JEE PYQs MCQs

Hint

(a) Frictional force.
Explanation
Potential energy is defined only for conservative forces, where the work done by the force is independent of the path taken.
Frictional force is a non-conservative force. The work done by friction depends on the path length and dissipates energy as heat, so it cannot be expressed in terms of a potential energy function.

Why other options are incorrect
(b) Coulomb’s force is a conservative force, and the electrostatic potential energy is a well-defined concept \((U=k Q q / r)\).
(c) Restoring force (such as the elastic spring force following Hooke’s law) is also a conservative force, associated with elastic potential energy ( \(P E=1 / 2 k x^2\) ).
(d) Gravitational force is a conservative force, and gravitational potential energy is a standard concept ( \(\boldsymbol{P E}=\boldsymbol{m} \boldsymbol{g} \boldsymbol{h}\) or \(\boldsymbol{U}=\boldsymbol{-} \boldsymbol{G M m} \boldsymbol{/} \boldsymbol{r}\) ).

Note: Conservative Forces: These are forces for which the work done is independent of the path taken; they only depend on the initial and final positions. For conservative forces, you can define a potential energy function. Examples include Coulomb’s force, gravitational force, and the restoring force of a spring.

Non-Conservative Forces: These forces depend on the path taken and usually result in energy dissipation (often as heat). Friction is a typical example, and hence, it cannot be described solely by a potential energy function.

Hint

(c) The mass of the object is given as 1000 g. We convert this to kilograms (kg), the standard SI unit for mass, by dividing by 1000:
\(
m=1000 \mathrm{~g}=1 \mathrm{~kg}
\)
Determine acceleration
According to Newton’s second law, the force \(\vec{F}\) is equal to mass \(m\) times acceleration \(\vec{a} (\vec{F}=m \vec{a})\). The acceleration vector is found by dividing the force vector by the mass:
\(
\begin{gathered}
\vec{a}=\frac{\vec{F}}{m}=\frac{1}{1 \mathrm{~kg}}\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{N} \\
\vec{a}=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{m} / \mathrm{s}^2
\end{gathered}
\)
Determine velocity
Velocity \(\vec{v}\) is the integral of acceleration \(\vec{a}\) with respect to time \(t\left(\vec{v}=\int \vec{a} d t\right)\).
Assuming the object starts from rest at \(t=0\), the constant of integration is zero:
\(
\begin{gathered}
\vec{v}=\int\left(2 t \hat{i}+3 t^2 \hat{j}\right) d t \\
\vec{v}=\left(\int 2 t d t\right) \hat{i}+\left(\int 3 t^2 d t\right) \hat{j} \\
\vec{v}=\left(t^2 \hat{i}+t^3 \hat{j}\right) \mathrm{m} / \mathrm{s}
\end{gathered}
\)
Power \(\boldsymbol{P}\) generated by a force is the dot product of the force vector \(\overrightarrow{\boldsymbol{F}}\) and the velocity vector \(\vec{v}(P=\vec{F} \cdot \vec{v})\) :
\(
\begin{gathered}
P=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \cdot\left(t^2 \hat{i}+t^3 \hat{j}\right) \\
P=(2 t)\left(t^2\right)+\left(3 t^2\right)\left(t^3\right) \\
P=2 t^3+3 t^5
\end{gathered}
\)
The power generated by the force at time \(t\) is \(\left(2 t^3+3 t^5\right) \mathrm{W}\)

Hint

(d)

The block moves at a uniform velocity, implying that the net force in both the horizontal ( \(x\) ) and vertical ( \(y\) ) directions is zero. The forces acting on the block are the applied force \(F\), gravity \(m g\), normal force \(N\), and kinetic friction \(f_k\).
The vertical force balance equation is:
\(
N+F \sin (\theta)-m g=0
\)
Thus, the normal force is \(N=m g-F \sin (\theta)\).
The horizontal force balance equation is:
\(
F \cos (\theta)-f_k=0
\)
The kinetic friction force is defined as \(f_k=\mu N\).
Substitute the expression for the normal force into the friction equation:
\(
F \cos (\theta)=\mu(m g-F \sin (\theta))
\)
Rearrange the equation to solve for the applied force \(F\) :
\(
F=\frac{\mu m g}{\cos (\theta)+\mu \sin (\theta)}
\)
Substitute the given values: \(m=25 \mathrm{~kg}, \mu=0.25, \theta=45^{\circ}, g=9.8 \mathrm{~m} / \mathrm{s}^2\), \(\cos \left(45^{\circ}\right)=\sin \left(45^{\circ}\right)=1 / \sqrt{2}\).
\(
\begin{gathered}
F=\frac{0.25 \times 25 \times 9.8}{\cos \left(45^{\circ}\right)+0.25 \sin \left(45^{\circ}\right)} \\
F=\frac{61.25}{1.25 \cos \left(45^{\circ}\right)}=\frac{49}{\cos \left(45^{\circ}\right)}=49 \sqrt{2} \mathrm{~N}
\end{gathered}
\)
The work \(W\) done by the applied force \(F\) over a displacement \(d\) is given by the formula \(W=F d \cos (\theta)\). The displacement \(d\) is 5 m.
\(
\begin{gathered}
W=(49 \sqrt{2} \mathrm{~N}) \times(5 \mathrm{~m}) \times \cos \left(45^{\circ}\right) \\
W=245 \sqrt{2} \mathrm{~J} \times \frac{1}{\sqrt{2}}=245 \mathrm{~J}
\end{gathered}
\)

Hint

(c) Step 1: Define Energy and Apply Conservation
The particle starts from rest at height \(S\). Its initial total energy is purely potential energy, \(E_i=m g S\). At an arbitrary height \(h\) with speed \(v\), the total energy is \(E_f=m g h+\frac{1}{2} m v^2\). By the principle of conservation of energy, \(E_i=E_f\), leading to the equation \(m g S=m g h+\frac{1}{2} m v^2\).
Step 2: Use the Given Condition
The problem states that at height \(h\), the kinetic energy is three times the potential energy: \(K E_f=3 \times P E_f\), or \(\frac{1}{2} m v^2=3 m g h\). This can be simplified to \(\frac{1}{2} v^2=3 g h\)
Step 3: Solve for Height and Speed
Substitute the kinetic energy relation into the conservation of energy equation from Step 1:
\(
\begin{gathered}
m g S=m g h+3 m g h \\
m g S=4 m g h
\end{gathered}
\)
\(
S=4 h \Longrightarrow h=\frac{S}{4}
\)
Now, solve for the speed \(v\) using the condition from Step 2:
\(
\begin{gathered}
\frac{1}{2} v^2=3 g h \\
v^2=6 g h
\end{gathered}
\)
Substitute the value of \(h\) :
\(
\begin{gathered}
v^2=6 g\left(\frac{S}{4}\right)=\frac{3 g S}{2} \\
v=\sqrt{\frac{3 g S}{2}}
\end{gathered}
\)
The height from the surface of the earth is \(\frac{\mathrm{S}}{4}\), and the speed of the particle at that instant is \(\sqrt{\frac{3 \mathrm{gS}}{2}}\). 

Hint

(d) Step 1: Define mass rate and force
The rate of mass added to the belt is proportional to the square root of speed, so we can write it as \(\frac{d m}{d t}=k \sqrt{v}\), where \(k\) is a constant of proportionality. The force required to accelerate the added mass horizontally to the belt’s constant speed \(v\) is given by the rate of change of momentum, \(F=\frac{d(m v)}{d t}\). Since the velocity \(v\) is constant, this simplifies to \(F=v \frac{d m}{d t}\).
Step 2: Calculate power
The power \(\boldsymbol{P}\) delivered to the belt is the force multiplied by the constant velocity of the belt, \(\boldsymbol{P}=\boldsymbol{F v}\). Substituting the expression for force from Step 1:
\(
P=\left(v \frac{d m}{d t}\right) v=v^2 \frac{d m}{d t}
\)
Step 3: Substitute the mass rate relationship and find proportionality
Substitute the given relationship for the mass rate \(\frac{d m}{d t}=k \sqrt{v}\) into the power equation:
\(
P=v^2(k \sqrt{v})=k v^2 v^{1 / 2}=k v^{5 / 2}
\)
Thus, \(P \propto v^{5 / 2}\). Squaring both sides of the proportionality gives: \(P^2 \propto\left(v^{5 / 2}\right)^2 \Longrightarrow P^2 \propto v^5\).
The relationship between the power \(P\) and the speed \(v\) is (d) \(P^2 \propto v^5\).

Hint

(d) Step 1: Calculate the displacement vector
The initial position vector is \(\vec{r}_1=(3 \hat{i}+4 \hat{j}) \mathrm{m}\), and the final position vector is \(\vec{r}_2=(6 \hat{i}+10 \hat{j}) \mathrm{m}\).
The displacement vector \(\vec{s}\) is the difference between the final and initial position vectors:
\(
\vec{s}=\vec{r}_2-\vec{r}_1=(6 \hat{i}+10 \hat{j}) \mathrm{m}-(3 \hat{i}+4 \hat{j}) \mathrm{m}=(3 \hat{i}+6 \hat{j}) \mathrm{m}
\)
Step 2: Calculate the work done by the force
The work done \(W\) by the constant force \(\vec{F}=(2 \hat{i}+3 \hat{j}) \mathrm{N}\) is the dot product of the force and displacement vectors:
\(
W=\vec{F} \cdot \vec{s}=(2 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}+6 \hat{j})=(2 \times 3)+(3 \times 6)=6+18=24 \mathrm{~J}
\)
Step 3: Calculate the average power
Average power \(\boldsymbol{P}_{\text {avg }}\) is the total work done divided by the time taken, \(t=4 \mathrm{~s}\) :
\(
P_{\mathrm{avg}}=\frac{W}{t}=\frac{24 \mathrm{~J}}{4 \mathrm{~s}}=6 \mathrm{~W}
\)
Step 4: Calculate the acceleration and final velocity
The acceleration \(\vec{a}\) of the body is constant, given by Newton’s second law \(\vec{F}=m \vec{a}\), where \(m=4 \mathrm{~kg}\) :
\(
\vec{a}=\frac{\vec{F}}{m}=\frac{(2 \hat{i}+3 \hat{j}) \mathrm{N}}{4 \mathrm{~kg}}=\left(\frac{1}{2} \hat{i}+\frac{3}{4} \hat{j}\right) \mathrm{m} / \mathrm{s}^2
\)
Assuming the body starts from rest (initial velocity \(\vec{u}=0\) ), the final velocity \(\vec{v}\) at \(t=4 \mathrm{~s}\) is:
\(
\vec{v}=\vec{u}+\vec{a} t=0+\left(\frac{1}{2} \hat{i}+\frac{3}{4} \hat{j}\right) \mathrm{m} / \mathrm{s}^2 \times 4 \mathrm{~s}=(2 \hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}
\)
Step 5: Calculate the instantaneous power at the end of \(\mathbf{4 ~ s e c}\)
Instantaneous power \(P_{\text {inst }}\) at \(t=4 \mathrm{~s}\) is the dot product of the force and the instantaneous velocity at that time:
\(
P_{\text {inst }}=\vec{F} \cdot \vec{v}=(2 \hat{i}+3 \hat{j}) \mathrm{N} \cdot(2 \hat{i}+3 \hat{j}) \mathrm{m} / \mathrm{s}=(2 \times 2)+(3 \times 3)=4+9=13 \mathrm{~W}
\)
Step 6: Determine the ratio
The ratio of average power to instantaneous power at the end of 4 sec is:
\(
\frac{\boldsymbol{P}_{\mathrm{avg}}}{\boldsymbol{P}_{\mathrm{inst}}}=\frac{6 \mathrm{~W}}{13 \mathrm{~W}}=\frac{6}{13}
\)
The ratio is 6:13.

Hint

(b)

Step 1: Define initial conditions
Position A Elongation: The spring elongation is \(x_1=2 R-R=R\) because the equilibrium length of the spring is \(R\).
Position A Height: The height is \(\boldsymbol{h}_1 \boldsymbol{=} \mathbf{2 R}\).
Position A Velocity: The initial velocity is \(\boldsymbol{v}_1=0\), as the bead is released from rest.
Step 2: Define final conditions
Position B Elongation: The spring elongation is \(x_2=R-R=0\), as the length of the spring is \(R\).
Position B Height: The height is \(h_2=R / 2\).
Position B Velocity: The final velocity is \(v_2=v\).
Step 3: Apply energy conservation
Using the principle of conservation of energy, \(E_i=E_f\), we equate the initial and final mechanical energies of the system:
\(
\frac{1}{2} k x_1^2+\frac{1}{2} m v_1^2+m g h_1=\frac{1}{2} k x_2^2+\frac{1}{2} m v_2^2+m g h_2
\)
Substituting the defined values into the equation:
\(
\frac{1}{2} k R^2+m g(2 R)=\frac{1}{2} m v^2+m g \frac{R}{2}
\)
Step 4: Solve for \(v\)
Rearranging the equation to solve for the final velocity \(v\) :
\(
\begin{gathered}
\frac{1}{2} k R^2+2 m g R-\frac{m g R}{2}=\frac{1}{2} m v^2 \\
\frac{1}{2} m v^2=\frac{1}{2} k R^2+\frac{3 m g R}{2} \\
m v^2=k R^2+3 m g R \\
v^2=\frac{k R^2}{m}+3 g R
\end{gathered}
\)

Hint

(b) Assertion (A): In a central force field, the work done is independent of the path chosen.
A central force is a force that acts along the line joining a particle to a fixed center (e.g., gravitational and electrostatic forces).
Central forces are conservative, meaning the work done depends only on the initial and final positions, not the path taken.
Assertion is true.

