JEE PYQs MCQs

Hint

(d) Step 1: Calculate the acceleration of the body
To find the acceleration, we use Newton’s second law, \(\vec{F}=m \vec{a}\). The force \(\vec{F}\) is 6 N directed along the positive z-axis, which can be written as \(\vec{F}=6 \hat{k} \mathrm{~N}\). The mass \(m\) is 2 kg.
The acceleration is given by:
\(
\vec{a}=\frac{\vec{F}}{m}=\frac{6 \hat{k}}{2}=3 \hat{k} \mathrm{~ms}^{-2}
\)
Step 2: Calculate the final velocity
The change in velocity \(\Delta \vec{v}\) is the product of acceleration \(\vec{a}\) and the time interval \(\Delta t\).
\(
\Delta \vec{v}=\vec{a} \Delta t=(3 \hat{k}) \times \frac{5}{3}=5 \hat{k} \mathrm{~ms}^{-1}
\)
The final velocity \(\vec{v}_{\text {out }}\) is the sum of the initial velocity \(\vec{v}_{\text {in }}\) and the change in velocity \(\Delta \vec{v}\). The initial velocity is given as \(\vec{v}_{\text {in }}=3 \hat{i}+4 \hat{j} \mathrm{~ms}^{-1}\).
\(
\vec{v}_{\text {out }}=\vec{v}_{\text {in }}+\Delta \vec{v}=(3 \hat{i}+4 \hat{j})+5 \hat{k}=3 \hat{i}+4 \hat{j}+5 \hat{k} \mathrm{~ms}^{-1}
\)
The velocity of the body when it emerges from the force field is \(3 \hat{i}+4 \hat{j}+5 \hat{k} \mathrm{~ms}^{-1}\).

Hint

(b) Step 1: Convert mass to standard units and find acceleration
First, convert the mass of the object from grams to kilograms:
\(
m=500 \mathrm{~g}=\frac{500}{1000} \mathrm{~kg}=0.5 \mathrm{~kg}
\)
Next, determine the acceleration of the object. Since the velocity \(v\) is a function of position \(x\), we use the chain rule to find the acceleration \(a\) :
\(
a=\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}
\)
The velocity is given as \(v=4 \sqrt{x}\). We need to find its derivative with respect to \(x\) :
\(
\frac{d v}{d x}=\frac{d}{d x}\left(4 x^{1 / 2}\right)=4 \cdot \frac{1}{2} x^{-1 / 2}=\frac{2}{\sqrt{x}}
\)
Now, substitute this back into the acceleration equation:
\(
a=v \frac{d v}{d x}=(4 \sqrt{x})\left(\frac{2}{\sqrt{x}}\right)=8 \mathrm{~m} / \mathrm{s}^2
\)
Step 2: Calculate the force
Using Newton’s Second Law, the force \(\boldsymbol{F}\) is the product of the object’s mass \(\boldsymbol{m}\) and its acceleration \(\boldsymbol{a}\) :
\(
F=m a
\)
Substitute the values for mass and acceleration:
\(
F=(0.5 \mathrm{~kg})\left(8 \mathrm{~m} / \mathrm{s}^2\right)=4 \mathrm{~N}
\)
The force acting on the object is \(\mathbf{4} \mathbf{N}\).

Hint

(d)

Step 1: Set up the equations of motion
First, we analyze the forces acting on the block. The force of gravity ( \(m g\) ) can be resolved into two components: one perpendicular to the inclined plane, \(m g \cos \theta\), and one parallel to it, \(m g \sin \theta\). The normal force ( \(N\) ) acts perpendicular to the plane, and the force of kinetic friction ( \(f_k\) ) acts parallel to the plane, opposing the motion.
The net force perpendicular to the plane is zero, so we have:
\(
N=m g \cos \theta
\)
The net force parallel to the plane is equal to the mass times the acceleration:
\(
m g \sin \theta-f_k=m a
\)
Step 2: Solve for the coefficient of kinetic friction
The force of kinetic friction is given by \(f_k=\mu_k N\), where \(\mu_k\) is the coefficient of kinetic friction. Substituting the expression for the normal force, we get \(f_k=\mu_k(m g \cos \theta)\)

Now we can substitute this into the equation for the parallel forces:
\(
m g \sin \theta-\mu_k(m g \cos \theta)=m a
\)
We can divide the entire equation by \(m g\) to simplify:
\(
\sin \theta-\mu_k \cos \theta=\frac{a}{g}
\)
We are given that the angle \(\theta=60^{\circ}\) and the acceleration \(a=\frac{g}{2}\). Plugging these values into the equation:
\(
\sin \left(60^{\circ}\right)-\mu_k \cos \left(60^{\circ}\right)=\frac{g / 2}{g}
\)
We know that \(\sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2}\) and \(\cos \left(60^{\circ}\right)=\frac{1}{2}\).
\(
\frac{\sqrt{3}}{2}-\mu_k\left(\frac{1}{2}\right)=\frac{1}{2}
\)
Now, we solve for \(\mu_k\) :
\(
\begin{aligned}
\mu_k\left(\frac{1}{2}\right) & =\frac{\sqrt{3}}{2}-\frac{1}{2} \\
\mu_k\left(\frac{1}{2}\right) & =\frac{\sqrt{3}-1}{2} \\
\mu_k & =\sqrt{3}-1
\end{aligned}
\)
The value of the coefficient of kinetic friction is \(\sqrt{3}-1\).

Hint

(d)

Vertical components of \(T_1\) and \(T_2\) balance the weight of the body.
\(
\therefore T_1 \sin \theta_1+T_2 \sin \theta_2=m g
\)
Given, \(\mathrm{T}_1=\sqrt{3} \mathrm{~T}_2\)
For \(\theta_1=60^{\circ} \& \theta_2=30^{\circ}\)
\(
\mathrm{T}_2\left[\sqrt{3} \sin \theta_1+\sin \theta_2\right]=\mathrm{mg}
\)
\(
\mathrm{T}_2=\frac{\mathrm{mg}}{2}
\)

Hint

(c) Step 1: Apply the Work-Energy Theorem
The Work-Energy Theorem states that the work done on an object by the net force is equal to the change in the object’s kinetic energy.
\(
W_{n e t}=\Delta K=K_f-K_i
\)
The work done by a variable force \(F(x)\) over a displacement from \(x_i\) to \(x_f\) is given by the integral:
\(
W=\int_{x_i}^{x_f} F(x) d x
\)
In this problem, the only force doing work is the retarding force, \(F_r=-k x\). We will use the given values to solve for the final speed, \(v_f\).
Step 2: Calculate the work done by the retarding force
The retarding force is \(F_r=-k x\), with \(k=10 \mathrm{~N} / \mathrm{m}\). The block moves from \(x_i=0.1 \mathrm{~m}\) to \(x_f=1.9 \mathrm{~m}\).
\(
\begin{gathered}
W_r=\int_{0.1}^{1.9}-10 x d x \\
W_r=-10\left[\frac{x^2}{2}\right]_{0.1}^{1.9} \\
W_r=-10\left(\frac{(1.9)^2}{2}-\frac{(0.1)^2}{2}\right) \\
W_r=-5(3.61-0.01) \\
W_r=-5(3.6)=-18 \mathrm{~J}
\end{gathered}
\)
Step 3: Solve for the final speed
Now, we set the work done equal to the change in kinetic energy.
\(
W_r=\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2
\)
We are given \(m=1 \mathrm{~kg}\) and \(v_i=10 \mathrm{~m} / \mathrm{s}\).
\(
-18=\frac{1}{2}(1) v_f^2-\frac{1}{2}(1)(10)^2
\)
\(
v_f=8 \mathrm{~m} / \mathrm{s}
\)
\(
\text { The final speed of the block is } 8 \mathrm{~m} / \mathrm{s} \text {. }
\)

Hint

(a) Using Lami’s theorem

\(
\begin{aligned}
& \frac{T_1}{\sin (90+30)}=\frac{T_2}{\sin (90+60)}=\frac{m g}{\sin 90} \\
& \Rightarrow T_1=m g \frac{\sqrt{3}}{2} \\
& T_2=\frac{M g}{2}=5 \\
& T_1=5 \sqrt{3}
\end{aligned}
\)

Alternate:

\(
T_1 \sin 60^{\circ}+T_2 \sin 30^{\circ}=m g \dots(i)
\)
\(
T_1 \cos 60^{\circ}=T_2 \cos 30^{\circ} \dots(ii)
\)
\(
T_1=\sqrt{3} T_2 \dots(iii)
\)
From (i) and (iii)
\(
\sqrt{3} T_2 \times \frac{\sqrt{3}}{2}+\frac{T_2}{2}=m g
\)
\(
T_2=\frac{m g}{2}=5 \mathrm{~N}
\)
and \(T_1=\frac{\sqrt{3} m g}{2}=5 \sqrt{3} \quad \mathrm{~N}\)

Hint

(b)

\(
\begin{aligned}
& F-m g=m a \\
& F=m a+m g \dots(i) \\
& F-(m-x) g=(m-x) 3 a \dots(ii)
\end{aligned}
\)
Put Eqn. (i) in Eqn.(ii)
\(
{ma}+{mg}-{mg}+{xg}=3 {ma}-3 {xa}
\)
\(
{x}=\frac{2 {ma}}{{~g}+3 {a}}
\)

Hint

(b)

\(
\begin{aligned}
&\text { AS system is in equilibrium, therefore }\\
&\begin{aligned}
& \mathrm{T}_1 \cos 45^{\circ}=\mathrm{mg} \\
& \mathrm{~T}_1 \sin 45^{\circ}=\mathrm{F} \\
& \therefore \tan 45^{\circ}=\frac{\mathrm{F}}{\mathrm{mg}} \\
& \Rightarrow \mathrm{~F}=\mathrm{mg} \\
& \mathrm{~F}=10 \mathrm{~N}
\end{aligned}
\end{aligned}
\)

Hint

(c)

Step 1:
In the above figure, point \(A\) corresponds to the end of the bar on the ground and point \(B\) corresponds to the end of the bar on the shoulder of the man. AB represents the length of the bar \(l\)
Here, the weight experienced by the man is the force \(F_2\). The corresponding reaction on the ground is \(F_1\).
Both \(F_1\) and \(F_2\) are directed upwards while the weight of the bar \(W\) is directed downwards.
The net forces acting on the system can be expressed as \(F_1+F_2=W\).
The distance AE will be equal to \(\frac{l}{2} \cos \theta\) and the distance AD represents the horizontal component of the length of the bar i.e., \(l \cos \theta\).
Step 2: Use the principle of moments to find the weight experienced by the man.
Here, the weight of the bar \(W\) tries to rotate the bar in the clockwise direction while the force \(F_2\) tries to rotate the bar in the anticlockwise direction.
The principle fo moments gives the sum of the moments as zero.
i.e., \(F_2(A D)-W(A E)=0\)
\(\Rightarrow F_2(A D)=W(A E) \dots(1)\)
Substituting for \(A E=\frac{l}{2} \cos \theta\) and \(A D=l \cos \theta\) in equation (1) we get,
\(F_2(l \cos \theta)=W\left(\frac{l}{2} \cos \theta\right)\)
On simplifying the above expression we get, \(F_2=\frac{W}{2}\)

Hint

(c) As the string is inextensible, both masses have the same acceleration \(a\). Also, the pulley is massless and frictionless, hence the tension at both ends of the string is the same. Suppose, the mass \(m_2\) is greater than mass \(m_1\) so the heavier mass \(m_2\) is accelerating downward and the lighter mass \(m_1\) is accelerating upwards. Therefore, by Newton’s 2nd law

\(
\begin{gathered}
T-m_1 g=m_1 a \dots(i)\\
m_2 g-T=m_2 a \dots(ii)
\end{gathered}
\)
After solving Eqs. (i) and (ii), we get
\(
a=\frac{\left(m_2-m_1\right)}{\left(m_1+m_2\right)} \cdot g=\frac{g}{8} \text { [given] }
\)
So, \(\frac{g}{8}=\frac{m_2\left(1-m_1 / m_2\right)}{m_2\left(1+m_1 / m_2\right)} \cdot g \dots(iii)\)
Let \(\frac{m_1}{m_2}=x\)
Thus, Eq. (iii) becomes
\(
\frac{1-x}{1+x}=\frac{1}{8} \text { or } x=\frac{7}{9} \text { or } \frac{m_2}{m_1}=\frac{9}{7}
\)
So, the ratio of the masses is \(9: 7\).

Hint

(a) Complete step-by-step answer:
First, we assume that time taken by an object of mass \(m\) to slide down on a smooth inclined plane is \(t\) and a rough inclined plane is \(t^{\prime}\). Since the object takes \(n\) times as much time to slide down rough incline as it takes to slide down a perfectly smooth incline, we have \(t^{\prime}=n t\)
Now we resolve the components acceleration in direction of motion and perpendicular to direction of motion.
For smooth incline, equation of motion:

In direction perpendicular to direction of motion, the normal reaction \(N\) when \(\theta\) is the angle of inclination
\(
N=m g \cos \theta
\)
Along the direction of motion, net force \(F_{n e t}\) when object accelerates with constant acceleration \(a\)
\(
\begin{aligned}
& F_{n e t}=m a \\
& m g \sin \theta=m a \Rightarrow a=g \sin \theta
\end{aligned}
\)
For rough incline, equation of motion:

In direction perpendicular to direction of motion
\(
N=m g \cos \theta
\)
Along the direction of motion net force when object accelerates with acceleration \(a^{\prime}\) is
\(
\begin{aligned}
& F_{n e t}^{\prime}=m a^{\prime} \\
& m g \sin \theta-f_k=m a^{\prime}
\end{aligned}
\)
Where kinetic friction \(f_k=\mu N\)
\(\mu\) is the coefficient of kinetic friction.
\(
\Rightarrow f_k=\mu m g \cos \theta
\)
Therefore we get,
\(
a^{\prime}=g(\sin \theta-\mu \cos \theta)
\)
Since distance travelled by object is same in both cases, we have
\(
\begin{aligned}
& s=s^{\prime} \\
& (g \sin \theta) t^2=g(\sin \theta-\mu \cos \theta)\left(t^{\prime}\right)^2
\end{aligned}
\)
On simplifying this equation and substituting \(t^{\prime}=n t\) and \(\theta=45^{\circ}\) we get
\(
t^2=\left(1-\frac{\mu \cos 45^{\circ}}{\sin 45^{\circ}}\right)(n t)^2
\)
On solving the above equation for coefficient of kinetic friction \(\mu\) we get
\(
\mu=1-\frac{1}{n^2}
\)

Hint

(c) Step 1: Convert units and identify given values
First, convert the mass of the cricket ball from grams to kilograms to use standard SI units.
\(\operatorname{Mass}(m): 150 \mathrm{~g}=0.15 \mathrm{~kg}\)
Initial velocity \(\left(v_i\right): 20 \mathrm{~m} / \mathrm{s}\)
Final velocity \(\left(v_f\right): 0 \mathrm{~m} / \mathrm{s}\) (since the ball is caught)
Time interval \((\Delta t): 0.1 \mathrm{~s}\)
Step 2: Calculate the change in momentum
The change in momentum ( \(\Delta p\) ) is the product of the mass and the change in velocity. The formula for momentum is \(p=m v\).
\(
\begin{gathered}
\Delta p=m\left(v_f-v_i\right) \\
\Delta p=0.15 \mathrm{~kg}(0 \mathrm{~m} / \mathrm{s}-20 \mathrm{~m} / \mathrm{s}) \\
\Delta p=0.15 \times(-20) \mathrm{kg} \cdot \mathrm{~m} / \mathrm{s} \\
\Delta p=-3 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}
\end{gathered}
\)
The magnitude of the change in momentum is \(3 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}\).
Step 3: Calculate the magnitude of the force
According to Newton’s Second Law, the average force ( \(\boldsymbol{F}\) ) is the rate of change of momentum.
\(
\begin{gathered}
F=\frac{|\Delta p|}{\Delta t} \\
F=\frac{3 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}}{0.1 \mathrm{~s}} \\
F=30 \mathrm{~N}
\end{gathered}
\)
The magnitude of the force exerted by the ball on the player’s hand is 30 N.

