Laws Of Motion – NEET UG PhysicsPractice Questions & PYQs

At the peak of a curved hill, a roller coaster car follows a circular path, which requires a centripetal force directed toward the center of curvature. Weightlessness occurs precisely when the contact force between the rider and the seat vanishes, meaning that the gravitational force alone must account for the entire centripetal force needed to maintain the car's circular trajectory.

Equating the gravitational force to the required centripetal force gives the condition $mg = \frac{mv^2}{R}$. Canceling the mass from both sides simplifies the relationship to $v = \sqrt{Rg}$. Substituting the given radius of 20 m and a gravitational acceleration of 10 m/s^2, the speed must equal $\sqrt{20 \times 10}$, which calculates to approximately 14.1 m/s. This value lies between 14 m/s and 15 m/s.

To ensure the resultant force acts only along the $y$-direction, the net sum of all force components along the $x$-axis must be zero. Resolving the three forces acting at $0^\circ$, $60^\circ$, and $180^\circ$ into their $x$-components yields $1 \cos 0^\circ = 1 \, \text{N}$, $1 \cos 60^\circ = 0.5 \, \text{N}$, and $1 \cos 180^\circ = -1 \, \text{N}$, respectively. Summing these values gives a net $x$-component of $\sum F_x = 1 + 0.5 - 1 = 0.5 \, \text{N}$. To make the resultant force act strictly along the $y$-axis, an additional force $\vec{F}_{\text{add}}$ must be applied such that its $x$-component counteracts the existing net $x$-force, requiring $F_{\text{add},x} = -0.5 \, \text{N}$. Given the magnitude formula $|\vec{F}_{\text{add}}| = \sqrt{F_x^2 + F_y^2}$, the minimum magnitude is achieved by setting the $y$-component to zero, resulting in a required force of $0.5 \, \text{N}$.

When a block slides along a horizontal surface, kinetic friction acts as a constant force opposing the motion, with a magnitude defined as $f = \mu mg$. Newton's second law allows us to calculate the resulting retardation as $a = f/m = \mu g$. Because the block eventually comes to a complete stop, we apply the kinematic equation $v = u - at$, where the initial velocity $u$ is $V$ and the final velocity $v$ is zero. Substituting these values gives $0 = V - (\mu g)t$, which leads to the expression for time:

$t = \frac{V}{\mu g}$

Force is defined as the rate of change of momentum, where the change involves only the velocity component perpendicular to the wall, as the parallel component remains unaffected by the impact. With the ball hitting the wall at an angle of 30 degrees relative to the surface, the perpendicular component of its velocity is given by the product of the speed and the sine of the angle.

During the collision, this perpendicular velocity component reverses direction, creating a total change in momentum equal to twice the magnitude of the initial perpendicular momentum. The average force can be calculated using the impulse equation:
$F = \frac{2mv \sin 30^{\circ}}{t}$
By substituting the values of mass 0.5 kg, speed 12 m/s, contact time 0.25 s, and the sine value of 0.5 into the equation, we get:
$F = \frac{2 \times 0.5 \times 12 \times 0.5}{0.25}$
This calculation results in an average force of 24 N.

For a block to remain stationary relative to an accelerating wedge, the forces acting on the block in the non-inertial frame of the wedge must be in equilibrium. By resolving the gravitational force $mg$ and the horizontal pseudo force $ma$ into components parallel and perpendicular to the incline, the condition for the block not slipping is defined by $mg \sin\theta = ma \cos\theta$. The total normal reaction force $N$ exerted by the wedge is the sum of the perpendicular components of both gravity and the pseudo force, leading to the expression $N = mg \cos\theta + ma \sin\theta$. Substituting the acceleration derived from the equilibrium condition, $a = g \tan\theta$, into this expression simplifies through trigonometric identities to the final value:

$N = \frac{mg}{\cos\theta}$

For a system connected by a string over a pulley to remain in equilibrium, the forces acting on the objects must perfectly balance. The hanging block B exerts a downward force equal to its weight, which creates tension in the string that pulls block A across the table. Simultaneously, block A experiences a normal force equal to its own weight, which allows the table to generate a maximum static friction force that resists this tension.

Equating these forces allows us to determine the maximum mass for block B before sliding occurs. The tension in the string is given by $T = m_B g$, while the limiting frictional force is defined by $f_s = \mu_s m_A g$. Setting the tension equal to the frictional force yields the balance condition $m_B g = \mu_s m_A g$. By canceling the gravity constant from both sides of the equation, the mass of block B is determined by the relationship $m_B = \mu_s m_A$. Using the coefficient of 0.2 and the mass of 2 kg, the maximum mass of block B is 0.4 kg.