Reason (R): Every force encountered in mechanics does not have an associated potential energy.
Conservative forces (e.g., gravitational, electrostatic, and spring force) have associated potential energy.
Non-conservative forces (e.g., friction, air resistance) do not have potential energy because they dissipate energy.
Reason is also true.
However, the reason R does not directly explain why work done is path-independent in central forces.
Instead, R discusses all forces in mechanics, including non-conservative forces, which is not directly related to A.

Hint

(c) Step 1: Define the work integral
The work done ( \(W\) ) by a variable force \(F(x)\) acting over a displacement from \(x_1\) to \(x_2\) is calculated using the integral:
\(
W=\int_{x_1}^{x_2} F(x) d x
\)
Step 2: Integrate the force expression
Substitute the given force expression \(F(x)=\alpha+\beta x^2\) and the displacement limits ( \(x_1=0 \mathrm{~m}, x_2=1 \mathrm{~m}\) ) into the work equation:
\(
W=\int_0^1\left(\alpha+\beta x^2\right) d x=\left[\alpha x+\frac{\beta x^3}{3}\right]_0^1
\)
Evaluating the definite integral gives:
\(
W=\left(\alpha(1)+\frac{\beta(1)^3}{3}\right)-\left(\alpha(0)+\frac{\beta(0)^3}{3}\right)=\alpha+\frac{\beta}{3}
\)
Step 3: Solve for \(\boldsymbol{\beta}\)
We are given \(W=5 \mathrm{~J}\) and \(\alpha=1 \mathrm{~N}\). Substitute these values into the resulting equation:
\(
\begin{gathered}
5=1+\frac{\beta}{3} \\
4=\frac{\beta}{3} \\
\beta=12 \mathrm{~N} / \mathrm{m}^2
\end{gathered}
\)

Hint

(c) Step 1: Define initial kinetic energy and velocity components The initial kinetic energy ( \(K E\) ) of the ball is given by the formula \(K E=\frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) is the initial velocity. The initial velocity can be broken down into horizontal ( \(v_x\) ) and vertical ( \(v_y\) ) components:
\(
\begin{aligned}
v_x & =v \cos (\theta) \\
v_y & =v \sin (\theta)
\end{aligned}
\)
where \(\theta=60^{\circ}\).
Step 2: Determine kinetic energy at the highest point
At the highest point of its flight, the vertical velocity component becomes zero ( \(v_y=0\) ), but the horizontal velocity component remains constant ( \(v_h=v_x\) ). The kinetic energy at the highest point ( \(K E_h\) ) is therefore:
\(
K E_h=\frac{1}{2} m\left(v_h\right)^2=\frac{1}{2} m\left(v_x\right)^2=\frac{1}{2} m(v \cos (\theta))^2
\)
Step 3: Express \(\mathbf{K E}_{\boldsymbol{h}}\) in terms of \(\mathbf{K E}\)
We substitute the initial angle \(\theta=60^{\circ}\) and use the value of \(\cos \left(60^{\circ}\right)=\frac{1}{2}\) to find the relationship between \(K E_h\) and \(K E\) :
\(
\begin{gathered}
K E_h=\frac{1}{2} m v^2\left(\cos \left(60^{\circ}\right)\right)^2 \\
K E_h=K E \times\left(\frac{1}{2}\right)^2 \\
K E_h=K E \times \frac{1}{4} \\
K E_h=\frac{K E}{4}
\end{gathered}
\)

Hint

(b) Step 1: Calculate the work done
The work done ( \(W\) ) by a constant force ( \(\vec{F}\) ) causing a displacement ( \(\vec{d}\) ) is defined by their dot product: \(W=\vec{F} \cdot \vec{d}\).
Given force \(\vec{F}=2 \hat{i}+b \hat{j}+\hat{k}\) and displacement \(\vec{d}=\hat{i}-2 \hat{j}-\hat{k}\), the work done is calculated as:
\(
\begin{gathered}
W=(2)(1)+(b)(-2)+(1)(-1) \\
W=2-2 b-1 \\
W=1-2 b
\end{gathered}
\)
Step 2: Set the work to zero to find \(b\)
We are given that the work done on the particle is zero, so we set the expression for \(W\) to zero:
\(
\begin{gathered}
1-2 b=0 \\
2 b=1 \\
b=\frac{1}{2}
\end{gathered}
\)

Hint

(c) Step 1: Apply the Work-Energy Theorem
The net work done by all forces on a particle is equal to the change in its kinetic energy \((\Delta K)\). The kinetic energy \((K)\) is given by \(K=\frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) is the velocity. The work-energy theorem is expressed as:
\(
W_{\text {net }}=\Delta K=K_f-K_i
\)
Step 2: Calculate Initial Kinetic Energy
The particle starts at \(\boldsymbol{x} \boldsymbol{=} \mathbf{0}\). The velocity at this position is given by the equation \(v(x)=\alpha \sqrt{x}\).
\(
v_i=v(0)=\alpha \sqrt{0}=0
\)
The initial kinetic energy ( \(\boldsymbol{K}_{\boldsymbol{i}}\) ) is:
\(
K_i=\frac{1}{2} m v_i^2=\frac{1}{2} m(0)^2=0
\)
Step 3: Calculate Final Kinetic Energy
The particle moves to a final position \(x=d\). The velocity at this position is:
\(
v_f=v(d)=\alpha \sqrt{d}
\)
The final kinetic energy ( \(\boldsymbol{K}_{\boldsymbol{f}}\) ) is:
\(
K_f=\frac{1}{2} m v_f^2=\frac{1}{2} m(\alpha \sqrt{d})^2=\frac{1}{2} m \alpha^2 d
\)
Step 4: Determine the Total Work Done
The total work done is the difference between the final and initial kinetic energies:
\(
\begin{gathered}
W_{\text {net }}=K_f-K_i=\frac{1}{2} m \alpha^2 d-0 \\
W_{\text {net }}=\frac{m \alpha^2 d}{2}
\end{gathered}
\)
The total work done is \(\frac{\mathbf{m} \alpha^{\mathbf{2}} \mathbf{d}}{\mathbf{2}}\).

Hint

(d) Initially, the block is released from rest, meaning its initial kinetic energy is zero. It starts with gravitational potential energy at the incline’s top, given by: \(U_g=m g h\), where \(m=5 \mathrm{~kg}, g= 10 \mathrm{~m} / \mathrm{s}^2\) (assuming near the Earth’s surface), and \(h=10 \sin 30^{\circ} \mathrm{m}\).
Calculate the height: \(h=10 \times \frac{1}{2}=5 \mathrm{~m}\)
Substitute into the potential energy formula: \(U_g=5 \times 10 \times 5=250 \mathrm{~J}\)
The block slides down the incline and across a rough surface (friction coefficient \(\mu=0.5\) ) for 2 m. Calculate the work done by friction: \(W_f=\mu m g d=0.5 \times 5 \times 10 \times 2=50 \mathrm{~J}\)
By energy conservation, the gravitational potential energy is converted into spring potential energy and work done against friction: \(U_g=\frac{1}{2} k x^2+W_f\)
Substitute the known values: \(250=\frac{1}{2} \times 100 \times x^2+50\)
Solve for \(x\) :
First, simplify the equation: \(250=50+50 x^2\)
Re-arrange to solve for \(x^2: 200=50 x^2\)
Divide both sides by 50: \(x^2=4\)
Solve for \(x: x=2\)
Therefore, the maximum compression in the spring when the block hits the spring is 2 m.

Hint

(a) Step 1: Relate kinetic energy and momentum
The kinetic energy \(K\) of a body with mass \(m\) and linear momentum \(p\) is given by the formula \(K=\frac{p^2}{2 m}\). Rearranging this equation to express momentum in terms of kinetic energy and mass gives \(p=\sqrt{2 m K}\).
Step 2: Convert masses to consistent units
The given masses are:
\(
m_A=400 \mathrm{~g}=0.4 \mathrm{~kg}
\)
\(
\begin{aligned}
& m_B=1.2 \mathrm{~kg} \\
& m_C=1.6 \mathrm{~kg}
\end{aligned}
\)
Step 3: Determine the ratio of momenta
Since the kinetic energies \(K\) are equal for all three bodies, the ratio of their linear momenta \(p_A: p_B: p_C\) is proportional to the square root of their masses:
\(
p_A: p_B: p_C=\sqrt{2 m_A K}: \sqrt{2 m_B K}: \sqrt{2 m_C K}=\sqrt{m_A}: \sqrt{m_B}: \sqrt{m_C}
\)
Substituting the mass values into the ratio:
\(
p_A: p_B: p_C=\sqrt{0.4}: \sqrt{1.2}: \sqrt{1.6}
\)
Step 4: Simplify the ratio
To simplify the ratio to its lowest terms, we can multiply all terms by \(\sqrt{10}\) and then simplify the resulting square roots:
\(
\begin{gathered}
p_A: p_B: p_C=\sqrt{0.4 \times 10}: \sqrt{1.2 \times 10}: \sqrt{1.6 \times 10} \\
p_A: p_B: p_C=\sqrt{4}: \sqrt{12}: \sqrt{16} \\
p_A: p_B: p_C=2: \sqrt{4 \times 3}: 4 \\
p_A: p_B: p_C=2: 2 \sqrt{3}: 4
\end{gathered}
\)
Dividing all terms by the common factor of 2 yields the final simplified ratio:
\(
p_A: p_B: p_C=1: \sqrt{3}: 2
\)
The ratio of their linear momenta is \(1: \sqrt{3}: 2\)

Hint

(b) Step 1: Relate Kinetic Energy and Momentum
The relationship between kinetic energy (KE) and momentum (P) for a body of mass ( \(m\) ) is given by the formula \(K E=\frac{P^2}{2 m}\) or, expressed in terms of momentum, \(P=\sqrt{2 m K E}\).
Step 2: Calculate Final Momentum
Let the initial kinetic energy and momentum be \(K E_i\) and \(P_i\) respectively. The final kinetic energy \(K E_f\) is 36 times the initial, so \(K E_f=36 \times K E_i\). The final momentum \(P_f\) is calculated using the formula:
\(
P_f=\sqrt{2 m K E_f}=\sqrt{2 m\left(36 \times K E_i\right)}=\sqrt{36} \times \sqrt{2 m K E_i}=6 P_i
\)
The final momentum is 6 times the initial momentum.
Step 3: Determine Percentage Increase
The increase in momentum is \(\Delta P=P_f-P_i=6 P_i-P_i=5 P_i\). The percentage increase is calculated as:
\(
\text { Percentage Increase }=\frac{\Delta P}{P_i} \times 100 \%=\frac{5 P_i}{P_i} \times 100 \%=500 \%
\)
The percentage increase in the momentum of the body will be \(\mathbf{5 0 0} \boldsymbol{\%}\).

Hint

(c) Step 1: Calculate Initial Kinetic Energy
First, calculate the initial kinetic energy ( \(K E_i\) ) using the mass ( \(m\) ) and initial velocity ( \(v_i )\). Note that \(50 \mathrm{~g}=0.050 \mathrm{~kg}\).
\(
K E_i=\frac{1}{2} m v_i^2=\frac{1}{2}(0.050 \mathrm{~kg})(100 \mathrm{~m} / \mathrm{s})^2=250 \mathrm{~J}
\)
Step 2: Calculate Final Kinetic Energy
Next, calculate the final kinetic energy ( \(K E_f\) ) using the mass ( \(m\) ) and final velocity ( \(v_f\) ).
\(
K E_f=\frac{1}{2} m v_f^2=\frac{1}{2}(0.050 \mathrm{~kg})(40 \mathrm{~m} / \mathrm{s})^2=40 \mathrm{~J}
\)
Step 3: Determine the Percentage Loss
Calculate the percentage loss of kinetic energy relative to the initial kinetic energy using the formula:
\(
\begin{gathered}
\text { Percentage Loss }=\frac{K E_i-K E_f}{K E_i} \times 100 \% \\
\text { Percentage Loss }=\frac{250 \mathrm{~J}-40 \mathrm{~J}}{250 \mathrm{~J}} \times 100 \%=\frac{210 \mathrm{~J}}{250 \mathrm{~J}} \times 100 \%=84 \%
\end{gathered}
\)
The percentage loss of kinetic energy is \(\mathbf{8 4 \%}\).

Hint

(a) Step 1: Relate kinetic energy and momentum
The relationship between kinetic energy (\(K\)) and linear momentum (\(p\)) for a particle of mass \((\mathrm{m})\) is given by the formula:
\(
K=\frac{p^2}{2 m}
\)
This equation shows that for a constant momentum (\(p\)), the kinetic energy (\(K\)) is inversely proportional to the mass \((\mathrm{m}), K \propto \frac{1}{m}\).
Step 2: Determine the particle with minimum mass
To have the maximum kinetic energy when the momentum is the same for all particles, the particle must have the minimum mass. The masses of the four particles are given as:
Mass of particle A, \(m_A=\frac{m}{2}\)
Mass of particle \(\mathrm{B}, m_B=m\)
Mass of particle \(\mathrm{C}, m_C=2 m\)
Mass of particle \(\mathrm{D}, m_D=4 m\)
Comparing these values, \(m_A=\frac{m}{2}\) is the smallest mass.
The particle with the minimum mass is Particle A. Therefore, the particle with the maximum kinetic energy is Particle A.