Hint

(a) Let’s restate the problem carefully:
Weight of the body \(=200 \mathrm{~N}\)
Mass of chain \(=10 \mathrm{~kg}\)
\(g=10 \mathrm{~m} / \mathrm{s}^2\)
We are to find the force with which the branch pulls the chain i.e., the tension at the top of the chain.
Step 1: Weight of the chain
\(
W_{\text {chain }}=m g=10 \times 10=100 \mathrm{~N}
\)
Step 2: What does the branch have to support?
The branch supports:
1. The entire weight of the chain ( 100 N )
2. The weight of the body (200 N), which is transmitted through the chain
Therefore, the total force the branch must exert (the tension at the top of the chain) is:
\(
T_{\text {top }}=200+100=300 \mathrm{~N}
\)

Hint

(c)

\(
\begin{aligned}
&T-m_1 g=m_1 a \dots(i)\\
&m_2 g-T=m_2 a \dots(ii)\\
&\text { After solving Eqs. (i) and (ii), we get }
\end{aligned}
\)
The acceleration of the system is given by:
\(
a=\frac{m_2-m_1}{m_1+m_2} g
\)
Given \(a=\frac{g}{\sqrt{2}}\), substitute in the equation:
\(
\frac{g}{\sqrt{2}}=\frac{m_2-m_1}{m_1+m_2} g
\)
Simplifying:
\(
\frac{1}{\sqrt{2}}=\frac{m_2-m_1}{m_1+m_2}
\)
Cross-multiplying:
\(
\sqrt{2}\left(m_2-m_1\right)=m_1+m_2
\)
Rearranging:
\(
m_1(\sqrt{2}+1)=m_2(\sqrt{2}-1)
\)
The ratio of masses is:
\(
\frac{m_1}{m_2}=\frac{\sqrt{2}-1}{\sqrt{2}+1}
\)

Hint

(a) Step 1: Find the force vector \(\vec{F}\)
The force \(\vec{F}\) is the time derivative of the linear momentum \(\vec{p}\). The linear momentum is given as \(\vec{p}(t)=\hat{i} \cos (\mathrm{kt})-\hat{j} \sin (\mathrm{kt})\).
\(
\vec{F}=\frac{d \vec{p}}{d t}
\)
Differentiating the components of \(\overrightarrow{\boldsymbol{p}}\) with respect to time \(t\) :
\(
\begin{aligned}
\vec{F} & =\frac{d}{d t}(\hat{i} \cos (k t)-\hat{j} \sin (k t)) \\
\vec{F} & =\hat{i} \frac{d}{d t}(\cos (k t))-\hat{j} \frac{d}{d t}(\sin (k t))
\end{aligned}
\)
Using the chain rule, we get:
\(
\begin{gathered}
\vec{F}=\hat{i}(-\sin (k t) \cdot k)-\hat{j}(\cos (k t) \cdot k) \\
\vec{F}=-k \sin (k t) \hat{i}-k \cos (k t) \hat{j}
\end{gathered}
\)
Step 2: Calculate the angle between \(\overrightarrow{\boldsymbol{F}}\) and \(\overrightarrow{\boldsymbol{p}}\)
The angle \(\boldsymbol{\theta}\) between two vectors can be found using the dot product formula:
\(
\vec{F} \cdot \vec{p}=|\vec{F}||\vec{p}| \cos (\theta)
\)
Let’s first calculate the dot product \(\vec{F} \cdot \vec{p}\) :
\(
\begin{gathered}
\vec{F} \cdot \vec{p}=(-k \sin (k t))(\cos (k t))+(-k \cos (k t))(-\sin (k t)) \\
\vec{F} \cdot \vec{p}=-k \sin (k t) \cos (k t)+k \sin (k t) \cos (k t) \\
\vec{F} \cdot \vec{p}=0
\end{gathered}
\)
Since the dot product of \(\vec{F}\) and \(\vec{p}\) is zero, the vectors are orthogonal. This means the angle \(\theta\) between them is \(90^{\circ}\) or \(\frac{\pi}{2}\) radians.

Hint

(b)

Step 1: Calculate the Normal Force
The normal force ( \(\boldsymbol{N}\) ) on a horizontal surface is equal to the gravitational force, which is the mass \((m)\) multiplied by the acceleration due to gravity \((g)\). We assume the standard value for \(g\) is \(9.8 \mathrm{~m} / \mathrm{s}^2\).
\(
\begin{gathered}
N=m g \\
N=(50 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right) \\
N=490 \mathrm{~N}
\end{gathered}
\)
Step 2: Calculate the Force of Kinetic Friction
The force of kinetic friction \(\left(\boldsymbol{F}_{\boldsymbol{k}}\right)\) is the product of the coefficient of kinetic friction \(\left(\boldsymbol{\mu}_{\boldsymbol{k}}\right)\) and the normal force ( \(N\) ).
\(
\begin{gathered}
F_k=\mu_k N \\
F_k=(0.3)(490 \mathrm{~N}) \\
F_k=147 \mathrm{~N}
\end{gathered}
\)
The force of kinetic friction is 147 N.

Hint

(b)

Step 1: Calculate the total mass of the system
The system consists of the wooden block and the iron cylinder.
The mass of the wooden block is \(m_b=5 \mathrm{~kg}\).
The mass of the iron cylinder is \(m_c=25 \mathrm{~kg}\).
The total mass of the system is the sum of their masses:
\(
M=m_b+m_c=5 \mathrm{~kg}+25 \mathrm{~kg}=30 \mathrm{~kg}
\)
Step 2: Apply Newton’s Second Law
According to Newton’s Second Law, the net force acting on the system is equal to its total mass multiplied by its acceleration ( \(F_{\text {net }}=m a\) ).
The forces acting on the system are:
1. The total weight of the system acting downwards, \(W=m g\).
2. The normal force from the floor acting upwards, \(N\).
Since the system is accelerating downwards with an acceleration \(a=0.1 \mathrm{~m} / \mathrm{s}^2\), we can write the equation of motion as:
\(
F_{n e t}=W-N=M a
\)
Using \(g=9.8 \mathrm{~m} / \mathrm{s}^2\) :
\(
(30 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)-N=(30 \mathrm{~kg})\left(0.1 \mathrm{~m} / \mathrm{s}^2\right)
\)
\(
\begin{gathered}
N=294 \mathrm{~N}-3 \mathrm{~N} \\
N=291 \mathrm{~N}
\end{gathered}
\)

Hint

(a)

Step 1: Analyze the forces acting on the brick
The brick is on the verge of sliding down the inclined surface. At this point, the component of gravity pulling the brick down the slope is exactly balanced by the maximum force of static friction.
The force of gravity acting on the brick is \(\boldsymbol{F}_{\boldsymbol{g}}=\boldsymbol{m}\).
The component of gravity parallel to the inclined surface is \(F_{\|}=m g \sin \theta\).
The component of gravity perpendicular to the inclined surface is \(F_{\perp}=m g \cos \theta\).
The normal force is equal in magnitude to the perpendicular component of gravity, so \(N=m g \cos \theta\).
The maximum force of static friction is given by \(f_{s, \max }=\mu_s N=\mu_s m g \cos \theta\).
Step 2: Set up the equation for the moment the brick begins to slide
When the brick begins to slide, the parallel component of gravity is equal to the maximum static friction force.
\(
\begin{gathered}
F_{\|}=f_{s, \max } \\
m g \sin \theta=\mu_s m g \cos \theta
\end{gathered}
\)
Step 3: Solve for the coefficient of static friction
We can cancel out \(m g\) from both sides of the equation and then solve for \(\mu_s\) :
\(
\begin{gathered}
\sin \theta=\mu_s \cos \theta \\
\mu_s=\frac{\sin \theta}{\cos \theta}=\tan \theta
\end{gathered}
\)
Given that the angle of inclination \(\theta\) is \(45^{\circ}\), we can find the value of \(\mu_s\) :
\(
\begin{gathered}
\mu_s=\tan \left(45^{\circ}\right) \\
\mu_s=1
\end{gathered}
\)

Hint

(b) Force-time relationship: The force applied to the block increases linearly with time, meaning \(F=k t\) where \(k\) is a constant.
Newton’s Second Law: \(\boldsymbol{F}=m a\), which relates force to mass and acceleration.
Result: By solving for \(a\) in \(F=m a\), we get \(a=\frac{F}{m}=\frac{k t}{m}\), which shows that acceleration is directly proportional to time (will vary linearly with time).

Hint

(a) Step 1: Calculate the net force
To find the net force ( \(\vec{F}_{\text {net }}\) ) acting on the body, add the two force vectors \(\vec{F}_1\) and \(\vec{F}_2\).
\(
\begin{gathered}
\vec{F}_{n e t}=\vec{F}_1+\vec{F}_2 \\
\vec{F}_{n e t}=(5 \hat{i}+8 \hat{j}+7 \hat{k})+(3 \hat{i}-4 \hat{j}-3 \hat{k}) \\
\vec{F}_{n e t}=(5+3) \hat{i}+(8-4) \hat{j}+(7-3) \hat{k} \\
\vec{F}_{n e t}=8 \hat{i}+4 \hat{j}+4 \hat{k}
\end{gathered}
\)
Step 2: Calculate the acceleration
Use Newton’s Second Law of Motion, \(\vec{F}_{\text {net }}=m \vec{a}\), to find the acceleration ( \(\vec{a}\) ) by dividing the net force by the mass ( \(m=4 \mathrm{~kg}\) ).
\(
\begin{gathered}
\vec{a}=\frac{\vec{F}_{\text {net }}}{m} \\
\vec{a}=\frac{8 \hat{i}+4 \hat{j}+4 \hat{k}}{4} \\
\vec{a}=\frac{8}{4} \hat{i}+\frac{4}{4} \hat{j}+\frac{4}{4} \hat{k} \\
\vec{a}=2 \hat{i}+1 \hat{j}+1 \hat{k} \\
\vec{a}=2 \hat{i}+\hat{j}+\hat{k}
\end{gathered}
\)

Hint

(a) Step 1: Identify given values and convert to SI units
The mass of the ball needs to be converted from grams to kilograms to be in SI units.
Mass, \(m=120 \mathrm{~g}=0.12 \mathrm{~kg}\)
Initial velocity, \(u=25 \mathrm{~m} / \mathrm{s}\)
Final velocity, \(v=0 \mathrm{~m} / \mathrm{s}\) (since the ball is caught and comes to a stop)
Time taken, \(\Delta t=0.1 \mathrm{~s}\)
Step 2: Calculate the change in momentum
The force is related to the change in momentum over time, as described by Newton’s Second Law: \(F=\frac{\Delta p}{\Delta t}\), where \(\Delta p\) is the change in momentum \((p=m v)\).
First, we calculate the initial and final momentum.
Initial momentum: \(p_{\text {initial }}=m u=(0.12 \mathrm{~kg})(25 \mathrm{~m} / \mathrm{s})=3 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}\)
Final momentum: \(p_{\text {final }}=m v=(0.12 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})=0 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}\)
Next, we find the change in momentum, \(\Delta p\) :
\(\Delta p=p_{\text {final }}-p_{\text {initial }}=0-3=-3 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}\)
Step 3: Calculate the magnitude of the force
Now we use the formula for force:
\(F=\frac{\Delta p}{\Delta t}=\frac{-3 \mathrm{~kg} \cdot \mathrm{~m} / \mathrm{s}}{0.1 \mathrm{~s}}=-30 \mathrm{~N}\)
The question asks for the magnitude of the force, so we take the absolute value.
\(|F|=|-30|=30 \mathrm{~N}\) (Here, negative sign shows the direction of force is opposite to motion.)
The magnitude of the force exerted by the ball on the hand of the player is 30 N.

