Laws Of Motion – NEET UG PhysicsPractice Questions & PYQs

Impulse represents the change in momentum and is defined as the integral of force with respect to time. Since the force acting on the bullet varies, the total impulse delivered until the bullet exits the barrel is found by integrating the force function $F = 600 - 2 \times 10^5 t$ over the duration for which the force is applied. The time at which the bullet leaves the barrel is found by setting the force to zero:
$t = \frac{600}{2 \times 10^5} = 3 \times 10^{-3} \text{ s}$
Integrating the force equation over this duration yields the total impulse:
$I = \int_{0}^{3 \times 10^{-3}} (600 - 2 \times 10^5 t) \, dt$
Evaluating this definite integral results in the expression $600t - 10^5 t^2$. Calculating this at the upper limit of $3 \times 10^{-3} \text{ s}$ produces $600(3 \times 10^{-3}) - 10^5(3 \times 10^{-3})^2$, which simplifies to $1.8 - 0.9$, resulting in a final impulse of $0.9 \text{ N}\cdot\text{s}$.

Since the block moves down the inclined plane at a constant speed, the net force acting along the slope must be zero because there is no acceleration. The downward component of the gravitational force pulling the block along the plane is $mg \sin \theta$, while the kinetic friction opposing this motion is given by $\mu_k mg \cos \theta$. Because these two opposing forces must balance each other exactly to maintain a steady speed, we have $mg \sin \theta = \mu_k mg \cos \theta$. Dividing both sides of this equation by $mg \cos \theta$ reveals that the coefficient of kinetic friction is equivalent to $\tan \theta$.

When a mass is suspended by a thread, the tension depends on the net force acting upon it according to Newton's second law. Moving the mass upward with acceleration requires the thread to overcome both gravity and the force of acceleration, resulting in a tension defined by $T_1 = m(g+a)$. Conversely, lowering the mass with the same acceleration allows gravity to assist the motion, reducing the required tension to $T_2 = m(g-a)$.

Dividing the upward tension by the downward tension cancels the mass, leaving a relationship dependent only on acceleration and gravity. Using 9.8 m/s² for gravity and 4.9 m/s² for the acceleration, the ratio is:
$\frac{T_1}{T_2} = \frac{g+a}{g-a} = \frac{9.8 + 4.9}{9.8 - 4.9}$
This simplifies to 14.7 / 4.9, resulting in a ratio of 3:1.

Newton's second law states that the acceleration of an object is determined by the net force acting upon it divided by its mass. When two forces act perpendicularly to one another, the magnitude of the resulting net force is found by using the Pythagorean theorem, treating the individual forces as legs of a right triangle. Calculating the net force provides $F_{\text{net}} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{N}$. Dividing this total force by the given mass of $5\ \text{kg}$ results in an acceleration of $2\ \text{m s}^{-2}$. To find the direction of this acceleration relative to the $8\ \text{N}$ force, we calculate the angle $\theta$ by taking the tangent, which is the ratio of the perpendicular force component to the reference component. This yields $\tan \theta = \frac{6}{8} = \frac{3}{4}$, indicating that the acceleration occurs at an angle of $\tan^{-1}\left(\frac{3}{4}\right)$ with respect to the $8\ \text{N}$ force.

Newton's first law of motion establishes the concept of inertia, asserting that an object will persist in its state of rest or uniform motion unless an external influence alters that condition. By defining force as the specific agent capable of changing an object's state, the law separates the external action from the body's intrinsic properties. This foundational isolation of force as a distinct, external influence effectively defines the physical independence of force.

Centrifugal force arises in a rotating reference frame, acting as an apparent force directed away from the center of rotation. When milk undergoes the rapid circular motion of churning, this force acts on the particles within the mixture, pushing the denser components outward toward the perimeter of the container. Because cream is less dense than the rest of the liquid, it is displaced toward the center, which allows the lighter cream to be effectively separated from the denser milk.

The thrust generated by a rocket is fundamentally described by Newton's second law, which relates force to the rate of change of momentum. By ejecting fuel at high speed, the rocket experiences a reaction force determined by the product of the mass flow rate and the exhaust velocity.

