu_s \).

Motion Down the Plane:
If the angle of inclination \( \theta \) is greater than \( \theta_s \), the object starts sliding down. In this case, kinetic friction \( f_k = \mu_k N = \mu_k mg \cos \theta \) acts up the slope. The net force causing acceleration \( a \) down the plane is:
\( F_{net} = mg \sin \theta - f_k = mg \sin \theta - \mu_k mg \cos \theta = ma \)
\( a = g(\sin \theta - \mu_k \cos \theta) \)

If the object moves with constant velocity down the plane, then acceleration \( a=0 \), which implies \( mg \sin \theta = \mu_k mg \cos \theta \), or \( \tan \theta = \mu_k \). This is called the angle of sliding (\( \theta_k \)).

The motion of a body on an inclined plane depends on the interplay between gravity and friction, which are affected by the slope angle and the surface properties.
In simple words: When an object is on a sloped surface, gravity tries to pull it down the slope. Friction tries to hold it in place or slow it down. Whether the object slides, stays put, or moves at a steady speed depends on how steep the slope is and how much friction there is.

🎯 Exam Tip: Always resolve the weight \( mg \) into components parallel (\( mg \sin \theta \)) and perpendicular (\( mg \cos \theta \)) to the inclined plane first; this is crucial for setting up the force equations correctly.

 

Question 7. Explain how will you find out the direction of static friction?
Answer: Static friction is the force that stops an object from moving when a force is applied to it. It acts between a stationary object and the surface it rests on. The direction of static friction is always *opposite* to the direction in which the object *tends to move*. For example, if you push a box to the right, the static friction acts to the left, preventing the box from sliding. This force adjusts itself to match the applied force up to a certain maximum point. Static friction plays a crucial role in preventing everyday objects from slipping or sliding unintentionally.
In simple words: Static friction prevents an object from moving. Its direction is always against the way you try to push or pull the object.

🎯 Exam Tip: The key point for static friction is that it acts *only* when there's an attempt to move an object, and its direction is always opposite to the potential motion.

 

Question 14. Explain Inertial and Non-inertial Frames of Reference.
Answer:
A **Frame of Reference** is a coordinate system used to describe the motion of a body. It provides a point of view from which observations are made.

An **Inertial Frame of Reference** is a frame that is either at rest or moving with a constant velocity relative to another inertial frame. In such a frame, Newton's first and second laws of motion are valid, meaning an object not acted upon by external forces will maintain its state of motion (either rest or constant velocity). These frames are unaccelerated. For example, our Earth (for many practical purposes), a space shuttle moving at a constant velocity, or a rocket flying at a constant speed are considered inertial frames. These frames are essential because the fundamental laws of physics are simplest in them.

A **Non-inertial Frame of Reference** is a frame that is accelerating relative to an inertial frame. In a non-inertial frame, Newton's laws of motion do not hold true without introducing fictitious forces (like centrifugal force or Coriolis force) to explain observed motions. Examples include a rotating carousel or a speeding-up car.
In simple words: A frame of reference is a way to describe motion. An inertial frame is steady or moves smoothly, and Newton's laws work perfectly there. A non-inertial frame is speeding up, slowing down, or turning, and needs extra "fake" forces to make Newton's laws work.

🎯 Exam Tip: Remember that Newton's laws are most naturally applied in inertial frames. When working in non-inertial frames, be prepared to introduce pseudo-forces to account for the frame's acceleration.

Rbse Class 11 Physics Chapter 4 Numerical Questions

 

Question 1. Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary. It is given that the coefficient of static friction between the box and the floor of the train is 0.13. (g = 9.8m/s²)
Answer:
To find the maximum acceleration of the train, we consider the forces acting on the box. The box stays stationary on the floor due to the static frictional force. This static friction provides the necessary force for the box to accelerate with the train.
Given:
Coefficient of static friction, \( \mu_s = 0.13 \)
Acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \)

For the box to remain stationary, the static frictional force (\( f_s \)) must be equal to or greater than the force required to accelerate the box (\( ma \)). At maximum acceleration (\( a_{max} \)), the static frictional force is at its maximum value, also known as limiting friction.
The maximum static frictional force is given by \( f_s = \mu_s N \), where \( N \) is the normal force. Since the box is on a horizontal surface, the normal force \( N \) balances the weight of the box, so \( N = mg \).
Therefore, \( f_s = \mu_s mg \).