As the monkey climbs upward with acceleration $a$, the rope experiences a tension force $T$ determined by the relationship $T = m(g + a)$, where $m$ represents the monkey's mass. Because the rope can only support a maximum load equivalent to 25 kg, the maximum allowable tension is $25g$. Equating the forces acting on the rope to this breaking limit provides the equation $20(g + a) = 25g$. Simplifying this to isolate $a$ yields $20a = 5g$, and substituting $g = 10 \text{ m/s}^2$ results in $a = \frac{5 \times 10}{20}$, which equals $2.5 \text{ m/s}^2$.

A weighing scale in an elevator measures the normal force exerted by the floor on the person, which manifests as their apparent weight. When an elevator accelerates upwards, the floor must exert additional force to counteract gravity while also providing the necessary upward acceleration for the person. Consequently, the apparent weight becomes higher than the person's true weight.

The total force is determined by the equation $R = m(g + a)$, where the person's mass is 80 kg, the acceleration due to gravity is 10 m/s^2, and the lift's upward acceleration is 5 m/s^2. Substituting these values into the formula yields the resulting force:

$R = 80(10 + 5) = 1200 \text{ N}$

When a lift accelerates in the upward direction, the tension in the supporting string must overcome gravity while also providing the force required to create that acceleration, leading to an increased tension compared to a stationary state. Based on Newton's second law of motion, the relationship between these forces is $T - mg = ma$, which rearranges to the formula $T = m(g + a)$. By substituting the values of 1000 kg for mass, 9.8 m/s² for gravity, and 1 m/s² for acceleration, the calculation becomes $T = 1000 \times (9.8 + 1)$. Solving this provides the final tension of 10,800 N.

Movement of the block depends on the net force, which is the applied force minus the opposing frictional force. The normal force $N = mg$ is 100 N, making the kinetic friction $f = \mu N$ equal to 50 N. Subtracting this friction from the 100 N horizontal pull leaves a net force of 50 N. Using Newton's second law $a = \frac{F_{net}}{m}$, we find the acceleration is 5 m/s².

The conservation of linear momentum dictates that because the stationary bomb possesses zero initial momentum, the vector sum of the momenta of the three fragments produced by the explosion must also be zero. Consequently, the momentum of the third, larger fragment must perfectly oppose the combined momentum of the two smaller fragments.

With a mass ratio of 1:1:3 and a total mass of 1 kg, the masses of the fragments are 1/5 kg, 1/5 kg, and 3/5 kg. Since the two smaller fragments of mass 1/5 kg move perpendicularly at 30 m/s, their respective momenta are $p_1 = 6$ kg m/s and $p_2 = 6$ kg m/s. The third fragment, with mass 3/5 kg, must carry momentum $\vec{p}_3$ such that $\vec{p}_3 = -(p_1 \hat{i} + p_2 \hat{j})$. Dividing this momentum by the mass 3/5 kg provides the velocity vector for the third part:
$\vec{v}_3 = \frac{-(6\hat{i} + 6\hat{j})}{3/5} = -10\hat{i} - 10\hat{j}$
The magnitude of the velocity is determined using the Pythagorean theorem, resulting in $v_3 = \sqrt{(-10)^2 + (-10)^2} = 10\sqrt{2}$ m/s.

Newton's second law of motion dictates that the force required to stop a moving object is equivalent to the change in momentum divided by the duration of the catch. First, convert the mass into standard SI units as $0.15 \text{ kg}$ and determine the total change in momentum as the product of this mass and the initial velocity. Using the relationship
$F = \frac{m \times v}{\Delta t}$
substitute the known values to find $F = \frac{0.15 \times 20}{0.1}$, which results in a net force of $30 \text{ N}$.

Efficiency in a pulley system reflects how effectively the input effort is converted into useful work, determined by the ratio of mechanical advantage to velocity ratio. Start by calculating the mechanical advantage, which is the load of 750 N divided by the effort of 250 N, resulting in 3. Next, find the velocity ratio by dividing the distance moved by the effort, 12 m, by the distance moved by the load, 3 m, which yields 4. Combining these values into the efficiency formula $\eta = \frac{\text{Mechanical Advantage}}{\text{Velocity Ratio}} \times 100$ results in a final value of $\frac{3}{4} \times 100 = 75\%$.

Friction acts to oppose the tendency of an object to slide when subject to an accelerating reference frame, such as a moving truck. The maximum static friction the surface can provide is determined by the formula $f_{L} = \mu N = \mu mg$. Substituting the given values into this expression, we find that $f_{L} = 0.6 \times 1 \times 10 = 6 \text{ N}$. The truck's acceleration creates an effective pseudo force on the block given by $F = ma$, which equals 5 N. Because this applied pseudo force is lower than the limiting friction, the block does not slide, and the actual static friction exerted is simply equal to the applied force, resulting in 5 N.

