Laws Of Motion – NEET UG PhysicsPractice Questions & PYQs

The centripetal force governing circular motion is given by the expression $F = \frac{mv^2}{r}$. Setting the centripetal force of the lighter stone with mass $m$ and radius $r$ equal to that of the heavier stone with mass $2m$ and radius $r/2$ leads to the equation:
$\frac{m v_1^2}{r} = \frac{2m v_2^2}{r/2}$
Simplifying the right-hand side results in $\frac{m v_1^2}{r} = \frac{4m v_2^2}{r}$, and canceling the common mass and radius terms yields $v_1^2 = 4 v_2^2$. Taking the square root shows that the velocity of the lighter stone is $v_1 = 2 v_2$, identifying that the multiplier $n$ must be 2.

Newton's second law serves as the guiding principle here, relating the net forces acting on each block to their common acceleration. For the hanging block $m_2$, gravity pulls downward while tension resists, creating the relation $m_2 g - T = m_2 a$, whereas block $m_1$ on the table is pulled forward by tension against the opposing kinetic friction force $\mu_k m_1 g$, yielding $T - \mu_k m_1 g = m_1 a$. Summing these equations eliminates the tension and reveals the system acceleration as $a = \frac{(m_2 - \mu_k m_1)g}{m_1 + m_2}$.

Substituting this expression for the acceleration back into the force balance for the hanging mass, $T = m_2(g - a)$, provides the tension in the string. Expanding the equation $T = m_2 \left( g - \frac{(m_2 - \mu_k m_1)g}{m_1 + m_2} \right)$ through algebraic simplification leads directly to the final expression $T = \frac{m_1 m_2 (1 + \mu_k) g}{m_1 + m_2}$.

When an object sits on an inclined plane, the gravitational force component acting down the slope is balanced by static friction until the angle of inclination reaches the critical point where slipping occurs. At this threshold of 30 degrees, the coefficient of static friction is equivalent to the tangent of the angle, which calculates to 0.6. This fundamental relationship captures the maximum grip the plank holds before the box begins to move.

After the box breaks free and starts sliding, the force of kinetic friction opposes its downward motion, resulting in constant acceleration. Using the kinematic relation $s = ut + \frac{1}{2}at^2$ with an initial velocity of zero, a distance of 4.0 meters, and a time of 4.0 seconds, the acceleration is found to be 0.5 meters per second squared. The net force causing this acceleration is described by the equation $a = g \sin \theta - \mu_k g \cos \theta$, where the angle is 30 degrees. Substituting the known acceleration and gravitational values into this force balance equation isolates the coefficient of kinetic friction, yielding a value of 0.5.

Newton's Second Law describes how the net force acting on a system dictates its acceleration, which in this case is determined by the gravitational pull on the hanging mass minus the resistive force of friction on the masses resting on the table. The driving force for the system is the weight of mass $m_1$, represented as $m_1g$, while the total opposing frictional force is $\mu(m_2+m_3)g$. Because all three masses are connected and move together, the system shares a common acceleration calculated by dividing the net force by the total mass $m_1+m_2+m_3$. By substituting the given equality where $m_1 = m_2 = m_3 = m$, the expression becomes $\frac{mg - \mu(m+m)g}{3m}$, which simplifies to an acceleration of $\frac{g(1-2\mu)}{3}$.

The motion of the balloon is governed by Newton's second law, which relates the net force to the mass and acceleration of the object. Since the upward buoyancy force exerted on the balloon remains constant, we determine this force from the initial downward descent, expressed as $mg - B = ma$, which simplifies to $B = m(g - a)$.

Once mass $m_0$ is removed, the balloon moves upward with acceleration $a$, resulting in the equation $B - (m - m_0)g = (m - m_0)a$. Substituting the expression for buoyancy into this equation leads to $m(g - a) - (m - m_0)g = (m - m_0)a$, which simplifies to $m_0(g + a) = 2ma$, yielding the required mass to be removed as $\frac{2ma}{g + a}$.

The impulse experienced by an object corresponds to the change in its momentum, and on a force-time graph, this value is determined by calculating the net area under the curve. Because force is applied over distinct time intervals, we calculate the area for each segment individually, treating areas below the horizontal axis as negative.

For the first interval between 0 and 2 seconds, the graph forms a triangle with a base of 2 and a height of 6, yielding an area of $\frac{1}{2} \times 2 \times 6 = 6$. The second interval, from 2 to 4 seconds, shows a force of -3 acting over 2 seconds, resulting in an area of $(-3) \times 2 = -6$. Finally, for the interval between 4 and 8 seconds, a constant force of 3 acts over 4 seconds, contributing an area of $3 \times 4 = 12$. Adding these components together, $\Delta p = 6 - 6 + 12 = 12$, which gives a total net area of 12, representing the total change in momentum for the particle in Ns.

