Newtonian Mechanics MCQ Quiz - Objective Question with Answer for Newtonian Mechanics - Download Free PDF

Calculation:

Firstly, check how much friction is:

f = μN = 0.5 × 10 × 9.8 × (√3 / 2) = 4.9 √3 × 5

f = 24.5 √3

So the required force is:

F = μN + 10g sin 30°

F = 24.5 √3 + 5 × 9.8

F = 91.4 N

Hence, the correct option is (B).

Correct Answer: (i) m√ (8ag / 3)

2mg - T = 2m × a ....(i)

T - mg = m × a ....(ii)

On solving eq. (i) & (ii), a = g / 3

v_b = ^√ u² + 2as = ^√ 0 + 2 ^{(g / 3)} a = ^√ (2ag / 3)

s = ut + 1/2at², a = 0 + 1/2(g / 3) t², t = ^√ (6a / g) = 3v / g

t = 2v / g

Calcultion:

F – mg = ma

F = ma + mg

F – (m – x)g = (m – x) 3a

Put F

Ma + mg – mg + xg = 3ma – 3xa

\(x=\frac{2 m a}{g+3 a}\)

Calculation:

Analyze the forces on the system

The 4 kg mass exerts a downward force due to gravity:

F = m × g = 4 × 10 = 40 N

T' = 2T           (1)

Kx = 2T'         (2)

T = 40 N         (3)

From 1, 2, 3 we get

160 = (8 × 10³) × x

Solve for extension

x = 160 / (8 × 10³)

x = 20 × 10^-3 m = 2 cm 

Thus, the extension in the spring is: 2 cm

The Correct Option is B

Calculation:

To solve for the ratio \( \frac{T_2}{T_1}\) , we use Kepler's third law, which states that the square of the orbital period T is proportional to the cube of the semi-major axis (a) of the orbit:

\(T^2 \propto a^3\)

The semi-major axis (a) for an elliptical orbit is the average of the minimum and maximum distances from the sun:

\(a = \frac{r_{\text{min}} + r_{\text{max}}}{2}\)

Step 1: Semi-major axes of \(P_1\) and \(P_2\):

For \(P_1\):

\(a_1 = \frac{R + 3R}{2} = 2R\)

For \(P_2\):

\(a_2 = \frac{2R + 4R}{2} = 3R\)

From Kepler's law:

\(\frac{T_2^2}{T_1^2} = \frac{a_2^3}{a_1^3}\)

Substitute \(a_1 = 2R\) and \(a_2 = 3R\):

\(\frac{T_2^2}{T_1^2} = \frac{(3R)^3}{(2R)^3} = \frac{27R^3}{8R^3} = \frac{27}{8}\)

Take the square root to find \(\frac{T_2}{T_1} \):

\(\frac{T_2}{T_1} = \sqrt{\frac{27}{8}} = \frac{3}{2} \sqrt{\frac{3}{2}}\)

Thus, option '1' is correct.

Translatory motion:

It is that motion in which a body, which is not a point mass body is moving such that all its constituent particles move simultaneously along parallel straight lines and all its constituent particles shift through an equal distance in a given interval of time.

For example, A body slipping along the inclined plane has translatory motion.

From above it is clear that the motion of a body as a whole from one point to another is called translation motion.

Additional Information

Types of motions:

• Rectilinear motion: It is that motion in which a particle or point mass body is moving along a straight line.
• Oscillatory or vibratory motion: It is that motion in which a body moves to and fro or back and forth repeatedly about a fixed point (called mean position) in a definite interval of time.
  • For example The motion of the pendulum of the wall clock is oscillatory motion.
• Brownian motion: The continuous zigzag movement of the colloidal particles in the dispersion medium in a colloidal solution is called the Brownian motion.
  • The cause of the Brownian movement is the collision of molecules of the dispersion medium on the colloidal particles.
  • The size of the particles and viscosity affects the Brownian movement.

CONCEPT:

Aristotle's Law of motion:

• According to Aristotelian law of motion, an external force is necessary to keep a body moving with uniform velocity.

EXPLANATION:

• Aristotle's views were proved wrong by Galileo Galicia (1564-1642) about two thousand years later on.
• It was observed that external forces were necessary to counter the opposing forces of friction to keep bodies in uniform motion.
• If there were no friction, no external force would be needed to maintain the state of uniform motion of a body.
• Thus, he did not take into account friction. Therefore option 3 is correct.

Concept:

• Centre of mass frame of reference: It is the isolated system of particles in which the total linear momentum of the particles of the system is zero.
• Momentum: The product of mass and velocity is called the momentum of the body.

Additional Information 

There are two types of frame of reference:

• Inertial frame of reference: The frame of reference having zero acceleration is called the Inertial frame of reference.
  • This frame of reference will be either at rest or will be moving with a constant velocity.
  • Newton’s law is valid in this frame of reference.
• The non-inertial frame of reference: The frame of reference having non-zero acceleration is called a non-inertial frame of reference.
  • Newton’s law is not valid in this frame of reference.
  • For example: If we are observing an object from a freely falling object then this will be a non-inertial frame of reference because the freely falling body has some acceleration.

CONCEPT:

• Rotatory Motion: Rotational motion or circular motion, is a physical motion that happens when an object rotates or spins on an axis. Example: Rotation of a fan.
• Spin Motion: Motion of an object as it rotates around the axis through its center of mass. Example: Rotation of a top.
• Projectile motion: Projectile motion is the motion of an object projected into the air, under only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. Example: a cannonball fired at some angle.
• Translatory motion: Motion with a fixed direction of
  the net force.All the particles of a body executing translatory motion move in the same direction traversing parallel paths. Example: the motion of bus or train.

