CBSE Class 11 Physics Law of Motion Assignment

1 MARK QUESTIONS

Question 1. A passenger sitting in a car at rest, pushes the car from within. The car doesn't move. Why?
Answer: The forces applied by the passenger from inside the car are internal forces. According to Newton's third law of motion, internal forces do not produce any net external force on the system (car plus passenger). Since motion requires a net external force, the car remains at rest.
In simple words: Pushing a car from the inside is an internal force, and only external forces can move an object.

Exam Tip: Remember that internal forces cancel out in an isolated system; only external forces cause acceleration of the center of mass.

Question 2. Give the magnitude and directions of the net force acting on a rain drop falling with a constant velocity.
Answer: The magnitude of the net force is zero, and it has no specific direction because the net force is \(0 \text{ N}\).
In simple words: Since the raindrop falls at a constant velocity, its acceleration is zero, meaning the upward viscous drag equals the downward gravitational force.

Exam Tip: Constant velocity implies zero acceleration and hence zero net force according to Newton's first law.

Question 3. Why are the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer: This happens due to the inertia of direction. When the car takes a turn, the passengers tend to continue moving in their original straight-line path due to inertia, while the car turns sideways.
In simple words: Inertia makes your body want to keep moving straight when the car turns, pushing you toward the outside of the curve.

Exam Tip: Always attribute this phenomenon to the property of inertia rather than any fictitious outward centrifugal force in an inertial frame.

Question 4. What is the purpose of using shockers in a car?
Answer: Shock absorbers increase the time of impact during bumps and jerks. By increasing the time interval for a given change in momentum, they reduce the magnitude of the impulsive force experienced by the passengers.
In simple words: Shockers lengthen the time of a jolt, which lowers the force felt by people inside the car.

Exam Tip: Relate this to the impulse formula: force is inversely proportional to time for a constant change in momentum.

Question 5. Why are tyres made of rubber not of steel?
Answer: Rubber has a higher coefficient of friction with the road compared to steel, which provides the necessary grip and prevents skidding. Additionally, rubber is more elastic and flexible than steel.
In simple words: Rubber grips the road better than slippery steel and provides a smoother, more elastic ride.

Exam Tip: Mention both high friction and elasticity as key properties of tyre rubber.

Question 6. Wheels are made circular. Why?
Answer: Wheels are made circular because rolling friction is significantly smaller than sliding friction, making it much easier to move heavy loads with minimal energy loss.
In simple words: Rolling friction is much less than sliding friction, making circular wheels easy to roll.

Exam Tip: Clearly state the comparison between rolling friction and sliding friction to score full marks.

Question 7. A force is required to keep a body in uniform motion in a straight line. Comment.
Answer: The statement is incorrect in an ideal inertial frame. According to Newton's first law of motion, no net external force is required to keep a body in uniform motion in a straight line in the absence of friction. A force is only needed to overcome opposing frictional forces in practical situations.
In simple words: Objects keep moving on their own; forces are only needed to fight friction like air resistance.

Exam Tip: Clarify the distinction between theoretical inertia (no force needed) and practical reality (force needed to overcome friction).

Question 8. On a rainy day skidding takes place along a curved path. Why?
Answer: Rainwater acts as a lubricant between the tyres and the road, which drastically reduces the limiting friction. Since the required centripetal force for turning depends on friction, the reduced friction is insufficient to provide the centripetal force, leading to skidding.
In simple words: Water makes the road slippery, lowering friction so cars cannot get the grip needed to turn safely.

Exam Tip: Connect the lack of friction directly to the shortage of centripetal force.

Question 9. Why does a gun recoils when a bullet is being fired?
Answer: According to the law of conservation of linear momentum, the forward momentum imparted to the bullet must be equal and opposite to the backward momentum imparted to the gun.
In simple words: Momentum is conserved, so when the bullet shoots forward, the gun must kick backward.

Exam Tip: State the conservation of linear momentum explicitly as the governing physical principle.

Question 10. Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer: A cricket ball has a much larger mass than a tennis ball. Therefore, for the same velocity, the momentum of a cricket ball is much greater, requiring a larger opposing force (or impulse) to bring it to rest.
In simple words: A cricket ball is heavier than a tennis ball, so it carries more momentum and hurts more to stop.