Newton's Second Law establishes that the net force on any object is the product of its mass and acceleration, expressed by the equation:

$F_{\text{net}} = ma$

Because the blocks are moving upward at a constant speed $v$, there is no change in velocity, which means the acceleration of the entire system is zero. Therefore, for the block with mass $2m$, the net force is determined by multiplying its mass by an acceleration of zero, which results in a net force of zero.

This problem is solved using the law of conservation of linear momentum, which dictates that in an isolated system, the total initial momentum must equal the total final momentum. Because the person and the rifle start at rest, the initial momentum of the system is zero. When the rifle is fired, the bullets gain forward momentum, forcing the person and the rifle to recoil backward with an equal and opposite amount of momentum to keep the total system momentum at zero.

To determine the recoil velocity, first calculate the total momentum of the bullets. With each of the 10 bullets having a mass of 0.01 kg and a velocity of 800 m/s, the total forward momentum generated is calculated as $P = n \cdot m \cdot u$. Using the given values, $10 \times 0.01 \times 800$, we find the total forward momentum is 80 kg m/s. This must be balanced by the recoil of the person and the rifle. Their combined mass of 100 kg must acquire an equal backward momentum, represented by the equation $100 \cdot v = -80$. Solving for $v$, we find the recoil velocity is -0.8 m/s.

The average force exerted on the person is determined by the rate of change of momentum, expressed as $F = \frac{\Delta p}{\Delta t}$. The total momentum imparted to the bullets over the 5 s interval is 80 kg m/s. By dividing this total change in momentum by the time elapsed, we get $F = \frac{80}{5}$, which results in an average force of 16 N.

The work-energy theorem provides the most direct approach here, as it equates the net work done on an object to its change in kinetic energy. Since the block begins at rest and returns to rest at the bottom, the total change in kinetic energy is zero, meaning the total work done by gravity must be perfectly balanced by the total work done by friction. Gravity acts throughout the entire incline length $L$, resulting in work equal to $W_g = mg \sin \theta L$.

Conversely, friction only opposes the motion along the lower half of the plane, which spans a distance of $\frac{L}{2}$. Given the coefficient of friction $\mu$ and the normal force $mg \cos \theta$, the frictional force magnitude is $\mu mg \cos \theta$. The negative work performed by this force is $W_f = -\mu mg \cos \theta \left(\frac{L}{2}\right)$. Setting the sum of these works to zero gives the equation $mg \sin \theta L - \mu mg \cos \theta \left(\frac{L}{2}\right) = 0$. Dividing both sides by the common factors $mg$ and $L$ simplifies the relationship to $\sin \theta = \frac{\mu}{2} \cos \theta$. Rearranging this expression to solve for $\mu$ yields $\mu = \frac{2 \sin \theta}{\cos \theta}$, which is equivalent to $\mu = 2 \tan \theta$.

When a vehicle travels along a circular path, a pendulum bob suspended inside it experiences a horizontal centrifugal force in addition to the vertical force of gravity. In the reference frame of the moving car, the bob shifts until it reaches an equilibrium position where the resultant of these two perpendicular forces aligns with the tension in the string. The angle $\theta$ that the string makes with the vertical is determined by the ratio of the centrifugal force to the weight of the bob, leading to the relationship $\tan \theta = \frac{v^2}{rg}$.

Substituting the given values, where the constant speed $v$ is 10 m/s, the track radius $r$ is 10 m, and the acceleration due to gravity $g$ is 10 m/s$^2$, we find:
$\tan \theta = \frac{10^2}{10 \cdot 10} = 1$
Since the tangent of the angle equals 1, the wire makes an angle of $\pi/4$ with the vertical. The length of the wire does not affect this equilibrium angle in horizontal circular motion.

Since the rock is initially stationary, the law of conservation of linear momentum dictates that the vector sum of the momenta of all fragments must be zero, expressed as $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$. The momentum of the first part is 12 kg ms⁻¹, and the momentum of the second part is 16 kg ms⁻¹, with both acting perpendicular to each other. The third piece must balance this combined momentum, with a magnitude calculated as $p_3 = \sqrt{12^2 + 16^2} = 20$ kg ms⁻¹. By applying the relation $m_3 = p_3 / v_3$ with the known speed of 4 ms⁻¹, we find that the mass is 5 kg.

The motion of a car around a level circular curve is governed by static friction, which provides the necessary centripetal force to maintain the path. By equating the maximum static frictional force $\mu_s N$, where the normal force $N = mg$, to the required centripetal force $\frac{mv^2}{R}$, we establish the relationship $\frac{mv^2}{R} \le \mu_s mg$. Simplifying this inequality to isolate the velocity demonstrates that the maximum speed possible without skidding is $v = \sqrt{\mu_sRg}$.

When a car travels along a frictionless banked curve, the centripetal force required for circular motion is generated solely by the horizontal component of the normal force exerted by the road. This physical requirement dictates that the speed of the vehicle must be perfectly matched to the banking angle and the radius of the turn to maintain a steady path, a relationship described by $\tan \theta = \frac{v^2}{Rg}$.