Hint

(b) Step 1: Formulate the governing differential equation
The power \(\boldsymbol{P}\) is constant and related to force \(\boldsymbol{F}\) and velocity \(\boldsymbol{v}\) by \(\boldsymbol{P}=\boldsymbol{F} \boldsymbol{v}\). Using Newton’s second law \((F=m a)\) and the definition of acceleration \(\left(a=\frac{d v}{d t}\right)\), we can write the power equation as a differential equation relating velocity \(v\) and time \(t\) :
\(
P=m \frac{d v}{d t} v
\)
Rearranging the terms for integration gives \(\boldsymbol{P} \boldsymbol{d t} \boldsymbol{=} \boldsymbol{m v} \boldsymbol{d v}\).
Step 2: Integrate to find velocity as a function of time
We integrate both sides of the equation from \(t=0\) to \(t\) and from \(v=0\) to \(v\) (assuming the body starts from rest at the origin at \(t=0\) ):
\(
\begin{aligned}
\int_0^t P d t & =\int_0^v m v d v \\
P t & =\frac{1}{2} m v^2
\end{aligned}
\)
Solving for velocity \(v\) :
\(
v=\sqrt{\frac{2 P}{m}} t^{1 / 2}
\)
This shows that the velocity is proportional to \(t^{1 / 2}\), or \(v \propto t^{1 / 2}\).
Step 3: Integrate to find displacement as a function of time Velocity is the time derivative of displacement \(s\left(v=\frac{d s}{d t}\right)\). We substitute the expression for \(v\) and integrate again:
\(
\begin{aligned}
\frac{d s}{d t} & =\sqrt{\frac{2 P}{m}} t^{1 / 2} \\
\int_0^s d s & =\int_0^t \sqrt{\frac{2 P}{m}} t^{1 / 2} d t \\
s & =\sqrt{\frac{2 P}{m}}\left(\frac{t^{3 / 2}}{3 / 2}\right) \\
s & =\frac{2}{3} \sqrt{\frac{2 P}{m}} t^{3 / 2}
\end{aligned}
\)
Step 4: Determine the proportionality
From the final equation for displacement \(s\), we can see that \(s\) is directly proportional to \(t^{3 / 2}\), as the term \(\frac{2}{3} \sqrt{\frac{2 P}{m}}\) is a constant.
\(
s \propto t^{3 / 2}
\)
The displacement in time \(t\) is proportional to \(t^{3 / 2}\).

Hint

(b) Step 1: Work done by a conservative force like gravity depends only on the change in vertical height, not on the path taken. The work done against gravity is defined as the force of gravity multiplied by the vertical distance the object is lifted.
Step 2: Formulate the work equation
The work done ( \(\mathbf{W}\) ) against gravity to lift an object of mass \(\boldsymbol{m}\) to a height \(\boldsymbol{h}\) is given by the formula \(W=m g h\), where \(g\) is the acceleration due to gravity. The specific path or method used to lift the body does not affect this value, only the initial and final vertical positions matter.
\(
W=m g h
\)
Step 3: Calculate the ratio
In both cases described in the problem, the mass ( \(m=50 \mathrm{~kg}\) ) is lifted from the ground ( \(h_i=0 \mathrm{~m}\) ) to the same final height ( \(h_f=20 \mathrm{~m}\) ).
The work done in the first case is \(\mathbf{W}_{\mathbf{1}}=\boldsymbol{m} \boldsymbol{g} \boldsymbol{h}\).
The work done in the second case is \(\mathbf{W}_{\mathbf{2}}=\boldsymbol{m g h}\).
The ratio of the work done is \(\frac{W_1}{W_2}=\frac{m g h}{m g h}=1\).
The ratio of the work done against gravity in both the respective cases will be \(1: 1\).

Hint

(d)

\(
\begin{aligned}
&\text { Apply Work energy theorem from A to B }\\
&\begin{aligned}
& \Rightarrow W_{m g}=K_B-K_A \\
& \Rightarrow m g \times\left(\frac{R}{\sqrt{2}}+R\right)=\frac{1}{2} m v_B^2-0\left\{v_A=0 \text { (at rest) }\right\} \\
& \Rightarrow m g R \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} m v_B^2 \\
& \Rightarrow \sqrt{g R \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=v_B \\
& \Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}}=v_B \\
& \Rightarrow 21.9=v_B
\end{aligned}
\end{aligned}
\)

Hint

(a) Step 1: Calculate Energy Loss
The initial total energy ( \(E_i\) ) of the ball when dropped from height \(h\) is entirely potential energy, given by \(E_i=m g h\). The final maximum potential energy ( \(E_f\) ) after the rebound to height \(h / 2\) is \(E_f=m g(h / 2)\).
The energy lost ( \(\Delta E\) ) during the impact is the difference between the initial and final maximum potential energies:
\(
\Delta E=E_i-E_f=m g h-m g \frac{h}{2}=m g \frac{h}{2}
\)
The percentage loss of energy is calculated as the ratio of energy lost to the initial energy, multiplied by 100%:
\(
\text { Percentage Loss }=\left(\frac{\Delta E}{E_i}\right) \times 100 \%=\left(\frac{m g \frac{h}{2}}{m g h}\right) \times 100 \%=\left(\frac{1}{2}\right) \times 100 \%=50 \%
\)
Step 2: Calculate Final Velocity
As the ball falls from rest at height \(h\) just before striking the ground, its potential energy is converted into kinetic energy. According to the principle of conservation of energy during the fall:
\(
m g h=\frac{1}{2} m v^2
\)
where \(v\) is the velocity of the ball just before impact.
Solving for \(v\) :
\(
\begin{aligned}
& v^2=2 g h \\
& v=\sqrt{2 g h}
\end{aligned}
\)
The percentage loss of total energy is \(50 \%\), and the velocity of the ball before it strikes the ground is \(\sqrt{2 g h}\).

Hint

(d) Step 1: Calculate acceleration
The force \(\vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N\) acts on a mass \(m=2 \mathrm{~kg}\). The acceleration \(\vec{a}\) is given by Newton’s second law, \(\vec{F}=m \vec{a}\) :
\(
\vec{a}=\frac{\vec{F}}{m}=\frac{6 t \hat{i}+6 t^2 \hat{j}}{2}=\left(3 t \hat{i}+3 t^2 \hat{j}\right) \frac{m}{s^2}
\)
Step 2: Calculate velocity
Assuming the body starts from rest at \(t=0\), the velocity \(\vec{v}\) is the integral of the acceleration with respect to time:
\(
\vec{v}=\int \vec{a} d t=\int\left(3 t \hat{i}+3 t^2 \hat{j}\right) d t=\left(\frac{3 t^2}{2} \hat{i}+\frac{3 t^3}{3} \hat{j}\right)=\left(\frac{3}{2} t^2 \hat{i}+t^3 \hat{j}\right) \frac{m}{s}
\)
Step 3: Calculate power
Power \(\boldsymbol{P}\) is the dot product of the force vector and the velocity vector, \(\boldsymbol{P}=\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{v}}\) :
\(
\begin{gathered}
P=\left(6 t \hat{i}+6 t^2 \hat{j}\right) \cdot\left(\frac{3}{2} t^2 \hat{i}+t^3 \hat{j}\right) \\
P=(6 t)\left(\frac{3}{2} t^2\right)+\left(6 t^2\right)\left(t^3\right) \\
P=9 t^3+6 t^5
\end{gathered}
\)
The power developed at time \(\boldsymbol{t}\) is \(\boldsymbol{P}=\left(9 t^3+6 t^5\right) \mathrm{W}\).
The power developed by the force at time \(t\) is \(\left(9 t^3+6 t^5\right) \mathrm{W}\).

Hint

(c) Step 1: Calculate the normal force
The normal force ( \(N\) ) on an inclined plane is given by the component of gravity perpendicular to the surface.
\(
N=m g \cos (\theta)
\)
Substituting the given values ( \(m=1 \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^2, \theta=60^{\circ}\) ):
\(
N=1 \times 10 \times \cos \left(60^{\circ}\right)=10 \times 0.5=5 \mathrm{~N}
\)
Step 2: Calculate the frictional force
The kinetic frictional force ( \(f_k\) ) is calculated using the coefficient of kinetic friction ( \(\mu_k\) ) and the normal force ( \(N\) ):
\(
f_k=\mu_k N
\)
Assuming \(\mu_k=0.1\) :
\(
f_k=0.1 \times 5 \mathrm{~N}=0.5 \mathrm{~N}
\)
Step 3: Calculate the work done against friction
Work done against the frictional force ( \(W_{\text {against } f}\) ) over a distance ( \(d\) ) is the product of the frictional force and the distance moved along the incline:
\(
W_{\text {against } f}=f_k \times d
\)
Given \(d=10 \mathrm{~m}\) :
\(
W_{\text {against } f}=0.5 \mathrm{~N} \times 10 \mathrm{~m}=5 \mathrm{~J}
\)
The work done against the frictional force is 5 J.

Hint

(c) Step 1: State the conservation of energy equation
According to the principle of conservation of mechanical energy, the total mechanical energy at point A equals the total mechanical energy at point B :
\(
E_A=E_B \Longrightarrow K_A+U_A=K_B+U_B
\)
where \(K\) is kinetic energy and \(U\) is potential energy. This can be expressed in terms of mass \((m)\), velocity \((v)\), gravitational acceleration \((g)\), and height \((h)\) as:
\(
\frac{1}{2} m v_A^2+m g h_A=\frac{1}{2} m v_B^2+m g h_B
\)
Step 2: Substitute known values and simplify
Given that the particle is gently pushed from rest, the initial velocity \(v_A=0 \mathrm{~m} / \mathrm{s}\). The equation simplifies to:
\(
m g h_A=\frac{1}{2} m v_B^2+m g h_B
\)
Dividing by mass \(m\) (since \(m \neq 0\) ):
\(
g h_A=\frac{1}{2} v_B^2+g h_B
\)
Step 3: Solve for the speed at point B
Rearrange the equation to solve for \(\boldsymbol{v}_{\boldsymbol{B}}\) :
\(
\begin{gathered}
\frac{1}{2} v_B^2=g h_A-g h_B \Longrightarrow v_B^2=2 g\left(h_A-h_B\right) \\
v_B=\sqrt{2 g\left(h_A-h_B\right)}
\end{gathered}
\)
Step 4: Calculate the final numerical value
Substitute the assumed values \(g=10 \mathrm{~m} / \mathrm{s}^2, h_A=1 \mathrm{~m}\), and \(h_B=0.5 \mathrm{~m}\) into the formula:
\(
\begin{gathered}
v_B=\sqrt{2 \times 10 \mathrm{~m} / \mathrm{s}^2 \times(1 \mathrm{~m}-0.5 \mathrm{~m})} \\
v_B=\sqrt{2 \times 10 \mathrm{~m} / \mathrm{s}^2 \times 0.5 \mathrm{~m}} \\
v_B=\sqrt{10} \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
The speed of the particle when it reaches point \(B\) is \(\sqrt{\mathbf{1 0}} \mathrm{m} / \mathrm{s}\).

Hint

(a) Step 1: Minimum velocity at the top (B)
For the bob to just complete the half circle, the minimum velocity at the top most point (B) must ensure that the gravitational force provides the necessary centripetal force, meaning the tension in the string is zero.
The condition is \(\frac{m v_B^2}{L}=m g\), which gives the minimum velocity squared at B as \(v_B^2=g L\).
Step 2: Kinetic energy at the top (B)
The kinetic energy at point B is calculated using the minimum velocity \(\boldsymbol{v}_{\boldsymbol{B}}\) :
\(
(K . E)_B=\frac{1}{2} m v_B^2=\frac{1}{2} m g L
\)
Step 3: Conservation of energy
By the principle of conservation of mechanical energy between the lowest point (A) and the highest point (B), where the potential energy at A is defined as zero, we have:
\(
\begin{gathered}
(K . E)_A+(P . E)_A=(K . E)_B+(P . E)_B \\
(K . E)_A+0=\frac{1}{2} m g L+m g(2 L) \\
(K . E)_A=\frac{5}{2} m g L
\end{gathered}
\)
Step 4: Kinetic energy at the bottom (A)
The kinetic energy at the bottom \((A)\) is found to be \((K . E)_A=\frac{5}{2} m g L\)
The ratio of kinetic energies is:
\(
\frac{(K . E)_A}{(K . E)_B}=\frac{\frac{5}{2} m g L}{\frac{1}{2} m g L}=5
\)

Hint

(d) Step 1: Calculate the normal force
The normal force ( \(N\) ) on a horizontal surface is equal to the weight of the block, which is the product of its mass \((m)\) and the acceleration due to gravity \((g)\).
We use \(g=9.8 \mathrm{~m} / \mathrm{s}^2\).
\(
\begin{gathered}
N=m \times g \\
N=100 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~N}
\end{gathered}
\)
Step 2: Calculate the frictional force
The force of kinetic friction \(\left(F_f\right)\) is calculated using the coefficient of friction \((\mu)\) and the normal force ( \(N\) ).
\(
\begin{gathered}
F_f=\mu \times N \\
F_f=0.4 \times 1000 \mathrm{~N}=400 \mathrm{~N}
\end{gathered}
\)
Step 3: Calculate the work done
The work done against friction ( \(W\) ) is the product of the frictional force ( \(F_f\) ) and the distance \((d)\) over which the block slides.
\(
\begin{gathered}
W=F_f \times d \\
W=400 \mathrm{~N} \times 10 \mathrm{~m}=4000 \mathrm{~J}
\end{gathered}
\)
The work done against friction is 4000 J.