Hint

(d) Step 1: Identify given values and forces
Mass of the trolley \(\left(m_1\right)=20 \mathrm{~kg}\)
Mass of the hanging block \(\left(m_2\right)=6 \mathrm{~kg}\)
Coefficient of kinetic friction \(\left(\boldsymbol{\mu}_{\boldsymbol{k}}\right)=\mathbf{0 . 0 4}\)
Acceleration due to gravity \((g)=10 \mathrm{~ms}^{-2}\)
The forces acting on the system are tension in the string ( \(\boldsymbol{T}\) ), the gravitational force on the block \(\left(m_2 g\right)\), and the kinetic friction force on the trolley \(\left(f_k\right)\)
Step 2: Apply Newton’s Second Law
We can set up two equations of motion for the system, one for the trolley and one for the block, since they move together with the same acceleration (\(a\)).
1. For the trolley: The net horizontal force is the tension minus the friction. The friction force is \(f_k=\mu_k N\), where the normal force \(N\) equals the trolley’s weight ( \(m_1 g\) ).
\(
T-f_k=m_1 a \quad \Longrightarrow \quad T-\mu_k m_1 g=m_1 a
\)
Plugging in the values:
\(
\begin{aligned}
& T-(0.04)(20)(10)=20 a \\
& T-8=20 a \quad \dots(1)
\end{aligned}
\)
2. For the block: The net vertical force is the gravitational force on the block minus the tension.
\(
m_2 g-T=m_2 a
\)
Plugging in the values:
\(
\begin{aligned}
& (6)(10)-T=6 a \\
& 60-T=6 a \quad \dots(2)
\end{aligned}
\)
Step 3: Solve the system of equations
Now, we can solve the two equations simultaneously for the acceleration \(a\). From Equation 2, we can express the tension \(T\) as \(T=60-6 a\). Substituting this into Equation 1:
\(
\begin{gathered}
(60-6 a)-8=20 a \\
52=26 a \\
a=\frac{52}{26} \\
a=2
\end{gathered}
\)
The acceleration of the system is \(2 \mathrm{~ms}^{-2}\).

Hint

(a)

Case-I:
\(
\begin{aligned}
& f_{\max }=\mu N \\
& f_{\max }=2.5 \sqrt{3} \\
& f_{\max }=\frac{5 \sqrt{3}}{2} \\
& F_2=25-\frac{5 \sqrt{3}}{2}
\end{aligned}
\)
Case-II:
\(
\begin{aligned}
& F_1=25+\frac{5 \sqrt{3}}{2} \\
& \left|F_1\right|-\left|F_2\right|=\frac{10 \sqrt{3}}{2} \\
& =5 \sqrt{3}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&T-m_1 g=m_1 a \dots(i)\\
&m_2 g-T=m_2 a \dots(ii)\\
&\text { After solving Eqs. (i) and (ii), we get }
\end{aligned}
\)
The acceleration of the system is given by:
\(
a=\frac{m_2-m_1}{m_1+m_2} g
\)
\(
\begin{aligned}
& a=\frac{\left(m_1-m_2\right) g}{\left(m_1+m_2\right)}=\frac{g}{8} \\
& 8 m_1-8 m_2=m_1+m_2 \\
& 7 m_1=9 m_2 \\
& \frac{m_1}{m_2}=\frac{9}{7}
\end{aligned}
\)

Hint

(a)

For \(M\) block:
\(
\begin{aligned}
& 10 \mathrm{~g} \sin 53^{\circ}-\mu(10 \mathrm{~g}) \cos 53^{\circ}-\mathrm{T}=10 \times 2 \\
& \mathrm{~T}=80-15-20 \\
& \mathrm{~T}=45 \mathrm{~N}
\end{aligned}
\)
For \(m\) block:
\(
\begin{aligned}
& \mathrm{T}-\mathrm{mg} \sin 37^{\circ}-\mu \mathrm{mg} \cos 37^{\circ}=\mathrm{m} \times 2 \\
& 45=10 \mathrm{~m} \\
& \mathrm{~m}=4.5 \mathrm{~kg}
\end{aligned}
\)

Hint

(d) Step 1: Find the angle of the tangent to the curve
The condition for a block to be on the verge of slipping on an inclined surface is that the tangent of the angle of inclination, \(\theta\), is equal to the coefficient of static friction, \(\mu\). That is, \(\tan \theta=\mu\). The angle of inclination at any point on the curve is given by the slope of the tangent line, which is the first derivative of the function describing the curve.
The equation of the curve is given by \(y=\frac{x^2}{4}\).
First, find the derivative of the function with respect to \(x\) :
\(
\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x^2}{4}\right)=\frac{2 x}{4}=\frac{x}{2}
\)
The slope of the tangent is given by \(m_{\text {slope }}=\frac{d y}{d x}=\tan \theta\).
Step 2: Relate the slope to the coefficient of friction
The maximum angle without slipping occurs when the tangent of the angle of the surface is equal to the coefficient of static friction.
\(
\tan \theta=\mu
\)
Given the coefficient of friction \(\boldsymbol{\mu} \boldsymbol{=} \mathbf{0 . 5}\), we can set the slope equal to this value:
\(
\frac{x}{2}=0.5
\)
Step 3: Solve for the coordinate \(\boldsymbol{x}\)
Solve the equation from Step 2 for \(x\) :
\(
x=2 \times 0.5=1
\)
This is the \(x\)-coordinate where the block is on the verge of slipping.
Step 4: Calculate the maximum height \(y\)
Substitute the value of \(x=1\) back into the original equation for the curve to find the corresponding height \(y\) :
\(
y=\frac{x^2}{4}=\frac{(1)^2}{4}=\frac{1}{4} \mathrm{~m}
\)
The maximum height at which the block can be placed without slipping is \(\frac{1}{4} \mathrm{~m}\)

Hint

(a)

Step 1: Calculate the total mass of all three blocks
\(
m_{\text {total }}=m_A+m_B+m_C=5+3+2=10 \mathrm{~kg}
\)
Step 2: Calculate the acceleration of the system
Using Newton’s second law, \(F=m_{\text {total }} \times a\)
\(
a=\frac{F}{m_{\text {total }}}=\frac{80}{10}=8 \mathrm{~m} / \mathrm{s}^2
\)
\(
\mathrm{T}_1=5 \times 8=40
\)
\(
\mathrm{T}_2-\mathrm{T}_1=3 \times 8 \Rightarrow \mathrm{~T}_2=64
\)

Hint

(d) 

\(
\begin{aligned}
& 40-2 T=4 a \\
& T-10=4 a \Rightarrow 20=12 a \\
& \Rightarrow a=\frac{5}{3} \Rightarrow 2 a=\frac{g}{3}
\end{aligned}
\)

Hint

(a) Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.
This statement is incorrect. The limiting force of static friction is given by the formula \(F_{s, \max }=\mu_s N\), where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.
The value of \(\boldsymbol{\mu}_{\boldsymbol{s}}\) depends on the nature and materials of the surfaces in contact.
The limiting force of static friction is, to a good approximation, independent of the apparent area of contact (provided the normal force remains the same).

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.
This statement is correct. The force of kinetic friction is given by the formula \(F_k=\mu_k N\), where \(\mu_k\) is the coefficient of kinetic friction.
Similar to static friction, the kinetic friction force is generally independent of the area of contact.
The value of \(\mu_k\) depends on the materials and nature of the surfaces in contact. \(\theta\)
Therefore, Statement I is false, and Statement II is true.

Hint

(b) Step 1: Differentiate the position vector to find the velocity vector
The position vector is given by \(\vec{r}=\left(10 t \hat{i}+15 t^2 \hat{j}+7 \hat{k}\right)\). To find the velocity vector \(\vec{v}\), we differentiate the position vector with respect to time \(t\)
\(
\vec{v}=\frac{d \vec{r}}{d t}=\frac{d}{d t}\left(10 t \hat{i}+15 t^2 \hat{j}+7 \hat{k}\right)=10 \hat{i}+30 t \hat{j}
\)
Step 2: Differentiate the velocity vector to find the acceleration vector
To find the acceleration vector \(\vec{a}\), we differentiate the velocity vector with respect to time \(t\).
\(
\vec{a}=\frac{d \vec{v}}{d t}=\frac{d}{d t}(10 \hat{i}+30 t \hat{j})=0 \hat{i}+30 \hat{j}
\)
The acceleration vector is therefore \(\vec{a}=30 \hat{j} \mathrm{~m} / \mathrm{s}^2\).
Step 3: Determine the direction of the net force
According to Newton’s Second Law of Motion, the net force \(\vec{F}_{\text {net }}\) is directly proportional to the acceleration \(\vec{a}\left(\vec{F}_{\text {net }}=m \vec{a}\right)\). This means the direction of the net force is the same as the direction of the acceleration.
The acceleration vector \(\vec{a}\) is \(30 \hat{j}\), which has a magnitude only in the positive \(y\) direction.

Hint

(c) Step 1: Calculate the resultant of forces \(\boldsymbol{F}_{\mathbf{2}}\) and \(\boldsymbol{F}_{\mathbf{3}}\)
The forces \(F_2\) and \(F_3\) are applied perpendicularly. We can find the magnitude of their resultant force, \(\mathrm{F}_{23}\), using the Pythagorean theorem.
\(
F_{23}=\sqrt{F_2^2+F_3^2}
\)
Given \(F_2=8 \mathrm{~N}\) and \(F_3=6 \mathrm{~N}\) :
\(
F_{23}=\sqrt{(8 \mathrm{~N})^2+(6 \mathrm{~N})^2}=\sqrt{64 \mathrm{~N}^2+36 \mathrm{~N}^2}=\sqrt{100 \mathrm{~N}^2}=10 \mathrm{~N}
\)
Step 2: Determine the net force after \(F_1\) is removed
Initially, with all three forces acting, the particle is at rest. This means the net force is zero. The resultant of \(\mathbf{F}_{\mathbf{2}}\) and \(\mathbf{F}_{\mathbf{3}}\) is \(10 \mathbf{N}\), and since the particle is at rest, the third force \(\mathbf{F}_{\mathbf{1}}\) must have an equal magnitude ( 10 N ) and act in the opposite direction to the resultant of \(\mathrm{F}_2\) and \(\mathrm{F}_3\).
When force \(\mathbf{F}_{\mathbf{1}}\) is removed, the remaining net force on the particle is the resultant of \(\mathbf{F}_{\mathbf{2}}\) and \(\mathbf{F}_{\mathbf{3}}\), which we calculated in Step 1.
\(
F_{n e t}=F_{23}=10 \mathrm{~N}
\)
Step 3: Calculate the acceleration of the particle
Using Newton’s second law of motion, the acceleration (\(a\)) of the particle can be found by dividing the net force ( \(F_{n e t}\) ) by the mass ( \(m\) ).
\(
a=\frac{F_{\text {net }}}{m}
\)
Given \(F_{\text {net }}=10 \mathrm{~N}\) and \(m=5 \mathrm{~kg}\) :
\(
a=\frac{10 \mathrm{~N}}{5 \mathrm{~kg}}=2 \mathrm{~ms}^{-2}
\)

Hint

(c) Step 1: Find the acceleration of the body
First, convert the mass from grams to kilograms:
\(
m=500 \mathrm{~g}=0.5 \mathrm{~kg}
\)
The velocity of the body is given by the relation \(v=10 \sqrt{x}\).
To find the acceleration, we need to differentiate the velocity with respect to time, which can be done using the chain rule since velocity is a function of displacement \(x\) :
\(
a=\frac{d v}{d t}=\frac{d v}{d x} \cdot \frac{d x}{d t}
\)
Since \(\frac{d x}{d t}=v\), the acceleration can be written as:
\(
a=v \frac{d v}{d x}
\)
Now, we find \(\frac{d v}{d x}\) :
\(
\frac{d v}{d x}=\frac{d}{d x}\left(10 x^{1 / 2}\right)=10 \cdot \frac{1}{2} x^{1 / 2-1}=5 x^{-1 / 2}=\frac{5}{\sqrt{x}}
\)
Substitute this back into the acceleration equation:
\(
a=v \frac{d v}{d x}=(10 \sqrt{x})\left(\frac{5}{\sqrt{x}}\right)=50
\)
The acceleration of the body is constant and equal to \(50 \mathrm{~m} / \mathrm{s}^2\).
Step 2: Calculate the force using Newton’s Second Law
According to Newton’s Second Law of Motion, the force acting on the body is given by:
\(
F=m a
\)
Substitute the values of mass and acceleration:
\(
F=(0.5 \mathrm{~kg})\left(50 \mathrm{~m} / \mathrm{s}^2\right)=25 \mathrm{~N}
\)
The force acting on the body is 25 N.

Hint

(d) Step 1: Find the acceleration
The acceleration \(\vec{a}\) is the time derivative of the velocity \(\vec{v}\).
Given the velocity \(\vec{v}(t)=\left(2 t \hat{i}+3 t^2 \hat{j}\right) \mathrm{ms}^{-1}\), the acceleration is:
\(
\vec{a}(t)=\frac{d \vec{v}}{d t}=\frac{d}{d t}\left(2 t \hat{i}+3 t^2 \hat{j}\right)=(2 \hat{i}+6 t \hat{j}) \mathrm{ms}^{-2}
\)
Step 2: Calculate acceleration at \(t=1 \mathrm{~s}\)
Substitute \(t=1 \mathrm{~s}\) into the acceleration equation:
\(
\vec{a}(1)=(2 \hat{i}+6(1) \hat{j})=(2 \hat{i}+6 \hat{j}) \mathrm{ms}^{-2}
\)
Step 3: Use Newton’s second law to find the force
According to Newton’s second law, the force \(\vec{F}\) is the product of mass \(m\) and acceleration \(\vec{a}\). The mass is given as 500 g, which is 0.5 kg.
\(
\begin{gathered}
\vec{F}=m \vec{a} \\
\vec{F}=0.5 \mathrm{~kg} \times(2 \hat{i}+6 \hat{j}) \mathrm{ms}^{-2}=(1 \hat{i}+3 \hat{j}) \mathrm{N}
\end{gathered}
\)
Step 4: Compare the calculated force with the given force
The calculated force is \((1 \hat{i}+3 \hat{j}) \mathrm{N}\). The problem states that the force is \((\hat{i}+x \hat{j}) \mathrm{N}\). By comparing the components of the two vectors, we find:
\(
\begin{aligned}
& \hat{i}: 1=1 \\
& \hat{j}: 3=x
\end{aligned}
\)
Thus, the value of \(x\) is 3.