EXPLANATION:

• The motion of an object is said to be translatory if the position of the object is changing with respect to a fixed point or object. So the motion of train is an example of translatory motion.

So option 4 is correct. 

Translatory motion:

It is that motion in which a body, which is not a point mass body is moving such that all its constituent particles move simultaneously along parallel straight lines and all its constituent particles shift through an equal distance in a given interval of time.

For example, A body slipping along the inclined plane has translatory motion.

From above it is clear that the motion of a body as a whole from one point to another is called translation motion.

Additional Information

Types of motions:

• Rectilinear motion: It is that motion in which a particle or point mass body is moving along a straight line.
• Oscillatory or vibratory motion: It is that motion in which a body moves to and fro or back and forth repeatedly about a fixed point (called mean position) in a definite interval of time.
  • For example The motion of the pendulum of the wall clock is oscillatory motion.
• Brownian motion: The continuous zigzag movement of the colloidal particles in the dispersion medium in a colloidal solution is called the Brownian motion.
  • The cause of the Brownian movement is the collision of molecules of the dispersion medium on the colloidal particles.
  • The size of the particles and viscosity affects the Brownian movement.

CONCEPT:

Aristotle's Law of motion:

• According to Aristotelian law of motion, an external force is necessary to keep a body moving with uniform velocity.

EXPLANATION:

• Aristotle's views were proved wrong by Galileo Galicia (1564-1642) about two thousand years later on.
• It was observed that external forces were necessary to counter the opposing forces of friction to keep bodies in uniform motion.
• If there were no friction, no external force would be needed to maintain the state of uniform motion of a body.
• Thus, he did not take into account friction. Therefore option 3 is correct.

Calculation:

To solve for the ratio \( \frac{T_2}{T_1}\) , we use Kepler's third law, which states that the square of the orbital period T is proportional to the cube of the semi-major axis (a) of the orbit:

\(T^2 \propto a^3\)

The semi-major axis (a) for an elliptical orbit is the average of the minimum and maximum distances from the sun:

\(a = \frac{r_{\text{min}} + r_{\text{max}}}{2}\)

Step 1: Semi-major axes of \(P_1\) and \(P_2\):

For \(P_1\):

\(a_1 = \frac{R + 3R}{2} = 2R\)

For \(P_2\):

\(a_2 = \frac{2R + 4R}{2} = 3R\)

From Kepler's law:

\(\frac{T_2^2}{T_1^2} = \frac{a_2^3}{a_1^3}\)

Substitute \(a_1 = 2R\) and \(a_2 = 3R\):

\(\frac{T_2^2}{T_1^2} = \frac{(3R)^3}{(2R)^3} = \frac{27R^3}{8R^3} = \frac{27}{8}\)

Take the square root to find \(\frac{T_2}{T_1} \):

\(\frac{T_2}{T_1} = \sqrt{\frac{27}{8}} = \frac{3}{2} \sqrt{\frac{3}{2}}\)

Thus, option '1' is correct.

Concept:

• Centre of mass frame of reference: It is the isolated system of particles in which the total linear momentum of the particles of the system is zero.
• Momentum: The product of mass and velocity is called the momentum of the body.

Additional Information 

There are two types of frame of reference:

• Inertial frame of reference: The frame of reference having zero acceleration is called the Inertial frame of reference.
  • This frame of reference will be either at rest or will be moving with a constant velocity.
  • Newton’s law is valid in this frame of reference.
• The non-inertial frame of reference: The frame of reference having non-zero acceleration is called a non-inertial frame of reference.
  • Newton’s law is not valid in this frame of reference.
  • For example: If we are observing an object from a freely falling object then this will be a non-inertial frame of reference because the freely falling body has some acceleration.

Explanation:

The Torque about the point O is given by :

\(\vec\tau =\vec{r}\times\vec{F} \)

The distance travel along the inclined plane is

 \(s=\frac{1}{2}g\sin\frac{\pi}{6}\cdot (2)^2 \\ \quad = 10 \text{ m}\)

Then \(r_{\perp}\) will be 

\(\tan60^\circ =\frac{r_{\perp}}{s} \\ \sqrt{3}\cdot s=r_{\perp} \\ 10 \sqrt{3}=r_{\perp}\)

The torque when the object is at s= 10 m after 2 sec will be :

\(\vec\tau =\vec{r}\times\vec{F} \\ \quad =10\sqrt{3}\times \frac{g}{2} \\ \quad =50\sqrt{3} = 86.6 \\ \quad \approx 87\)

The correct answer is 87.

Explanation:
The average power P is defined as the work done W divided by the time taken T . In this case, the work done is equal to the potential energy gained by the person, given by Mgh , where M is the mass, g is the acceleration due to gravity, and h is the height reached. The time taken for the jump can be determined from the initial velocity v_0 using the equation of motion:

\( h = \frac{1}{2} g t^2\)

Therefore, the average power is:
\( P = \frac{W}{T} = \frac{Mgh}{\frac{1}{2}gt^2} = \frac{2Mgh}{t}\)

To find the time taken, we first calculate the initial velocity v₀ . Using the equation of motion:
\( v_0^2 = 2gh\\ v_0 = \sqrt{2gh}\)

Using the equation of motion:
\( h = v_0 t - \frac{1}{2} g t^2 \\ t = \frac{2h}{v_0} = \frac{2h}{\sqrt{2gh}} = \sqrt{\frac{2h}{g}}\)

Substituting the expression for t into the average power equation:

\( P = \frac{2Mgh}{\sqrt{\frac{2h}{g}}} = Mg \sqrt{2gh}\)

The average power is:
\( P = 72.7 \times 9.8 \times \sqrt{\frac{2 \times 0.91}{9.8}} \approx 4510 \, \text{W}\)

The correct answer is 4510 W.