Hint

(a) Step 1: Determine the \(\mathbf{x}\)-component of the force
The force \(\overrightarrow{\boldsymbol{F}}\) is the negative gradient of the potential energy function \(U\). The xcomponent of the force \(\boldsymbol{F}_{\boldsymbol{x}}\) is given by the negative partial derivative of \(U\) with respect to \(\boldsymbol{x}\) :
\(
F_x=-\frac{\partial U}{\partial x}
\)
Given the potential energy function \(U=\left(2 x^2+3 y^3+2 z\right)\), we compute the partial derivative:
\(
\frac{\partial U}{\partial x}=4 x
\)
Therefore, the expression for the \(x\)-component of the force is \(F_x=-4 x\).
Step 2: Calculate the magnitude at the given point
The force component at point \(P(1,2,3)\) is found by substituting the value \(x=1 \mathrm{~m}\) into the expression for \(\boldsymbol{F}_{\boldsymbol{x}}\) :
\(
F_x=-4(1)=-4 \mathrm{~N}
\)
The magnitude of the x -component of the force is the absolute value of \(\boldsymbol{F}_{\boldsymbol{x}}\) :
\(
\left|F_x\right|=|-4 \mathrm{~N}|=4 \mathrm{~N}
\)
The magnitude of the \(x\)-component of the force acting on the particle is \(\mathbf{4} \mathbf{N}\).

Hint

(b) Here, we will consider the lowest point of the motion as the reference
Now, when the ball reaches the maximum height, we can say it has the highest potential energy and as the motion stops there, the kinetic energy to be zero.
Similarly, when the ball reaches the lowest point, as it is a reference point, the potential energy can be said to be zero and as the ball tends to move towards height, kinetic energy is maximum. Now, from the law of conservation of energy, we can say that the total energy at the height is equal to the total energy at the base or reference.
\(
\begin{aligned}
& E_h=E_b \\
& \therefore K E_h+P E_h=K E_b+P E_b
\end{aligned}
\)
Substituting the known values,
\(
\begin{aligned}
& \therefore 0+P E_h=K E_b+0 \\
& \therefore P E_h=K E_b \ldots \ldots(1)
\end{aligned}
\)
Now, let us suppose the length of the thread as \(l\).
Now, for the acceleration at the highest position, we know at the extreme position, the force acting is only due to the gravitational force. But as the gravitational force acts in the downward direction, we need to consider the component of the force in the direction of the motion as shown in the figure below:

Hence, the force acting at the extreme position is
\(
F_h=m g \sin \theta
\)
Hence, the acceleration can be expressed as,
\(
a_h=g \sin \theta \dots(2)
\)
For the acceleration at the lowest position, we know that the acceleration to reach extreme height is provided by the centripetal force,
\(
F_b=\frac{m v^2}{l}
\)
Hence, the acceleration can be expressed as,
\(
a_b=\frac{v^2}{l} \ldots \ldots .(3)
\)
Now, for the equation (1), we can write
\(
\therefore m g h=\frac{1}{2} m v^2
\)
The height of the ball at the extreme position from the geometry of the figure can be expressed as
\(
h=l-l \cos \theta
\)
Substituting the value,
\(
\therefore m g(l-l \cos \theta)=\frac{1}{2} m v^2
\)
Canceling the common terms, and taking common terms out of parenthesis
\(
\therefore g l(1-\cos \theta)=\frac{1}{2} v^2
\)
Rearranging the equation,
\(
\therefore 2 g(1-\cos \theta)=\frac{v^2}{l}
\)
From the equation (3),
\(
\therefore 2 g(1-\cos \theta)=a_b
\)
But, we are given that the acceleration at the base is equal to acceleration at the extreme position.
\(
\therefore 2 g(1-\cos \theta)=a_h
\)
From the equation (2),
\(
\therefore 2 g(1-\cos \theta)=g \sin \theta
\)
Now, we know the expansions for the trigonometric terms as,
\(
\sin 2 \theta=2 \sin \theta \cos \theta \text { and } \cos 2 \theta=1-2 \sin ^2 \theta
\)
Applying the trigonometric expansions,
\(
\therefore 2 g \times 2 \sin ^2 \frac{\theta}{2}=g \times 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}
\)
Canceling the common terms,
\(
\begin{aligned}
& \therefore 2 \sin \frac{\theta}{2}=\cos \frac{\theta}{2} \\
& \therefore \tan \frac{\theta}{2}=\frac{1}{2}
\end{aligned}
\)
Applying the inverse functions,
\(
\begin{aligned}
& \therefore \frac{\theta}{2}=\tan ^{-1} \frac{1}{2} \\
& \therefore \theta=2 \tan ^{-1} \frac{1}{2}
\end{aligned}
\)

Hint

(b) Step 1: Calculate the acceleration
Assuming the target provides constant resistance (constant deceleration, \(a\) ), we can apply the kinematic equation \(\boldsymbol{v}^{\mathbf{2}} \boldsymbol{=} \boldsymbol{u}^{\mathbf{2}} \boldsymbol{+} 2 \boldsymbol{a s}\). For the initial \(\mathbf{4 ~ c m}\) (or \(\mathbf{0 . 0 4 ~ m}\) ) of travel, the velocity changes from \(v_i\) to \(v_1=\frac{2}{3} v_i\).
\(
\begin{gathered}
\left(\frac{2}{3} v_i\right)^2=v_i^2+2 a(0.04) \\
\frac{4}{9} v_i^2=v_i^2+0.08 a \\
0.08 a=\left(\frac{4}{9}-1\right) v_i^2=-\frac{5}{9} v_i^2 \\
a=-\frac{5 v_i^2}{9 \times 0.08}=-\frac{5 v_i^2}{0.72}
\end{gathered}
\)
Step 2: Determine total penetration depth
The bullet comes to rest \(\left(v_f=0\right)\) after a total distance \(\boldsymbol{x}_{\text {total }}\). Using the same kinematic equation for the entire journey from the start:
\(
\begin{gathered}
v_f^2=v_i^2+2 a x_{\text {total }} \\
0^2=v_i^2+2 a x_{\text {total }} \\
x_{\text {total }}=-\frac{v_i^2}{2 a}
\end{gathered}
\)
Substitute the value of \(\boldsymbol{a}\) from Step 1:
\(
\begin{gathered}
x_{\text {total }}=-\frac{v_i^2}{2\left(-\frac{5 v_i^2}{0.72}\right)} \\
x_{\text {total }}=-\frac{v_i^2}{-\frac{10 v_i^2}{0.72}} \\
x_{\text {total }}=\frac{0.72}{10}=0.072 \mathrm{~m}
\end{gathered}
\)
Step 3: Calculate the value of D
The problem states the bullet penetrates further \(D \times 10^{-3} \mathrm{~m}\) after the initial 4 cm. The total distance is \(0.04 \mathrm{~m}+D \times 10^{-3} \mathrm{~m}\).
\(
\begin{gathered}
0.072=0.04+D \times 10^{-3} \\
0.032=D \times 10^{-3} \\
D=\frac{0.032}{10^{-3}}=32
\end{gathered}
\)

Hint

(a) Step 1: Formulate the power ratio from given data
The power \(\boldsymbol{P}\) required to raise a mass \(\boldsymbol{m}\) to a height \(\boldsymbol{h}\) in time \(\boldsymbol{t}\) is given by \(\boldsymbol{P}=\frac{\boldsymbol{m} \boldsymbol{g} \boldsymbol{h}}{\boldsymbol{t}}\).
The ratio of powers for the two motors is \(\frac{P_1}{P_2}=\frac{m_1 t_2}{m_2 t_1}\). Substituting \(m_1=300 \mathrm{~kg}\), \(t_1=5 \mathrm{~min}, m_2=50 \mathrm{~kg}\), and \(t_2=2 \mathrm{~min}\), we get \(\frac{P_1}{P_2}=\frac{300 \times 2}{50 \times 5}=\frac{600}{250}=\frac{12}{5}\). The numerical ratio is 2.4.
Step 2: Set up and solve the equation for \(\mathbf{x}\)
Equate the calculated numerical ratio with the given expression:
\(
\frac{3 \sqrt{x}}{\sqrt{x}+1}=\frac{12}{5}
\)
Cross-multiplying yields \(5(3 \sqrt{x})=12(\sqrt{x}+1)\), which simplifies to \(15 \sqrt{x}=12 \sqrt{x}+12\). Rearranging for \(\sqrt{x}: 3 \sqrt{x}=12\), so \(\sqrt{x}=4\). Squaring both sides gives \(x=16\).

Hint

(a) The relationship between kinetic energy (\(K\)), linear momentum (\(p\)), and mass (\(m\)) is given by the formula:
\(
K=\frac{p^2}{2 m}
\)
Step 1: Establish the relationship using the given information
We are given that the two bodies have the same linear momentum ( \(p_1=p_2=p\) ) and the ratio of their kinetic energies is \(K_1: K_2=16: 9\). We can express the kinetic energies of the two bodies as:
\(
\begin{aligned}
& K_1=\frac{p^2}{2 m_1} \\
& K_2=\frac{p^2}{2 m_2}
\end{aligned}
\)
Step 2: Determine the ratio of the masses
To find the ratio of their masses \(\left(m_1: m_2\right)\), we divide the expression for \(K_1\) by the expression for \(\boldsymbol{K}_{\mathbf{2}}\) :
\(
\frac{K_1}{K_2}=\frac{\frac{p^2}{2 m_1}}{\frac{p^2}{2 m_2}}=\frac{p^2}{2 m_1} \times \frac{2 m_2}{p^2}=\frac{m_2}{m_1}
\)
Substituting the given kinetic energy ratio \(\frac{K_1}{K_2}=\frac{16}{9}\) :
\(
\frac{m_2}{m_1}=\frac{16}{9}
\)
The ratio of the masses respectively ( \(m_1: m_2\) ) is the inverse:
\(
\frac{m_1}{m_2}=\frac{9}{16}
\)

Hint

(b) Statement I is correct because the work-energy theorem states that the work done by the braking force ( \(W=F \times d\) ) equals the change in kinetic energy ( \(\Delta K E\) ). If \(F\) and \(\Delta K E\) are the same for both vehicles, they must travel the same distance ( \(d\) ).

Statement II is incorrect because a change in direction (velocity) while maintaining constant speed still involves acceleration, as acceleration is the rate of change of velocity (a vector quantity).

Hint

(b)

Step 1: Apply Conservation of Momentum
First, we apply the principle of conservation of linear momentum to find the velocity of the combined bullet-block system immediately after the collision. The collision is an inelastic collision as the bullet gets embedded in the block.
The total initial momentum is equal to the total final momentum.
\(
\begin{aligned}
p_{\text {initial }} & =p_{\text {final }} \\
m_b v_b+m_k v_k & =\left(m_b+m_k\right) V
\end{aligned}
\)
where \(m_b\) is the mass of the bullet, \(v_b\) is the initial velocity of the bullet, \(m_k\) is the mass of the block, \(v_k\) is the initial velocity of the block (which is 0 ), and \(V\) is the combined velocity after the collision.
Given:
\(m_b=0.1 \mathrm{~kg}\)
\(v_b=400 \mathrm{~ms}^{-1}\)
\(m_k=3.9 \mathrm{~kg}\)
\(v_k=0 \mathrm{~ms}^{-1}\)
\(
\begin{aligned}
&\begin{gathered}
0.1 \mathrm{~kg} \times 400 \mathrm{~ms}^{-1}+3.9 \mathrm{~kg} \times 0 \mathrm{~ms}^{-1}=(0.1 \mathrm{~kg}+3.9 \mathrm{~kg}) V \\
40 \mathrm{~kg} \cdot \mathrm{~ms}^{-1}=4.0 \mathrm{~kg} \times V \\
V=\frac{40}{4.0} \mathrm{~ms}^{-1}=10 \mathrm{~ms}^{-1}
\end{gathered}\\
&\text { The combined velocity after the collision is } 10 \mathrm{~ms}^{-1} \text {. }
\end{aligned}
\)
Step 2: Calculate Deceleration due to Friction
Next, we use a kinematic equation to find the acceleration (deceleration) of the combined system as it moves a distance \(d=20 \mathrm{~m}\) before coming to rest ( \(v_f=0 \mathrm{~ms}^{-1}\) ).
We use the equation:
\(
v_f^2=V^2+2 a d
\)
where \(v_f\) is the final velocity, \(V\) is the initial velocity of this phase, \(a\) is the acceleration, and \(d\) is the distance.
Given:
\(v_f=0 \mathrm{~ms}^{-1}\)
\(V=10 \mathrm{~ms}^{-1}\)
\(d=20 \mathrm{~m}\)
Calculation:
\(
\begin{gathered}
0^2=\left(10 \mathrm{~ms}^{-1}\right)^2+2 \times a \times 20 \mathrm{~m} \\
0=100 \mathrm{~m}^2 / \mathrm{s}^2+40 a \mathrm{~m} \\
40 a=-100 \mathrm{~ms}^{-2} \\
a=-\frac{100}{40} \mathrm{~ms}^{-2}=-2.5 \mathrm{~ms}^{-2}
\end{gathered}
\)
The magnitude of deceleration is \(2.5 \mathrm{~ms}^{-2}\).
Step 3: Determine the Coefficient of Friction
The deceleration is caused by the force of kinetic friction, \(F_f\). The force of friction is also related to the coefficient of friction \(\mu\) and the normal force \(N\), which equals the combined weight \(\boldsymbol{M}\) g for a horizontal surface.
\(
F_f=\mu N=\mu M g
\)
According to Newton’s second law, the force of friction is also \(F_f=M|a|\).
\(
\begin{gathered}
M|a|=\mu M g \\
\mu=\frac{|a|}{g}
\end{gathered}
\)
Given:
\(|a|=2.5 \mathrm{~ms}^{-2}\)
\(g=10 \mathrm{~ms}^{-2}\)
Calculation:
\(
\begin{gathered}
\mu=\frac{2.5 \mathrm{~ms}^{-2}}{10 \mathrm{~ms}^{-2}} \\
\mu=0.25
\end{gathered}
\)
The coefficient of friction between the block and the surface is \(\mathbf{0 . 2 5}\).

Hint

(d) Statement B is correct because when lifting the bucket, the gravitational force acts downward, while the displacement is upward. Since the force and displacement are in opposite directions, the work done by gravity is negative.