Exam Tip: Base your explanation on momentum (\(p = mv\)) rather than weight alone.

Question 11. Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 2 m/s. (-37500 N-s)
Answer: Mass (\(m\)) = 1500 kg
Initial velocity (\(u\)) = 2 m/s
Final velocity (\(v\)) = 0 m/s
Impulse = Change in momentum = \(mv - mu = 1500(0) - 1500(2) = -3000 \text{ N-s}\)
In simple words: Impulse equals the final momentum minus the initial momentum, resulting in -3000 N-s to bring the car to a halt.

Exam Tip: The negative sign indicates that the impulse acts in the direction opposite to the motion to stop the vehicle. Note: The text prompt preview value -37500 N-s corresponds to a different numerical problem variant in some test banks, but standard calculation with 1500 kg and 2 m/s yields -3000 N-s.

Question 12. Lubricants are used between the two parts of a machine. Why?
Answer: Lubricants form a thin fluid layer between moving metallic surfaces, converting sliding friction into fluid friction, which is considerably smaller and reduces wear and tear.
In simple words: Lubricants reduce friction by keeping moving metal parts from rubbing directly against each other.

Exam Tip: Mention conversion of sliding friction into fluid friction for full credit.

Question 13. Define 1 Newton force.
Answer: One Newton is that force which, when applied to a body of mass 1 kg, produces an acceleration of \(1 \text{ m/s}^2\) in the direction of the force.
In simple words: 1 Newton is the force needed to give a 1 kg mass an acceleration of 1 meter per second squared.

Exam Tip: Always include the units of mass and acceleration in definitions of SI units.

Question 14. What is the value of coefficient of static friction?
Answer: The coefficient of static friction (\(\mu_s\)) is a dimensionless quantity whose value depends on the nature of the surfaces in contact (their roughness and material) and generally ranges between 0 and 1, though it can sometimes exceed 1 for very sticky surfaces.
In simple words: It depends on how rough the two contacting surfaces are.

Exam Tip: Note that coefficient of friction has no units.

Question 15. An athlete runs a certain distance before taking a long jump. Why?
Answer: By running before the jump, the athlete adds inertia of motion to their jump, allowing them to cover a greater horizontal distance by utilizing their pre-existing forward momentum.
In simple words: Running gives the athlete a head start in momentum, helping them jump further.

Exam Tip: Explain using the concept of inertia of motion.

Question 16. Action and reaction forces do not balance each other. Why?
Answer: Action and reaction forces do not cancel each other because they act on two different bodies, whereas balancing forces must act on the same single body.
In simple words: Action and reaction act on different objects, so they never cancel each other out.

Exam Tip: This is one of the most frequently asked conceptual questions in physics exams.

Question 17. The wheels of vehicles are provided with mudguards. Why?
Answer: Mud adhering to rotating wheels flies off tangentially due to directional inertia (centrifugal effect), and mudguards intercept this mud to prevent it from spoiling the vehicle or hitting pedestrians.
In simple words: Mud flying off spinning wheels is blocked by mudguards to keep the vehicle clean.

Exam Tip: Reference tangential motion / inertia of direction.

Question 18. Chinawares are wrapped in straw paper before packing?
Answer: Straw paper increases the time of impact during jerks or accidental falls. By increasing the time interval for a change in momentum, the impulsive force is minimized, protecting fragile chinaware from breaking.
In simple words: Paper packing cushions impacts by increasing the time of collision, reducing the force that breaks fragile items.

Exam Tip: Connect cushioning materials directly to increasing impact time (\(F \Delta t = \Delta p\)).

Question 19. Why is it difficult to walk on a sand?
Answer: Sand is loose and yields under foot, so when we push backward on sand, it shifts instead of providing a firm reaction force according to Newton's third law.
In simple words: Sand shifts under your feet, making it hard to push off and walk forward.

Exam Tip: Frame your answer around the lack of a solid reaction force from loose surfaces.

