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

Solution:

F₃ = F₃(p, Q, t)

∂F₃/∂p = -q,    ∂F₃/∂Q = -P

⇒ Q = log(1 + q^1/2 cos p) ⇒ e^Q = 1 + q^1/2 cos p

⇒ (e^Q - 1) / cos p = q^1/2 ⇒ q = ((e^Q - 1)²) / cos² p

⇒ P = 2(1 + q^1/2 cos p) q^1/2 sin p ⇒ P = 2e^Q q^1/2 sin p ⇒ 2e^Q(e^Q - 1) tan p

⇒ ∂F₃/∂p = -q = -((e^Q - 1)² / cos² p) ⇒ F₃ = -∫(e^Q - 1)² sec²p dp

⇒ F₃ = - (e^Q - 1)² tan p + f₁(Q)   ........A

∂F₃/∂Q = -P = -2(e^2Q - e^Q) tan p

F₃ = -2 ∫ (e^2Q - e^Q) tan p + f₂(p)

= - (e^Q - 1)² tan p + tan p + f₂(p)   ....(B)

Equating A and B: f₁(Q) = 0,   f₂(p) = -tan p

So F₃ = - (e^Q - 1)² tan p

Solution:

H = ( p – qA )² / 2m + qφ

⇒ H = |p|² / 2m + q² |A|² / 2m – (q / m) p · A

Use p = p_xi + p_yj + p_zk

and A = A_xi + A_yj + A_zk

∂H/∂p_x = (p_x – qA_x) / m = ẋ

∂H/∂p_y = (p_y – qA_y) / m = ẏ

∂H/∂p_z = (p_z – qA_z) / m = ż

Let us find {x, ẏ}:

{x, ẏ} = ∂x/∂x ∂ẏ/∂p_x – ∂x/∂p_x ∂ẏ/∂x + ∂x/∂y ∂ẏ/∂p_y – ∂x/∂p_y ∂ẏ/∂y

0 = – (1/m)(q) ( ∂A_y/∂x – ∂A_x/∂y ) = (q/m²)( ∇ × A )_z = (q/m²) B_k

⇒ {v_i, v_j} = (q/m²) B_k ε_ij ⇒ ε_ijk {v_i, v_j} = (q/m²) B_l ε_ijk ε_ij

= (q/m²) B_l 2δ_kl = 2qB_k / m²

Solution:

L = (15/2) m x² + 6mxẏ + 3my² − mg(x + 2y)

⇒  (∂L/∂ẋ) − ∂L/∂x = 0 ⇔ 15mẍ + 6mÿ + mg = 0 ........(1)

⇒  (∂L/∂ẏ) − ∂L/∂y = 0 ⇔ 6mẍ + 6mÿ + 2mg = 0 .......(2)

Use operation 2(1) − (2)

24mẍ + 6ẏ = 0 ⇒ d/dt (4x + y) = 0 ⇒ 4ẋ + ẏ = 0 ⇒ 12ẋ + 3ẏ = c

The correct answer is Option 1.Key Points

• When an object undergoes acceleration, it means there is a change in its velocity. This change in velocity can occur either in terms of speed, direction, or both.
• A force always acts on it:
  • This statement is generally true.
  • According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma).
  • So, if there's acceleration, there must be a force acting on the object.

Additional Information

• Acceleration is a fundamental concept in physics that describes the rate of change of velocity with respect to time.
• Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. Therefore, any change in speed, direction, or both constitutes acceleration.
• The formula for acceleration (a) is a = F/m where a is acceleration. F is the net force acting on an object and m is the mass of the object.
• This formula states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
• In simpler terms, if you apply a force to an object, it will accelerate, and the acceleration will be larger if the force is stronger or if the object has less mass.
• Acceleration can occur in various forms:
  • Linear Acceleration: Change in speed in a straight line.
  • Angular Acceleration: Change in rotational speed or direction.
  • Centripetal Acceleration: Acceleration directed towards the center of a circular path.
• Acceleration can be positive or negative:
  • Positive Acceleration: Speeding up in the positive direction.
  • Negative Acceleration (Deceleration): Slowing down or moving in the opposite direction.
• Gravity is a common force causing acceleration. Near the Earth's surface, objects in free fall experience acceleration due to gravity, denoted as g (approximately 9.8 m/s²).

Concept:

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory.