Hint

(a)

\(
\begin{aligned}
&\begin{gathered}
\mathrm{mg}-\mathrm{F} \sin 30^{\circ}=\mathrm{N} \\
\mathrm{f}=\mu(\mathrm{N}) \quad=\mu \mathrm{mg}-\mathrm{F} \sin 30^{\circ} \\
\mu=0.25 \text { given }
\end{gathered}\\
&\text { To start motion, } \mathrm{F} \cos 30^{\circ} \geq \mu \mathrm{N}\\
&\begin{gathered}
\mathrm{F} \cos 30^{\circ} \geq \mu \mathrm{mg}-\mathrm{F} \sin 30^{\circ} \\
\mathrm{F} \frac{\sqrt{3}}{2} \geq \frac{1}{4} 100-\mathrm{F} \frac{1}{2}, \quad \mathrm{~F} \frac{\sqrt{3}}{2}+\mathrm{F} \frac{1}{8} \geq \frac{100}{4} \\
\mathrm{~F} \frac{4 \sqrt{3}+1}{8} \geq 25, \mathrm{~F}=\frac{25 \times 8}{4 \sqrt{3}+1}=25.2 \mathrm{~N}
\end{gathered}
\end{aligned}
\)

Hint

(b) As we know that impulse is given by
\(
I=\Delta P=F \times \Delta t
\)
or \(I=\) Area of \(F-t\) graph
(a) \(\frac{1}{2} \times 1 \times 0.5=\frac{1}{4} \mathrm{~N} . \mathrm{s}\)
(b) \(0.5 \times 2=1 \mathrm{~N} . \mathrm{s}\) (maximum)
(c) \(\frac{1}{2} \times 1 \times 0.75=\frac{3}{8} \mathrm{~N} . \mathrm{s}\)
(d) \(\frac{1}{2} \times 2 \times 0.5=\frac{1}{2} \mathrm{~N} . \mathrm{s}\)

Hint

(d) Step 1: Calculate the acceleration of the block
First, we need to find the acceleration (a) of the block using the given kinematic information. The block starts from rest, so its initial velocity \((u)\) is 0. We can use the following kinematic equation:
\(
s=u t+\frac{1}{2} a t^2
\)
Plugging in the given values for distance ( \(s=50 \mathrm{~m}\) ) and time ( \(t=10 \mathrm{~s}\) ), we get:
\(
\begin{gathered}
50=(0)(10)+\frac{1}{2} a(10)^2 \\
50=\frac{1}{2} a(100) \\
50=50 a \\
a=\frac{50}{50}=1 \mathrm{~ms}^{-2}
\end{gathered}
\)
Step 2: Apply Newton’s Second Law to find the coefficient of kinetic friction
Next, we apply Newton’s Second Law of Motion ( \(\Sigma F=m a\) ) to the block in the horizontal direction. The net force ( \(\boldsymbol{\Sigma} \boldsymbol{F}\) ) acting on the block is the applied force ( \(\boldsymbol{F}_{\text {app }}\) ) minus the force of kinetic friction \(\left(\boldsymbol{F}_{\boldsymbol{k}}\right)\).
\(
F_{n e t}=F_{a p p}-F_k=m a
\)
Now, we can rearrange the equation to solve for \(\mu_k\) :
\(
\begin{gathered}
F_{a p p}-m a=\mu_k m g \\
\mu_k=\frac{F_{a p p}-m a}{m g}
\end{gathered}
\)
Finally, we plug in the known values: mass ( \(m=5 \mathrm{~kg}\) ), applied force ( \(F_{a p p}=30 \mathrm{~N}\) ), acceleration ( \(a=1 \mathrm{~ms}^{-2}\) ), and gravity ( \(g=10 \mathrm{~ms}^{-2}\) ).
\(
\begin{gathered}
\mu_k=\frac{30-(5)(1)}{(5)(10)} \\
\mu_k=\frac{30-5}{50} \\
\mu_k=\frac{25}{50} \\
\mu_k=0.5
\end{gathered}
\)
The coefficient of kinetic friction is \(\mathbf{0 . 5}\).

Hint

(d) Step 1: Calculate the acceleration
To find the acceleration of the body, we can use the kinematic equation \(v=u+a t\). The body starts with an initial speed of \(u=20 \mathrm{~m} / \mathrm{s}\) and comes to a stop ( \(v=0\) ) in \(t=5 \mathrm{~s}\).
\(
\begin{gathered}
v=u+a t \\
0=20+a(5) \\
5 a=-20 \\
a=-4 \mathrm{~m} / \mathrm{s}^2
\end{gathered}
\)
The magnitude of the deceleration is \(4 \mathrm{~m} / \mathrm{s}^2\).
Step 2: Find the force of friction
According to Newton’s second law, the net force on the body is equal to its mass times its acceleration ( \(F=m a\) ). In this case, the only horizontal force acting on the body is the force of friction, which causes the deceleration. 0
\(
\begin{gathered}
F_f=m a \\
F_f=(10 \mathrm{~kg})\left(4 \mathrm{~m} / \mathrm{s}^2\right) \\
F_f=40 \mathrm{~N}
\end{gathered}
\)
Step 3: Relate friction to the coefficient of friction
The force of friction ( \(F_f\) ) is related to the coefficient of friction ( \(\mu\) ) and the normal force ( \(N)\) by the formula \(F_f=\mu N\). Since the body is on a horizontal surface, the normal force is equal to the gravitational force ( \(N=m g\) ).
\(
\begin{gathered}
N=m g \\
N=(10 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^2\right) \\
N=100 \mathrm{~N}
\end{gathered}
\)
Now we can solve for the coefficient of friction ( \(\mu\) ).
\(
\begin{gathered}
F_f=\mu N \\
40=\mu(100) \\
\mu=\frac{40}{100} \\
\mu=0.4
\end{gathered}
\)

Hint

(a)

\(
\begin{aligned}
\mathrm{N} & =\mathrm{mg}+\mathrm{F} \sin 30^{\circ} \\
& =700+200 \times \frac{1}{2}=800 \text { newton }
\end{aligned}
\)

Hint

(d)

\(
\frac{\sqrt{3} \mathrm{~g}}{\sin 120^{\circ}}=\frac{\mathrm{T}}{\sin 90^{\circ}} \Rightarrow \mathrm{T}=\frac{\sqrt{3} \mathrm{~g}}{\sqrt{3} / 2}=20 N
\)

Hint

(c) Step 1: Calculate the velocity of the body after the force ceases
The force ceases and the body moves at a constant velocity for the next 10 s, covering a distance of 50 m . The constant velocity can be calculated using the formula \(v=d / t\).
\(
v=\frac{50 \mathrm{~m}}{10 \mathrm{~s}}=5 \mathrm{~m} / \mathrm{s}
\)
Step 2: Calculate the acceleration of the body
During the first 20 s , the body starts from rest ( \(v_0=0\) ) and accelerates to a final velocity of \(5 \mathrm{~m} / \mathrm{s}\) (from Step 1). The acceleration can be calculated using the formula \(a=\frac{v_f-v_0}{t}\)
\(
a=\frac{5 \mathrm{~m} / \mathrm{s}-0 \mathrm{~m} / \mathrm{s}}{20 \mathrm{~s}}=0.25 \mathrm{~m} / \mathrm{s}^2
\)
Step 3: Calculate the value of the force
The force acting on the body can be found using Newton’s second law of motion, \(F=m a\), where \(m\) is the mass ( 20 kg ) and \(a\) is the acceleration ( \(0.25 \mathrm{~m} / \mathrm{s}^2\) ) calculated in the previous step.
\(
F=(20 \mathrm{~kg})\left(0.25 \mathrm{~m} / \mathrm{s}^2\right)=5 \mathrm{~N}
\)

Hint

(a) Step 1: Smooth Case Analysis
The acceleration of an object on a frictionless inclined plane is given by \(\boldsymbol{a}=\boldsymbol{g} \boldsymbol{\operatorname { s i n }} \boldsymbol{\theta}\). For \(\theta=45^{\circ}\), the acceleration is \(a=g \sin 45^{\circ}=\frac{g}{\sqrt{2}}\). Using the kinematic equation \(L=\frac{1}{2} a t^2\), where \(L\) is the distance traveled, the time \(t_1\) is found by rearranging the equation:
\(
t_1=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 L}{g / \sqrt{2}}}=\sqrt{\frac{2 \sqrt{2} L}{g}}
\)
Step 2: Rough Case Analysis
The acceleration of an object on a rough inclined plane is given by \(a=g \sin \theta-\mu g \cos \theta\). For \(\theta=45^{\circ}\), the acceleration is:
\(
a=g \sin 45^{\circ}-\mu g \cos 45^{\circ}=\frac{g}{\sqrt{2}}-\mu \frac{g}{\sqrt{2}}=\frac{g}{\sqrt{2}}(1-\mu)
\)
Using the same kinematic equation as in the smooth case, the time \(t_2\) is:
\(
t_2=\sqrt{\frac{2 L}{a}}=\sqrt{\frac{2 L}{\frac{g}{\sqrt{2}}(1-\mu)}}=\sqrt{\frac{2 \sqrt{2} L}{g(1-\mu)}}
\)
Step 3: Deriving the Coefficient of Friction \(\boldsymbol{\mu}\)
The problem states that \(t_1=\frac{t_2}{n}\). By substituting the expressions for \(t_1\) and \(t_2\) from the previous steps, we get:
\(
\sqrt{\frac{2 \sqrt{2} L}{g}}=\frac{1}{n} \sqrt{\frac{2 \sqrt{2} L}{g(1-\mu)}}
\)
To solve for \(\mu\), first square both sides of the equation:
\(
\frac{2 \sqrt{2} L}{g}=\frac{1}{n^2} \frac{2 \sqrt{2} L}{g(1-\mu)}
\)
The term \(\frac{2 \sqrt{2} L}{g}\) appears on both sides, so it can be canceled out, leaving:
\(
1=\frac{1}{n^2(1-\mu)}
\)
Rearranging the equation to solve for \(\mu\) :
\(
\begin{aligned}
& n^2(1-\mu)=1 \\
& 1-\mu=\frac{1}{n^2} \\
& \mu=1-\frac{1}{n^2}
\end{aligned}
\)
The coefficient of kinetic friction is \(\boldsymbol{\mu}=\mathbf{1}-\frac{\mathbf{1}}{\mathbf{n}^{\mathbf{2}}}\).

Hint

(c)

\(
\begin{aligned}
&\mathrm{Mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{ma}\\
&\frac{g}{2}-\frac{\sqrt{3}}{2} \cdot \mu g=\frac{g}{4}\\
&\begin{aligned}
& \frac{\sqrt{3}}{2} \mu=\frac{1}{4} \\
& \mu=\frac{1}{2 \sqrt{3}}
\end{aligned}
\end{aligned}
\)

Hint

(a)

From Fig(a)

\(
\begin{aligned}
& \mathrm{F}_1=\mathrm{mg} \sin 45^{\circ}+\mathrm{f}=\mathrm{mg} \sin 45^{\circ}+\mu \mathrm{N} \\
& \mathrm{~F}_1=\frac{\mathrm{mg}}{\sqrt{2}}+\mu \mathrm{mg} \cos 45^{\circ} \\
& \mathrm{F}_1=\frac{\mathrm{mg}}{\sqrt{2}}(1+\mu)
\end{aligned}
\)

From Fig(b)

\(
\begin{aligned}
& F_2=m g \sin 45^{\circ}-f=m g \sin 45^{\circ}-\mu N \\
& =\frac{m g}{\sqrt{2}}(1-\mu) \\
& F_1=2 F_2 \\
& \frac{m g}{\sqrt{2}}(1+\mu)=2 \frac{m g}{\sqrt{2}}(1-\mu) \\
& 1+\mu=2-2 \mu \\
& \mu=1 / 3=0.33
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& 4 \mathrm{~g} \frac{\sqrt{3}}{2}-\mathrm{T}=4 \mathrm{a} \\
& \mathrm{~T}-\frac{\mathrm{g}}{2}=1 \mathrm{a} \\
& 2 \sqrt{3} \mathrm{~g}-\mathrm{T}=4\left(\mathrm{~T}-\frac{\mathrm{g}}{2}\right) \Rightarrow 5 \mathrm{~T}=(2 \sqrt{3}+2) \mathrm{g} \\
& \mathrm{~T}=\frac{10}{5}(2 \sqrt{3}+2) \Rightarrow \mathrm{T}=4(\sqrt{3}+1) \mathrm{N}
\end{aligned}
\)

Hint

(d)

Statement I is true because an elevator moving at a uniform speed (zero acceleration) has a net force of zero, meaning its weight (downward force) is balanced by the tension of its cable (upward force). This condition applies whether the uniform speed is upward or downward.

Statement II is false because when an elevator goes down with increasing speed (accelerating downward), the force exerted by the floor on the person (the normal force, or apparent weight) is less than their actual weight. This is because the net force must be downward to cause the downward acceleration. The equation for the normal force ( \(N\) ) is \(N=m(g-a)\), where \(m\) is the person’s mass, \(g\) is the acceleration due to gravity, and \(a\) is the magnitude of the downward acceleration. Since \(\boldsymbol{a}\) is a positive value, \(\boldsymbol{N}\) is less than the actual weight \(m g\). \(\theta\)

Therefore, the correct answer is that Statement I is true but Statement II is false.

Hint

(b)

For a system at rest
\(
\mathrm{m}_1 \mathrm{~g} \cos \theta=\mathrm{N} \dots(1)
\)
( \(\mathrm{N} \rightarrow\) Force by inclined plane on the block)
\(
\begin{aligned}
& \mathrm{T}=\mathrm{m}_2 \mathrm{~g}=30 \mathrm{~N} \dots(2) \\
& \mathrm{m}_1 \mathrm{~g} \sin \theta=\mathrm{T} \\
& 50 \sin \theta=30 \\
& \sin \theta=\frac{3}{5} \\
& \theta=37^{\circ} \\
& \mathrm{N}=\mathrm{m}_1 \mathrm{~g} \cos \theta \\
& =50 \times \cos 37^{\circ} \\
& \mathrm{N}=40 \mathrm{~N}
\end{aligned}
\)
The force exerted by the inclined plane on the body of mass \(m_1\) will be 40 N.

