Laws Of Motion – NEET UG PhysicsPractice Questions & PYQs

The maximum acceleration is governed by the limit of static friction, which must provide the necessary force to accelerate the box along with the trolley. When the trolley moves, inertia causes the box to tend toward slipping, but friction acts to keep it in place. The maximum force available is calculated as $f_{\text{max}} = \mu_s mg$, where $\mu_s$ is the coefficient of static friction, $m$ is the mass of the box, and $g$ is the acceleration due to gravity.

Because the box must move with the trolley, the required force for the box is $ma$. To prevent slipping, this required force must not exceed the maximum frictional limit, leading to the condition $ma \le \mu_s mg$. Since the mass of the 15 kg box cancels out on both sides, the calculation depends only on the friction coefficient of 0.12 and the gravitational constant of $10\ \text{m/s}^2$. Multiplying these values, $0.12 \times 10$ results in a maximum allowable acceleration of $1.2\ \text{m/s}^2$.

When an object slides down a surface of length $L$ starting from rest, the kinematics dictate that $L = \frac{1}{2} a t^2$, which implies that time is inversely proportional to the square root of acceleration, $t \propto \frac{1}{\sqrt{a}}$. On the smooth surface, the acceleration is $a_s = g \sin\theta$, whereas on the rough surface, kinetic friction reduces this to $a_r = g \sin\theta - \mu_k g \cos\theta$. Given that the time taken on the rough surface is twice the time taken on the smooth surface, the ratio of the accelerations must be the square of the inverse time ratio, resulting in $\frac{a_s}{a_r} = (\frac{t_r}{t_s})^2 = 2^2 = 4$. Substituting the angle $\theta = 45^{\circ}$ where $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$, the ratio becomes $\frac{g \sin 45^{\circ}}{g \sin 45^{\circ} - \mu_k g \cos 45^{\circ}} = \frac{1}{1 - \mu_k} = 4$. Solving this linear equation yields $1 - \mu_k = 0.25$, which results in $\mu_k = 0.75$.

When two blocks are pushed together, they move as a single system with the same acceleration. By applying Newton's Second Law to the combined mass of 2 kg and 3 kg under the influence of the 10 N force, the acceleration is found using $10 = (2 + 3) a$, which yields $a = 2 \text{ m/s}^2$.

Since block A pushes against block B, the force transmitted through the contact point is what drives block B. Focusing exclusively on block B, which has a mass of 3 kg and accelerates at 2 m/s$^2$, the interaction force is calculated as $F = 3 \times 2 = 6 \text{ N}$.

Newton's second law dictates that the acceleration of the box depends on the net force acting along the frictionless incline, which is the tension pulling upward minus the component of gravitational force pulling downward. The downward component of the weight is calculated as $mg \sin 30^\circ$, which equals $5 \times 10 \times 0.5 = 25$ newtons. Subtracting this from the upward tension of 30 newtons provides a net force of $30 - 25 = 5$ newtons. Substituting these values into $F_{net} = ma$ gives the equation $5 = 5a$, resulting in an acceleration of 1 m/s².

An object placed on a rough inclined plane remains in static equilibrium as long as the angle of inclination is less than or equal to the angle of repose, which is determined by the condition $\mu \geq \tan \theta$. Given that $\mu = 1/\sqrt{3}$ and the inclination angle is $30^\circ$, we observe that $\tan 30^\circ = 1/\sqrt{3}$, meaning the inclination angle is exactly equal to the angle of repose. Under these conditions, the component of gravity pulling the $2 \text{kg}$ block down the slope is fully offset by the maximum static friction opposing that motion. Since the block is at rest and possesses no acceleration, the net force acting on it must be zero.

Newton's second law dictates that the force acting on an object is proportional to the rate of change of its momentum, which requires the force to be in the same direction as the change in the velocity vector:
$\vec{F} = \frac{m(\vec{v}_f - \vec{v}_i)}{\Delta t}$
Since the player transitions from southward to eastward movement, the change in velocity involves subtracting the initial southward vector from the final eastward vector. This vector subtraction is equivalent to adding the eastward vector to a northward vector, resulting in a net direction that points toward the north-east.

For a body lying on the floor of a car to remain stationary while the car accelerates, static friction must provide the necessary force to oppose the tendency of the body to slide backwards. This tendency arises due to the pseudo-force, which is equal to $ma$ in the non-inertial frame of the car. The maximum value this friction can provide is limited by the product of the coefficient of static friction and the normal force, expressed as $f_{\max} = \mu_s mg$. Setting the pseudo-force equal to this maximum friction leads to the relation $a = \mu_s g$. Using the provided values of 0.15 for the coefficient and 10 for gravity, the limiting acceleration is 0.15 multiplied by 10, which equals 1.5 m/s^2.