According to Newton's second law, the force causing the box to accelerate is \( F = ma_{max} \).
At the maximum acceleration, these two forces are equal:
\( ma_{max} = \mu_s mg \)

We can cancel out the mass \( m \) from both sides:
\( a_{max} = \mu_s g \)

Now, we substitute the given values:
\( a_{max} = 0.13 \times 9.8 \)
\( a_{max} = 1.274 \text{ m/s}^2 \)
So, the maximum acceleration of the train for the box to remain stationary is approximately \( 1.27 \text{ m/s}^2 \). It's important that the friction is enough to keep the box from sliding.
In simple words: The train can speed up only as much as the friction between the box and the floor allows. We find this maximum speed-up by multiplying the friction number (coefficient) by gravity. This gives us the fastest the train can go without the box slipping.

🎯 Exam Tip: For problems involving static friction on an accelerating platform, remember that the maximum static friction is what provides the acceleration. The mass of the object will cancel out, simplifying the calculation.

 

Question 2. A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
Answer:
First, we need to convert the cyclist's speed from km/h to m/s:
\( v = 18 \text{ km/h} = 18 \times \frac{5}{18} \text{ m/s} = 5 \text{ m/s} \)

Given values:
Speed \( v = 5 \text{ m/s} \)
Radius of turn \( r = 3 \text{ m} \)
Coefficient of static friction \( \mu_s = 0.1 \)
Acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \)

For the cyclist to make the turn without slipping, the required centripetal force must be provided by the static friction between the tyres and the road. We calculate two values:
1. The **required centripetal force** (\( F_c \)) needed to keep the cyclist moving in a circle.
\( F_c = \frac{mv^2}{r} = \frac{m \times (5)^2}{3} = \frac{25m}{3} \approx 8.33m \)

2. The **maximum possible static frictional force** (\( F_f \)) that the road can provide.
\( F_f = \mu_s N \), where \( N \) is the normal force. On a level road, \( N = mg \).
\( F_f = \mu_s mg = 0.1 \times m \times 9.8 = 0.98m \)

Now, we compare the required force with the maximum available force:
Required centripetal force \( F_c \approx 8.33m \)
Maximum static frictional force \( F_f = 0.98m \)

Since the required centripetal force (\( 8.33m \)) is much greater than the maximum available static frictional force (\( 0.98m \)), the static friction is not enough to keep the cyclist on the circular path. Therefore, the cyclist will slip and fall down while taking the turn. This shows how crucial friction is for safe turning.
In simple words: The cyclist is going too fast for the turn. The push needed to stay in the circle is much bigger than what the tyres can get from friction. So, the cyclist will slide and fall.

🎯 Exam Tip: Always compare the *required* centripetal force with the *maximum available* frictional force. If the required force exceeds the available friction, slipping will occur.

 

Question 3. A bomb at rest explodes into three fragments. Two fragments fly off at right angles to each other. These two fragments: the one with mass 2 kg moves with a speed of 12 m/s, and the other one with mass 1 kg moves with 8 m/s. The speed of the third fragment is 20 m/s, then calculate its mass.
Answer:
This problem uses the principle of conservation of momentum. Since the bomb is initially at rest, its total momentum before the explosion is zero. After the explosion, the total momentum of all three fragments must still be zero.