Hint

(d)

Let the mass of complete chain be ‘ \(m\) ‘and mass of chain of length’ \(\ell\) ‘ and ‘ \(\mathrm{L}-\ell^{\prime}\) be \(\mathrm{m}_1\) and \(\mathrm{m}_2\) respectively
\(
\begin{aligned}
& \mathrm{m}_1+\mathrm{m}_2=\mathrm{m} \\
& \mathrm{~m}_1=\left(\frac{\mathrm{m}}{\mathrm{~L}}\right) \ell \\
& \mathrm{m}_2=\left(\frac{\mathrm{m}}{\mathrm{~L}}\right)(\mathrm{L}-\ell) \\
& \left(\mathrm{m}_2-\mathrm{m}_1\right)_{\mathrm{g}}=\left(\mathrm{m}_2+\mathrm{m}_1\right) \mathrm{a} \\
& \mathrm{a}=\left(\frac{\mathrm{m}_2-\mathrm{m}_1}{\mathrm{~m}_2+\mathrm{m}_1}\right) \mathrm{g} \\
& \frac{\mathrm{~g}}{2}=\frac{\left(\frac{\mathrm{m}}{\mathrm{~L}}\right)(\mathrm{L}-2 \ell)}{\mathrm{m}} \mathrm{~g}
\end{aligned}
\)
\(
\begin{aligned}
& \frac{1}{2}=\frac{\mathrm{L}-2 \ell}{\mathrm{~L}} \\
& \mathrm{~L}=2 \mathrm{~L}-4 \ell \\
& 4 \ell=\mathrm{L} \\
& \ell=\frac{\mathrm{L}}{4} \\
& \therefore \mathrm{x}=4
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \mathrm{N}=\mathrm{Mg} \cos \theta \\
& \mathrm{f}=\mathrm{Mg} \sin \theta \\
& \mathrm{R}=\sqrt{\mathrm{N}^2+\mathrm{f}^2} \\
& \mathrm{R}=\mathrm{Mg}
\end{aligned}
\)

Explanation:

The forces are:
Weight (Mg): Acts vertically downward.
Normal force ( \(\boldsymbol{N}\) ): Acts perpendicular to the incline, away from the surface.
Kinetic friction force ( \(f_k\) ): Acts parallel to the incline, opposing the motion.
We need to resolve the weight vector into components along our chosen coordinate system.
The component parallel to the incline is \(M g \sin \theta\).
The component perpendicular to the incline is \(M g \cos \theta . \theta\)
Step 2: Apply Newton’s first law
Since the block slides with constant velocity, its acceleration is zero. According to Newton’s first law, the net force on the block must be zero.
Along the \(\mathbf{y}\)-axis (perpendicular to the incline): The normal force and the perpendicular component of gravity are balanced.
\(
\begin{aligned}
& \sum_N F_y=N-M g \cos \theta=0 \\
& N=M g \cos \theta
\end{aligned}
\)
Along the \(\mathbf{x}\)-axis (parallel to the incline): The kinetic friction force and the parallel component of gravity are balanced.
\(
\sum F_x=M g \sin \theta-f_k=0
\)
\(
f_k=M g \sin \theta
\)
Step 3: Calculate the magnitude of the contact force
The contact force is the resultant vector of the normal force and the friction force. Since the normal force and the friction force are perpendicular to each other, the magnitude of the contact force ( \(\boldsymbol{F}_{\text {contact }}\) ) can be found using the Pythagorean theorem.
\(
F_{c o n t a c t}=\sqrt{N^2+f_k^2}
\)
Substitute the values for \(N\) and \(f_k\) we found in Step 2:
\(
\begin{aligned}
F_{\text {contact }} & =\sqrt{(M g \cos \theta)^2+(M g \sin \theta)^2} \\
F_{\text {contact }} & =\sqrt{M^2 g^2 \cos ^2 \theta+M^2 g^2 \sin ^2 \theta}
\end{aligned}
\)
Factor out the common term \(\boldsymbol{M}^2 \boldsymbol{g}^2\) :
\(
F_{\text {contact }}=\sqrt{M^2 g^2\left(\cos ^2 \theta+\sin ^2 \theta\right)}
\)
Using the trigonometric identity \(\sin ^2 \theta+\cos ^2 \theta=1\) :
\(
\begin{gathered}
F_{\text {contact }}=\sqrt{M^2 g^2(1)} \\
F_{\text {contact }}=M g
\end{gathered}
\)

Hint

(c)

\(
\begin{aligned}
& \mathrm{a}=\mathrm{g} \sin \theta \\
& \ell=\frac{1}{2} \mathrm{~g} \sin 30^{\circ}(2)^2 \\
& \ell=\frac{1}{2} \mathrm{~g} \sin 45^{\circ} \mathrm{t}^2 \\
& \left(\frac{1}{2}\right)(4)=\frac{1}{\sqrt{2}} \mathrm{t}^2 \Rightarrow \mathrm{t}=\sqrt{2 \sqrt{2}}=1.68
\end{aligned}
\)

Explanations:

Step 1: Determine the length of the incline
The motion of the block down a frictionless incline is governed by the acceleration component of gravity along the incline, which is \(a=g \sin (\theta)\). Since the lift is moving with a uniform velocity, there is no additional acceleration, so the effective gravity remains \(g\). The distance ‘ \(l\) ‘ is covered in time ‘ \(t\) ‘ with initial velocity \(\boldsymbol{u}=\mathbf{0}\). The equation of motion is \(l=\frac{1}{2} a t^2\).
For the first case, we have:
\(
\begin{gathered}
\theta_1=30^{\circ} \\
t_1=2 \mathrm{~s} \\
a_1=g \sin \left(30^{\circ}\right)=\frac{g}{2}
\end{gathered}
\)
The length of the incline is:
\(
l=\frac{1}{2} a_1 t_1^2=\frac{1}{2}\left(\frac{g}{2}\right)(2)^2=g \text { meters }
\)
Step 2: Calculate the time for the second incline angle
For the second case, the incline angle is changed to \(\theta_2=45^{\circ}\), and the length ‘\(l\)‘ remains the same.
The acceleration is now:
\(
a_2=g \sin \left(45^{\circ}\right)=g\left(\frac{1}{\sqrt{2}}\right)
\)
The time \(t_2\) is found using the same kinematic equation:
\(
l=\frac{1}{2} a_2 t_2^2
\)
We know \(l=g\) from the first step.
\(
\begin{aligned}
&\begin{gathered}
g=\frac{1}{2}\left(g \frac{1}{\sqrt{2}}\right) t_2^2 \\
1=\frac{1}{2 \sqrt{2}} t_2^2 \\
t_2^2=2 \sqrt{2} \\
t_2=\sqrt{2 \sqrt{2}}=\sqrt[4]{8} \mathrm{~s}
\end{gathered}\\
&\text { To find the approximate value: }\\
&t_2 \approx \sqrt{2 \times 1.414}=\sqrt{2.828} \approx 1.68 \mathrm{~s}
\end{aligned}
\)
The time taken by the block to slide down the incline will be approximately 1.68 s.

Hint

(b)

Retardation due to friction \(=\frac{\mu M g}{M}=\mu g\)
Now, \(s=\frac{v^2-u^2}{2 a}=\frac{0^2-2^2}{2 x-\mu g}=\frac{2}{0.4 \times 10}=0.5 \mathrm{~m}\)

Explanation: Step 1: Calculate the acceleration of the bag
The force of kinetic friction on the bag is given by \(f=\mu N\), where \(\mu\) is the coefficient of friction and \(N\) is the normal force. Since the bag is on a horizontal surface, the normal force is equal to its weight, \(N=m g\).
The force of friction provides the acceleration, so \(m a=f=\mu m g\).
Solving for acceleration, we get \(a=\mu \mathrm{g}\).
Given:
\(
\begin{aligned}
&\begin{aligned}
& \mu=0.4 \\
& g=10 \mathrm{~m} / \mathrm{s}^{-2}
\end{aligned}\\
&a=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^{-2}
\end{aligned}
\)
Step 2: Calculate the time the bag takes to reach the belt’s speed
The bag starts with an initial velocity of \(\boldsymbol{u}=\mathbf{0}\) and accelerates to the belt’s velocity of \(v=2 \mathrm{~m} / \mathrm{s}\). We can use the kinematic equation \(v=u+a t\).
Given:
\(v=2 \mathrm{~m} / \mathrm{s}\)
\(\boldsymbol{u}=0 \mathrm{~m} / \mathrm{s}\)
\(a=4 \mathrm{~m} / \mathrm{s}^{-2}\)
\(
\begin{gathered}
2=0+4 t \\
t=\frac{2}{4}=0.5 \mathrm{~s}
\end{gathered}
\)
Step 3: Calculate the distance traveled by the belt and the bag
The distance the conveyor belt travels during this time is \(s_{\text {belt }}=v_{\text {belt }} \times t\).
\(
s_{\text {belt }}=2 \mathrm{~m} / \mathrm{s} \times 0.5 \mathrm{~s}=1 \mathrm{~m}
\)
The distance the bag travels during this time is \(s_{\text {bag }}=u t+\frac{1}{2} a t^2\).
\(
s_{\text {bag }}=(0)(0.5)+\frac{1}{2}(4)(0.5)^2=0+2(0.25)=0.5 \mathrm{~m}
\)
Step 4: Calculate the distance the bag slips
The distance the bag slips relative to the belt is the difference between the distance the belt traveled and the distance the bag traveled.
\(
s_{\text {slip }}=s_{\text {belt }}-s_{\text {bag }}=1 \mathrm{~m}-0.5 \mathrm{~m}=0.5 \mathrm{~m}
\)
The distance traveled by the bag on the belt during slipping motion is 0.5 m.

Hint

(b)

Case-I:
\(
\begin{aligned}
a & =\frac{m_2 g-m_1 g}{m_1+m_2} \\
M_2=2 m_1 & \\
a_1 & =\frac{2 m_1 g-m_1 g}{3 m_1} \\
a_1 & =g / 3
\end{aligned}
\)
Case-II:
\(
\begin{gathered}
\mathrm{M}_2=3 \mathrm{~m}_1 \\
\mathrm{a}_2=\frac{3 \mathrm{~m}_1 \mathrm{~g}-\mathrm{m}_1 \mathrm{~g}}{4 \mathrm{~m}_1} \\
\mathrm{a}_2=\frac{\mathrm{g}}{2} \\
\frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\frac{\mathrm{g}}{3}}{\frac{\mathrm{~g}}{2}}=\frac{2}{3}
\end{gathered}
\)

Hint

(a)

If \(T\) is the tension in thread and a be acceleration of 100 kg block
For the 10 kg block kept on the 100 kg block (In the 100 kg block frame of reference,
\(
10 a-T=10 \times 2 \dots(1)
\)
For the 20 kg block in vertical direction,
\(
\begin{aligned}
& \mathrm{T}-20 \mathrm{~g}=20 \times 2 \\
& \Rightarrow \mathrm{~T}-20 \times 10=20 \times 2 \dots(2)
\end{aligned}
\)
Adding the above two equations,
\(
\begin{aligned}
& \Rightarrow 10 \mathrm{a}=3 \times 20+20 \times 10=260 \\
& \mathrm{a}=26 \mathrm{~ms}^{-2}, \mathrm{~T}=240 \mathrm{~N}
\end{aligned}
\)
The applied force F will be responsible for the horizontal acceleration of 100 kg and 20 kg block.
Therefore, ((Fig(b))
\(
\begin{aligned}
& \mathrm{F}-\mathrm{T}=120 \mathrm{a} \\
& \Rightarrow \mathrm{~F}=3360 \mathrm{~N}
\end{aligned}
\)

Hint

(c) 

When Monkey Moving downward:
Using Newton’s second law
\(
\begin{aligned}
& \mathrm{mg}-\mathrm{T}=\mathrm{ma}_1 \\
& \therefore 500-\mathrm{T}=50 \times 4 \\
& \Rightarrow \mathrm{~T}=300 \mathrm{~N}
\end{aligned}
\)

When Monkey Moving upward:
Using Newton’s second law of motion
\(
\begin{aligned}
& \mathrm{T}-\mathrm{mg}=\mathrm{ma}_2 \\
& \Rightarrow \mathrm{~T}-500=50 \times 5 \\
& \Rightarrow \mathrm{~T}=750 \mathrm{~N}
\end{aligned}
\)
Breaking strength of string \(=350 \mathrm{~N}\)
\(\therefore\) String will break while monkey is moving upward.

Hint

(a) Resultant of already applied forces
\(
F=5 \hat{i}-6 \hat{i}+7 \hat{j}-8 \hat{j}=-\hat{\mathrm{i}}-\hat{\mathrm{j}}
\)
Force required to balance \(=\hat{i}+\hat{j}\)
\(
F=\sqrt{1^2+1^2}=\sqrt{2} ; \tan \quad \theta=\frac{y}{x}=\frac{1}{1} ; \quad \theta=\tan ^{-1} 1=45^{\circ}
\)
Force required \(=\sqrt{2} \mathrm{~N}\) in magnitude at angle \(45^{\circ}\) with + ve \(x\)-axis

Hint

(d)

Step 1: Identify and analyze the forces acting on the block
The forces acting on the block are:
A horizontal force of 50 N pushing the block against the wall.
A normal force \(N\) from the wall, acting horizontally and opposite to the applied force.
The weight of the block \(W\), acting vertically downward.
The upward force \(F\).
A force of static friction \(f_{s^{\prime}}\), acting vertically downward to oppose the potential upward motion.
Step 2: Calculate the normal force and maximum static friction
Since the block is in horizontal equilibrium, the applied horizontal force is balanced by the normal force from the wall.
\(
N=50 \mathrm{~N}
\)
The maximum static friction force is given by the formula \(f_{s, \max }=\mu_s N\).
Given \(\mu_s=0.5\) and \(N=50 \mathrm{~N}\) :
\(
f_{s, \max }=(0.5)(50)=25 \mathrm{~N}
\)
Step 3: Calculate the weight of the block
The weight of the block is given by \(W=m g\).
Given \(m=2 \mathrm{~kg}\) and \(g=10 \mathrm{~ms}^{-2}\) :
\(
W=(2)(10)=20 \mathrm{~N}
\)

Step 4: Determine the maximum upward force \(F\)
For the block to be on the verge of moving upward, the sum of the upward forces must be equal to the sum of the downward forces. The upward forces are the applied force \(\boldsymbol{F}\). The downward forces are the weight of the block and the maximum static friction force.
\(
\begin{gathered}
F_{\text {upward }}=F_{\text {downward }} \\
F=W+f_{s, \max }
\end{gathered}
\)
Substituting the values calculated in the previous steps:
\(
\begin{gathered}
F=20 \mathrm{~N}+25 \mathrm{~N} \\
F=45 \mathrm{~N}
\end{gathered}
\)
The block will not move upward as long as \(F \leq 45 \mathrm{~N}\). Therefore, the maximum value of \(F\) is 45 N.
The maximum value of \(\boldsymbol{F}\) applied, so that the block does not move upward, will be \(\mathbf{4 5}\) N. 