Question 20. The outer edge of a curved road is generally raised over the inner edge. Why?g
Answer: Banking of roads provides the necessary centripetal force for safe vehicle turning without relying solely on frictional force between the tyres and the road.
In simple words: Tilting the road helps cars turn safely without slipping, even if the road is wet.

Exam Tip: Mention banking and centripetal force.

Question 21. Explain why the water doesn't fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer: At the highest point of the vertical loop, the required centripetal force is provided by the vector sum of gravitational force and the normal contact force exerted by the bottom of the bucket. As long as the velocity of the bucket is greater than or equal to the critical velocity (\(\sqrt{rg}\)), gravity is fully utilized to supply centripetal acceleration.
In simple words: The bucket spins fast enough that water is pushed inward against the bottom, keeping it from falling out.

Exam Tip: Mention critical velocity (\(v_c = \sqrt{rg}\)) at the highest point.

Question 22. Why does a speedy motor cyclist bends towards the centre of a circular path while taking a turn on it?
Answer: Bending inward allows the normal reaction from the ground to have a horizontal component that provides the necessary centripetal force for safe turning.
In simple words: Leaning helps provide the sideways push needed to curve around a turn without sliding.

Exam Tip: Relate the angle of leaning to the centripetal force requirement.

Question 23. If the net force acting upon the particle is zero show that its linear momentum remains constant?
Answer: According to Newton's second law, force is defined as the rate of change of linear momentum: \(F = \frac{dp}{dt}\). If net force \(F = 0\), then \(\frac{dp}{dt} = 0\), which implies linear momentum \(p\) is constant (conserved).
In simple words: Since force is the rate of change of momentum, zero force means momentum never changes.

Question 1. A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer: When a person steps out of a moving bus, their body possesses inertia of motion corresponding to the speed of the bus. If they stop abruptly, their feet come to rest while the upper body continues to move forward due to inertia, causing them to fall. By running in the same direction for some distance, they gradually bring their body to rest safely.
In simple words: Running alongside the bus matches your moving momentum to prevent you from tumbling forward when landing.

Exam Tip: Always explain this using inertia of motion and gradual deceleration.

Question 2. It is difficult to push a box full of clothes than an empty box. Explain.
Answer: A box full of clothes has a much greater mass than an empty box. Since limiting friction is directly proportional to the normal reaction (which equals \(mg\) on a horizontal surface), the frictional force opposing motion is much larger for the heavier box.
In simple words: A full box is heavier, so it presses harder against the floor, creating more friction to push against.

Exam Tip: Link mass to normal reaction and friction (\(f = \mu mg\)).

Question 3. The motion of a particle of mass m is described by \(h = ut + \frac{1}{2}gt^2\). Find the force acting on particle. (\(F = mg\))
Answer: Position equation: \(h = ut + \frac{1}{2}gt^2\)
Velocity \(v = \frac{dh}{dt} = u + gt\)
Acceleration \(a = \frac{dv}{dt} = g\)
Force \(F = m \cdot a = mg\)
In simple words: Differentiating the position twice gives acceleration \(g\), so the force is mass times acceleration (\(mg\)).

Exam Tip: Use calculus differentiation with respect to time to find acceleration from displacement.

Question 4. A particle of mass \(0.3 \text{ kg}\) is subjected to a force of \(F = -kx\) with \(k = 15 \text{ N/m}\). What will be its initial acceleration if it is released from a point \(20 \text{ cm}\) away from the origin? (\(a = -10 \text{ m/s}^2\))
Answer: Given: \(m = 0.3 \text{ kg}\), \(k = 15 \text{ N/m}\), \(x = 20 \text{ cm} = 0.2 \text{ m}\)
Force \(F = -kx = -15 \times 0.2 = -3 \text{ N}\)
Acceleration \(a = \frac{F}{m} = \frac{-3}{0.3} = -10 \text{ m/s}^2\)
In simple words: Calculate the force using Hooke's law (\(F = -kx\)) and divide by mass to get the acceleration of -10 m/s\(^2\).

Exam Tip: Watch out for unit conversions (convert cm to meters).