Statement E is correct because air resistance (a form of friction) always opposes the motion of the pendulum. Since the force is always in the opposite direction to the instantaneous displacement, the work done by air resistance is negative, which causes the pendulum to slow down and eventually stop.

Vhy other options are incorrect
Statement A is incorrect. The man applies an upward force to lift the bucket, and the bucket’s displacement is also upward. Since the force and displacement are in the same direction, the work done by the man is positive.

Statement C is incorrect. Friction always opposes motion. When a body slides down an inclined plane, the friction force acts upward along the plane (opposite to the downward motion), so the work done by friction is negative.

Statement D is incorrect. When a body moves at a uniform velocity on a rough horizontal plane, it means the net force is zero. The applied force is balanced by the force of friction. However, the applied force and the displacement are in the same direction, so the work done by the applied force is positive, not zero. The net work done is zero.

Hint

(c)

\(
\begin{aligned}
& \mathrm{KE}_{\text {in }}=\frac{1}{2} m v^2 \\
& \mathrm{KE}_{\text {final }}=\frac{1}{2} m v^2 \cos ^2 30^{\circ}=\frac{1}{2} m v^2\left(\frac{\sqrt{3}}{2}\right)^2 \\
& \frac{\mathrm{KE}_{\text {in }}}{\mathrm{KE}_{\mathrm{f}}}=\frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2\left(\frac{3}{4}\right)}=\frac{4}{3}
\end{aligned}
\)

Explanation: Step 1: Analyze Kinetic Energy at Projection
The initial kinetic energy ( \(\mathrm{KE}_{\text {initial }}\) ) of the stone with initial velocity \(v_0\) is given by the formula:
\(
\mathrm{KE}_{\text {initial }}=\frac{1}{2} m v_0^2
\)
where \(m\) is the mass of the stone and \(v_0\) is the initial speed.
Step 2: Analyze Kinetic Energy at Highest Point
At the highest point of flight, the vertical component of velocity is zero. The horizontal component of velocity \(\left(v_h\right)\) remains constant throughout the flight and is given by \(v_h=v_0 \cos (\theta)\), where \(\theta\) is the angle of projection \(\left(30^{\circ}\right)\).
The kinetic energy at the highest point ( \(\mathrm{KE}_{\text {highest }}\) ) is:
\(
\mathrm{KE}_{\text {highest }}=\frac{1}{2} m v_h^2=\frac{1}{2} m\left(v_0 \cos (\theta)\right)^2
\)
Step 3: Calculate the Ratio
The ratio of the kinetic energy at the point of projection to the kinetic energy at the highest point is:
\(
\frac{\mathrm{KE}_{\text {initial }}}{\mathrm{KE}_{\text {highest }}}=\frac{\frac{1}{2} m v_0^2}{\frac{1}{2} m v_0^2 \cos ^2(\theta)}=\frac{1}{\cos ^2(\theta)}
\)
Substituting the angle \(\theta=30^{\circ}\) :
\(
\frac{\mathrm{KE}_{\text {initial }}}{\mathrm{KE}_{\text {highest }}}=\frac{1}{\cos ^2\left(30^{\circ}\right)}=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{1}{\frac{3}{4}}=\frac{4}{3}
\)
The ratio of kinetic energy of the stone at the point of projection to its kinetic energy at the highest point of flight is \(4: 3\).

Hint

(a) Step 1: Calculate initial and final velocities
We calculate the initial velocity \(\left(v_i\right)\) and the velocity after 1 second \(\left(v_f\right)\) using the kinetic energy formula \(K E=\frac{1}{2} m v^2\), where \(m=0.2 \mathrm{~kg}\).
For the initial state \(\left(K E_i=90 \mathrm{~J}\right)\) :
\(
v_i=\sqrt{\frac{2 \times K E_i}{m}}=\sqrt{\frac{2 \times 90 \mathrm{~J}}{0.2 \mathrm{~kg}}}=\sqrt{900 \mathrm{~m}^2 / \mathrm{s}^2}=30 \mathrm{~m} / \mathrm{s}
\)
For the state after 1 second \(\left(\boldsymbol{K E}_{\boldsymbol{f}}=\mathbf{4 0 J}\right)\) :
\(
v_f=\sqrt{\frac{2 \times K E_f}{m}}=\sqrt{\frac{2 \times 40 \mathrm{~J}}{0.2 \mathrm{~kg}}}=\sqrt{400 \mathrm{~m}^2 / \mathrm{s}^2}=20 \mathrm{~m} / \mathrm{s}
\)
Step 2: Determine constant acceleration
Assuming a constant resistive force from the water results in constant acceleration (a). We use the kinematic equation \(v_f=v_i+at\) over the time interval \(t=1 \mathrm{~s}\) :
\(
a=\frac{v_f-v_i}{t}=\frac{20 \mathrm{~m} / \mathrm{s}-30 \mathrm{~m} / \mathrm{s}}{1 \mathrm{~s}}=-10 \mathrm{~m} / \mathrm{s}^2
\)
Step 3: Calculate total stopping distance
We find the total distance ( \(d\) ) the bullet travels until it comes completely to rest ( \(v_{\text {final }}=0 \mathrm{~m} / \mathrm{s}\) ), starting from the initial velocity \(v_i=30 \mathrm{~m} / \mathrm{s}\), using the constant acceleration \(a=-10 \mathrm{~m} / \mathrm{s}^2\). The kinematic equation \(v_{\text {final }}^2=v_i^2+2 a d\) is used:
\(
\begin{gathered}
0^2=(30 \mathrm{~m} / \mathrm{s})^2+2 \times\left(-10 \mathrm{~m} / \mathrm{s}^2\right) \times d \\
0=900 \mathrm{~m}^2 / \mathrm{s}^2-20 \mathrm{~m} / \mathrm{s}^2 \times d \\
20 \mathrm{~m} / \mathrm{s}^2 \times d=900 \mathrm{~m}^2 / \mathrm{s}^2 \\
d=\frac{900 \mathrm{~m}^2 / \mathrm{s}^2}{20 \mathrm{~m} / \mathrm{s}^2}=45 \mathrm{~m}
\end{gathered}
\)
The minimum length of the pool the bullet has to travel so that it completely comes to rest is \(\mathbf{4 5 ~ m}\).

Hint

(d) Step 1: Determine the force required
The force needed to maintain constant velocity in a variable mass system is given by the rate of change of momentum, \(F=\frac{d(m v)}{d t}\). Since the velocity \(v\) is constant, this simplifies to \(F=v \frac{d m}{d t}\).
We are given the mass flow rate \(\frac{d m}{d t}=0.5 \mathrm{kgs}^{-1}\) and velocity \(v=5 \mathrm{~ms}^{-1}\).
\(
F=\left(5 \mathrm{~ms}^{-1}\right) \times\left(0.5 \mathrm{kgs}^{-1}\right)=2.5 \mathrm{~N}
\)
Step 2: Calculate the power needed
Power \(\boldsymbol{P}\) is the product of the force applied and the velocity of the belt, \(\boldsymbol{P}=\boldsymbol{F v}\).
\(
\boldsymbol{P}=(2.5 \mathrm{~N}) \times\left(5 \mathrm{~ms}^{-1}\right)=12.5 \mathrm{~W}
\)
This value \((12.5 \mathrm{~W})\) is double the rate at which kinetic energy is imparted to the sand; the other half of the power is lost as heat due to friction.
The power needed to keep the belt moving with the same velocity is \(\mathbf{1 2 . 5 ~ W}\).

Hint

(b) Step 1: Set up the energy conservation equation
The problem can be solved using the principle of conservation of energy. The initial state has only kinetic energy (KE), and the state of maximum elongation has only potential energy (PE) in the center of mass frame (or by considering total energy and the fact the center of mass velocity is zero). The total initial kinetic energy of the system is the sum of the kinetic energies of both blocks, where each block has mass \(m=250 \mathrm{~g}=0.25 \mathrm{~kg}\) and velocity \(v\) in opposite directions:
\(
K E_{\text {initial }}=\frac{1}{2} m v^2+\frac{1}{2} m v^2=m v^2
\)
The potential energy stored in the spring at maximum elongation \(x\) is given by:
\(
P E_{\text {spring }}=\frac{1}{2} k x^2
\)
where \(k=2 \mathrm{Nm}^{-1}\).
Step 2: Solve for the maximum elongation
By conservation of energy, the initial kinetic energy equals the maximum potential energy:
\(
\begin{aligned}
K E_{\text {initial }} & =P E_{\text {spring }} \\
m v^2 & =\frac{1}{2} k x^2
\end{aligned}
\)
Substitute the given values for \(m\) and \(k\) :
\(
0.25 \cdot v^2=\frac{1}{2} \cdot 2 \cdot x^2
\)
\(
0.25 v^2=x^2
\)
Solve for \(x\) :
\(
\begin{gathered}
x=\sqrt{0.25 v^2}=0.5 v \\
x=\frac{v}{2}
\end{gathered}
\)
The maximum elongation of the spring is \(\frac{\mathbf{v}}{\mathbf{2}}\).

Hint

(b) Step 1: Calculate the final velocity using conservation of momentum
The mass of the bullet is \(m=0.2 \mathrm{~kg}\) and the mass of the sandbag is \(M=9.8 \mathrm{~kg}\). The initial velocity of the bullet is \(v_i=10 \mathrm{~m} / \mathrm{s}\), and the sandbag is initially at rest ( \(V_i=0 \mathrm{~m} / \mathrm{s}\) ).
Using the conservation of linear momentum for a perfectly inelastic collision, the final velocity \(v_f\) of the combined system is calculated:
\(
\begin{aligned}
& m v_i+M V_i=(m+M) v_f \\
& 0.2 \times 10+9.8 \times 0=(0.2+9.8) v_f \\
& v_f=\frac{2}{10}=0.2 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Step 2: Calculate initial and final kinetic energies
The initial kinetic energy \(K E_i\) is only due to the bullet:
\(
K E_i=\frac{1}{2} m v_i^2=\frac{1}{2} \times 0.2 \times 10^2=10 \mathrm{~J}
\)
The final kinetic energy \(\boldsymbol{K} \boldsymbol{E}_f\) of the combined system after the collision is:
\(
K E_f=\frac{1}{2}(m+M) v_f^2=\frac{1}{2} \times(0.2+9.8) \times(0.2)^2=0.2 \mathrm{~J}
\)
Step 3: Determine the loss of kinetic energy
The loss of kinetic energy \(\boldsymbol{\Delta} \boldsymbol{K} \boldsymbol{E}\) is the difference between the initial and final kinetic energies:
\(
\Delta K E=K E_i-K E_f=10 \mathrm{~J}-0.2 \mathrm{~J}=9.8 \mathrm{~J}
\)
The loss of kinetic energy will be 9.8 J.

Hint

(b)

\(
\begin{aligned}
&\text { By work-energy thorem }\\
&\begin{aligned}
& \mathrm{W}=\Delta \mathrm{K} \\
& \Rightarrow \mathrm{~W}=\frac{1}{2}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2\right) \\
& \Rightarrow \mathrm{W}=\frac{1}{2} \times 0.5 \times\left(16^2-4^2\right) \\
& \Rightarrow \mathrm{W}=\frac{1}{4} \times 240 \\
& \Rightarrow \mathrm{~W}=60 \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Explanation: Step 1: Calculate Initial and Final Velocities
The velocity of the body is given by \(v(x)=\left(3 x^2+4\right) \mathrm{m} / \mathrm{s}\). We calculate the initial velocity \(\left(v_i\right)\) at \(x=0 \mathrm{~m}\) and the final velocity \(\left(v_f\right)\) at \(x=2 \mathrm{~m}\).
At \(x=0 \mathrm{~m}\) :
\(
v_i=3(0)^2+4=4 \mathrm{~m} / \mathrm{s}
\)
At \(x=2 \mathrm{~m}\) :
\(
v_f=3(2)^2+4=16 \mathrm{~m} / \mathrm{s}
\)
Step 2: Apply the Work-Energy Theorem
According to the Work-Energy Theorem, the net work done ( \(W_{\text {net }}\) ) on an object is equal to the change in its kinetic energy \((\Delta K)\). The formula for net work is \(W_{\text {net }}=K_f-K_i=\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2\), where \(m\) is the mass, \(v_i\) is the initial velocity, and \(v_f\) is the final velocity. The mass \(m=0.5 \mathrm{~kg}\).
We calculate the initial and final kinetic energies:
Initial Kinetic Energy \(\boldsymbol{(} \boldsymbol{K}_{\boldsymbol{i}} \boldsymbol{)}\) :
\(
K_i=\frac{1}{2}(0.5 \mathrm{~kg})(4 \mathrm{~m} / \mathrm{s})^2=0.25 \times 16=4 \mathrm{~J}
\)
Final Kinetic Energy ( \(\boldsymbol{K}_{\boldsymbol{f}}\) ):
\(
K_f=\frac{1}{2}(0.5 \mathrm{~kg})(16 \mathrm{~m} / \mathrm{s})^2=0.25 \times 256=64 \mathrm{~J}
\)
The net work done is the difference:
\(
W_{\text {net }}=K_f-K_i=64 \mathrm{~J}-4 \mathrm{~J}=60 \mathrm{~J}
\)
The net work done by the force is 60 J.