Given values for the fragments:
For the first fragment:
Mass \( m_1 = 2 \text{ kg} \)
Velocity \( v_1 = 12 \text{ m/s} \)
Momentum \( p_1 = m_1 v_1 = 2 \times 12 = 24 \text{ kg m/s} \)

For the second fragment:
Mass \( m_2 = 1 \text{ kg} \)
Velocity \( v_2 = 8 \text{ m/s} \)
Momentum \( p_2 = m_2 v_2 = 1 \times 8 = 8 \text{ kg m/s} \)

For the third fragment:
Velocity \( v_3 = 20 \text{ m/s} \)
Mass \( m_3 = ? \)

The first two fragments fly off at right angles to each other. We can find their combined momentum (\( \vec{p}' \)) using the Pythagorean theorem, as their momentum vectors are perpendicular:
\[ p' = \sqrt{p_1^2 + p_2^2} \]
\[ p' = \sqrt{(24)^2 + (8)^2} \]
\[ p' = \sqrt{576 + 64} \]
\[ p' = \sqrt{640} \]
\[ p' \approx 25.298 \text{ kg m/s} \]

According to the conservation of momentum, the vector sum of all momenta after the explosion must be zero. This means the momentum of the third fragment (\( \vec{p}_3 \)) must be equal in magnitude and opposite in direction to the combined momentum of the first two fragments (\( \vec{p}' \)).
So, \( p_3 = p' \).
\( p_3 \approx 25.298 \text{ kg m/s} \)

We also know that \( p_3 = m_3 v_3 \). We can now find the mass of the third fragment:
\( 25.298 = m_3 \times 20 \)
\( m_3 = \frac{25.298}{20} \)
\( m_3 \approx 1.2649 \text{ kg} \)

Rounding to two decimal places, the mass of the third fragment is approximately \( 1.27 \text{ kg} \). The total momentum remains zero after the explosion, a fundamental concept in physics.
In simple words: A bomb at rest has zero total movement. When it breaks into three pieces, the movement of all pieces must still add up to zero. We found the combined movement of the first two pieces and then used that to calculate the mass of the third piece.

🎯 Exam Tip: Always remember that momentum is a vector quantity. For explosions or collisions where total momentum is conserved, use vector addition for momenta, especially when objects move at angles to each other.

 

Question 4. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find out the acceleration of the masses and the tension in the string when the masses are released.
Answer:
Given values:
Mass \( m_1 = 12 \text{ kg} \) (the heavier mass)
Mass \( m_2 = 8 \text{ kg} \) (the lighter mass)
Acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \) (we will use this value for acceleration, and then approximate to 10 m/s\(^2\) for tension to match common textbook examples).

When the masses are released, the heavier mass \( m_1 \) will move downwards, and the lighter mass \( m_2 \) will move upwards with the same acceleration, let's call it \( a \). Let \( T \) be the tension in the string.

We can write the equations of motion for each mass:
For mass \( m_1 \) (moving downwards):
\( m_1 g - T = m_1 a \) ........ (1)

For mass \( m_2 \) (moving upwards):
\( T - m_2 g = m_2 a \) ........ (2)

To find the acceleration \( a \), we add equation (1) and equation (2):
\( (m_1 g - T) + (T - m_2 g) = m_1 a + m_2 a \)
\( m_1 g - m_2 g = (m_1 + m_2) a \)
\( (m_1 - m_2) g = (m_1 + m_2) a \)

Now, we can solve for \( a \):
\( a = \frac{(m_1 - m_2) g}{m_1 + m_2} \)
Substitute the given values:
\( a = \frac{(12 - 8) \times 9.8}{12 + 8} \)
\( a = \frac{4 \times 9.8}{20} \)
\( a = \frac{39.2}{20} \)
\( a = 1.96 \text{ m/s}^2 \)
The acceleration is approximately \( 2 \text{ m/s}^2 \). This means both masses speed up or slow down by this amount each second.