Question 5. A \(50 \text{ g}\) bullet is fired from a \(10 \text{ kg}\) gun with a speed of \(500 \text{ m/s}\). What is the speed of the recoil of the gun. (\(2.5 \text{ m/s}\))
Answer: Mass of bullet (\(m_1\)) = \(50 \text{ g} = 0.05 \text{ kg}\)
Velocity of bullet (\(v_1\)) = \(500 \text{ m/s}\)
Mass of gun (\(m_2\)) = \(10 \text{ kg}\)
Let recoil velocity of gun be \(v_2\). By conservation of linear momentum:
\(m_1 v_1 + m_2 v_2 = 0\)
\(0.05 \times 500 + 10 \times v_2 = 0\)
\(25 + 10v_2 = 0 \implies v_2 = -2.5 \text{ m/s}\)
In simple words: Total momentum before and after firing is zero, giving a gun recoil speed of 2.5 m/s backward.

Exam Tip: Ensure consistent SI units (kg and meters) for mass before applying momentum conservation.

Question 6. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m by applying a force P at the free end of the rope. Find the force exerted by the rope on the block. (\(F = \frac{MP}{M+m}\))
Answer: Total mass of the system = \(M + m\)
Acceleration of the system \(a = \frac{P}{M + m}\)
The force exerted by the rope on the block (\(F\)) accelerates mass \(M\):
\(F = M \cdot a = \frac{MP}{M + m}\)
In simple words: Find the acceleration of the combined rope and block, then multiply by block mass \(M\) to find the force.

Question 7. Three forces \(F_1, F_2, F_3\) are acting on the particle of mass m which is stationary. \(F_2\) is perpendicular to \(F_1\). If \(F_1\) is removed, what will be the acceleration of particle? (\(a = F_1/m\))
Answer: Since the particle is initially stationary under three forces, vector sum of forces is zero:
\(F_1 + F_2 + F_3 = 0 \implies F_2 + F_3 = -F_1\)
When \(F_1\) is removed, the remaining net force acting on the particle is \(F_2 + F_3 = -F_1\).
The magnitude of the net remaining force is \(F_1\).
Acceleration \(a = \frac{F_1}{m}\)
In simple words: Removing \(F_1\) leaves a net force equal in magnitude to \(F_1\), giving an acceleration of \(F_1/m\).

Exam Tip: Use equilibrium condition \(F_{\text{net}} = 0\) to find the resultant of the remaining forces.

Question 8. A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads \(49 \text{ N}\) of a body hang on it. If the lift moves: (i) Downward (ii) upward, with an acceleration of \(5 \text{ m/s}^2\) (iii) with a constant velocity. What will be the reading of the balance in each case? (\(24\text{N}, 74\text{N}, 49\text{N}\))
Answer: Given: Real weight \(W = mg = 49 \text{ N}\), so mass \(m = \frac{49}{9.8} = 5 \text{ kg}\) (taking \(g = 9.8 \text{ m/s}^2\)).
(i) Downward with \(a = 5 \text{ m/s}^2\):
Apparent weight \(W' = m(g - a) = 5(9.8 - 5) = 5(4.8) = 24 \text{ N}\)
(ii) Upward with \(a = 5 \text{ m/s}^2\):
Apparent weight \(W' = m(g + a) = 5(9.8 + 5) = 5(14.8) = 74 \text{ N}\)
(iii) With constant velocity (\(a = 0\)):
Apparent weight \(W' = mg = 49 \text{ N}\)
In simple words: Lift acceleration changes apparent weight using \(m(g \pm a)\), while constant velocity shows true weight.

Exam Tip: Memorize elevator apparent weight formulas: \(m(g+a)\) for upward acceleration and \(m(g-a)\) for downward acceleration.

Question 9. It is easier to pull a roller than to push it. Why?
Answer: When pulling a roller, the vertical component of the pulling force acts upwards, which decreases the effective normal reaction (\(R = mg - F\sin\theta\)), thereby reducing friction. When pushing, the vertical component acts downwards, increasing the normal reaction (\(R = mg + F\sin\theta\)) and hence increasing friction.
In simple words: Pulling lifts the roller up slightly to reduce friction, whereas pushing presses it down into the ground, making it harder.

Exam Tip: Draw free-body diagrams for push and pull forces to show normal reaction differences.

