Mechanics

A car undergoes a uniform circular motion. The acceleration of the car is

[amp_mcq option1=”zero” option2=”a non-zero constant” option3=”a non-zero but not a constant” option4=”None of the above” correct=”option3″]

In uniform circular motion, the speed of the object (car) is constant, but its direction of motion is continuously changing. Velocity is a vector quantity (magnitude + direction). Since the direction of velocity is changing, the velocity itself is not constant. Acceleration is the rate of change of velocity. Therefore, there is a non-zero acceleration. The acceleration in uniform circular motion is the centripetal acceleration, which is directed towards the center of the circle. Its magnitude (a = v^2/r) is constant because the speed (v) and radius (r) are constant. However, the direction of the acceleration vector is continuously changing as the car moves around the circle. A vector quantity is constant only if both its magnitude and direction are constant. Since the direction changes, the acceleration vector is non-zero but not constant.

Acceleration is a vector quantity. For acceleration to be constant, both its magnitude and direction must be constant.

The acceleration in uniform circular motion is always perpendicular to the velocity vector and points towards the center of the circle. This causes the direction of velocity to change, resulting in circular motion, while the speed remains constant.

Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint; as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g = 10 m/s²)

[amp_mcq option1=”200 N” option2=”20 N” option3=”10 N” option4=”infinitely large” correct=”option4″]

When a mass is attached to the midpoint of a horizontally held rope, the rope sags downwards due to the weight of the mass (W = mg). The tension in the rope acts along the direction of the rope segments on either side of the mass. For the rope to be in equilibrium, the vector sum of the two tensions and the weight must be zero. If the rope is perfectly horizontal, the tension vectors would have only horizontal components (neglecting the mass of the rope itself). However, the weight of the attached mass acts vertically downwards. To balance this downward force, there must be an equal and opposite upward force provided by the vertical components of the tension in the rope. As the rope approaches a perfectly horizontal state, the angle (θ) it makes with the horizontal approaches zero. The vertical component of the tension in each half of the rope is T * sin(θ), where T is the tension. The total upward vertical force is 2 * T * sin(θ). To balance the weight W, 2 * T * sin(θ) = W. If the rope is to be perfectly straight and horizontal (θ = 0), then sin(θ) = sin(0) = 0. For 2 * T * sin(θ) to equal the non-zero weight W (20 kg * 10 m/s² = 200 N), the tension T must tend towards infinity as sin(θ) tends towards zero. Thus, an infinitely large tension is required to completely straighten the rope.

– Static equilibrium requires the net force in all directions to be zero.
– When a mass hangs from a rope, the weight acts vertically downwards.
– Tension in the rope must have a vertical component to balance the weight.
– For a nearly horizontal rope, the angle with the horizontal is very small.
– As the angle approaches zero, the sine of the angle approaches zero, requiring tension to approach infinity to provide a non-zero vertical force.

This scenario represents a classic physics problem illustrating the vector resolution of forces and the limit as an angle approaches zero. In any real-world scenario, the rope will always sag slightly, having a non-zero angle, because no rope can withstand infinite tension.

The position vector of a particle is r⃗ = 2t² x̂ + 3t ŷ + 4 ẑ. Then the instantaneous velocity v⃗ and acceleration a⃗ respectively lie

[amp_mcq option1=”on xy-plane and along z-direction” option2=”on yz-plane and along x-direction” option3=”on yz-plane and along y-direction” option4=”on xy-plane and along x-direction” correct=”option4″]

The position vector is given by r⃗ = 2t² x̂ + 3t ŷ + 4 ẑ.
The instantaneous velocity vector v⃗ is the first derivative of the position vector with respect to time:
v⃗ = dr⃗/dt = d/dt (2t² x̂ + 3t ŷ + 4 ẑ) = (d/dt 2t²) x̂ + (d/dt 3t) ŷ + (d/dt 4) ẑ = 4t x̂ + 3 ŷ + 0 ẑ = 4t x̂ + 3 ŷ.
A vector of the form Ax̂ + Bŷ has components only in the x and y directions. Such a vector lies in the xy-plane.
The instantaneous acceleration vector a⃗ is the first derivative of the velocity vector with respect to time:
a⃗ = dv⃗/dt = d/dt (4t x̂ + 3 ŷ) = (d/dt 4t) x̂ + (d/dt 3) ŷ = 4 x̂ + 0 ŷ = 4 x̂.
A vector of the form Ax̂ has a component only in the x direction. Such a vector lies along the x-direction.
Therefore, the instantaneous velocity v⃗ lies on the xy-plane, and the instantaneous acceleration a⃗ lies along the x-direction.