Newton's Second Law relates the average force exerted on a wall to the rate of change of an object's momentum, which is the impulse delivered during the contact time. Because the object reflects at the same angle it strikes, only the velocity component perpendicular to the wall changes direction, resulting in a total change in momentum of $2mv \sin \theta$, where $\theta$ is the angle made with the wall.

With a mass of $1 \text{ kg}$, velocity of $1 \text{ m s}^{-1}$, and an angle of $60^{\circ}$ relative to the wall surface, the perpendicular velocity component involves $\sin(60^{\circ})$. The force is determined by dividing the change in momentum by the contact time of $0.1 \text{ s}$:
$F = \frac{2mv \sin \theta}{\Delta t} = \frac{2(1)(1) \sin(60^{\circ})}{0.1}$
Substituting $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ into the equation gives $F = \frac{\sqrt{3}}{0.1}$, which simplifies to $10 \sqrt{3} \text{ N}$.

Newton's Second Law of Motion governs the behavior of linked bodies, stating that the net force on a system equals the total mass multiplied by the acceleration. To find the acceleration of the entire collection of blocks, one must calculate the effective driving force by subtracting all opposing forces-both external resistances and the parallel components of gravity-from the applied pulling force. With a total mass of 6 kg and a net driving force calculated as $60 - 18 - (6 \times 10 \times 0.5) = 12 \text{ N}$, the system accelerates up the incline at $a = 12 / 6 = 2 \text{ m/s}^2$.

Focusing on the 1 kg block allows us to isolate the internal contact force between the blocks. The forces acting on this mass in the direction of motion are the normal reaction $N$ pushing upward, the opposing 18 N resistance, and the component of weight pulling downward equal to $1 \times 10 \times 0.5 = 5 \text{ N}$. Applying the second law for this specific block yields the equation $N - 18 - 5 = 1 \times 2$. Solving this expression results in $N - 23 = 2$, which confirms the normal reaction force is 25 N.

Impulse is defined as the change in linear momentum, represented by the equation $I = \Delta p = p_f - p_i$. Since the ball returns to its original release height after the collision, the kinetic energy and speed must be identical immediately before and after the impact, although the direction of motion reverses.

Calculating the impact velocity involves the formula $v = \sqrt{2gh}$. Using the provided values of 10 meters per second squared for gravity and 10 meters for height, the velocity becomes $\sqrt{2 \times 10 \times 10}$, which equals $10\sqrt{2}$ meters per second. The initial momentum before impact is $-mv$ and the final momentum after impact is $+mv$. The total change in momentum is $mv - (-mv) = 2mv$. Substituting the mass of 0.15 kilograms and the velocity of $10\sqrt{2}$ meters per second, the impulse becomes $2 \times 0.15 \times 10\sqrt{2}$, resulting in $3\sqrt{2}$ kilogram meters per second. Using the approximation $\sqrt{2} \approx 1.414$, the magnitude of the impulse is approximately 4.2 kilogram meters per second.

The motion of this system is governed by Newton's second law, where the net force acting on each body equals its mass multiplied by its acceleration. For the $4 \text{kg}$ block, the tension pulling upward overcomes its weight, leading to the expression $T - 4g = 4a$. Because the $6 \text{kg}$ block is heavier, it accelerates downward against the tension, which gives the equation $6g - T = 6a$. Summing these two equations effectively removes the internal tension force, resulting in:

$2g = 10a$

Dividing both sides by $10$ shows that the acceleration of the system is $a = \frac{g}{5}$.

When three force vectors form a closed triangle PQR, they indicate that the resultant force acting on the object is zero, placing the particle in a state of equilibrium. Because the net force $F_{net} = 0$, the acceleration is likewise zero according to the relationship:
$F_{net} = m \frac{d v}{d t} = 0$
Consequently, the particle continues to move without any change in its motion, meaning the velocity remains constant.

Whenever an object breaks apart without external forces interfering, the principle of conservation of linear momentum dictates that the total initial momentum must remain equal to the combined final momentum of the fragments. Considering an object of mass $M$ with an initial velocity $\vec{v}_0 = 20 \hat{i}+25 \hat{j}-12 \hat{k}$, the total momentum is given by $M\vec{v}_0$. Upon splitting into two pieces with masses $\frac{M}{6}$ and $\frac{5M}{6}$, the smaller fragment moves with velocity $\vec{v}_1 = 100 \hat{i}+35 \hat{j}+8 \hat{k}$, while the larger fragment moves with an unknown velocity $\vec{v}_2$. Setting up the equation $M(20 \hat{i}+25 \hat{j}-12 \hat{k}) = \frac{M}{6}(100 \hat{i}+35 \hat{j}+8 \hat{k}) + \frac{5M}{6} \vec{v}_2$ allows for the cancellation of the total mass $M$. Multiplying the entire equation by six to clear the denominators simplifies the expression to $6(20 \hat{i}+25 \hat{j}-12 \hat{k}) = (100 \hat{i}+35 \hat{j}+8 \hat{k}) + 5 \vec{v}_2$, which expands to $120 \hat{i}+150 \hat{j}-72 \hat{k} = (100 \hat{i}+35 \hat{j}+8 \hat{k}) + 5 \vec{v}_2$. By subtracting the smaller fragment's momentum from the initial total, we get $5 \vec{v}_2 = (120-100)\hat{i} + (150-35)\hat{j} + (-72-8)\hat{k}$, resulting in $5 \vec{v}_2 = 20 \hat{i}+115 \hat{j}-80 \hat{k}$. Finally, dividing by five yields the velocity of the larger piece as $4 \hat{i}+23 \hat{j}-16 \hat{k}$.