Calculating the force in standard units involves first converting the ejection velocity of 60 km/s into 60000 m/s. Given that the fuel burns at a rate of 1 kg/s, the thrust is calculated using the expression $F = v \frac{dm}{dt}$. Substituting the known values leads to the following calculation:

$F = 60000 \times 1 = 60000$

This results in a total force of 60000 N acting on the rocket.

Newton's second law of motion states that the force acting on an object is equal to the product of its mass and acceleration, represented by the formula $F = ma$. To solve for the mass of the body, we can rearrange the relationship to $m = F / a$. By substituting the given force of 10 N and the acceleration of 1 m/s², the calculation becomes $m = 10 / 1$, which confirms the mass is 10 kg.

Imagine the chain split into two segments: one section resting on the table providing the necessary frictional resistance, and the other hanging off the edge generating the gravitational pull. For the chain to be on the verge of slipping, the weight of the hanging portion must be exactly balanced by the frictional force acting on the part of the chain resting on the table.

Letting $L$ represent the total length and $y$ the hanging portion, the equilibrium condition is expressed as $(\frac{M}{L}y)g = \mu(\frac{M}{L}(L-y))g$. Solving this equation for the ratio $\frac{y}{L}$ yields the relationship $\frac{y}{L} = \frac{\mu}{1 + \mu}$. Substituting $\mu = 0.25$ into this formula gives $\frac{0.25}{1.25} = \frac{1}{5}$, indicating that 20% of the chain can hang over the table edge.

For a rocket to launch vertically, the thrust generated by its engine must overcome both the downward pull of gravity and provide the force necessary to produce the desired upward acceleration. The total force required is given by the expression $M(g+a)$, which represents the sum of the rocket's weight and its inertial force. Since thrust is generated by the high-speed ejection of exhaust gas, it is defined by the product of the exhaust speed and the mass ejection rate, represented as $u \frac{dm}{dt}$. By equating these two expressions, the required burn rate is determined by the formula $\frac{dm}{dt} = \frac{M(g+a)}{u}$.

Substituting the provided values into this equation using 5000 kg for mass, 10 m/s^2 for gravity, 20 m/s^2 for acceleration, and 800 m/s for exhaust speed, we find the numerator by multiplying 5000 by the sum of 10 and 20. This results in a total force of 150,000 N, which must be divided by the exhaust speed of 800 m/s. Performing this division yields a fuel ejection rate of 187.5 kg/s.

The stopping distance of a vehicle depends on the deceleration provided by the frictional force acting against its motion. By using kinematic motion equations, we can calculate exactly how far a car travels before friction brings it to a complete standstill.

First, convert the speed from 72 km/h into SI units by multiplying by 5/18, which results in 20 m/s. The maximum deceleration the car experiences is provided by the static friction, calculated as $a = \mu g$. Substituting the given coefficient of friction 0.5 and gravity 10 m/s^2, the deceleration value becomes 5 m/s^2.

Using the kinematic equation $v^2 = u^2 + 2as$ where the final velocity is 0, the expression for stopping distance is rearranged to $s = \frac{-u^2}{2a}$. Substituting the initial velocity 20 m/s and the deceleration of -5 m/s^2, the calculation results in 40 m.

Rocket propulsion relies on the principle that the thrust force experienced by the craft is equal to the product of its velocity and the rate of fuel mass ejection, given by the relation $F = v \frac{dm}{dt}$. Since the rocket travels at 300 m/s and experiences a force of 210 N, substituting these values produces the equation $210 = 300 \frac{dm}{dt}$. Calculating the rate of fuel consumption requires dividing 210 by 300, which results in 0.7 kg/s.

The law of conservation of momentum dictates that since the gun and bullet start at rest, the total momentum of the system must remain zero after firing. This implies the bullet's forward momentum is countered by an equal and opposite backward momentum from the gun. First, determine the bullet's forward momentum by calculating $0.2 \times 5 = 1$ kg m/s. To maintain a net system momentum of zero, the gun-with a mass of 1 kg-must recoil such that the sum of the bullet and gun momenta is $1 + 1 \times v = 0$. Solving this reveals that the gun moves backward with a velocity of 1 m/s.