Hint

(d)

For 4 kg block:
\(
4 g-T=4 a \dots(i)
\)
For 40 kg block:
\(
\mathrm{T}-40 \mathrm{~g} \times 0.02=40 \mathrm{a} \dots(ii)
\)
Adding both eq.(i) & (ii)
\(
\begin{aligned}
& 40-8=44 a \\
& a=\frac{32}{44}=\frac{8}{11} m / s^2
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\text { Using Newton’s second law }\\
&\begin{aligned}
& m g-\frac{m g}{4}=m a \\
& \Rightarrow a=\frac{3 g}{4}
\end{aligned}
\end{aligned}
\)

Hint

(b)

\(
F=-k x=ma
\)
\(
\begin{gathered}
a=\frac{-k x}{2}=\frac{-12 x}{2}=-6 x \\
\frac{d v}{d x}=-6 x \\
\int_4^v v d v=-\int_{\frac{1}{2}}^{3 / 2} 6 x d x \\
\frac{v^2-4^2}{2}=-\frac{63^2}{2}-\frac{1^2}{2} \\
v^2-16=-6 \frac{9}{4}-\frac{1}{4} \\
v^2=16-6 \times 2=4 \\
v=2 m / s
\end{gathered}
\)

Hint

(c)

Step 1: Analyze the forces on the top block ( \(m\) )
The top block of mass \(\boldsymbol{m}\) is accelerated by the static friction force exerted by the bottom block of mass \(M\). The maximum static friction force is given by \(f_{s, \max }=\mu_s N\), where \(N\) is the normal force. In this case, the normal force is the weight of the top block, so \(N=m g\)
The maximum static friction force is:
\(
\begin{gathered}
f_{s, \max }=\mu_s m g \\
f_{s, \max }=(0.5)(2 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)=9.8 \mathrm{~N}
\end{gathered}
\)
The maximum acceleration ( \(a_{\max }\) ) of the top block without slipping is found using Newton’s second law:
\(
\begin{gathered}
f_{s, \max }=m a_{\max } \\
10 \mathrm{~N}=(2 \mathrm{~kg}) a_{\max } \\
a_{\max }=\frac{9.8 \mathrm{~N}}{2 \mathrm{~kg}}=4.9 \mathrm{~m} / \mathrm{s}^2
\end{gathered}
\)
Step 2: Analyze the forces on the combined system
For the blocks to move together without slipping, the entire system of mass \(m+M\) must accelerate at this maximum rate, \(a_{\text {max }}\). The total mass of the system is \(m+M=2 \mathrm{~kg}+8 \mathrm{~kg}=10 \mathrm{~kg}\). The external force \(F\) is applied to this combined mass. Since the table is smooth, there is no friction opposing the motion of the entire system.
Using Newton’s second law for the combined system:
\(
\begin{gathered}
F=(m+M) a_{\max } \\
F=(10 \mathrm{~kg})\left(4.9 \mathrm{~m} / \mathrm{s}^2\right)=49 \mathrm{~N}
\end{gathered}
\)
The maximum horizontal force \(F\) is 49 N.

Hint

(a) Step 1: Apply the constraint relation
The problem can be solved using the constraint relation for an inextensible string in a pulley system. This principle states that for a system of masses connected by an inextensible string, the sum of the products of the tension in each string segment and the acceleration of the mass it is connected to is zero. This is mathematically expressed as \(\sum_i \boldsymbol{T}_i \cdot a_i=0\).
Step 2: Determine the effective tension on each mass
Based on the standard figure for this problem, the tensions acting on each mass are as follows:
The mass \(m_1\) is connected to a pulley with four segments of the string passing over it, so the effective tension is \(4 T\).
The mass \(m_2\) is connected to a pulley with two segments of the string, so the effective tension is \(2 T\).
The masses \(m_3\) and \(m_4\) are each connected to a single string, so the effective tension for each is \(\boldsymbol{T}\).
Step 3: Formulate the equation
By applying the constraint relation \(\sum \vec{T}_i \cdot \vec{a}_i=0\) and considering the directions of the accelerations, the equation becomes:
\(
4 T a_1+2 T a_2+T a_3+T a_4=0
\)
Step 4: Simplify the equation
By factoring out the tension \(T\) (which is non-zero) and dividing the equation by it, we get the relationship between the accelerations:
\(
4 a_1+2 a_2+a_3+a_4=0
\)

Hint

(b) When the elevator moves downward with constant acceleration.

Explanation
Apparent Weight: The sensation of “weight” a person feels is actually the normal force exerted on them by the surface they are standing on, not their true weight which is constant (mass x gravity). This is called apparent weight.
Net Force: According to Newton’s Second Law, the net force ( \(\boldsymbol{F}_{\text {net }}\) ) acting on the person is the sum of the downward force of gravity ( \(m g\) ) and the upward normal force ( \(N\) ) from the floor. This net force is equal to the person’s mass times their acceleration ( \(m a\) ).
Downward Acceleration: When the elevator accelerates downward, the direction of acceleration is the same as the force of gravity. The normal force must be less than the gravitational force to allow for this downward acceleration ( \(N=m g-m a\) ). This reduced normal force causes the person to feel lighter, or experience a “weight loss”.

Why other options are incorrect
A. When the elevator moves upward with constant acceleration: In this case, the normal force must be greater than the force of gravity to produce an upward acceleration ( \(N=m g+m a\) ). This makes the person feel heavier.
C & D. When the elevator moves with uniform velocity: If the elevator is moving at a constant speed (uniform velocity), there is no acceleration ( \(a=0\) ). The normal force is equal to the person’s true weight ( \(N=m g\) ), so they feel their normal weight and experience no change.

Alternate:

\(
\begin{aligned}
&\begin{aligned}
& \mathrm{mg}-\mathrm{N}=\mathrm{ma} \\
& \Rightarrow \mathrm{~N}=\mathrm{mg}(\mathrm{~g}-\mathrm{a})
\end{aligned}\\
&\therefore \text { Person experiences weight loss, when acceleration of lift is downward. }
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { Let time taken to ascent is } t_1 \text { and that to descent is } t_2 \text {. Height will be same so }\\
&\begin{aligned}
& H=\frac{1}{2} \times 12 t_1^2=\frac{1}{2} \times 8 t_2^2 \\
& \Rightarrow \frac{t_1}{t_1}=\frac{\sqrt{2}}{\sqrt{3}}
\end{aligned}
\end{aligned}
\)

Hint

(b) Step 1: Calculate the force of friction
The normal force ( \(N\) ) is equal to the weight of the block ( \(m g\) ). The force of friction ( \(F_f\) ) is the product of the coefficient of friction ( \(\mu\) ) and the normal force.
\(
\begin{gathered}
N=m g \\
N=(10 \mathrm{~kg})\left(9.8 \mathrm{~ms}^{-2}\right)=98 \mathrm{~N} \\
F_f=\mu N \\
F_f=(0.5)(98 \mathrm{~N})=49 \mathrm{~N}
\end{gathered}
\)
Step 2: Calculate the acceleration
According to Newton’s second law, the net force on the block ( \(\boldsymbol{F}_{\boldsymbol{f}}\) ) is equal to its mass ( \(m\) ) times its acceleration ( \(a\) ). Since the force of friction opposes the motion, the acceleration is negative.
\(
\begin{gathered}
F_f=m a \\
49 \mathrm{~N}=(10 \mathrm{~kg}) a \\
a=\frac{49 \mathrm{~N}}{10 \mathrm{~kg}}=4.9 \mathrm{~ms}^{-2}
\end{gathered}
\)
The acceleration is \(-4.9 \mathrm{~ms}^{-2}\).
Step 3: Calculate the distance
Using the kinematic equation \(v^2=u^2+2 a s\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance. The final velocity \(v\) is \(0 \mathrm{~ms}^{-1}\) since the block comes to rest.
\(
\begin{gathered}
0^2=\left(9.8 \mathrm{~ms}^{-1}\right)^2+2\left(-4.9 \mathrm{~ms}^{-2}\right) s \\
0=96.04-9.8 s \\
9.8 s=96.04 \\
s=\frac{96.04}{9.8}=9.8 \mathrm{~m}
\end{gathered}
\)
The distance covered by the block before coming to rest is 9.8 m.

Hint

(d) Let acceleration of wedge is \(a_1\) and acceleration of block w.r.t. wedge is \(a_2\)

\(
\begin{aligned}
&\begin{aligned}
& N \cos 60^{\circ}=M a_1=16 a_1 \\
& \Rightarrow \mathrm{~N}=32 \mathrm{a}_1
\end{aligned}\\
&\text { F.B.D. of block w.r.t wedge }
\end{aligned}
\)

\(\perp\) to incline
\(
\begin{aligned}
& \mathrm{N}=8 \mathrm{~g} \cos 30^{\circ}-8 \mathrm{a}_1 \sin 30^{\circ} \Rightarrow 32 \mathrm{a}_1=4 \sqrt{3} \mathrm{~g}-4 \mathrm{a}_1 \\
& \Rightarrow \mathrm{a}_1=\frac{\sqrt{3}}{9} \mathrm{~g}
\end{aligned}
\)
Along incline
\(
\begin{aligned}
& 8 g \sin 30^{\circ}+8 \mathrm{a}_1 \cos 30^{\circ}=m \mathrm{a}_2=8 \mathrm{a}_2 \\
& \mathrm{a}_2=\mathrm{g} \times \frac{1}{2}+\frac{\sqrt{3}}{9} \mathrm{~g} \cdot \frac{\sqrt{3}}{2}=\frac{2 \mathrm{~g}}{3}
\end{aligned}
\)

Hint

(b) During upward motion

\(
F=m g \sin \theta=2N
\)
As velocity need to be constant, acceleration will be 0. So Net Force has to be 0.
\(
F_{net} =0
\)

\(
\mathrm{F}=2 \mathrm{~N}=(+\mathrm{ve}) \text { constant }
\)

During downward motion

\(
F=m g \sin \theta=-2N
\)
As velocity need to be constant, acceleration will be 0. So Net Force has to be 0.
\(
F_{net}=0
\)

\(
\Rightarrow \mathrm{F}=-2 \mathrm{~N}=(-\mathrm{ve}) \text { constant }
\)

Hint

(a)

\(
\begin{aligned}
& \overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2=-\overrightarrow{\mathrm{F}}_3 \\
& \text { Here } \overrightarrow{\mathrm{F}}_1+\overrightarrow{\mathrm{F}}_2+\overrightarrow{\mathrm{F}}_3=0 \\
& \text { Since } \overrightarrow{\mathrm{F}}_{\mathrm{net}}=0 \text { (equilibrium) }
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& \left(m_1 g-2 T\right)=m_1 a \dots(1) \\
& T-m_2 g=m_2(2 a) \\
& 2 T-2 m_2 g=4 m_2 a \dots(2) \\
& \text { Adding eqn (1) & eqn (2), we get }
& m_1 g-2 m_2 g=\left(m_1+4 m_2\right) a \\
& a=\frac{(8-4) g}{(8+8)}=\frac{4}{16} g=\frac{g}{4} \\
& a=\frac{10}{4} m / s^2 \\
& S=\frac{1}{2} a t^2 \\
& \frac{0.2 \times 2 \times 4}{10}=t^2 \\
& t=0.4 s e c
\end{aligned}
\)

Hint

(d)

Step 1: Identify the forces acting on the rocket
The net force on the rocket is the difference between the upward thrust and the downward force of gravity. The total mass of the rocket at any time is the initial mass ( \(m_0\) ) plus the remaining fuel. At the initial moment, the total mass is \(m=1000 \mathrm{~kg}\).
Thrust force ( \(\boldsymbol{F}_{\text {thrust }}\) ): This is the force created by the expulsion of exhaust gases. It is given by the product of the relative exhaust velocity and the rate of change of the rocket’s mass:
\(
F_{t h r u s t}=v_{r e l} \frac{d m}{d t}
\)
Gravitational force ( \(\mathrm{W}^{\prime}\) ): This is the weight of the rocket, which is its mass multiplied by the acceleration due to gravity:
\(
W=m g
\)
Net force ( \(\boldsymbol{F}_{\text {net }}\) ): According to Newton’s Second Law, the net force is equal to the mass of the rocket times its acceleration:
\(
F_{n e t}=m a
\)
Step 2: Set up the equation and solve for the rate of fuel burn
The net force is the thrust minus the gravitational force, so we can write the equation:
\(
\begin{aligned}
& F_{\text {net }}=F_{\text {thrust }}-W \\
& m a=v_{\text {rel }} \frac{d m}{d t}-m g
\end{aligned}
\)
\(
\begin{aligned}
&\text { We can rearrange this equation to solve for the rate of fuel burn, } \frac{d m}{d t} \text { : }\\
&\begin{gathered}
m a+m g=v_{\text {rel }} \frac{d m}{d t} \\
m(a+g)=v_{\text {rel }} \frac{d m}{d t} \\
\frac{d m}{d t}=\frac{m(a+g)}{v_{\text {rel }}}
\end{gathered}
\end{aligned}
\)
Step 3: Substitute the given values and calculate the result
Initial mass \((m)=1000 \mathrm{~kg}\)
Acceleration \((a)=20 \mathrm{~ms}^{-2}\)
Gravitational acceleration \((\mathrm{g})=10 \mathrm{~ms}^{-2}\)
Relative speed of gases \(\left(v_{r e l}\right)=500 \mathrm{~ms}^{-1}\)
Substituting these values into the equation:
\(
\begin{gathered}
\frac{d m}{d t}=\frac{1000 \mathrm{~kg}\left(20 \mathrm{~ms}^{-2}+10 \mathrm{~ms}^{-2}\right)}{500 \mathrm{~ms}^{-1}} \\
\frac{d m}{d t}=\frac{1000 \mathrm{~kg}^{-1}\left(30 \mathrm{~ms}^{-2}\right)}{500 \mathrm{~ms}^{-1}} \\
\frac{d m}{d t}=\frac{30000 \mathrm{~kg} \mathrm{~ms}^{-2}}{500 \mathrm{~ms}^{-1}} \\
\frac{d m}{d t}=60 \mathrm{~kg} \mathrm{~s}^{-1}
\end{gathered}
\)
The rate at which the fuel should be burnt is \(60 \mathbf{~ k g ~ s}^{-\mathbf{1}}\).