Rearranging this expression to solve for speed gives $v = \sqrt{Rg \tan \theta}$. Substituting the given values where the radius is 90 m, the acceleration due to gravity is 10 ms⁻², and the banking angle is 45{}^{\circ}, we calculate $v = \sqrt{90 × 10 × \tan 45^{\circ}}$. Because $\tan 45^{\circ}$ is 1, the calculation simplifies to the square root of 900, resulting in a speed of 30 ms⁻¹.

When a box is placed on a moving conveyor belt, the kinetic friction between the surfaces acts as a net force that accelerates the box relative to the belt. This force is determined by the coefficient of friction and the acceleration due to gravity, defined by the formula $f = \mu mg$, which leads to the acceleration $a = \mu g$.

Using the values provided, the acceleration is $a = 0.5 \times 10 = 5$ m/s². Since the box is initially at rest relative to the ground and the belt moves at a constant speed of 2 m/s, the box has an initial velocity of 2 m/s relative to the belt. As the belt accelerates the box, its velocity relative to the belt decreases to zero. Applying the kinematic equation $v^2 - u^2 = 2aS$ where $v = 0$, $u = 2$ m/s, and $a = -5$ m/s², we substitute these values to get $0^2 - 2^2 = 2(-5)S$, which simplifies to $S = 0.4$ m.

The relationship between impact velocity and drop height is derived from kinematic principles, which establish that velocity scales with the square root of height as $v = \sqrt{2gh}$. Because mass remains constant, momentum is directly proportional to the square root of the height, expressed as $P \propto \sqrt{h}$. Dropping the stone from a height that is 100% greater than the original means doubling the distance, changing the height from $h$ to $2h$.

By applying this proportionality, the new momentum becomes $P' = m\sqrt{2g(2h)}$, which simplifies to $P' = \sqrt{2}P$. Since $\sqrt{2}$ is approximately 1.414, the final momentum is 1.414 times the initial value. The percentage change is calculated as $\frac{P'-P}{P} \times 100$, which results in $(\sqrt{2}-1) \times 100$, yielding an increase of approximately 41%.

When a lift ascends with acceleration, the cable supporting the lift must exert enough tension to overcome both the downward gravitational pull on the system and provide the additional force required to accelerate the mass upward. By treating the lift and the passenger inside as a single combined body, the total mass becomes 1000 kg. Applying Newton's second law of motion, the net force acting on the system is represented by the equation $T - (M + m)g = (M + m)a$.

Rearranging this relationship to solve for the tension gives the formula $T = (M + m)(g + a)$. Substituting the given values into this expression results in $T = (1000)(10 + 1)$, which leads to a total tension of 11000 N.

Impulse is defined as the change in an object's linear momentum during an interaction. Because the body strikes the wall with momentum $MV$ and rebounds with the same speed in the opposite direction, its final momentum is represented as $-MV$. Calculating the change involves subtracting the initial momentum from the final momentum, expressed as $(-MV) - (MV) = -2MV$, which results \in an impulse magnitude of $2MV$.

To keep a block from sliding down the side of an accelerating cart, we analyze the situation from the cart's non-inertial reference frame, where the block experiences an outward pseudo force $F_{\text{fic}} = m \alpha$. This force pushes the block against the cart surface, generating a normal force $N = m \alpha$ and a resulting upward static friction force $f = \mu N = \mu m \alpha$. For the block to remain stationary, this frictional force must be at least equal to the block's weight, which is expressed by the inequality $\mu m \alpha \geq mg$. Simplifying this relationship shows that the cart must maintain a minimum acceleration of $\alpha \geq g/\mu$.

Newton's Second Law relates force, mass, and acceleration through $\vec{F} = m\vec{a}$. The magnitude of the given force is found by calculating the square root of the sum of the squares of its components, which results in $\sqrt{6^2 + (-8)^2 + 10^2} = 10\sqrt{2}\,\text{N}$. Dividing this force magnitude by the given acceleration of $1\,\text{m/s}^2$ gives the mass $m = \frac{10\sqrt{2}}{1} = 10\sqrt{2}\,\text{kg}$.

According to Newton's second law, the force required to change the state of motion is equivalent to the time rate of change of momentum, expressed as $F = \frac{dp}{dt} = \frac{d(mv)}{dt}$. In this scenario, because the conveyor belt maintains a constant velocity of $v$ m/s, the sand falling onto the belt must be accelerated from rest to match the belt's speed, which creates a constant change in momentum.

Since the velocity remains steady, the expression simplifies to the product of velocity and the rate of mass accumulation, represented as $F = v\frac{dm}{dt}$. Substituting the given rate of mass flow, which is $M$ kg/s, directly into this relationship confirms that the force necessary to sustain the motion is $Mv$ newton.