Next, to find the tension \( T \), we can use equation (2) and approximate \( a \approx 2 \text{ m/s}^2 \) and \( g \approx 10 \text{ m/s}^2 \) as often done in simplified physics problems:
\( T = m_2 g + m_2 a \)
\( T = m_2 (g + a) \)
\( T = 8 \text{ kg} \times (10 \text{ m/s}^2 + 2 \text{ m/s}^2) \)
\( T = 8 \times 12 \)
\( T = 96 \text{ N} \)
The tension in the string is \( 96 \text{ N} \). This tension is the force exerted by the string on each mass, pulling them upwards or downwards.
In simple words: The heavier mass pulls down, making both masses move. We use special physics rules to find how fast they speed up (acceleration) and how hard the string pulls (tension). The heavier mass goes down at 1.96 m/s\(^2\), and the string pulls with 96 N of force.

🎯 Exam Tip: When dealing with Atwood machine problems, clearly define the positive direction for each mass and draw free-body diagrams to correctly set up the force equations. Remember to sum these equations to eliminate tension when solving for acceleration first.

 

Question 5. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball when the mass of the ball is 0.15 kg?
Answer:
Impulse is the change in momentum of an object. The mass of the ball is \( m = 0.15 \text{ kg} \). The initial speed is \( 54 \text{ km/h} \).

First, convert the speed from km/h to m/s:
\( v = 54 \text{ km/h} = 54 \times \frac{5}{18} \text{ m/s} = 15 \text{ m/s} \)

The ball is deflected by an angle of 45° without a change in speed. This means the magnitude of the initial velocity (\( u \)) and final velocity (\( v \)) are both 15 m/s. Let's assume the deflection happens such that the momentum component along the axis of deflection changes direction, while the component perpendicular to it remains unchanged. In this scenario, the angle each velocity vector makes with the axis of change is \( \frac{45^\circ}{2} = 22.5^\circ \).

Let's consider the x-axis as the direction along which the momentum changes (the direction of the impulse). The initial and final momentum vectors are \( \vec{p}_i \) and \( \vec{p}_f \), each with magnitude \( mv \). The x-component of momentum changes from \( mv \cos(22.5^\circ) \) to \( -mv \cos(22.5^\circ) \). The y-component remains the same (effectively zero change, for impulse calculation along this axis).

Change in momentum in the x-direction (\( \Delta p_x \)):
\( \Delta p_x = p_{fx} - p_{ix} \)
\( \Delta p_x = (-mv \cos 22.5^\circ) - (mv \cos 22.5^\circ) \)
\( \Delta p_x = -2mv \cos 22.5^\circ \)

The impulse in the x-direction (\( I_x \)) is equal to this change in momentum:
\( I_x = -2mv \cos 22.5^\circ \)

Substitute the values:
\( I_x = -2 \times 0.15 \text{ kg} \times 15 \text{ m/s} \times \cos(22.5^\circ) \)
We know \( \cos(22.5^\circ) \approx 0.9239 \).
\( I_x = -2 \times 0.15 \times 15 \times 0.9239 \)
\( I_x = -4.15755 \text{ Ns} \)

The change in momentum in the y-direction (\( \Delta p_y \)) is zero because the vertical components cancel out:
\( \Delta p_y = 0 \)
So, the impulse in the y-direction (\( I_y \)) is 0.

The total impulse is the sum of impulses in x and y directions. Since the impulse is purely in the x-direction, its magnitude is the absolute value of \( I_x \).
Total Impulse \( |I| = \sqrt{I_x^2 + I_y^2} = \sqrt{(-4.15755)^2 + 0^2} \approx 4.15755 \text{ Ns} \)
Rounding this, the magnitude of the impulse imparted to the ball is approximately \( 4.2 \text{ Ns} \). The negative sign shows the impulse acts in the direction opposite to the initial momentum component along the chosen axis of deflection.
In simple words: When the ball hits the bat, its direction changes, but its speed stays the same. The "push" (impulse) given to the ball is found by how much its movement changes. Because of how it bounces off, the change in movement is about 4.2 Ns in the opposite direction.