Maintaining the block against the inner wall relies on the upward force of static friction being sufficient to counterbalance the downward pull of gravity. This frictional force is determined by the normal reaction, which acts as the centripetal force required to sustain the circular motion of the block, expressed as $N = mr\omega^2$.

For the block to remain stationary, the limiting condition requires the friction force $\mu N$ to equal the weight $mg$, leading to the equilibrium equation $\mu mr\omega^2 = mg$. Canceling the mass $m$ from both sides and isolating the angular velocity yields $\omega = \sqrt{\frac{g}{\mu r}}$. Substituting the given values $g = 10 \text{ m/s}^{2}$, $r = 1 \text{ m}$, and $\mu = 0.1$ into this relation results in $\omega = \sqrt{\frac{10}{0.1 \times 1}}$, which simplifies to 10 rad/s.

When viewed from the non-inertial frame of the wedge, the block remains stationary because the pseudo-force acting opposite to the wedge's acceleration balances the horizontal component of the normal force. Specifically, the vertical component $N \cos \theta$ must perfectly counteract the weight $mg$, while the horizontal component $N \sin \theta$ balances the pseudo-force $ma$. Dividing the equation $N \sin \theta = ma$ by $N \cos \theta = mg$ cancels the normal force to arrive at the relationship $a = g \tan \theta$.

Friction is determined by the relationship between the force of friction and the normal force, expressed by the equation $f = \mu N$. Since the coefficient of friction is derived from the ratio of two forces, $\mu = \frac{f}{N}$, the units of force in both the numerator and the denominator cancel out perfectly. Consequently, the coefficient of friction is a dimensionless value, which contradicts the suggestion that it possesses dimensions of length.

When the string is cut, we must account for the immediate response of the spring force versus the instantaneous disappearance of tension. In the moments before the cut, the lower block of mass $m$ is held in equilibrium by the tension in the string, establishing $T = mg$. Simultaneously, the upper block of mass $3m$ supports both the lower block and itself against the spring, requiring the spring to exert an upward force equal to $3mg + T$, or $4mg$.

Once the string snaps, the tension vanishes instantly, but the spring force remains constant at $4mg$ because springs cannot change their force state instantaneously. For the lower block, the only remaining force is its weight, resulting in a downward acceleration of
$a_B = \frac{mg}{m} = g$
For the upper block, the upward spring force of $4mg$ acts against its weight of $3mg$, creating a net upward force of $mg$. Dividing this net force by the mass of the block gives an acceleration of
$a_A = \frac{mg}{3m} = \frac{g}{3}$

When a particle undergoes uniform circular motion on a horizontal surface, the net force directed towards the center must be equal to the centripetal force. This physical requirement is defined by the equation
$F_c = \frac{mv^2}{l}$
where the tension in the string acts as the only horizontal force on the particle. Since this tension provides the entirety of the necessary centripetal force to maintain the circular path, the net radial force on the particle is simply $T$.

Impulse imparted by the wall equals the net change in the ball's momentum, calculated as the difference between its final and initial momentum vectors. Since the wall exerts force only perpendicular to its surface, the component of momentum parallel to the wall remains unchanged, while the perpendicular component reverses direction. The total change in momentum is determined by $J = 2mV \cos(60^{\circ})$, where the factor of two accounts for the reversal of the normal momentum component $mV \cos(60^{\circ})$. Substituting the trigonometric value $\cos(60^{\circ}) = 0.5$ into this expression yields $J = 2mV(0.5)$, which simplifies to a final impulse value of $mV$.

When a vehicle travels along a banked curve at its maximum safe speed, the inertia of the car creates a strong tendency to slide outward and upward along the incline, which causes static friction to act downward parallel to the road surface to oppose this motion. By resolving the normal reaction force and the frictional force into horizontal and vertical components, one can equate the net radial force to the required centripetal force and balance the vertical forces against gravity. Combining these equations leads to the expression for the maximum permissible speed on such a curve, which is $V_{\max} = \sqrt{gR \left( \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right)}$.