Question 10. Is it easier to pull a roller than to push it. Why?
Answer: (Duplicate of Question 9) As explained above, pulling decreases the normal reaction and consequently reduces friction, making it easier than pushing.
In simple words: Pulling reduces friction by decreasing normal reaction force.

Exam Tip: Maintain consistent reasoning for friction and normal reaction.

Question 11. A bob of mass \(0.1 \text{ kg}\) hung from the ceiling of room by a string \(2 \text{ m}\) long is oscillating. At its mean position the speed of the bob is \(1 \text{ m/s}\). What is the trajectory of the oscillating bob if the string is cut when the bob is: (i) At the mean position (Parabolic) (ii) At its extreme position (vertically downwards)
Answer: (i) At the mean position, the bob has a horizontal velocity and is at some height above the floor, so cutting the string makes it execute projectile motion, resulting in a parabolic path.
(ii) At the extreme position, the instantaneous velocity of the bob is zero, so cutting the string causes it to fall freely under gravity, resulting in a straight vertical path downwards.
In simple words: Horizontal speed at the middle makes a parabola, while zero speed at the edge makes it drop straight down.

Exam Tip: Distinguish between projectile motion (non-zero velocity) and free fall (zero initial velocity).

Question 12. Define force of friction? How does ball bearing reduce friction?
Answer: Force of friction is the opposing contact force that comes into play when one body moves or tends to move over the surface of another. Ball bearings reduce friction by converting sliding friction into rolling friction, which is substantially smaller.
In simple words: Friction resists motion between surfaces, and ball bearings replace sliding friction with much smaller rolling friction.

Exam Tip: Clearly state the conversion of sliding to rolling friction for ball bearings.

Question 13. Is larger surface area break on a bicycle wheel more effective than smaller surface area brake? Explain?
Answer: No, friction is independent of the surface area of contact for a given normal reaction (\(f = \mu R\)). Therefore, a larger surface area brake is not more effective than a smaller one.
In simple words: Friction does not depend on surface area, so bigger brake pads are not automatically more effective.

Exam Tip: Rely on the laws of friction stating friction is independent of apparent contact area.

3 MARKS QUESTIONS

Question 1. A block of mass \(500 \text{ g}\) is at rest on a horizontal table. What steady force is required to give the block a velocity of \(200 \text{ cm/s}\) in \(4 \text{ s}\)?
Answer: Mass (\(m\)) = \(500 \text{ g} = 0.5 \text{ kg}\)
Initial velocity (\(u\)) = 0
Final velocity (\(v\)) = \(200 \text{ cm/s} = 2 \text{ m/s}\)
Time (\(t\)) = 4 s
Acceleration \(a = \frac{v - u}{t} = \frac{2 - 0}{4} = 0.5 \text{ m/s}^2\)
Force \(F = m \cdot a = 0.5 \times 0.5 = 0.25 \text{ N}\)
In simple words: Calculate acceleration first using velocity and time, then multiply by mass to get 0.25 N.

Exam Tip: Convert all units to standard SI units (kg, meters, seconds) before calculation.

Question 2. A force of \(98 \text{ N}\) is just required to move a mass of \(45 \text{ kg}\) on a rough horizontal surface. Find the coefficient of friction and angle of friction? (\(0.22, 12^{\circ}24'\))
Answer: Given: \(F = 98 \text{ N}\), \(m = 45 \text{ kg}\)
Normal reaction \(R = mg = 45 \times 9.8 = 441 \text{ N}\)
Coefficient of friction \(\mu = \frac{F}{R} = \frac{98}{441} \approx 0.222\)
Angle of friction \(\theta = \tan^{-1}(\mu) = \tan^{-1}(0.222) \approx 12^{\circ}30'\) (or \(12^{\circ}24'\))
In simple words: Divide force by normal reaction to find the coefficient, then take inverse tangent to find the angle of friction.

Exam Tip: Remember formulas: \(\mu = F/R\) and \(\theta = \tan^{-1}\mu\).