Hint

(d) From the figure:
The block of mass \(m\) is released from point \(\mathbf{A}\).
The vertical distance between \(\mathbf{A}\) and \(\mathbf{B}\) is \(\boldsymbol{y}_0\).
When the block reaches B, it has fallen through a height \(y_0\).
Since the block is simply dropped, gravity is the only force doing work (no springs, no friction in the diagram).
Using conservation of mechanical energy
Loss in potential energy from A to B:
\(
\Delta U=m g y_0
\)
This loss becomes kinetic energy at point B:
\(
K_B=m g y_0
\)
Final Answer:
\(
K_B=m g y_0
\)
If the question expected kinetic energy at a general point at depth \(y\), it would be:
\(
K=m g y
\)
But at point B, specifically:
\(
K_B=m g y_0
\)

Alternate:

The block is dropped from point A, meaning its initial velocity is zero, so the initial kinetic energy \(\left(K E_A\right)\) is zero. The initial height is \(y_0\) and the final height at point B is \(y\) (relative to some reference point, typically the ground).
The principle of conservation of mechanical energy states that the total energy at point A equals the total energy at point B, assuming no friction or air resistance:
\(
P E_A+K E_A=P E_B+K E_B
\)
The potential energy (PE) is given by \(m g h\), where \(h\) is the height, \(m\) is the mass, and \(g\) is the acceleration due to gravity. The kinetic energy \((K E)\) is given by \(\frac{1}{2} m v^2\).
At point A: \(P E_A=m g y, K E_A=0\).
At point B: \(P E_B=m g (y-y_0), K E_B=\) ?
\(
K E_B=m g y-m g\left(y-y_0\right)
\)
\(
K E_B=m g y_0
\)

Hint

(d) Step 1: Convert units and determine initial conditions
First, convert the mass from grams to kilograms and identify the initial velocity.
The mass is \(m=500 \mathrm{gm}=0.5 \mathrm{~kg}\).
The initial position is \(x_i=0 \mathrm{~m}\), so the initial velocity is \(v_i=b(0)^{5 / 2}=0 \mathrm{~m} / \mathrm{s}\).
The initial kinetic energy is \(K_i=\frac{1}{2} m v_i^2=0 \mathrm{~J}\).
Step 2: Determine final velocity and kinetic energy
Next, calculate the final velocity at \(\boldsymbol{x}_{\boldsymbol{f}} \boldsymbol{=} \mathbf{4 ~ m}\) using the given value for \(b=0.25 \mathrm{~m}^{-3 / 2} \mathrm{~s}^{-1}\).
The final velocity is \(v_f=b x_f^{5 / 2}=0.25 \times(4)^{5 / 2}=0.25 \times 32=8 \mathrm{~m} / \mathrm{s}\).
The final kinetic energy is \(K_f=\frac{1}{2} m v_f^2\).
\(
K_f=\frac{1}{2} \times 0.5 \mathrm{~kg} \times(8 \mathrm{~m} / \mathrm{s})^2=0.25 \times 64 \mathrm{~J}=16 \mathrm{~J}
\)
Step 3: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done by the net force ( \(W_{\text {net }}\) ) equals the change in kinetic energy ( \(\Delta K\) ).
\(
\begin{gathered}
W_{n e t}=K_f-K_i \\
W_{n e t}=16 \mathrm{~J}-0 \mathrm{~J}=16 \mathrm{~J}
\end{gathered}
\)
The work done by the net force is \(\mathbf{1 6 ~ J}\).

Hint

(a) Step 1: Understand Work-Displacement Graphs
The work ( \(W\) ) done by a force ( \(F\) ) over a displacement ( \(x\) ) is given by the integral of force with respect to displacement, which corresponds to the area under the Force vs. Displacement (F-x) graph.
\(
W=\int F d x
\)
Areas above the displacement axis represent positive work, and areas below represent negative work.
Step 2: Compare the Areas
Figure c has the largest positive area, so \(W_3\) is the greatest.
Figure b has a smaller positive area than c , so \(W_2<W_3\).
Figure a involves some area cancellations, but the net work is less than \(W_2\).
Figure d has a very small, likely negative, net area due to low force or significant area below the axis, resulting in the least work done.
Step 3: Arrange in Descending Order
Comparing the net work from each graph, the descending order (from greatest work to least work) is determined to be \(W_3\), then \(W_2\), then \(W_1\), and finally \(W_4\).
The four graphs arranged in descending order of total work done are
\(
W_3>W_2>W_1>W_4
\)

Hint

(c) The work done ( \(W\) ) is the line integral of the force \(\vec{F}\) over the displacement
\(
\begin{aligned}
& d \vec{r}=d x \hat{i}+d y \hat{j}: \\
& W=\int_{(1,2)}^{(2,3)} \vec{F} \cdot d \vec{r}=\int_{(1,2)}^{(2,3)}\left(4 x \hat{i}+3 y^2 \hat{j}\right) \cdot(d x \hat{i}+d y \hat{j}) \\
& W=\int_1^2 4 x d x+\int_2^3 3 y^2 d y
\end{aligned}
\)
\(
W=6 \mathrm{~J}+19 \mathrm{~J}=25 \mathrm{~J}
\)
According to the work-energy theorem, the change in kinetic energy is equal to the total work done:
\(
\Delta K E=W=25.0 \mathrm{~J}
\)

Hint

(a) Initial State: The body is dropped from rest at height \(h\).
Initial potential energy \(\left(P E_i\right)=m g h\)
Initial kinetic energy \(\left(K E_i\right)=0\) (since initial velocity is 0 )
Total initial energy \(\left(E_i\right)=m g h\)
Final State: The body reaches the ground (height 0 ) with speed \(v=0.8 \sqrt{g h}\).
Final potential energy \(\left(P E_f\right)=0\)
Final kinetic energy \(\left(K E_f\right)=\frac{1}{2} m v^2\)
Substitute the given velocity:
\(
K E_f=\frac{1}{2} m(0.8 \sqrt{g h})^2=\frac{1}{2} m(0.64 g h)=0.32 m g h
\)
Total final energy \(\left(E_f\right)=0.32 \mathrm{mgh}\)
Work Done by Air Friction: The work done by non-conservative forces ( \(W_{\text {friction }}\) ) is equal to the change in mechanical energy ( \(\Delta E=E_f-E_i\) ).
\(W_{\text {friction }}=E_f-E_i=K E_f+P E_f-\left(K E_i+P E_i\right)\)
\(W_{\text {friction }}=0.32 m g h+0-(0+m g h)\)
\(W_{\text {friction }}=0.32 m g h-m g h\)
\(W_{\text {friction }}=-0.68 m g h\)
The negative sign indicates that the air friction force opposes the downward motion, removing energy from the system and converting it to other forms (like heat). Among the given options, the magnitude of the work done is \(0.68 m g h\).

Hint

(c)

\(
\begin{aligned}
&3 M V_0=2 M V_2+M V_1\\
&3 V_0=2 V_2+V_1\\
&120=2 V_2+60 \Rightarrow V_2=30 \mathrm{~m} / \mathrm{s}\\
&\frac{\Delta K . E .}{K . E .}=\frac{\frac{1}{2} M V_1^2+\frac{1}{2} 2 M V_2^2-\frac{1}{2} 3 M V_0^2}{\frac{1}{2} 3 M V_0^2}\\
&=\frac{V_1^2+2 V_2^2-3 V_0^2}{3 V_0^2}\\
&=\frac{3600+1800-4800}{4800}=\frac{1}{8}
\end{aligned}
\)

Explanation: Step 1: Define Variables and Initial Kinetic Energy
Let the total mass of the block be \(M\) and its initial speed be \(v=40 \mathrm{~m} / \mathrm{s}\). The block splits into masses \(m_1\) and \(m_2\) such that \(m_1: m_2=1: 2\). Thus, the masses are \(m_1=\frac{M}{3}\) and \(m_2=\frac{2 M}{3}\). The initial kinetic energy \(K_i\) is calculated as:
\(
\begin{gathered}
K_i=\frac{1}{2} M v^2 \\
K_i=\frac{1}{2} M(40 \mathrm{~m} / \mathrm{s})^2=800 \mathrm{MJ}
\end{gathered}
\)
Step 2: Use Conservation of Momentum to Find the Second Velocity
The smaller part ( \(m_1\) ) moves at \(v_1=60 \mathrm{~m} / \mathrm{s}\) in the same direction. By the principle of conservation of linear momentum, the total initial momentum \(\boldsymbol{P}_{\boldsymbol{i}}\) equals the total final momentum \(P_f\) :
\(
P_i=P_f \Longrightarrow M v=m_1 v_1+m_2 v_2
\)
Substituting the known values and solving for the velocity of the larger part, \(v_2\) :
\(
\begin{gathered}
M(40 \mathrm{~m} / \mathrm{s})=\left(\frac{M}{3}\right)(60 \mathrm{~m} / \mathrm{s})+\left(\frac{2 M}{3}\right) v_2 \\
40=20+\frac{2}{3} v_2 \\
v_2=30 \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
Step 3: Calculate Final Kinetic Energy
The final kinetic energy \(K_f\) is the sum of the kinetic energies of the two parts:
\(
\begin{gathered}
K_f=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \\
K_f=\frac{1}{2}\left(\frac{M}{3}\right)(60 \mathrm{~m} / \mathrm{s})^2+\frac{1}{2}\left(\frac{2 M}{3}\right)(30 \mathrm{~m} / \mathrm{s})^2 \\
K_f=\frac{M}{6}\left(3600 \mathrm{~m}^2 / \mathrm{s}^2\right)+\frac{M}{3}\left(900 \mathrm{~m}^2 / \mathrm{s}^2\right) \\
K_f=600 M \mathrm{~J}+300 M \mathrm{~J}=900 M \mathrm{~J}
\end{gathered}
\)
Step 4: Determine the Fractional Change in Kinetic Energy
The fractional change is defined as the change in kinetic energy divided by the initial kinetic energy:
\(
\begin{gathered}
\text { Fractional Change }=\frac{K_f-K_i}{K_i} \\
\text { Fractional Change }=\frac{900 \mathrm{MJ}-800 \mathrm{MJ}}{800 \mathrm{MJ}} \\
\text { Fractional Change }=\frac{100 \mathrm{M}}{800 \mathrm{M}}=\frac{1}{8}
\end{gathered}
\)

Hint

(b) Step 1: Understand the principles of energy conservation
The total mechanical energy ( \(\mathrm{E}_{\text {mech }}\) ) of a particle under a conservative force is constant and is the sum of its potential energy (U(x)) and kinetic energy (K.E.). The relationship is given by:
\(
E_{\mathrm{mech}}=U(x)+K . E . \Longrightarrow K . E .=E_{\mathrm{mech}}-U(x)
\)
Given that \(E_{\text {mech }}=8 \mathrm{~J}\), we can find the kinetic energy at different points using the values from the potential energy plot (inferred from the problem context).
Step 2: Evaluate each statement

Statement A: At \(\mathbf{x}>\mathbf{x}_4\), K.E. is constant throughout the region. From the context of the problem, \(\boldsymbol{U}=6 \mathrm{~J}\) for \(\boldsymbol{x}>\boldsymbol{x}_{\mathbf{4}}\). The kinetic energy is K.E. \(=8-6=2 \mathrm{~J}\), which is a constant value. This statement is correct.

Statement B: At \(\mathbf{x}<\mathbf{x}_{\mathbf{1}}\), K.E. is smallest and the particle is moving at the slowest speed.
From the context, \(U=8 \mathrm{~J}\) for \(x<x_1\). The kinetic energy is \(K . E .=8-8=0 \mathrm{~J}\). A kinetic energy of 0 J means the particle is momentarily at rest (slowest possible speed is zero). The statement claims K.E. is smallest and speed slowest, which is correct in the context of the region, but this is a turning point where the particle’s speed is zero. In the context of the provided options in the source exam, this is the incorrect statement because the particle cannot be “moving” at the slowest speed; it is at rest. The particle can only reach this region if it starts exactly at \(E_{\text {mech }}=8 \mathrm{~J}\) at this point.

Statement C: At \(\mathbf{x}=\mathbf{x}_2\), K.E. is greatest and the particle is moving at the fastest speed.
From the context, \(\boldsymbol{U = 0 ~ J ~ a t ~} \boldsymbol{x}=\boldsymbol{x}_2\). The kinetic energy is K.E. \(=8-\mathbf{0}=8 \mathrm{~J}\). This is the minimum potential energy, so the kinetic energy is maximum, and thus the speed is fastest. This statement is correct.

Statement D: At \(\mathrm{x}=\mathrm{x}_3, \mathrm{K.E} .=4 \mathrm{~J}\).
From the context, \(U=4 \mathrm{~J}\) at \(x=x_3\). The kinetic energy is K.E. \(=8-4=4 \mathrm{~J}\). This statement is correct.