– Velocity is the time derivative of the position vector.
– Acceleration is the time derivative of the velocity vector.
– A vector is in a plane if it has non-zero components only in the directions defining that plane.
– A vector is along an axis if it has a non-zero component only in the direction of that axis.

In this case, the position vector has a constant z-component (4). This means the particle is always located on the plane z=4, which is parallel to the xy-plane. The motion is confined to this plane z=4. The velocity vector (4t x̂ + 3 ŷ) correctly reflects motion in the x and y directions within this plane. The acceleration vector (4 x̂) indicates that the acceleration is constant and directed purely in the positive x-direction.

A particle moves with uniform acceleration along a straight line from rest. The percentage increase in displacement during sixth second compared to that in fifth second is about

[amp_mcq option1=”11%” option2=”22%” option3=”33%” option4=”44%” correct=”option2″]

For a particle starting from rest (initial velocity u = 0) and moving with uniform acceleration ‘a’, the displacement in the n-th second (S_n) is given by the formula S_n = u + a(n – 1/2). Since u = 0, S_n = a(n – 1/2).
Displacement in the fifth second (n=5): S₅ = a(5 – 1/2) = a(4.5).
Displacement in the sixth second (n=6): S₆ = a(6 – 1/2) = a(5.5).
The percentage increase in displacement during the sixth second compared to the fifth second is calculated as [(S₆ – S₅) / S₅] * 100%.
Percentage Increase = [(a(5.5) – a(4.5)) / a(4.5)] * 100% = [a(5.5 – 4.5) / a(4.5)] * 100% = [a(1) / a(4.5)] * 100% = (1 / 4.5) * 100% = (10 / 45) * 100% = (2 / 9) * 100% ≈ 0.2222 * 100% ≈ 22.22%.
The closest option is 22%.

– Displacement in the n-th second formula: S_n = u + a(n – 1/2).
– Uniform acceleration from rest means u=0.
– Percentage increase calculation involves finding the difference, dividing by the original value, and multiplying by 100.

The total displacement after ‘n’ seconds is given by S = ut + (1/2)at². For a body starting from rest (u=0), S = (1/2)at². The displacement in the n-th second is the difference between the total displacement after n seconds and the total displacement after (n-1) seconds: S_n = S(n) – S(n-1) = (1/2)an² – (1/2)a(n-1)² = (1/2)a [n² – (n² – 2n + 1)] = (1/2)a [n² – n² + 2n – 1] = (1/2)a (2n – 1) = a(n – 1/2). This confirms the formula used.

A person throws an object on a horizontal frictionless plane surface. It is noticed that there are two forces acting on this object—(i) gravitational pull and (ii) normal reaction of the surface. According to the third law of motion, the net resultant force is zero. Which one of the following can be said for the motion of the object?

[amp_mcq option1=”The object will move with acceleration.” option2=”The object will move with deceleration.” option3=”The object will move with constant speed, but varying direction.” option4=”The object will move with constant velocity.” correct=”option4″]

On a horizontal frictionless plane, the vertical forces (gravitational pull and normal reaction) are equal in magnitude and opposite in direction, resulting in zero net force in the vertical direction. If the object is thrown, it has an initial horizontal velocity. Since the plane is frictionless, there is no horizontal force (ignoring air resistance).

According to Newton’s first law of motion, an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force. With zero net force acting on the object in both vertical and horizontal directions, its velocity (magnitude and direction) remains constant.

The third law of motion states that for every action, there is an equal and opposite reaction. While it ensures that forces exist in pairs (e.g., Earth pulls the object down, object pulls Earth up; surface pushes object up, object pushes surface down), it doesn’t imply the *net resultant force on the object itself* is zero unless the forces acting *on that object* happen to be balanced. In this scenario, the forces *on the object* (gravity and normal reaction) are balanced vertically, and there are no horizontal forces, resulting in zero net force on the object, leading to constant velocity.

A person is standing on a frictionless horizontal ground. How can he move by a certain distance on this ground?

[amp_mcq option1=”By sneezing” option2=”By jumping” option3=”By running” option4=”By rolling” correct=”option1″]

The correct option is A) By sneezing.

On a frictionless horizontal surface, a person cannot propel themselves forward by pushing against the ground (like walking, running, or jumping) because there is no friction to provide an external reaction force in the horizontal direction. However, by sneezing, the person expels air in one direction. According to Newton’s third law of motion, this action creates an equal and opposite reaction force on the person’s body, propelling them in the opposite direction of the expelled air.

This is a classic physics problem illustrating Newton’s laws of motion, particularly the conservation of momentum. In the absence of external forces (friction), the total momentum of the person-air system remains constant. If the person expels air with a certain momentum, their body must acquire an equal and opposite momentum to conserve the total momentum (initially zero).