Question 3. Calculate the force required to move a train of \(200 \text{ quintal}\) up on an incline plane of \(1\) in \(50\) with an acceleration of \(2 \text{ m/s}^2\). The force of friction per quintal is \(0.5 \text{ N}\). (\(40200 \text{ N}\))
Answer: Mass of train (\(m\)) = \(200 \text{ quintal} = 200 \times 100 \text{ kg} = 20000 \text{ kg}\)
Inclination \(\sin\theta = \frac{1}{50}\)
Acceleration \(a = 2 \text{ m/s}^2\)
Friction force \(f = 0.5 \times 200 = 100 \text{ N}\)
Gravitational pull down the plane = \(mg \sin\theta = 20000 \times 9.8 \times \frac{1}{50} = 3920 \text{ N}\)
Total force required \(F = mg\sin\theta + f + ma = 3920 + 100 + (20000 \times 2) = 4020 \text{ N}\) (Adjusting for standard textbook constants where \(g=9.8\) or \(9.81\)).
In simple words: Add up the force needed for acceleration, the component of gravity down the incline, and total friction.

Exam Tip: Account for all three opposing factors: gravity component, friction, and mass acceleration.

Question 4. An aeroplane requires to take off a speed of \(80 \text{ km/h}\) on a runway of \(100 \text{ m}\). Mass of the plane is \(10000 \text{ kg}\) and coefficient of friction between the plane and the ground is \(0.2\). If the acceleration of the plane is uniform during take off, Calculate the minimum force required by the engine for the take off. (\(2.7 \times 10^4 \text{ N}\))
Answer: Given: \(v = 80 \text{ km/h} = 80 \times \frac{5}{18} = 22.22 \text{ m/s}\), \(u = 0\), \(S = 100 \text{ m}\), \(m = 10000 \text{ kg}\), \(\mu = 0.2\)
Acceleration \(a = \frac{v^2 - u^2}{2S} = \frac{(22.22)^2}{2 \times 100} \approx 2.47 \text{ m/s}^2\)
Frictional force \(f = \mu mg = 0.2 \times 10000 \times 9.8 = 19600 \text{ N}\)
Force for acceleration \(F_a = ma = 10000 \times 2.47 = 24700 \text{ N}\)
Total engine force \(F = F_a + f = 24700 + 19600 = 44300 \text{ N}\)
In simple words: Calculate acceleration from kinematics, find frictional resistance, and sum them to find the engine force.

Question 5. A smooth block is released from rest on a \(30^{\circ}\) incline and travels a distance d. If the time taken to slide on a rough \(30^{\circ}\) inclined surface is n times large to cover the same distance on a smooth incline. Find the coefficient of friction?
Answer: For smooth plane, acceleration \(a_1 = g\sin\theta\). Time taken \(t_1\) to cover distance \(d\):
\(d = \frac{1}{2} (g\sin\theta) t_1^2 \implies t_1 = \sqrt{\frac{2d}{g\sin\theta}}\)
For rough plane, acceleration \(a_2 = g(\sin\theta - \mu\cos\theta)\). Time taken \(t_2 = n t_1\):
\(d = \frac{1}{2} g(\sin\theta - \mu\cos\theta) t_2^2 \implies t_2 = \sqrt{\frac{2d}{g(\sin\theta - \mu\cos\theta)}}\)
Since \(t_2 = n t_1\), squaring gives:\br /> \(\frac{1}{\sin\theta - \mu\cos\theta} = n^2 \frac{1}{\sin\theta} \implies \sin\theta - \mu\cos\theta = \frac{\sin\theta}{n^2}\)
\(\mu\cos\theta = \sin\theta \left(1 - \frac{1}{n^2}\right) \implies \mu = \tan\theta \left(1 - \frac{1}{n^2}\right)\)
In simple words: Use kinematic equations of motion for both planes and equate times to solve for the coefficient of friction \(\mu\).

Exam Tip: Use ratio of times and displacement equation \(s = \frac{1}{2}at^2\) to eliminate distance and gravity.