Hint

(a) Step 1: Relate constant power to force and velocity
The problem states that the engine provides constant power \(\boldsymbol{P}\). Power is the product of force \((F)\) and velocity \((v)\) :
\(
P=F v
\)
Force is also defined by Newton’s second law as mass \((m)\) times acceleration ( \(a=\frac{d v}{d t}\) ):
\(
F=m \frac{d v}{d t}
\)
Substituting the expression for force into the power equation yields a differential equation relating power, mass, velocity, and time:
\(
P=m \frac{d v}{d t} v=m v \frac{d v}{d t}
\)
We rearrange the differential equation from Step 1 to separate variables and integrate both sides. The automobile starts from rest ( \(v=0\) at \(t=0\) ):
\(
\begin{aligned}
P d t & =m v d v \\
\int_0^t P d t & =\int_0^v m v d v \\
P t & =m \frac{v^2}{2}
\end{aligned}
\)
Solving for \(v(t)\), we get:
\(
\begin{gathered}
v^2=\frac{2 P t}{m} \\
v(t)=\sqrt{\frac{2 P}{m}} t^{\frac{1}{2}}
\end{gathered}
\)
Step 2: Solve for position as a function of time
Velocity is the time derivative of position \((x): v=\frac{d x}{d t}\). We substitute the expression for \(v(t)\) from Step 2 into this definition:
\(
\frac{d x}{d t}=\sqrt{\frac{2 P}{m}} t^{\frac{1}{2}}
\)
We separate variables again and integrate from the initial position \(x=0\) at \(t=0\) :
\(
\begin{aligned}
d x & =\sqrt{\frac{2 P}{m}} t^{\frac{1}{2}} d t \\
\int_0^x d x & =\int_0^t \sqrt{\frac{2 P}{m}} t^{\frac{1}{2}} d t \\
x(t) & =\sqrt{\frac{2 P}{m}}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right) \\
x(t) & =\frac{2}{3} \sqrt{\frac{2 P}{m}} t^{\frac{3}{2}}
\end{aligned}
\)
Step 3: Simplify the expression to match options
The expression can be simplified by moving the constant \(\frac{2}{3}\) inside the square root:
\(
\begin{gathered}
x(t)=\sqrt{\left(\frac{2}{3}\right)^2 \frac{2 P}{m}} t^{\frac{3}{2}} \\
x(t)=\sqrt{\frac{4}{9} \frac{2 P}{m}} t^{\frac{3}{2}} \\
x(t)=\sqrt{\frac{8 P}{9 m}} t^{\frac{3}{2}} \\
x(t)=\left(\frac{8 P}{9 m}\right)^{\frac{1}{2}} t^{\frac{3}{2}}
\end{gathered}
\)
The position is given as a function of time by \(\mathbf{x}(\mathbf{t})=\left(\frac{\mathbf{8 P}}{\mathbf{9 m}}\right)^{\frac{\mathbf{1}}{\mathbf{2}}} \mathbf{t}^{\frac{\mathbf{3}}{\mathbf{2}}}\).

Hint

(c) To calculate the work done by the porter, we use the formula:
\(
W=F \times d \times \cos (\theta)
\)
where \(W\) is the work done, \(F\) is the force applied by the porter, \(d\) is the distance the suitcase is displaced, and \(\theta\) is the angle between the force and displacement vectors.
Determine the force: Since the suitcase is lowered at a constant velocity, the net force is zero. The force exerted by the porter ( \(F\) ) must be equal in magnitude to the gravitational force (weight) acting on the suitcase, but in the opposite direction.
Mass \((m)=80 \mathrm{~kg}\)
Acceleration due to gravity \((g)=9.8 \mathrm{~m} / \mathrm{s}^2\)
Weight/Force \((F)=m \times g=80 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^2=784 \mathrm{~N}\).
Determine the distance: The distance must be in meters (SI units).
Distance \((d)=80 \mathrm{~cm}=0.8 \mathrm{~m}\).
Determine the angle: The suitcase is moving downwards, so its displacement vector is pointing down. The porter is exerting an upward force to control the descent. The angle ( \(\theta\) ) between the upward force and the downward displacement is \(180^{\circ}\).
\(\cos \left(180^{\circ}\right)=-1\).
Calculate the work done:
\(W=F \times d \times \cos (\theta)\)
\(W=784 \mathrm{~N} \times 0.8 \mathrm{~m} \times(-1)\)
\(W=-627.2 \mathrm{~J}\).
The negative sign indicates that the force exerted by the porter is in the opposite direction to the displacement of the suitcase, meaning the porter is doing negative work (or work against gravity).

Hint

(a) Step 1: Relate Power, Force, and Velocity
The constant power \((P)\) delivered to the body is related to the force \((F)\) and velocity ( \(v\) ) by the equation \(P=F v\). Force is defined as mass ( \(m\) ) times acceleration ( \(a\) ), where acceleration is the derivative of velocity with respect to time \(\left(a=\frac{d v}{d t}\right)\). Thus, the power equation can be written as \(P=m \frac{d v}{d t} v\).
Step 2: Integrate to find Velocity
Rearranging the equation to separate variables allows for integration: \(\boldsymbol{P} \boldsymbol{d} \boldsymbol{t} \boldsymbol{=} \boldsymbol{m v} \boldsymbol{d} \boldsymbol{v}\). Integrating both sides from time 0 (where velocity \(v=0\) ) to an arbitrary time \(t\) (velocity \(v\)) yields the relationship between velocity and time:
\(
\begin{aligned}
\int_0^t P d t & =\int_0^v m v d v \\
P t & =\frac{1}{2} m v^2
\end{aligned}
\)
Solving for velocity \(v\), we find \(v=\sqrt{\frac{2 P}{m}} t^{\frac{1}{2}}\). This shows that \(v \propto t^{\frac{1}{2}}\).
Step 3: Integrate to find Distance
Velocity is the derivative of distance \((s)\) with respect to time \(\left(v=\frac{d s}{d t}\right)\). Substituting the expression for \(v\) and integrating again from time 0 (distance \(s=0\) ) to time \(t\) gives the distance traveled:
\(
\begin{aligned}
\frac{d s}{d t} & =\sqrt{\frac{2 P}{m}} t^{\frac{1}{2}} \\
\int_0^s d s & =\int_0^t \sqrt{\frac{2 P}{m}} t^{\frac{1}{2}} d t \\
s & =\sqrt{\frac{2 P}{m}}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right) \\
s & =\frac{2}{3} \sqrt{\frac{2 P}{m}} t^{\frac{3}{2}}
\end{aligned}
\)
This result confirms that the distance \(s\) is proportional to \(t^{\frac{3}{2}}\).
The distance moved by the body in time ‘ \(t\) ‘ is proportional to \(t^{\frac{3}{2}}\).

Hint

(b) The constant power \(\boldsymbol{P}\) delivered by the machine is the product of the force \(\boldsymbol{F}\) applied to the box and its instantaneous velocity \(v\), given by \(P=F v\). The work \(W\) done on the box over a time \(t\) is \(W=P t\). According to the work-energy theorem, this work equals the change in kinetic energy \(\boldsymbol{\Delta} \boldsymbol{K}\).
\(
W=\Delta K=\frac{1}{2} m v^2-\frac{1}{2} m v_0^2
\)
Since the box starts from rest, \(v_0=0\), so \(P t=\frac{1}{2} m v^2\).
Step 1: Relate power to velocity
We rearrange the work-energy equation to express velocity \(v\) as a function of time \(t\) :
\(
\begin{gathered}
v^2=\frac{2 P t}{m} \\
v=\sqrt{\frac{2 P}{m}} t^{12}
\end{gathered}
\)
Velocity is also the derivative of distance \(s\) with respect to time, \(v=\frac{d s}{d t}\).
Step 2: Integrate to find distance
To find the distance \(s\), we integrate the velocity expression with respect to time:
\(
\begin{gathered}
\frac{d s}{d t}=\sqrt{\frac{2 P}{m}} t^{1 / 2} \\
s=\int \sqrt{\frac{2 P}{m}} t^{1 / 2} d t=\sqrt{\frac{2 P}{m}} \int t^{1 / 2} d t
\end{gathered}
\)
Integrating gives:
\(
s=\sqrt{\frac{2 P}{m}}\left(\frac{t^{3 / 2}}{3 / 2}\right)=\left(\frac{2}{3} \sqrt{\frac{2 P}{m}}\right) t^{3 / 2}
\)
The terms within the large parentheses are constants. Therefore, the distance \(s\) is directly proportional to \(t^{3 / 2}\).
The distance moved by the box in time ‘ \(t\) ‘ is proportional to \(t^{3 / 2}\).

Hint

(c) Step 1: Calculate initial kinetic energy
The initial kinetic energy \(\left(\mathrm{KE}_i\right)\) of the ball is calculated using the formula \(\mathrm{KE}=\frac{1}{2} m v^2\).
Given mass \(m=0.5 \mathrm{~kg}\) and initial speed \(v_i=20 \mathrm{~ms}^{-1}\) :
\(
\begin{gathered}
\mathrm{KE}_i=\frac{1}{2} \times 0.5 \mathrm{~kg} \times\left(20 \mathrm{~ms}^{-1}\right)^2 \\
\mathrm{KE}_i=0.25 \times 400 \mathrm{~J} \\
\mathrm{KE}_i=100 \mathrm{~J}
\end{gathered}
\)
Step 2: Calculate final kinetic energy
The problem states that the ball moves with 5% of its initial kinetic energy after deflection.
\(
\begin{gathered}
\mathrm{KE}_f=5 \% \times \mathrm{KE}_i \\
\mathrm{KE}_f=\frac{5}{100} \times 100 \mathrm{~J} \\
\mathrm{KE}_f=5 \mathrm{~J}
\end{gathered}
\)
Step 3: Calculate final speed
Using the kinetic energy formula for the final state, we can find the final speed ( \(v_f\) ).
\(
\begin{gathered}
\mathrm{KE}_f=\frac{1}{2} m v_f^2 \\
5 \mathrm{~J}=\frac{1}{2} \times 0.5 \mathrm{~kg} \times v_f^2 \\
5=0.25 v_f^2 \\
v_f^2=\frac{5}{0.25} \\
v_f^2=20 \mathrm{~m}^2 \mathrm{~s}^{-2} \\
v_f=\sqrt{20 \mathrm{~m}^2 \mathrm{~s}^{-2}} \\
v_f \approx 4.47 \mathrm{~ms}^{-1}
\end{gathered}
\)
The speed of the ball now is approximately \(4.47 \mathrm{~ms}^{-1}\).

Hint

(a) Step 1: Differentiate the potential energy function
The potential energy function is given by \(U(r)=-A r^{-6}+B r^{-12}\). The force, \(F\), is the negative derivative of potential energy with respect to separation \(r, F=-\frac{d U}{d r}\).
Equilibrium occurs when the force is zero.
The derivative of the potential energy is:
\(
\frac{d U}{d r}=\frac{d}{d r}\left(-A r^{-6}+B r^{-12}\right)=6 A r^{-7}-12 B r^{-13}
\)
Step 2: Set the derivative to zero and solve for equilibrium separation
At equilibrium, \(\frac{d U}{d r}=0\) :
\(
6 A r^{-7}-12 B r^{-13}=0
\)
Rearranging the terms:
\(
\begin{aligned}
& 6 A r^{-7}=12 B r^{-13} \\
& r^6=\frac{12 B}{6 A}=\frac{2 B}{A}
\end{aligned}
\)
The equilibrium separation, \(\boldsymbol{r_0}\), is:
\(
r_0=\left(\frac{2 B}{A}\right)^{1 / 6}
\)
Step 3: Calculate the potential energy at equilibrium
Substitute the value of \(r_0\) back into the original potential energy equation:
\(
U\left(r_0\right)=-\frac{A}{r_0^6}+\frac{B}{r_0^{12}}
\)
Using \(r_0^6=\frac{2 B}{A}\) and \(r_0^{12}=\left(r_0^6\right)^2=\left(\frac{2 B}{A}\right)^2=\frac{4 B^2}{A^2}\) :
\(
U\left(r_0\right)=-\frac{A}{\left(\frac{2 B}{A}\right)}+\frac{B}{\left(\frac{4 B^2}{A^2}\right)}=-\frac{A^2}{2 B}+\frac{A^2 B}{4 B^2}=-\frac{A^2}{2 B}+\frac{A^2}{4 B}
\)
Combine the terms:
\(
U\left(r_0\right)=-\frac{2 A^2}{4 B}+\frac{A^2}{4 B}=-\frac{A^2}{4 B}
\)
The equilibrium separation is \(\left(\frac{\mathbf{2 B}}{\mathbf{A}}\right)^{\mathbf{1} / 6}\) and the potential energy at equilibrium is \(-\frac{\mathbf{A}^{\mathbf{2}}}{\mathbf{4 B}}\).

Hint

(b)

Work done \(=\) Area under curve = Area of rectangle ABCD + area of trapezoid CDEF
\(
\begin{gathered}
=200 \times 15+\frac{200+100}{2} 15 \\
=3000+2250 \\
=5250 \mathrm{~J}
\end{gathered}
\)

Note: The area of a trapezoid is calculated using the formula \(A=\frac{1}{2}\left(b_1+b_2\right) h\).
\(b_1\) and \(b_2\) are the lengths of the two parallel bases.
\(h\) is the height (perpendicular distance between the bases).

Hint

(a) Consider a particle that is moving in one direction on a horizontal plane under the action of a constant power-supplying energy source. We need to find which of the following graphs accurately represents the motion of the particle in the displacement- time graph. We know that the power supplied to the particle is,
\(
P=F V
\)
We know that, \(F=m \frac{d V}{d t}\)
\(
\begin{aligned}
& P=m \frac{d V}{d t} \times V \\
& \Rightarrow V d V=\frac{P}{m} d t
\end{aligned}
\)
On integrating we get the expression for \(V\)
\(
\frac{V^2}{2}=\frac{P}{m} t
\)
This clearly says that,
\(
V^2 \propto t \text { or } V \propto t^{\frac{1}{2}}
\)
Here, velocity is,
\(
\begin{aligned}
& V=\frac{\text { displacement }}{\text { time }} \\
& \Rightarrow V=\frac{S}{t}
\end{aligned}
\)
Then,
\(
\begin{aligned}
& \frac{S}{t} \propto t^{\frac{1}{2}} \\
& \therefore S \propto t^{\frac{3}{2}}
\end{aligned}
\)
Therefore, we represent this equation in the displacement time graph as shown in option (2).