Calculation:

v = \(\sqrt{2gh} \) and v = e\(\sqrt{2gh} \)

0 = (ev)² - 2gh₁

h₁ = \({e^2 × 2gh\over 2g}\) = e²h

Similarly, h₂ = e⁴h

H = h + 2h₁ + 2h₂ +...∞ 

= h + 2(e²h + e⁴h + ... ∞)

= h + 2e²h(\({1\over 1-e^2}\))

= h × (\({1+ e^2 \over 1-e^2}\))

The coefficient of restitution between the ball and the floor is 0.5.

e = 0.5

H = 5h/3

The correct answer is option (2).

Concept:

We are using the angular momentum formula which is \(\overrightarrow{L}=I\overrightarrow{\omega}\) and then using this formula for \(x,y,z,\) planes.

By using matrices for values of \(L,\omega\) and \(I\) we get the value of magnitude and direction of angular momentum.

Explanation:

 A circular disc is in rotation with the origin as center in \(xy\) plane.

Given,

• \(\omega=\omega_0(\hat j+\hat k)\)  where \(\omega\) is angular velocity
• \(I_0\) is the moment of inertia

We are using formula for angular momentum

• \(\overrightarrow{L}=I\overrightarrow{\omega}\)
• \(I_{xx}=I_{yy}=I_0\)

For denoting angular momentum in \(x,y,z,\)planes we will use vector notations for angular momentum \(L\),

• \(\begin{bmatrix}L_x\\[0.3em]L_y\\[0.3em]L_z\end{bmatrix}\)\(=\begin{bmatrix}I_{xx} & I_{xy} & I_{xz} \\ I_{yx} &I_{yy} & I_{yz} \\[0.3em] I_{zx} & I_{zy} & I_{zz}\end{bmatrix}\)\(\begin{bmatrix}\omega_x\\[0.3em]\omega_y\\[0.3em]\omega_z\end{bmatrix}\)
• \(I_{xx}=I_{yy}=I_0\)

Using the perpendicular axis theorem,

• \(I_{xx}+I_{yy}=I_{zz}=I_0+I_0=2I_0\)
• \(\omega=\omega_0(\hat j+\hat k)\)

putting values of \(I\) and \(\omega\) in matrix equation, we get

• \(\begin{bmatrix}L_x\\[0.3em]L_y\\[0.3em]L_z\end{bmatrix}\)\(=\begin{bmatrix}I_0 & 0 & 0 \\ 0 &I_0 & 0 \\[0.3em] 0 & 0 & 2I_0\end{bmatrix}\)\(\begin{bmatrix}0\\[0.3em]\omega_0\\[0.3em]\omega_0\end{bmatrix}\)

By multiplication above matrices, we get

• \(\begin{bmatrix}L_x\\[0.3em]L_y\\[0.3em]L_z\end{bmatrix}\)\(=\)\(\begin{bmatrix}0\\[0.3em]I_0\omega_0\\[0.3em]2I_0\omega_0\end{bmatrix}\)
• \(\overrightarrow{L}=\overrightarrow{L_0}\hat j+2\overrightarrow{L_0}\hat k\)

This is the magnitude of angular momentum. The direction of angular momentum is \((\hat j +\hat 2k)\).

Concept:

 We are using Kepler's law here which states that the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.

• \(\frac{dA}{dT}=\frac{L}{2m}=\frac{A} {T}=constant\)

Explanation:

Using Kepler's second law, 

• \(\frac{dA}{dT}=\frac{L}{2m}=\frac{A} {T}=constant\)
• \(A_1=\frac {\pi ab}{2}+2\times\frac{1}{2}\times b\times c\)

eccentricity(e)\(=\frac {c}{a}\)=>\(c=ea\)

• \(A_1=\frac {\pi ab}{2}+eba=ab(\frac {\pi} {2}+e)\)
• \(A_2=\pi ab-ab(\frac {\pi}{2}+e)=ab(\frac {\pi} {2}-e)\)

Now,

• \(\frac{T_1}{T_2}=\frac{A_1}{A_2}\)
• \(\frac{T_1}{T_2}=\frac {ab(\frac {\pi}{2}+e)} {ab(\frac {\pi}{2}-e)}=\frac {(\frac {\pi}{2}+e)} {(\frac {\pi}{2}-e)}\)
• \(T_1=\frac {\frac{\pi }{2}+e}{\pi}, T_2=\frac {\frac{\pi }{2}-e}{\pi}\)
• \(T_1-T_2=\frac{2e}{\pi}=\frac {2\times0.0167\times 365}{3.14}\approx3.9 days.\)