Newton's second law relates the magnitude of an applied force to mass and acceleration through the equation $|\vec{F}| = ma$. Since mass is a scalar quantity, determining it requires finding the magnitude of the force vector $\vec{F} = 6 \hat{i} - 8 \hat{j} + 10 \hat{k}$ before dividing that result by the acceleration. The magnitude is calculated by taking the square root of the sum of the squares of the individual components.

Performing this calculation involves finding the square root of the sum of 36, 64, and 100, which simplifies to $\sqrt{200}$. Dividing this value by the acceleration of 1 m/s squared yields the mass of the body as $10\sqrt{2}$ kg.

Newton's second law of motion establishes that the net force acting on a body is the product of its mass and acceleration, which in this scenario requires balancing the upward tension against the downward force of gravity. Given the mass of 2000 kg and the standard gravitational acceleration of 10 meters per second squared, the weight of the lift is 20000 N. Using the equation $T - Mg = Ma$ we substitute the known values to get $28000 - 20000 = 2000a$, which simplifies to $8000 = 2000a$ and results \in $a = 4$ meters per second squared. Since the tension force pulling upward is greater than the gravitational force pulling downward, the acceleration is directed upwards.

When a body initially at rest undergoes an explosion, the law of conservation of linear momentum states that the vector sum of the momenta of all fragments must be zero. Because the total mass of 5 kg is distributed in a ratio of 1:1:3, the three fragments have masses of 1 kg, 1 kg, and 3 kg. The two equal fragments each move with a speed of 21 m/s in mutually perpendicular directions.

Their individual momentum magnitudes are each 21 kg m/s, and their resultant momentum vector is determined by the Pythagorean relationship $\sqrt{21^2 + 21^2} = 21\sqrt{2}$ kg m/s. Since the system's initial momentum is zero, the heaviest fragment must carry a momentum equal in magnitude to this resultant to balance the system. By setting the momentum of the heaviest fragment to this value, we have the equation $3v = 21\sqrt{2}$ kg m/s. Solving this for the velocity of the third fragment yields $v = 7\sqrt{2}$ m/s.

The average force exerted on the wall is determined by the impulse-momentum theorem, which states that the force equals the rate of change of momentum during the impact. When an object strikes a wall at an angle and reflects at the same angle, the momentum component parallel to the wall remains unchanged, while the perpendicular component reverses direction. This creates a net change in momentum equal to twice the initial perpendicular component of the momentum, calculated as $\Delta p = 2mv \sin 60^{\circ}$.

Given a mass of $3 \text{ kg}$, an impact velocity of $10 \text{ m/s}$, and an angle of $60^{\circ}$, the total change in momentum is $2 \times 3 \times 10 \times \sin 60^{\circ} = 30\sqrt{3} \text{ kg}\cdot\text{m/s}$. By dividing this change in momentum by the impact duration of $0.2 \text{ s}$, the average force is found using $F = \frac{\Delta p}{\Delta t} = \frac{30\sqrt{3}}{0.2}$, which yields $150 \sqrt{3} \text{ N}$.

When an object slides down a smooth inclined plane of angle $\theta$, its acceleration is determined solely by the component of gravity acting down the slope, given by $a = g \sin \theta$. In the presence of friction, this acceleration is reduced by the frictional force, resulting in a net acceleration of $a' = g \sin \theta - \mu g \cos \theta$. Since the distance $s$ covered from rest follows the kinematic relationship $s = \frac{1}{2}at^2$, the time taken to slide down the incline is inversely proportional to the square root of the acceleration, $t \propto \frac{1}{\sqrt{a}}$.

Given that the time taken with friction is twice the time taken without friction, we have $t_{friction} = 2t_{smooth}$. Squaring both sides yields $t_{friction}^2 = 4t_{smooth}^2$, which implies that the acceleration without friction must be four times the acceleration with friction, or $a = 4a'$. Substituting the expressions for acceleration into this relationship gives $g \sin \theta = 4(g \sin \theta - \mu g \cos \theta)$. Dividing both sides by $g$ and rearranging the terms leads to $\sin \theta = 4 \sin \theta - 4 \mu \cos \theta$, which simplifies to $4 \mu \cos \theta = 3 \sin \theta$. Isolating $\mu$ results in $\mu = \frac{3}{4} \tan \theta$. With $\theta = 45^{\circ}$ and $\tan 45^{\circ} = 1$, the coefficient of friction calculates to $0.75$.