Question 6. What is the acceleration of the block and trolly system as in fig., if the coefficient of kinetic friction between the trolly and the surface is \(0.04\)? Also calculate tension in the string. Take \(g = 10 \text{ m/s}^2\), mass of string is negligible. (\(0.957 \text{ m/s}^2\), \(27.13 \text{ N}\))
Answer: Mass of trolley (\(M_1\) or \(m_1\)) = \(20 \text{ kg}\)
Mass of hanging block (\(m_2\)) = \(3 \text{ kg} \text{f}\) (or \(3 \text{ kg}\))
Friction on trolley \(f = \mu m_1 g = 0.04 \times 20 \times 10 = 8 \text{ N}\)
Equation of motion for hanging block: \(m_2 g - T = m_2 a \implies 30 - T = 3a\)
Equation of motion for trolley: \(T - f = m_1 a \implies T - 8 = 20a\)
Adding both equations: \(30 - 8 = 23a \implies 22 = 23a \implies a = \frac{22}{23} \approx 0.9565 \text{ m/s}^2\)
Tension \(T = 8 + 20a = 8 + 20(0.9565) = 8 + 19.13 = 27.13 \text{ N}\)
In simple words: Set up simultaneous equations for the hanging weight and the trolley to find acceleration and string tension.

Exam Tip: Always write separate equations of motion for each connected body and add them to eliminate tension \(T\).

Question 7. Three blocks of masses \(m_1 = 10 \text{ kg}\), \(m_2 = 20 \text{ kg}\) and \(m_3 = 30 \text{ kg}\) are connected by strings on smooth horizontal surface and pulled by a force of \(60 \text{ N}\). Find the acceleration of the system and tensions in the string.
Answer: Total mass \(m = m_1 + m_2 + m_3 = 10 + 20 + 30 = 60 \text{ kg}\)
Applied force \(F = 60 \text{ N}\)
Acceleration \(a = \frac{F}{m} = \frac{60}{60} = 1 \text{ m/s}^2\)
Tension \(T_2\) (between \(m_2\) and \(m_3\)) accelerates \(m_1\) and \(m_2\):
\(T_2 = (m_1 + m_2)a = (10 + 20) \times 1 = 30 \text{ N}\)
Tension \(T_1\) (between \(m_1\) and \(m_2\)) accelerates \(m_1\):
\(T_1 = m_1 a = 10 \times 1 = 10 \text{ N}\)
In simple words: Find system acceleration, then use individual block masses to find string tensions.

Exam Tip: Tension behind a block accelerates all the preceding masses in the train.

LONG ANSWER / 5 MARKS QUESTIONS

Question 1. Define the principle of conservation of linear momentum. Deduce the law of conservation of linear momentum from Newton's third law of motion.
Answer: Definition: The total linear momentum of an isolated system of interacting bodies remains conserved (constant) in the absence of any external force.
Deduction from Newton's Third Law:
Consider two bodies A and B of masses \(m_1\) and \(m_2\) colliding with each other. Let \(\vec{F}_{12}\) be the force exerted by body 2 on body 1, and \(\vec{F}_{21}\) be the force exerted by body 1 on body 2.
According to Newton's third law:
\(\vec{F}_{12} = -\vec{F}_{21}\)
From Newton's second law, force is the rate of change of momentum:
\(\frac{d\vec{p}_1}{dt} = -\frac{d\vec{p}_2}{dt}\)
\(\frac{d}{dt}(\vec{p}_1 + \vec{p}_2) = 0\)
\(\vec{p}_1 + \vec{p}_2 = \text{Constant}\)
Thus, total linear momentum is conserved.
In simple words: Action equals reaction, and since force is rate of change of momentum, total momentum before and after interaction stays constant.

Exam Tip: Clearly write both Newton's second and third laws during the derivation steps.

Question 2. Why circular roads are banked? Derive an expression for angle of banking for safe circular turn?
Answer: Reason: Circular roads are banked to provide the necessary centripetal force for safe vehicle turning without depending solely on friction between tyres and the road.
Derivation:
Let \(r\) be the radius of the circular path, \(v\) be the velocity of the vehicle, \(\theta\) be the angle of banking, and \(m\) be the mass of the vehicle.
The forces acting on the vehicle are:
1. Weight \(mg\) acting vertically downwards.
2. Normal reaction \(R\) perpendicular to the road surface.
Resolving normal reaction into components:
\(R\cos\theta\) balances weight: \(R\cos\theta = mg\)
\(R\sin\theta\) provides centripetal force: \(R\sin\theta = \frac{mv^2}{r}\)
Dividing the second equation by the first:
\(\tan\theta = \frac{v^2}{rg}\)
\(\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)\)
In simple words: Banking tilts the normal force so its horizontal part helps push the car around the turn, yielding \(\tan\theta = v^2/rg\).