Explanation:

• Given the transformations \(X=\frac{α p}{x}\) and \(P = βx²\), we have to check them against the requirements of canonical transformations.
• The main requirement, among others, is that the Poisson bracket {X, P} = 1.
• In terms of partial derivatives, the Poisson bracket can be written as: \({X, P} = (\frac{∂X}{∂x})(\frac{∂P}{∂p}) - (\frac{∂X}{∂p})(\frac{∂P}{∂x})\)
• Substituting X and P from the problem statement, we get: \({X, P} = (\frac{∂(αp/x)}{∂x})(\frac{∂(βx²)}{∂p}) - (\frac{∂(αp/x)}{∂p})(\frac{∂(βx²)}{∂x})\)
• Computing partial derivatives, we get:\({X, P} = (\frac{-αp}{x²})(0) - (\frac{α}{x})(2βx) = -2αβ \)
• Setting the above equation to 1: \(-2αβ = 1\)
• Finally, \(1+2αβ = 0\)

Concept:

The Lagranges equation of motion of a system is given by

\({d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0\)

Calculation:

The Lagrangian L depends on 

L(x,y,\(̇ x\),\(̇ y\))

\({d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0\)

\({d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0 \)

L^' = L(x,y,\(\dot x\),\(\dot{y}\))

\({d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = ​​{d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0\)

⇒\(\dot{y} = c_1\)

Similarly \(\dot{x} = c_2\)

The correct answer is option (2).

Explanation:

We will first write the velocity vector in polar co-ordinates and write the radial velocity and radial acceleration and by differentiating we get the desired solution.

Given,-\(r=r_0 e^{β t} \text { and } \frac{d \theta}{d t}=ω\)     ----------------1

Velocity in polar co-ordinates is given by the sum of radial and transverse velocity.

\(v=v_r+v_t=\dot r \hat{r}+r\dot \theta \hat{\theta}\)

Radial velocity \(v_r=\dot r\)

Radial acceleration \(a_r=(\ddot {r}-r\dot \theta^2)\)

Now, \(r=r_0 e^{β t} \) and \(\dot r=r_0 \beta e^{β t} \)

\(\frac{d v_r}{d t}=\)\(\beta ^2r_0 \beta e^{β t} \) => \(\frac{d v_r}{d t}\)\(>0\)

\(\ddot r=r_0 \beta^2 e^{β t} \) and \(\ddot \theta^2=\omega^2\)

\(a_r=\)\(\beta ^2r_0 \beta e^{β t} \) \(-r\omega^2\)\(=\)\(\beta ^2r_0 \ e^{β t} \)\(-r_0 \omega^2e^{β t}\)\(=\)\(r_0 \ e^{β t} (\beta^2-\omega^2)\)

If \(\beta=\omega\), \(a_r=0\) for some choices of parameters.

So, the correct answer is  -\(\frac{d v_r}{d t}>0\) however, \(a_r=0\) for some choices of parameters.

Concept:

The Hamiltonian of a system specifies its total energy—i.e., the sum of its kinetic energy (that of motion) and its potential energy (that of position)—in terms of the Lagrangian function derived in earlier studies of dynamics and of the position and momentum of each of the particles.

Calculation:

H = q1p1 - q2p2 + \(aq_1^2\) , where a > 0 is a constant.

f = (q₁q₂ + λp₁p₂)

\({df \over dt}\) = [f,H] + \({\partial f \over \partial t }\)

\({\partial f \over \partial t }\) = 0 ⇒ \({df \over dt}\) = [f,H] = 0

[f,H] = \([{\partial f \over \partial q_1}{\partial H \over \partial p_1}-{\partial f \over \partial p_1}{\partial H \over \partial q_1}]+[{\partial f \over \partial q_2}{\partial H \over \partial p_2}-{\partial f \over \partial p_2}{\partial H \over \partial q_2}]= 0\)

q₂ . q₁ - λ p₂ (p₁ + 2aq₁) + q₁(-q₂) - λ p₁(-p₂) = 0

∴ λ = 0

The correct answer is option (1).