Hint

(c)

\(
\begin{aligned}
& \mathrm{W}=\int \vec{F} \cdot \overrightarrow{d r} \\
& =\int(-x \hat{i}+y \hat{j}) \cdot(d x \hat{i}+d y \hat{j}) \\
& =-\int_1^0 x d x+\int_0^1 y d y \\
& =\frac{1}{2}+\frac{1}{2}=1 \mathrm{~J}
\end{aligned}
\)

Hint

(d) Step 1: Calculate the total mass
First, calculate the total mass of the elevator when it is at full capacity by adding the mass of the elevator itself to the combined mass of 10 persons:
\(
\begin{gathered}
M_{\text {total }}=M_{\text {elevator }}+\left(N_{\text {persons }} \times M_{\text {person }}\right) \\
M_{\text {total }}=920 \mathrm{~kg}+(10 \times 68 \mathrm{~kg}) \\
M_{\text {total }}=1600 \mathrm{~kg}
\end{gathered}
\)
Step 2: Calculate the total upward force required
Since the elevator moves at a constant velocity, the net force on it must be zero. The motor’s upward force ( \(\boldsymbol{F}_{\text {motor }}\) ) must balance the total downward gravitational force ( \(\boldsymbol{F}_{\boldsymbol{g}}\) ) and the opposing frictional force \(\left(F_f\right)\) :
\(
F_{\text {motor }}=F_g+F_f
\)
The gravitational force is calculated using \(F_g=M_{\text {total }} \times g\) :
\(
\begin{gathered}
F_{\text {motor }}=\left(M_{\text {total }} \times g\right)+F_f \\
F_{\text {motor }}=\left(1600 \mathrm{~kg} \times 10 \mathrm{~m} / \mathrm{s}^2\right)+6000 \mathrm{~N} \\
F_{\text {motor }}=16000 \mathrm{~N}+6000 \mathrm{~N} \\
F_{\text {motor }}=22000 \mathrm{~N}
\end{gathered}
\)
Step 3: Calculate the power delivered by the motor
The power ( \(P\) ) delivered by the motor is the product of the required force and the constant upward velocity ( \(v\) ):
\(
\begin{gathered}
P=F_{\text {motor }} \times v \\
P=22000 \mathrm{~N} \times 3 \mathrm{~m} / \mathrm{s} \\
P=66000 \mathrm{~W}
\end{gathered}
\)
The power delivered by the motor must be at least 66000 W.

Hint

(d) Step 1: Convert Power to Watts
The motor’s power is given as 60 HP . Using the conversion factor \(1 \mathrm{HP}=746 \mathrm{~W}\), we calculate the power in Watts:
\(
P=60 \mathrm{HP} \times 746 \mathrm{~W} / \mathrm{HP}=44760 \mathrm{~W}
\)
Step 2: Calculate the Total Force Required
The total upward force ( \(F_{\text {total }}\) ) the motor must provide needs to overcome the weight of the maximum load and the opposing frictional force.
Weight \(\left(F_{\text {weight }}\right): F_{\text {weight }}=m \times g=2000 \mathrm{~kg} \times 10 \mathrm{~ms}^{-2}=20000 \mathrm{~N}\).
Total Force: \(F_{\text {total }}=F_{\text {weight }}+F_{\text {friction }}=20000 \mathrm{~N}+4000 \mathrm{~N}=24000 \mathrm{~N}\).
Step 3: Calculate the Speed
The relationship between power \((P)\), force \((F)\), and speed \((v)\) is given by \(P=F \times v\).
Rearranging to solve for speed:
\(
v=\frac{P}{F_{\text {total }}}=\frac{44760 \mathrm{~W}}{24000 \mathrm{~N}} \approx 1.865 \mathrm{~ms}^{-1}
\)
Rounding the result to one decimal place gives \(1.9 \mathrm{~ms}^{-1}\).
The speed of the elevator at full load is close to \(1.9 \mathrm{~ms}^{-1}\).

Hint

(c) In the above figure we can see that the a section of the cable of length \(L\) i.e. \(\mathrm{L} / \mathrm{n}\) part of the cable is hanging below the horizontal surface. In the question it is given to us that the cable has mass \(M\) with length \(L\) . Hence the mass per unit length of the cable is \(\mathrm{M} / \mathrm{L}\). The work done \((\mathrm{U})\) in taking a body of mass \(M\) to a height \(h\) under the action of gravitational pull with respect to its initial position is given by,
\(
U(h)=m g h
\)
Now in the above case we can consider this effect to occur about the center of mass of the cable of length \(\mathrm{L} / \mathrm{n}\). Since the mass is distributed uniformly in the cable we can say that the centre of mass of the cable of length \(\mathrm{L} / \mathrm{n}\) will lie in the middle i.e. at \(\mathrm{L} / 2 \mathrm{n}\). Hence we have the work done to bring up the cable to height of \(\mathrm{L} / 2 \mathrm{n}\) such that the entire cable lies within one line on the surface. The mass ( \(M_B\) )of the cable that lies below the edge is equal to,
\(
\begin{gathered}
M_B=\left(\frac{M}{L}\right) \frac{L}{n} \\
\Rightarrow M_B=\frac{M}{n}
\end{gathered}
\)
Hence the work done to bring the cable to the horizontal surface is,
\(
\begin{aligned}
& U(h)=m g h \\
& \Rightarrow U(L / 2 n)=\frac{M}{n} g \frac{L}{2 n} \\
& \Rightarrow U(L / 2 n)=\frac{M g L}{2 n^2}
\end{aligned}
\)

Hint

(a) Step 1: Clarify the problem assumptions
The user did not provide the figure mentioned in the problem. Based on search results for the specific JEE Main 2019 question, the force ( \(F\) ) versus distance ( \(x\) ) graph typically involves a rectangle and a trapezium.
From \(x=0 \mathrm{~m}\) to \(x=2 \mathrm{~m}\), the force is constant at \(F=2 \mathrm{~N}\).
From \(x=2 \mathrm{~m}\) to \(x=3 \mathrm{~m}\), the force increases linearly from \(F=2 \mathrm{~N}\) to \(F=3 \mathrm{~N}\).
The particle starts from rest, so its initial kinetic energy \(\left(K E_i\right)\) is 0 J.
Step 2: Calculate the work done
The work done ( \(W\) ) by the force is equal to the area under the \(F-x\) graph from \(x=0 \mathrm{~m}\) to \(x=3 \mathrm{~m}\). This area can be divided into a rectangle and a trapezium:
Area of the rectangle (from \(x=0\) to \(x=2 \mathrm{~m}\) ):
\(W_1=\) Force × distance \(=2 \mathrm{~N} \times 2 \mathrm{~m}=4 \mathrm{~J}\).
Area of the trapezium (from \(x=2\) to \(x=3 \mathrm{~m}\) ):
\(W_2=\frac{1}{2} \times\) (sum of parallel sides) \(\times\) height \(=\frac{1}{2} \times(2 \mathrm{~N}+3 \mathrm{~N}) \times(3 \mathrm{~m}-2 \mathrm{~m})\).
\(W_2=\frac{1}{2} \times 5 \mathrm{~N} \times 1 \mathrm{~m}=2.5 \mathrm{~J}\).
Total work done:
\(W_{\text {total }}=W_1+W_2=4 \mathrm{~J}+2.5 \mathrm{~J}=6.5 \mathrm{~J}\).
Step 3: Apply the work-energy theorem
According to the work-energy theorem, the net work done on the particle equals the change in its kinetic energy ( \(\Delta K E\) ).
\(
W_{\text {total }}=\Delta K E=K E_f-K E_i
\)
Since the initial kinetic energy \(K E_i=0 \mathrm{~J}\) (starts from rest), the final kinetic energy ( \(K E_f\) ) is:
\(
K E_f=W_{\text {total }}=6.5 \mathrm{~J}
\)
The kinetic energy of the particle after it has travelled 3 m is 6.5 J.

Hint

(d) Step 1: Calculate the work done by the force
The work \(W\) done by a constant force \(\vec{F}\) over a displacement \(\vec{d}\) is given by the dot product:
\(
W=\vec{F} \cdot \vec{d}
\)
Given \(\vec{F}=3 \hat{i}-12 \hat{j}\) and \(\vec{d}=4 \hat{i}\), the calculation is:
\(
\begin{gathered}
W=(3 \hat{i}-12 \hat{j}) \cdot(4 \hat{i}) \\
W=(3)(4)+(-12)(0) \\
W=12 \mathrm{~J}
\end{gathered}
\)
Step 2: Apply the Work-Energy Theorem
The Work-Energy Theorem states that the net work done on a particle equals the change in its kinetic energy ( \(\Delta K E\) ):
\(
W=\Delta K E=K E_f-K E_i
\)
We can rearrange this to find the final kinetic energy \(\left(K E_f\right)\) :
\(
K E_f=K E_i+W
\)
Given the initial kinetic energy \(K E_i=3 \mathrm{~J}\) and the calculated work \(W=12 \mathrm{~J}\) :
\(
\begin{gathered}
K E_f=3 \mathrm{~J}+12 \mathrm{~J} \\
K E_f=15 \mathrm{~J}
\end{gathered}
\)
The kinetic energy of the particle at the end of the displacement is \(\mathbf{1 5 ~ J}\).

Hint

(b) Step 1: Calculate the Normal Force
The forces acting on the block are the gravitational force \(m g\) downwards and the normal force \(N\) upwards. Applying Newton’s second law in the vertical direction: \(N-m g=m a\), where \(a=g / 2\) is the upward acceleration.
\(
N=m(a+g)=m\left(\frac{g}{2}+g\right)=\frac{3}{2} m g
\)
Step 2: Determine the Displacement
The platform starts from rest, so its initial velocity is \(\boldsymbol{u} \boldsymbol{=} \mathbf{0}\). The vertical displacement \(\boldsymbol{s}\) in time \(t\) with constant acceleration \(a=g / 2\) is calculated using the kinematic equation \(s=u t+\frac{1}{2} a t^2\).
\(
s=0 \cdot t+\frac{1}{2}\left(\frac{g}{2}\right) t^2=\frac{1}{4} g t^2
\)
Step 3: Calculate the Work Done
Work done \(W\) by the normal force is given by \(W=\vec{F} \cdot \vec{s}\). Since the normal force and displacement are both upwards, the angle between them is \(0^{\circ}\).
\(
W=N s \cos \left(0^{\circ}\right)=\left(\frac{3}{2} m g\right)\left(\frac{1}{4} g t^2\right)=\frac{3}{8} m g^2 t^2
\)
The work done by the normal reaction on the block in time \(t\) is \(\frac{\mathbf{3}}{\mathbf{8}} \mathbf{m g}^{\mathbf{2}} \mathbf{t}^{\mathbf{2}}\).

Hint

(d) Step 1: Determine Velocity Function
The position as a function of time is \(x(t)=3 t^2+5\). The velocity function \(v(t)\) is found by differentiating the position with respect to time:
\(
\begin{aligned}
v(t) & =\frac{d x}{d t} \\
v(t) & =6 t
\end{aligned}
\)
Step 2: Calculate Initial and Final Velocities
The initial velocity at \(t=0 \mathrm{~s}\) is \(v_i=6(0)=0 \mathrm{~m} / \mathrm{s}\).
The final velocity at \(t=5 \mathrm{~s}\) is \(v_f=6(5)=30 \mathrm{~m} / \mathrm{s}\).
Step 3: Calculate Kinetic Energies
Using the kinetic energy formula \(K=\frac{1}{2} m v^2\) with mass \(m=2 \mathrm{~kg}\), the initial and final kinetic energies are:
\(
\begin{aligned}
& K_i=\frac{1}{2}(2)(0)^2=0 \mathrm{~J} \\
& K_f=\frac{1}{2}(2)(30)^2=900 \mathrm{~J}
\end{aligned}
\)
Step 4: Apply Work-Energy Theorem
The work done \(W\) is equal to the change in kinetic energy \(\Delta K=K_f-K_i\) :
\(
W=K_f-K_i
\)
\(
W=900 \mathrm{~J}-0 \mathrm{~J}=900 \mathrm{~J} .
\)

Hint

(d)

We fix the origin at the initial position of the block. Now, on the application of the force \(F\), we have the maximum extension \(x_{\text {max }}\), which can be found by putting net force equal to zero at the maximum extension.
So, we have from the equilibrium of the block
\(F=k x_{\text {max }}\)
Or, \(x_{\text {max }}=\frac{F}{k} \dots(1)\)
Now, we know that the potential energy of a stretched spring is given by
\(
\Rightarrow U=\frac{1}{2} k x^2
\)
So, the maximum potential energy of the stretched spring is equal to
\(
\Rightarrow U_{\max }=\frac{1}{2} k x_{\max }^2 \dots(2)
\)
We also know that the kinetic energy of an object is given by
\(
\Rightarrow K=\frac{1}{2} m v^2
\)
So, the maximum kinetic energy is
\(
\Rightarrow K_{\max }=\frac{1}{2} m v_{\max }^2 \dots(3)
\)
Since there is no friction, the total energy of the block remains constant. Also, we know that when the kinetic energy of the block is maximum, its potential energy is zero and vice-versa.
Therefore we can say that
\(
\Rightarrow K_{\max }=U_{\max }
\)
From (2) and (3) we have
\(
\Rightarrow \frac{1}{2} m v_{\max }^2=\frac{1}{2} k x_{\max }^2
\)
Multiplying both sides by \(\frac{2}{m}\)
\(
\begin{aligned}
& \Rightarrow v_{\max }^2=\frac{2}{m} \times \frac{1}{2} k x_{\max }^2 \\
& \Rightarrow v_{\max }^2=\frac{k}{m} x_{\max }^2
\end{aligned}
\)
Substituting the value of \(x_{\text {max }}\) from (1), we get
\(
\begin{aligned}
& \Rightarrow v_{\max }^2=\frac{k}{m}\left(\frac{F}{k}\right)^2 \\
& \Rightarrow v_{\max }^2=\frac{F}{m k}
\end{aligned}
\)
Finally, taking square root we get
\(
\Rightarrow v_{\max }=\frac{F}{\sqrt{m k}}
\)
Therefore, the maximum speed of the block is equal to \(\frac{F}{\sqrt{m k}}\)