A car moving with a speed of 12 m/s is subjected to brakes which produces a deceleration of 6 m/s². The car takes 2 s to stop after the application of brakes. What is the distance covered by the car after the application of brakes?

[amp_mcq option1=”12 m” option2=”24 m” option3=”36 m” option4=”48 m” correct=”option1″]

Given:
Initial velocity, $u = 12$ m/s
Deceleration, $a = -6$ m/s² (negative because it’s deceleration)
Final velocity, $v = 0$ m/s (the car stops)
Time, $t = 2$ s
We need to find the distance covered, $s$.
We can use the equation of motion $s = ut + (1/2)at^2$.
Substituting the given values:
$s = (12 \, \text{m/s}) \times (2 \, \text{s}) + (1/2) \times (-6 \, \text{m/s}^2) \times (2 \, \text{s})^2$
$s = 24 \, \text{m} + (1/2) \times (-6) \times 4 \, \text{m}$
$s = 24 \, \text{m} – (3) \times 4 \, \text{m}$
$s = 24 \, \text{m} – 12 \, \text{m}$
$s = 12 \, \text{m}$
Alternatively, using $v^2 = u^2 + 2as$:
$0^2 = (12)^2 + 2 \times (-6) \times s$
$0 = 144 – 12s$
$12s = 144$
$s = 144 / 12 = 12$ m.

– Use the appropriate kinematic equations for uniformly accelerated motion.
– Deceleration is negative acceleration.

The kinematic equations for constant acceleration are:
1. $v = u + at$
2. $s = ut + (1/2)at^2$
3. $v^2 = u^2 + 2as$
4. $s = (u+v)t/2$
These equations are valid only when acceleration is constant.

The area under the velocity-time graph for a particle moving in a straight line with uniform acceleration gives

[amp_mcq option1=”its average velocity” option2=”its net displacement” option3=”the distance travelled by it” option4=”its average speed” correct=”option2″]

For a particle moving in a straight line, the area under the velocity-time graph represents the displacement of the particle during the given time interval. If the velocity is plotted on the y-axis and time on the x-axis, the area under the curve between two time points gives the change in position (displacement) over that period. This is valid even if the velocity is non-uniform or changes direction.

The area under the velocity-time graph represents the net displacement.

If the particle moves in only one direction (velocity does not change sign), the magnitude of the displacement is equal to the distance traveled. However, if the velocity-time graph goes below the time axis (indicating motion in the opposite direction), the area below the axis is considered negative displacement. The net displacement is the sum of the signed areas. The total distance traveled would be the sum of the magnitudes of the areas. The question specifies uniform acceleration, which means the graph is a straight line, but it doesn’t preclude the velocity from changing direction (e.g., starting with negative velocity and accelerating in the positive direction).

An object weighs 9 N on the surface of the Earth. What would be its weight, when measured on the surface of a planet where the acceleration due to gravity is 9 times that on the Earth ?

[amp_mcq option1=”The weight would remain the same” option2=”The weight would be equal to 1 N” option3=”The weight would become 9 times” option4=”The weight will be reduced to $\frac{1}{9}$ N” correct=”option3″]

The weight would become 9 times.

Weight is defined as the force of gravity acting on an object, given by the formula W = m * g, where W is weight, m is mass, and g is the acceleration due to gravity. The mass of the object remains constant regardless of location. If the acceleration due to gravity (g) on the planet is 9 times that on Earth (g_planet = 9 * g_earth), then the weight on the planet will be W_planet = m * g_planet = m * (9 * g_earth) = 9 * (m * g_earth) = 9 * W_earth. Thus, the weight on the planet becomes 9 times the weight on Earth.

The initial weight on Earth is 9 N. The new weight on the planet would be 9 * 9 N = 81 N. Option C accurately states the relationship between the new weight and the original weight.

What happens to the gravitational force between two objects if the mass of one object is doubled and the distance between them is also doubled ?

[amp_mcq option1=”The force would remain the same” option2=”The force would be doubled” option3=”The force would be halved” option4=”The force would increase by a factor of 4″ correct=”option3″]

According to Newton’s Law of Gravitation, the force is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. Doubling one mass increases the force by a factor of 2. Doubling the distance increases the distance squared by a factor of 4 (2^2), which decreases the force by a factor of 4. The combined effect is a force that is (2 * 1) / 4 = 1/2 times the original force.

Newton’s Law of Gravitation is given by F = G * (m1 * m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.

If the mass of one object is doubled (m1′ = 2m1) and the distance is doubled (r’ = 2r), the new force F’ is given by F’ = G * (m1′ * m2) / (r’)^2 = G * (2m1 * m2) / (2r)^2 = G * (2m1 * m2) / (4r^2) = (2/4) * G * (m1 * m2) / r^2 = (1/2) * F. Thus, the force would be halved.