Concept:

We will use the relationship of Hamiltonian and Lagrangian which is given by

• \(H=\sum p_i \dot{q_i}-L\)
• For a two-particle system, we can write the above equation as \(H=p_1\dot{q_1}+p_2\dot {q_2}-L\)

Explanation:

Given, \(H=p_1p_2+q_1q_2\)

• \(H=p_1\dot{q_1}+p_2\dot {q_2}-L\)
• \(H=p_1p_2+q_1q_2\) (given)

substitute value of H, we get,

• \(L=p_1\dot{q_1}+p_2\dot {q_2}-p_1p_2-q_1q_2\)--------------------------------1
• Using Hamilton equations, \(\dot p=\frac{-\partial H}{\partial q}, \dot{q}=\frac{\partial H}{\partial p}\)
• \(\dot q_1=\frac {\partial H}{\partial p_1}=p_2\) and \(\dot q_2=\frac {\partial H}{\partial p_2}=p_1\)
• Put the value of \(p_1\) and \(p_2\) in equation 1, to get the value of Lagrangian
• \(L=p_1\dot{q_1}+p_2\dot {q_2}-p_1p_2-q_1q_2\) \(=\)\(L=\dot q_2\dot{q_1}+\dot q_1\dot {q_2}-\dot q_1\dot q_2-q_1q_2\)
• \(L=\dot q_2 \dot q_1-q_1q_2\)

So, the correct answer is \(L=\dot q_2 \dot q_1-q_1q_2\).

Concept:

We will first write lagrangian for a given condition then we use the equation for normal nodes which is given by \(|\hat V-\omega^2\hat T|=0\)

Explanation:

Given m are two identical masses, k is spring constant and \(x_1\) and \(x_2\)are displacement of spring first and spring second respectively.

• \(T=\frac {1}{2}mx_1^2+\frac{1}{2}mx_2^2\)
• \(V=\frac {1}{2}k[x_1-0]^2+\frac{1}{2}[x_2-x_1]^2\)
• \(V=\frac{1}{2}kx_1^2+\frac{1}{2}kx_1^2+\frac{1}{2}kx_2^2-kx_1x_2\)
• \(V=kx_1^2+\frac{1}{2}kx_2^2-\frac{kx_1x_2}{2}-\frac{kx_1 x_2}{2}\)
• Using matrix we can write operators of \(V\) and \(T\) as,
• \(\hat T=\begin{bmatrix} \frac{m}{2} & 0 \\[0.3em] 0 & \frac{m}{2} \\[0.3em] \end{bmatrix} \) and \(\hat V=\begin{bmatrix} k & \frac{-k}{2} \\[0.3em] \frac{-k}{2} & \frac{k}{2} \\[0.3em] \end{bmatrix} \)
• Put these values in equation-\(|\hat V-\omega^2\hat T|=0\) to get the ratio of normal modes of frequencies,
• \(|\hat V-\omega^2\hat T|=0\)
• \(k|\begin{bmatrix} (1-\frac{\omega^2}{2}) & \frac{-1}{2} \\[0.3em] \frac{-1}{2} &( \frac{1}{2}-\frac{\omega^2}{2}) \\[0.3em] \end{bmatrix} |=0\)
• \(\det \begin{bmatrix} (1-\frac{\omega^2}{2}) & \frac{-1}{2} \\[0.3em] \frac{-1}{2} & (\frac{1}{2}-\frac{\omega^2}{2}) \\[0.3em] \end{bmatrix} =0\)
• \((1-\frac{\omega ^2}{2})(\frac{1}{2}-\frac {\omega^2}{2})=0\)
• \(\frac{1}{2}-\frac{\omega^2}{2}-\frac {\omega^2}{4}+\frac{\omega^4}{4}-\frac{1}{4}=0\)
• \(2-2\omega^2-\omega^2+\omega^4-1=0 \)
• \(\omega^4-3\omega^2+1=0\)
• Solution of above equation is \(\omega^2=\frac {3\pm \sqrt5}{2}\)
• \(\therefore \omega_+=\sqrt \frac{3+\sqrt 5}{2}, \omega_-=\sqrt\frac{3-\sqrt {5}}{2}\)
• ratio of both frequencies \(=\frac {\omega_-}{\omega_+}=\frac{\sqrt {3-\sqrt 5}}{\sqrt{3+\sqrt 5}}\)

So, the correct answer is  \(=\frac {\omega_-}{\omega_+}=\frac{\sqrt {3-\sqrt 5}}{\sqrt{3+\sqrt 5}}\)