To lift a rocket vertically against gravity, the upward thrust produced by ejecting exhaust gases must be equal to the downward force of the rocket's weight. This thrust is determined by the product of the exhaust speed and the rate of mass ejection, expressed as $v \frac{dm}{dt}$, which must balance the gravitational force $mg$. Rearranging this relationship to solve for the ejection rate gives the formula $\frac{dm}{dt} = \frac{mg}{v}$. By substituting the rocket mass of 600 kg, a gravitational acceleration of 10 m/s^2, and an exhaust speed of 1000 m/s, the calculation $\frac{600 \times 10}{1000}$ yields a mass ejection rate of 6 kg/s.

Newton's second law states that the net force on the monkey is the difference between its gravitational weight and the tension of the branch, expressed as $mg - T = ma$. Because the branch is only strong enough to support $75\%$ of the monkey's weight, we identify the maximum tension as $T = \frac{3}{4}mg$ to determine the minimum safe acceleration. Substituting this limit into the equation of motion provides:
$mg - \frac{3}{4}mg = ma$
Simplifying the left side gives $\frac{1}{4}mg = ma$, which confirms that the minimum acceleration the monkey must maintain is $a = \frac{g}{4}$.

Impulse is defined as the net change in momentum experienced by a particle, which occurs when an external force acts on it. Since momentum is the product of mass and velocity, calculating the change simply involves finding the difference between the final momentum and the initial momentum. Mathematically, this is expressed as:

$Impulse = m(v_2 - v_1)$

This relationship demonstrates that the impulse is equivalent to the mass of the particle multiplied by the change in its velocity vector.

Analyzing the motion from the non-inertial frame of the wedge requires accounting for the pseudo force $Ma$ acting on the mass in the direction opposite to the wedge's acceleration. If the wedge is accelerated to the left, this pseudo force acts horizontally to the right, generating a component $Ma \cos \theta$ directed up the incline, while the gravitational force exerts a component $Mg \sin \theta$ pulling the mass down the slope.

To ensure the mass remains stationary relative to the wedge, these two forces must be perfectly balanced, which is represented by the equation $Ma \cos \theta = Mg \sin \theta$. Simplifying this expression by canceling common terms leads to the result $a = g \tan \theta$. Consequently, applying an acceleration of $a = g \tan \theta$ to the wedge toward the left provides the exact pseudo force necessary to keep the mass in equilibrium.

Newton's second law states that the acceleration of an object is determined by the ratio of force to mass, expressed as $a = F/m$. With a force of 6 N and a mass of 1 kg, the acceleration is calculated as $a = 6 / 1 = 6 \text{ m/s}^2$. Using the kinematic relationship $v = u + at$ to describe the motion from an initial velocity of 0 m/s to a final velocity of 30 m/s, we can set up the equation $30 = 0 + 6t$. Solving this for time gives a duration of 5 seconds.

When a vehicle navigates a banked circular track, minimizing wear on the tires requires driving at the specific design speed where the normal reaction force alone supplies the necessary centripetal acceleration, eliminating the need for frictional assistance. This ideal state is governed by the relationship $\tan \theta = \frac{v^2}{rg}$, where $\theta$ represents the banking angle.

By substituting the specified speed 10 m/s, the radius 50 m, and the acceleration due to gravity 10 m/s^2 into the equation, the expression becomes $\tan \theta = \frac{100}{500}$. Simplifying this calculation leads to $\tan \theta = \frac{1}{5}$. Consequently, the banking angle is determined to be $\theta = \tan^{-1}\left(\frac{1}{5}\right)$.

The movement of these two suspended masses is governed by Newton's Second Law, where the net force on the system drives the shared acceleration of both objects. Because the 10 kg mass is pulled downwards by gravity with a force of $10g$ and the 5 kg mass is similarly pulled downwards with a force of $5g$, the net force driving the system is the difference between these weights, $10g - 5g$. Since the total mass being accelerated is the sum of both, or 15 kg, we find the acceleration using the equation $a = \frac{10g - 5g}{15} = \frac{5g}{15} = \frac{g}{3}$