🎯 Exam Tip: For deflection problems, resolve initial and final momenta into components along and perpendicular to the axis of impulse. The impulse is usually perpendicular to the surface of impact, and it is the change in the momentum component along this direction.

 

Question 6. 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 takes to stop?
Answer:
Given values:
Retarding force \( F = -50 \text{ N} \) (negative because it slows the body down)
Mass of the body \( m = 20 \text{ kg} \)
Initial speed \( u = 15 \text{ m/s} \)
Final speed \( v = 0 \text{ m/s} \) (because the body stops)

First, we need to find the acceleration (\( a \)) of the body using Newton's second law of motion, \( F = ma \):
\( a = \frac{F}{m} \)
\( a = \frac{-50 \text{ N}}{20 \text{ kg}} \)
\( a = -2.5 \text{ m/s}^2 \)
The negative sign indicates that this is a deceleration or a retarding acceleration, meaning the body is slowing down.

Next, we use a kinematic equation to find the time (\( t \)) it takes for the body to stop:
\( v = u + at \)
Substitute the known values:
\( 0 = 15 \text{ m/s} + (-2.5 \text{ m/s}^2) t \)
Rearrange the equation to solve for \( t \):
\( 2.5 t = 15 \)
\( t = \frac{15}{2.5} \)
\( t = 6 \text{ s} \)
So, the body takes 6 seconds to stop. This calculation helps understand how quickly objects come to rest under a constant opposing force.
In simple words: A push-back force slows down a moving object. We first figure out how fast it's slowing down (acceleration). Then, we use that to find out exactly how many seconds it will take for the object to stop completely. It takes 6 seconds.

🎯 Exam Tip: Remember to use consistent signs for force and acceleration (e.g., if initial velocity is positive, a retarding force/acceleration should be negative). Always clearly state the kinematic equation used for time-related problems.

 

Question 7. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m/s, then what is the recoil speed of the gun?
Answer:
This problem uses the principle of conservation of momentum. Before the gun is fired, both the gun and the shell are at rest, so the total initial momentum of the system is zero. After firing, the total momentum must still be zero, meaning the momentum of the shell in one direction must be balanced by the momentum of the gun in the opposite direction.

Given values:
Mass of the shell \( m_{shell} = 0.020 \text{ kg} \)
Mass of the gun \( m_{gun} = 100 \text{ kg} \)
Muzzle speed of the shell \( v_{shell} = 80 \text{ m/s} \)
Initial velocities of both gun and shell are 0 m/s.
Let \( v_{gun} \) be the recoil speed of the gun.

According to the law of conservation of momentum:
Total initial momentum = Total final momentum
\( m_{shell} u_{shell} + m_{gun} u_{gun} = m_{shell} v_{shell} + m_{gun} v_{gun} \)
\( 0 + 0 = m_{shell} v_{shell} + m_{gun} v_{gun} \)
\( 0 = (0.020 \text{ kg} \times 80 \text{ m/s}) + (100 \text{ kg} \times v_{gun}) \)
\( 0 = 1.6 + 100 v_{gun} \)
\( 100 v_{gun} = -1.6 \)
\( v_{gun} = -\frac{1.6}{100} \)
\( v_{gun} = -0.016 \text{ m/s} \)
The recoil speed of the gun is \( 0.016 \text{ m/s} \). The negative sign indicates that the gun moves in the direction opposite to the shell. This is a clear demonstration of Newton's third law of motion in action.
In simple words: Before firing, the gun and bullet are still, so their total movement is zero. When the gun fires, the bullet shoots forward, and to keep the total movement zero, the gun kicks backward with a small speed. That small kick is the recoil speed.

🎯 Exam Tip: For problems involving conservation of momentum, always define a positive direction and remember that the total momentum before an event (like firing a gun or an explosion) equals the total momentum after the event.