Hint

(c)

\(
\begin{aligned}
&\text { At } {t}=0, {u}=0\\
&\begin{aligned}
& a=\frac{F_0}{M}-\frac{F_0}{M T^2}(t-T)^2=\frac{d v}{d t} \\
& \int_0^v d v=\int_{t=0}^{2 T}\left(\frac{F_0}{M}-\frac{F_0}{M T^2}(t-T)^2\right) d t \\
& V=\left[\frac{F_0}{M} t\right]_0^{2 T}-\frac{F_0}{M T^2}\left[\frac{t^3}{3}-t^2 T+T^2 t\right]_0^{2 T} \\
& \Rightarrow V=\frac{4 F_0 T}{3 M}
\end{aligned}
\end{aligned}
\)

Hint

(c) Step 1: Calculate the acceleration vector
According to Newton’s second law, the acceleration vector \(\vec{a}\) is given by \(\vec{a}=\frac{\vec{F}}{m}\).
Given the force vector \(\vec{F}=(40 \hat{i}+10 \hat{j}) N\) and mass \(m=5 \mathrm{~kg}\), we can find the acceleration:
\(
\vec{a}=\frac{40 \hat{i}+10 \hat{j}}{5}=(8 \hat{i}+2 \hat{j}) \frac{m}{\mathrm{~s}^2}
\)
Step 2: Calculate the position vector
The position vector \(\vec{r}\) of a body starting from rest ( \(\vec{v}_0=0\) ) and an initial position of \(\overrightarrow{r_0}=0\) is given by the kinematic equation:
\(
\vec{r}=\vec{r}_0+\vec{v}_0 t+\frac{1}{2} \vec{a} t^2=0+0+\frac{1}{2} \vec{a} t^2
\)
Substituting the acceleration vector \(\vec{a}=(8 \hat{i}+2 \hat{j}) \frac{m}{s^2}\) and time \(t=10 \mathrm{~s}\) :
\(
\begin{gathered}
\vec{r}=\frac{1}{2}(8 \hat{i}+2 \hat{j})(10)^2=\frac{1}{2}(8 \hat{i}+2 \hat{j})(100) \\
\vec{r}=(4 \hat{i}+1 \hat{j})(100)=(400 \hat{i}+100 \hat{j}) m
\end{gathered}
\)
The position vector of the body at time \(t=10 \mathrm{~s}\) will be \((400 \hat{i}+100 \hat{j}) m\).

Hint

(a)

\(M=\) collective mass of block and all three iron cylinders \(=10+3 \times 20=70 \mathrm{~kg}\)
\(\mathrm{a}=\) acceleration of block \(=0.2 \mathrm{~ms}^{-2}\)
\(\mathrm{g}=10 \mathrm{~ms}^{-2}\) and \(\mu_{\mathrm{s}}=0.2\)
and \(\mathrm{R}=\) normal reaction
Force along vertical axis \(\mathrm{Mg}-\mathrm{R}=\mathrm{Ma}\)
\(
\begin{aligned}
& \therefore 70 \mathrm{~g}-\mathrm{R}=70 \times 0.2 \\
& \Rightarrow \mathrm{R}=70 \times 10-14 \\
& =700-14=686 \mathrm{~N}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\Rightarrow \mathrm{N}=\mathrm{mg}-\mathrm{F} \sin \theta \dots(1)\\
&\text { Also, } F \cos \theta-\mu_k N=m a \dots(2)\\
&\text { Substituting the value of } \mathrm{N} \text { from eq. (1) in eq. (2) }\\
&\begin{aligned}
& \Rightarrow \mathrm{F} \cos \theta-\mu_{\mathrm{k}}(\mathrm{mg}-\mathrm{F} \sin \theta)=\mathrm{m} \cdot \mathrm{a} \\
& \Rightarrow \mathrm{a}=\frac{F}{m} \cos \theta-\mu_{\mathrm{k}}\left(g-\frac{F}{m} \sin \theta\right)
\end{aligned}
\end{aligned}
\)

Hint

(d)

As we can see from the diagram that the only force here is due to the force applied on B. But due to F there will be some tension T in the spring and this tension is for both A and B as both are attached to spring.
Let us consider the body of mass A , it is given that when force is applied on mass B then mass A has acceleration, \(a\) and the only force which is experienced by mass \(A\) is due to tension in spring.
Hence we can write
\(
T=M a \dots(i)
\)
Here \(M\) is the mass of \(A\).
Now for the mass B there will be force F applied on it and the tension in the spring, if the acceleration of B is \(\mathrm{a}^{\prime}\), then by balancing the forces we get
\(
F-T=M a^{\prime}
\)
Substituting value of \(T\) from equation (i) we get
\(
\begin{aligned}
& F-M a=M a^{\prime} \\
& \Rightarrow a^{\prime}=\frac{F-M a}{M} \\
& \Rightarrow a^{\prime}=\frac{F}{M}-a
\end{aligned}
\)
Hence acceleration of B in terms of force, mass and acceleration of A is \(\frac{F}{M}-a\).

Hint

(a) Step 1: Setting up the differential equation
The damping force is given as \(\boldsymbol{F}=m a=-\alpha x^2\). From this, the acceleration is \(a=-\frac{\alpha}{m} x^2\). Using the kinematic identity \(a=v \frac{d v}{d x}\), we get:
\(
v \frac{d v}{d x}=-\frac{\alpha}{m} x^2
\)
Step 2: Integration
Separating the variables gives \(v d v=-\frac{\alpha}{m} x^2 d x\). To find the position \(x\) where the body comes to a stop ( \(v=0\) ), we integrate with the limits from the initial velocity \(v_0\) at position \(x=0\) to the final velocity 0 at position \(x\).
\(
\begin{aligned}
\int_{v_0}^0 v d v & =\int_0^x-\frac{\alpha}{m} x^2 d x \\
{\left[\frac{v^2}{2}\right]_{v_0}^0 } & =-\frac{\alpha}{m}\left[\frac{x^3}{3}\right]_0^x \\
\left(0-\frac{v_0^2}{2}\right) & =-\frac{\alpha}{m}\left(\frac{x^3}{3}-0\right) \\
-\frac{v_0^2}{2} & =-\frac{\alpha}{3 m} x^3
\end{aligned}
\)
Solving for \(x\), we find the maximum distance the object travels is:
\(
\begin{gathered}
x^3=\frac{v_0^2}{2} \cdot \frac{3 m}{\alpha} \\
x=\left(\frac{3 m v_0^2}{2 \alpha}\right)^{1 / 3}
\end{gathered}
\)

Hint

(b) Given \(\vec{F}=k\left(v_y \hat{i}+v_x \hat{j}\right)\)
\(
\begin{aligned}
& \Rightarrow \overrightarrow{m a}=k\left(v_y \hat{i}+v_x \hat{j}\right) \\
& \Rightarrow \vec{a}=\frac{k}{m}\left(v_y \hat{i}+v_x \hat{j}\right)
\end{aligned}
\)
Also \(\frac{d v_x}{d t}=\frac{k}{m} v_x\)
and \(\frac{d v_y}{d t}=\frac{k}{m} v_y\)
\(
\begin{aligned}
& \frac{d v_x}{d v_y}=\frac{v_y}{v_x} \\
& \Rightarrow \int v_x d v_x=\int v_y d v_y \\
& \Rightarrow v_x^2=v_y^2+C \\
& \Rightarrow v_x^2-v_y^2=C=\text { Constant }
\end{aligned}
\)
\(
\begin{aligned}
& \vec{v} \times \vec{a} \\
& =\left(v_x \hat{i}+v_y \hat{j}\right) \times \frac{k}{m}\left(v_y \hat{i}+v_x \hat{j}\right) \\
& =\left(v_x^2 \widehat{k}-v_y^2 \widehat{k}\right) \times \frac{k}{m} \\
& =\left(v_x^2-v_y^2\right) \times \frac{k}{m} \widehat{k} \\
& =\text { Constant }
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\text { At a point where slipping just starts, }\\
&\begin{aligned}
f= & m g \sin \theta \\
\Rightarrow \quad & \mu m g \cos \theta=m g \sin \theta \\
\tan \theta=\frac{3}{4}
\end{aligned}
\end{aligned}
\)

\(
\begin{aligned}
& h=N-N \cos \theta \\
& =\mathrm{N}-\mathrm{N}\left(\frac{4}{5}\right)=\frac{\mathrm{N}}{5} \\
& \mathrm{~h}=\frac{\mathrm{N}}{5}=0.2 \mathrm{~m}
\end{aligned}
\)

Hint

(c) Step 1:
The momentum of the spaceship is given by \(p=M(t) v(t)\). According to Newton’s second law, the external force on the system is the rate of change of momentum. Since there are no external forces acting on the spaceship (the dust is stationary and being “swept up,” which is an internal process for the system of spaceship+dust), the total external force, \(F_{\text {ext }}\), is zero.
\(
F_{e x t}=\frac{d p}{d t}=\frac{d(M v)}{d t}=0
\)
Step 2: Differentiate the momentum equation
Using the product rule for differentiation, we expand the expression for the time derivative of momentum:
\(
\frac{d(M v)}{d t}=M \frac{d v}{d t}+v \frac{d M}{d t}
\)
Since \(a=\frac{d v}{d t}\) (instantaneous acceleration) and \(\frac{d M}{d t}=b v^2(t)\) (given rate of mass increase), we can substitute these into the equation.
\(
\begin{aligned}
& M(t) a(t)+v(t)\left(\frac{d M}{d t}\right)=0 \\
& M(t) a(t)+v(t)\left(b v^2(t)\right)=0
\end{aligned}
\)
Step 3: Solve for acceleration
Now, we can rearrange the equation to solve for the acceleration, \(a(t)\).
\(
\begin{gathered}
M(t) a(t)=-v(t)\left(b v^2(t)\right) \\
M(t) a(t)=-b v^3(t) \\
a(t)=-\frac{b v^3(t)}{M(t)}
\end{gathered}
\)
The instantaneous acceleration of the satellite is \(-\frac{\mathbf{b v}^{\mathbf{3}}}{\mathbf{M}(\mathbf{t})}\).

Hint

(c) Step 1: Set up the equation of motion
The forces acting on the ball as it travels upward are gravity ( \(m g\) ) and the resistive force ( \(m k v^2\) ). Both forces act in the downward direction. According to Newton’s second law, the net force is equal to mass times acceleration. Let the upward direction be positive. The net force is:
\(
F_{n e t}=-m g-m k v^2=-m\left(g+k v^2\right)
\)
The acceleration \(a\) can be expressed as \(a=v \frac{d v}{d y}\), where \(v\) is the velocity and \(y\) is the height. Substituting this into the force equation gives:
\(
\begin{aligned}
m v \frac{d v}{d y} & =-m\left(g+k v^2\right) \\
v \frac{d v}{d y} & =-\left(g+k v^2\right)
\end{aligned}
\)
Step 2: Separate variables and integrate
To find the maximum height, \(\boldsymbol{H}\), we can separate the variables \(v\) and \(y\) and integrate. The velocity changes from its initial value \(\boldsymbol{u}\) at \(\boldsymbol{y}=\mathbf{0}\) to \(\mathbf{0}\) at the maximum height \(\boldsymbol{H}\).
\(
\frac{v}{g+k v^2} d v=-d y
\)
Integrating both sides with the appropriate limits:
\(
\int_u^0 \frac{v}{g+k v^2} d v=\int_0^H-d y
\)
Step 3: Solve the integral
To solve the integral on the left side, we can use a substitution. Let \(\mathrm{z}=g+k v^2\). Then the differential is \(d \mathrm{z}=2 k v d v\), which means \(v d v=\frac{1}{2 k} d z\).
The limits for \(\boldsymbol{z}\) are:
When \(v=u, \mathrm{z}=g+k u^2\).
When \(v=0, \mathrm{z}=\mathrm{g}\).
The integral becomes:
\(
\begin{gathered}
\int_{g+k u^2}^g \frac{1}{z}\left(\frac{1}{2 k}\right) d z=-\int_0^H d y \\
\frac{1}{2 k}[\ln |z|]_{g+k u^2}^g=-[y]_0^H \\
\frac{1}{2 k}\left(\ln (g)-\ln \left(g+k u^2\right)\right)=-H
\end{gathered}
\)
Using the properties of logarithms, \(\ln (a)-\ln (b)=\ln (a / b)\) :
\(
\frac{1}{2 k} \ln \left(\frac{g}{g+k u^2}\right)=-H
\)
To get a positive height, we can multiply both sides by -1 and use the property \(-\ln (x)=\ln (1 / x)\) :
\(
\begin{gathered}
H=-\frac{1}{2 k} \ln \left(\frac{g}{g+k u^2}\right)=\frac{1}{2 k} \ln \left(\frac{g+k u^2}{g}\right) \\
H=\frac{1}{2 k} \ln \left(1+\frac{k u^2}{g}\right)
\end{gathered}
\)
The maximum height attained by the ball is \(\frac{1}{2 k} \ln \left(1+\frac{k u^2}{g}\right)\).

Hint

(a)

For equilibrium,
\(
\mathrm{T} \sin 45^{\circ}=\mathrm{F} \dots(1)
\)
\(
\text { and } \mathrm{T} \cos 45^{\circ}=10 \mathrm{~g} \dots(2)
\)
Equation (1)/(2)
we get \(\mathrm{F}=10 \mathrm{~g}\)
\(
\text { = } 100 \mathrm{~N}
\)

Hint

(b) Step 1: Establish the relationship between spring constant and length
The force constant \(\boldsymbol{k}\) of a spring is inversely proportional to its unstretched length \(\boldsymbol{l}\). This means the product of the force constant and the length is a constant for a given spring material. Let’s call this constant \(C\).
\(
k l=C
\)
For the original spring, we have \(k l=C\).
After cutting, the two new pieces will have force constants \(k_1\) and \(k_2\) and lengths \(l_1\) and \(l_2\). The constant \(C\) remains the same for both pieces.
\(
k_1 l_1=C \quad \text { and } \quad k_2 l_2=C
\)
Step 2: Express the force constants in terms of the constant and lengths
From the equations in Step 1, we can express the new force constants as:
\(
k_1=\frac{C}{l_1} \quad \text { and } \quad k_2=\frac{C}{l_2}
\)
Step 3: Determine the ratio of the force constants
To find the ratio \(k_1 / k_2\), we divide the two expressions from Step 2:
\(
\frac{k_1}{k_2}=\frac{C / l_1}{C / l_2}=\frac{l_2}{l_1}
\)
The problem states that \(l_1=n l_2\). We can substitute this into the ratio:
\(
\frac{k_1}{k_2}=\frac{l_2}{n l_2}=\frac{1}{n}
\)
The ratio of the corresponding force constants, \(k_1 / k_2\), will be \(\frac{1}{n}\).