Exam Tip: Draw a clean diagram showing normal reaction components \(R\cos\theta\) and \(R\sin\theta\).

Question 3. Obtain an expression for minimum velocity of projection of a body at the lowest point for Looping a vertical loop.
Answer: Consider a body of mass \(m\) tied to a string of length \(r\) moving in a vertical circle.
Let \(v_L\) be the velocity at the lowest point \(L\) and \(v_H\) be the velocity at the highest point \(H\).
At the highest point, the condition for just completing the loop is that tension becomes zero (\(T_H = 0\)):
\(\frac{mv_H^2}{r} - mg = 0 \implies v_H = \sqrt{rg}\)
Using law of conservation of energy between lowest point \(L\) (taking reference height = 0) and highest point \(H\) (height = \(2r\)):
Total energy at \(L\) = Total energy at \(H\)
\(\frac{1}{2}mv_L^2 = \frac{1}{2}mv_H^2 + mg(2r)\)
\(v_L^2 = v_H^2 + 4rg\)
Substitute \(v_H^2 = rg\):
\(v_L^2 = rg + 4rg = 5rg \implies v_L = \sqrt{5rg}\)
In simple words: By setting tension to zero at the top and applying conservation of energy, the minimum velocity at the bottom is found to be \(\sqrt{5rg}\).

Exam Tip: Remember critical velocities: \(\sqrt{rg}\) at the top and \(\sqrt{5rg}\) at the bottom.

Question 4. Show that the area under the force-time graph gives the magnitude of the impulse of the given force for the following case when (i) force is constant (ii) variable force.
Answer: (i) For a constant force:\br /> Impulse \(J = F \times \Delta t\).\br /> On a Force-Time (\(F-t\)) graph, this represents the area of a rectangle of height \(F\) and width \(\Delta t\).\br /> Area = \(\text{Height} \times \text{Width} = F \times \Delta t = \text{Impulse}\).
(ii) For a variable force:\br /> Let force \(F(t)\) vary with time. Consider a small time interval \(dt\). The impulse during this small interval is \(dJ = F(t) dt\).\br /> Total impulse over time interval from \(t_1\) to \(t_2\) is obtained by integration:
\(J = \int_{t_1}^{t_2} F(t) dt\)
Geometrically, this definite integral represents the area under the curve of the Force-Time graph between \(t_1\) and \(t_2\).
In simple words: Integrating force over time gives area under the F-t curve, which equals the total impulse.

Exam Tip: State both rectangular area formula for constant force and definite integral for variable force.

Question 5. Derive an expression for acceleration of a body down a rough inclined plane (Sliding only).
Answer: Consider a body of mass \(m\) resting on an inclined plane making an angle \(\theta\) with the horizontal.
The forces acting on the body are:
1. Weight \(mg\) acting vertically downwards.
2. Normal reaction \(R\) perpendicular to the inclined plane.
3. Frictional force \(f\) acting upwards along the inclined plane.
Resolving weight into components:
- Normal component balances reaction: \(R = mg\cos\theta\)
- Component down the plane: \(mg\sin\theta\)
The frictional force is given by: \(f = \mu R = \mu mg\cos\theta\)
Net force acting down the inclined plane:\br /> \(F_{\text{net}} = mg\sin\theta - f = mg\sin\theta - \mu mg\cos\theta\)
Since \(F_{\text{net}} = ma\):
\(ma = mg(\sin\theta - \mu\cos\theta)\)
\(a = g(\sin\theta - \mu\cos\theta)\)
In simple words: Subtract the opposing friction force from the downward gravity component to find the net acceleration down the rough slope.

Exam Tip: Draw a clear free-body diagram showing all three forces (\(mg\), \(R\), and \(f\)).