 

Question 8. The motion of a particle of mass 0.1 kg is described by \( y = 0.3 t+\frac{9.8}{2} t^{2} \). Find out the force acting on the particle.
Answer:
Given:
Mass of the particle \( m = 0.1 \text{ kg} \)
Equation describing the position of the particle along the y-axis: \( y(t) = 0.3t + \frac{9.8}{2} t^2 \)

To find the force acting on the particle, we first need to find its acceleration. We can do this by taking derivatives of the position equation with respect to time.

1. **Find the velocity (\( v \)) by differentiating the position equation (\( y \)) with respect to time (\( t \)):**
\( v(t) = \frac{dy}{dt} = \frac{d}{dt} \left(0.3t + \frac{9.8}{2} t^2\right) \)
\( v(t) = \frac{d}{dt} (0.3t + 4.9t^2) \)
\( v(t) = 0.3 + 2 \times 4.9t \)
\( v(t) = 0.3 + 9.8t \text{ m/s} \)

2. **Find the acceleration (\( a \)) by differentiating the velocity equation (\( v \)) with respect to time (\( t \)):**
\( a(t) = \frac{dv}{dt} = \frac{d}{dt} (0.3 + 9.8t) \)
\( a(t) = 0 + 9.8 \)
\( a = 9.8 \text{ m/s}^2 \)
This shows that the particle is moving with a constant acceleration of \( 9.8 \text{ m/s}^2 \), which is equal to the acceleration due to gravity.

3. **Calculate the force (\( F \)) acting on the particle using Newton's second law (\( F = ma \)):**
\( F = 0.1 \text{ kg} \times 9.8 \text{ m/s}^2 \)
\( F = 0.98 \text{ N} \)
The force acting on the particle is \( 0.98 \text{ N} \). This constant force is likely the gravitational force acting on the particle.
In simple words: We are given an equation that tells us where the particle is at any time. We use this to find its speed, then its acceleration. Once we have the acceleration, we can multiply it by the particle's mass to find the force acting on it, which comes out to 0.98 N.

🎯 Exam Tip: Remember that velocity is the first derivative of position, and acceleration is the second derivative of position (or first derivative of velocity). Once acceleration is found, Newton's second law \( F=ma \) can directly give the force.

 

Question 9. 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 frequency of \( \frac{2}{3} \text{ Hz} \). What is the tension in the string? What is the maximum speed with which it can be whirled if the breaking tension of the string is 200 N?
Answer:
Given values:
Mass of the stone \( m = 0.25 \text{ kg} \)
Radius of the circle \( r = 1.5 \text{ m} \)
Frequency of whirling \( n = \frac{2}{3} \text{ Hz} \)
Maximum tension the string can withstand \( T_{max} = 200 \text{ N} \)
We will use \( \pi^2 \approx 9.86 \) for calculations as hinted by the source.

**Part 1: Calculate the tension in the string.**
The tension in the string provides the centripetal force required for circular motion. The formula for centripetal force (\( F_c \)) is given by \( F_c = \frac{mv^2}{r} \).
Since \( v = 2\pi r n \) (where \( n \) is frequency), we can substitute this into the formula for tension \( T \):
\( T = m \frac{(2\pi r n)^2}{r} \)
\( T = m \frac{4\pi^2 r^2 n^2}{r} \)
\( T = 4\pi^2 m r n^2 \)

Substitute the given values into the formula:
\( T = 4 \times 9.86 \times 0.25 \text{ kg} \times 1.5 \text{ m} \times \left(\frac{2}{3}\right)^2 \)
\( T = 4 \times 9.86 \times 0.25 \times 1.5 \times \frac{4}{9} \)
\( T = 9.86 \times 1.5 \times \frac{4}{9} \)
\( T = \frac{59.16}{9} \)
\( T \approx 6.57 \text{ N} \)
The tension in the string when whirled at this frequency is approximately \( 6.57 \text{ N} \).