Hint

(b) 

Case I:
\(
\begin{aligned}
& \mathrm{N}_1=\mathrm{mg}+10=60 \\
& (\mathrm{fs})_{\max }=0.2 \times(60)=12 \mathrm{~N}
\end{aligned}
\)
Case-II:
\(
\begin{aligned}
& \mathrm{N}_2=\mathrm{mg}-10 \\
& \quad=50-10=40 \mathrm{~N} \\
& (\mathrm{fs})_{\text {max }}=0.2 \times 40=8 \mathrm{~N}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{array}{rl|rl}
\mathrm{a}_{\mathrm{A}} & =\frac{10 \sqrt{3}-12}{5} & & \mathrm{a}_{\mathrm{B}}=\frac{10 \sqrt{3}-8}{5} \\
& =\frac{17.32-12}{5} & & =\frac{17.32-8}{5} \\
\mathrm{a}_{\mathrm{A}} & =\frac{5.32}{5} & & \mathrm{a}_{\mathrm{B}}=\frac{9.32}{5}
\end{array}\\
&\text { difference between acceleration }\\
&\begin{aligned}
a_B-a_A & =\frac{1}{5}(9.32-5.32)=\frac{4}{5} \\
\Delta a & =0.8 m / s^2
\end{aligned}
\end{aligned}
\)

Hint

(b) You have been given the following data
\(
\begin{aligned}
& \mu=0.2 \\
& m_A=1 \mathrm{~kg} \\
& m_B=3 \mathrm{~kg}
\end{aligned}
\)
Acceleration \(\mathrm{a}=\mu g=2 m s^{-2}\)
So you can calculate frictional force using the following formula:
\(
\text { Frictional force }=\mu m g=0.2 \times 4 \times 10=8 N
\)
Let us assume F is the force applied on the system,
\(
\begin{aligned}
& \Rightarrow F-\mu m g=m a \\
& F-8=4 \times 2 \\
& F=16N
\end{aligned}
\)

Explanation: Step 1: Analyze the motion of block A
For block A not to slide over block B, both blocks must have the same acceleration, \(a\). The force causing this acceleration on block A is the static friction force, \(f_s\), from block B. The maximum static friction force is given by \(f_{s, \max }=\mu_{A B} N_A\), where \(\mu_{A B}\) is the coefficient of friction between A and B , and \(N_A\) is the normal force on A.
From the free-body diagram of block A, the vertical forces are the weight \(m_A g\) and the normal force from \(\mathrm{B}, N_A\). Since there is no vertical motion, \(N_A=m_A g\).
The maximum acceleration, \(a_{\text {max }}\), of block A without sliding is given by Newton’s second law in the horizontal direction:
\(
\begin{gathered}
f_{s, \max }=m_A a_{\max } \\
\mu_{A B} m_A g=m_A a_{\max } \\
a_{\max }=\mu_{A B} g
\end{gathered}
\)
Given \(\mu_{A B}=0.2\) and \(g=10 \mathrm{~m} / \mathrm{s}^2\) :
\(
a_{\max }=(0.2)\left(10 \mathrm{~m} / \mathrm{s}^2\right)=2 \mathrm{~m} / \mathrm{s}^2
\)
This is the maximum acceleration the system can have before block A starts to slip.
Step 2: Analyze the motion of the combined system
Now, consider blocks A and B as a single system with a total mass \(m_{\text {total }}=m_A+m_B\). The external horizontal forces acting on this system are the applied force \(F\) and the kinetic friction force between block B and the table, \(f_{\text {table }}\).
The normal force on the combined system from the table is \(N_{\text {total }}=\left(m_A+m_B\right) g\). The kinetic friction force is \(f_{\text {table }}=\mu_{\text {table }} N_{\text {total }}=\mu_{\text {table }}\left(m_A+m_B\right) g\). Given \(\mu_{\text {table }}=0.2, m_A=1 \mathrm{~kg}, m_B=3 \mathrm{~kg}\), and \(g=10 \mathrm{~m} / \mathrm{s}^2\) :
\(
f_{\text {table }}=(0.2)(1 \mathrm{~kg}+3 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^2\right)=(0.2)(4 \mathrm{~kg})\left(10 \mathrm{~m} / \mathrm{s}^2\right)=8 \mathrm{~N}
\)
The maximum applied force \(F_{\text {max }}\) corresponds to the maximum acceleration \(a_{\text {max }}\) calculated in the previous step. Applying Newton’s second law to the combined system:
\(
\begin{gathered}
F_{\max }-f_{\text {table }}=m_{\text {total }} a_{\max } \\
F_{\max }-f_{\text {table }}=\left(m_A+m_B\right) a_{\max } \\
F_{\max }=\left(m_A+m_B\right) a_{\max }+f_{\text {table }}
\end{gathered}
\)
Using the values calculated:
\(
\begin{gathered}
F_{\max }=(1 \mathrm{~kg}+3 \mathrm{~kg})\left(2 \mathrm{~m} / \mathrm{s}^2\right)+8 \mathrm{~N} \\
F_{\max }=(4 \mathrm{~kg})\left(2 \mathrm{~m} / \mathrm{s}^2\right)+8 \mathrm{~N}=8 \mathrm{~N}+8 \mathrm{~N}=16 \mathrm{~N}
\end{gathered}
\)

Hint

(c) Given, resistance offered by the wall
\(
=F=-25 \times 10^{-2} \mathrm{~N}
\)
So, deacceleration of bullet,
\(
\begin{aligned}
a=\frac{F}{m}=\frac{-2.5 \times 10^{-2}}{20 \times 10^{-3}} & =-\frac{5}{4} \mathrm{~ms}^{-2} \\
& \left(\because m=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\right)
\end{aligned}
\)
Now, using the equation of motion,
\(
v^2-u^2=2 a s
\)
We have,
\(
\begin{aligned}
& v^2=1+2\left(-\frac{5}{4}\right)\left(20 \times 10^{-2}\right) \\
& \left(\because u=1 \mathrm{~ms}^{-1} \text { and } s=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\right) \\
& \Rightarrow v^2=\frac{1}{2} \\
& \therefore v=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

(a) Step 1: Set up the equation of motion
The forces acting on the ball as it travels upward are gravity and the drag force. Both forces act in the downward direction, opposing the upward motion of the ball.
Force of gravity: \(F_g=m g\) (downward)
Drag force: \(F_d=m \gamma u^2\) (downward, since \(u\) is upward)
According to Newton’s second law, the net force is equal to mass times acceleration ( \(\boldsymbol{F}_{\text {net }}=\boldsymbol{m} \boldsymbol{a}\) ). Since both forces are in the same direction, we can sum their magnitudes. Taking the upward direction as positive, the net force is:
\(
F_{n e t}=-m g-m \gamma u^2
\)
The acceleration is given by \(a=\frac{d u}{d t}\), so we have:
\(
m \frac{d u}{d t}=-m g-m \gamma u^2
\)
Dividing by \(m\) :
\(
\frac{d u}{d t}=-g-\gamma u^2=-\left(g+\gamma u^2\right)
\)
Step 2: Separate variables and integrate to find time
To find the time taken for the ball to reach its zenith, we need to integrate the equation of motion. At the zenith, the velocity of the ball is \(\boldsymbol{u}=\mathbf{0}\). The initial velocity is \(\boldsymbol{V}_{\mathbf{0}}\). We can separate the variables \(u\) and \(t\) :
\(
d t=-\frac{d u}{g+\gamma u^2}
\)
To find the total time \(\boldsymbol{T}\) to reach the zenith, we integrate from the initial conditions ( \(t=0, u=V_0\) ) to the final conditions ( \(t=T, u=0\) ):
\(
\int_0^T d t=-\int_{V_0}^0 \frac{d u}{g+\gamma u^2}
\)
Step 3: Solve the integral
We can rewrite the integral on the right side by factoring out \(g\) :
\(
\int_0^T d t=-\int_{V_0}^0 \frac{d u}{g\left(1+\frac{\gamma}{g} u^2\right)}
\)
Let’s use a substitution to simplify the integral. Let \(k^2=\frac{\gamma}{g}\), which means \(k=\sqrt{\frac{\gamma}{g}}\). The integral becomes:
\(
T=-\frac{1}{g} \int_{V_0}^0 \frac{d u}{1+k^2 u^2}
\)
We know that the integral of \(\frac{1}{1+x^2}\) is \(\arctan (x)\). The form of our integral is \(\int \frac{d x}{a^2+b^2 x^2}=\frac{1}{a b} \arctan \left(\frac{b x}{a}\right)\). In our case, \(a=1\) and \(b=k\).
\(
\begin{gathered}
T=-\frac{1}{g}\left[\frac{1}{k} \arctan (k u)\right]_{V_0}^0 \\
T=-\frac{1}{g}\left[\frac{1}{k} \arctan (0)-\frac{1}{k} \arctan \left(k V_0\right)\right] \\
T=-\frac{1}{g k}\left[0-\arctan \left(\sqrt{\frac{\gamma}{g}} V_0\right)\right] \\
T=\frac{1}{g k} \arctan \left(\sqrt{\frac{\gamma}{g}} V_0\right)
\end{gathered}
\)
Substitute \(k=\sqrt{\frac{\gamma}{g}}\) back into the equation:
\(
\begin{aligned}
T & =\frac{1}{g_{\sqrt{g}}} \arctan \left(V_0 \sqrt{\frac{\gamma}{g}}\right) \\
T & =\frac{1}{\sqrt{g \gamma}} \arctan \left(V_0 \sqrt{\frac{\gamma}{g}}\right)
\end{aligned}
\)
The time taken by the ball to rise to its zenith is \(\boldsymbol{T}=\frac{1}{\sqrt{g r}} \arctan \left(\boldsymbol{V}_{\mathbf{0}} \sqrt{\frac{\boldsymbol{\gamma}}{\boldsymbol{g}}}\right)\).

Hint

(c)

\(
\begin{aligned}
& \sum \mathrm{F}=0 \\
& 2+\mathrm{mg} \sin 30=\mu \mathrm{mg} \cos 30^{\circ} \dots(i) \\
& 10=\mathrm{mg} \sin 30+\mu \mathrm{mg} \cos 30^{\circ} \dots(ii)
\end{aligned}
\)
Adding eq (i) and (ii), we get
\(
12=2 \mu \mathrm{mg} \cos 30
\)
\(6=\mu \mathrm{mg} \cos 30 \dots(iii)\)
\(
4 \text { = mgsin30 } \dots(iv)
\)
From eq (iii) and (iv), we get
\(
3 / 2=\mu \times \cot 30^{\circ}
\)
\(
3 / 2=\mu \times \sqrt{3}
\)
\(
\mu=\frac{\sqrt{3}}{2}
\)

Hint

(d) Step 1: Set up the initial resultant force equation
The magnitude of the resultant force, R, of two forces P and Q with an angle \(\theta\) between them is given by the formula:
\(
R^2=P^2+Q^2+2 P Q \cos \theta
\)
Given that \(P=2 F\) and \(Q=3 F\), the initial resultant force squared is:
\(
\begin{gathered}
R^2=(2 F)^2+(3 F)^2+2(2 F)(3 F) \cos \theta \\
R^2=4 F^2+9 F^2+12 F^2 \cos \theta \\
R^2=13 F^2+12 F^2 \cos \theta
\end{gathered}
\)
Step 2: Set up the second resultant force equation
When force \(Q\) is doubled, its new magnitude is \(Q^{\prime}=2 Q=2(3 F)=6 F\). The new resultant force, \(R^{\prime}\), is also doubled, so \(R^{\prime}=2 R\). Substituting these into the resultant force formula:
\(
\begin{gathered}
\left(R^{\prime}\right)^2=P^2+\left(Q^{\prime}\right)^2+2 P\left(Q^{\prime}\right) \cos \theta \\
(2 R)^2=(2 F)^2+(6 F)^2+2(2 F)(6 F) \cos \theta \\
4 R^2=4 F^2+36 F^2+24 F^2 \cos \theta \\
4 R^2=40 F^2+24 F^2 \cos \theta
\end{gathered}
\)
Step 3: Solve for \(\cos \theta\)
Now we have a system of two equations. Substitute the expression for \(R^2\) from Step 1 into the equation from Step 2:
\(
4\left(13 F^2+12 F^2 \cos \theta\right)=40 F^2+24 F^2 \cos \theta
\)
Divide the entire equation by \(F^2\) (since \(F \neq 0\) ):
\(
\begin{gathered}
4(13+12 \cos \theta)=40+24 \cos \theta \\
52+48 \cos \theta=40+24 \cos \theta
\end{gathered}
\)
Rearrange the terms to solve for \(\cos \theta\) :
\(
\begin{gathered}
48 \cos \theta-24 \cos \theta=40-52 \\
24 \cos \theta=-12 \\
\cos \theta=-\frac{12}{24}=-\frac{1}{2}
\end{gathered}
\)
The angle \(\theta\) whose cosine is \(-1 / 2\) is \(120^{\circ}\).

Hint

(d)

\(
\begin{aligned}
& \tan 45^{\circ}=\frac{F}{m g} \\
& \therefore \mathrm{~F}=\mathrm{mg} \\
& =10 \times 10 \\
& =100 \mathrm{~N}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& \text { Limiting friction } f_s=\mu m g \cos 45^{\circ} \\
& =0.6 \times 10 \times 10 \times \frac{1}{\sqrt{2}}=30 \sqrt{2} \mathrm{~N}=42.43 \mathrm{~N}
\end{aligned}\\
&\text { When block starts to slide downward, the downward force on the block is }\\
&\begin{aligned}
& F=3+m g \sin 45^0=3+10 \times 10 \times \frac{1}{\sqrt{2}}=3+50 \sqrt{2}=73.71 \mathrm{~N}>f_{\mathrm{s}} \\
& \text { Block will not move if } P=F-f_s=73.71-42.43=31.38 \mathrm{~N} \simeq 32 \mathrm{~N}
\end{aligned}
\end{aligned}
\)