**Part 2: Calculate the maximum speed.**
The maximum speed (\( v_{max} \)) at which the stone can be whirled is limited by the maximum tension (\( T_{max} \)) the string can withstand before breaking.
Using the centripetal force formula:
\( T_{max} = \frac{m v_{max}^2}{r} \)

Rearrange the formula to solve for \( v_{max} \):
\( v_{max}^2 = \frac{T_{max} r}{m} \)
\( v_{max} = \sqrt{\frac{T_{max} r}{m}} \)

Substitute the values:
\( v_{max} = \sqrt{\frac{200 \text{ N} \times 1.5 \text{ m}}{0.25 \text{ kg}}} \)
\( v_{max} = \sqrt{\frac{300}{0.25}} \)
\( v_{max} = \sqrt{1200} \)
\( v_{max} \approx 34.64 \text{ m/s} \)
The maximum speed at which the stone can be whirled without breaking the string is approximately \( 34.64 \text{ m/s} \). Whirling the stone faster than this speed would cause the string to break. This highlights the importance of string strength in circular motion.
In simple words: First, we find the pull on the string (tension) when the stone spins at a given rate. Then, knowing how much pull the string can handle before breaking (max tension), we calculate the fastest speed the stone can spin. The tension is about 6.57 N, and the maximum speed is about 34.64 m/s.

🎯 Exam Tip: Pay attention to the units (Hz for frequency vs. m/s for speed) and ensure you use the correct formula for centripetal force. For maximum speed, equate the maximum tension to the required centripetal force.

 

Question 10. A body of mass 5 kg is acted upon by two perpendicular forces 8N and 6N. Give the magnitude and direction of acceleration of the body.
Answer:
Given values:
Mass of the body \( m = 5 \text{ kg} \)
First force \( F_1 = 8 \text{ N} \)
Second force \( F_2 = 6 \text{ N} \)
The two forces are perpendicular to each other.

**Part 1: Find the magnitude of the acceleration.**
First, we need to find the net force (\( F_{net} \)) acting on the body. Since the two forces are perpendicular, we can find their resultant using the Pythagorean theorem:
\[ F_{net} = \sqrt{F_1^2 + F_2^2} \]
\[ F_{net} = \sqrt{(8 \text{ N})^2 + (6 \text{ N})^2} \]
\[ F_{net} = \sqrt{64 + 36} \]
\[ F_{net} = \sqrt{100} \]
\[ F_{net} = 10 \text{ N} \]

Now, we use Newton's second law of motion, \( F_{net} = ma \), to find the acceleration (\( a \)):
\( a = \frac{F_{net}}{m} \)
\( a = \frac{10 \text{ N}}{5 \text{ kg}} \)
\( a = 2 \text{ m/s}^2 \)
The magnitude of the acceleration of the body is \( 2 \text{ m/s}^2 \). This means its velocity changes by 2 m/s every second in the direction of the net force.

**Part 2: Find the direction of the acceleration.**
The direction of the acceleration is the same as the direction of the net force. Let \( \theta \) be the angle that the resultant force (and thus acceleration) makes with the \( F_1 \) force.
We can use trigonometry:
\[ \tan \theta = \frac{F_2}{F_1} \]
\[ \tan \theta = \frac{6 \text{ N}}{8 \text{ N}} \]
\[ \tan \theta = 0.75 \]

To find \( \theta \), we take the inverse tangent:
\( \theta = \tan^{-1}(0.75) \)
\( \theta \approx 36.87^\circ \)
The direction of the acceleration is approximately \( 36.87^\circ \) with respect to the 8 N force. Understanding how forces combine is crucial for predicting motion.
In simple words: Two forces push the body at a right angle. We first find the total push by combining these forces. Then, we divide the total push by the body's weight to find how fast it speeds up (acceleration). The body speeds up at 2 m/s\(^2\), and its direction is about 37 degrees from the 8 N force.

🎯 Exam Tip: For perpendicular forces, treat them as components of a right triangle to find the resultant force using the Pythagorean theorem. Use tangent to find the direction relative to one of the forces.