Past NEET Papers

Hint

As the three forces are represented by three sides of a triangle forming a closed loop in the same order, then they will be in equilibrium and the net force is zero.
\(
\begin{aligned}
\Rightarrow \mathbf{F}_{\text {net }} &=\mathbf{F}_{P Q}+\mathbf{F}_{Q R}+\mathbf{F}_{R P}=0 \\
\mathbf{F}_{\text {net }} &=m \times \mathbf{a}=m \frac{d \mathbf{v}}{d t}=0 \Rightarrow \frac{d v}{d t}=0
\end{aligned}
\)
or \(\mathbf{v}=\) constant
Therefore, the velocity of the particle remains constant.

Hint

(d) Change in momentum = Area of F-t graph
\(\Delta p=\int \mathrm{Fdt}\)
\(=\) area of \(\Delta-area\) of \(\square+area\) of \(\square\)
\(
=\frac{1}{2} \times 2 \times 6-3 \times 2+4 \times 3=12 \mathrm{~N}\mathrm{s}
\)
Note: According to Newton’s 2nd law. \(F \propto \frac{d \vec{p}}{d t} \Rightarrow F\) \(=K \frac{d \vec{p}}{d t}\) or, \(F=\frac{d \vec{p}}{d t}[\because k=1, in S.I \& CGS system]\) and \(d \vec{p}=\vec{F} d t\)

Hint

(c) Momentum \(P=m v=m \sqrt{2 g h}\) (\(\therefore\ v^{2}=u^{2}+2 g h\); Here \(\left.u=0\right)\)
When the stone hits the ground its momentum,
\(P=m \sqrt{2 g h}\)
when same stone dropped from \(2 h\) then momentum,
\(
P^{\prime}=m \sqrt{2 g(2 h)}=\sqrt{2} P
\)
\(50 \%\) change in momentum, \(\frac{\left(P^{\prime}-P\right)}{P} \times 100 \%\)

\(
\begin{aligned}
&\Rightarrow \quad\left(\frac{\sqrt{2} P-P}{P}\right) \times 100 \%=(\sqrt{2}-1) \times 100 \% \\
&\Rightarrow \quad(1.414-1) \times 100 \% \\
&\Rightarrow \quad 414 \times 100 \%=41.4 \%
\end{aligned}
\)
Which is changed by \(41 \%\) of the initial.

Hint

(a) Impulse experienced by the body = change in momentum
\(
=\mathrm{MV}-(-\mathrm{MV}) =2 \mathrm{MV}
\)

Hint

(c) \(\vec{F}=6 \hat{i}-8 \hat{j}+10 \hat{k}\),
\(
|\overrightarrow{\mathrm{F}}|=\sqrt{36+64+100}=10 \sqrt{2} \mathrm{~N}
\)

\(\left(\because \mathrm{F}=\sqrt{\mathrm{F}_{\mathrm{x}}^{2}+\mathrm{F}_{\mathrm{y}}^{2}+\mathrm{F}_{\mathrm{z}}^{2}}\right)\)

Given, \(a=1 \mathrm{~ms}^{-2}\)
\(
\because \mathrm{F}=\mathrm{ma} \Rightarrow \mathrm{m}=\frac{10 \sqrt{2}}{1}=10 \sqrt{2} \mathrm{~kg}
\)

Hint

(a) \(\mathrm{F}=\frac{\mathrm{d}(\mathrm{Mv})}{\mathrm{dt}}=\mathrm{M} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dM}}{\mathrm{dt}}\); Given \(\mathrm{v}\) is constant,
\(
\therefore \mathrm{F}=\mathrm{v} \frac{\mathrm{dM}}{\mathrm{dt}} \text { But } \frac{\mathrm{dM}}{\mathrm{dt}}=\mathrm{M  kg} / \mathrm{s}
\)
\(
\therefore \mathrm{F}=\mathrm{Mv} \text { newton. }
\)

Hint

The magnitude of both velocity \(v_{1}\) and \(v_{2}\) are equal. Components of velocity along the wall are equal and are in the same direction.
But the components of velocity perpendicular to the wall are equal but opposite in direction.
So, the net change in momentum in the perpendicular direction,
\(
\Delta \mathrm{P}=m v_{1} \sin 30^{\circ}-\left(-m v_{2} \sin 30^{\circ}\right)
\)
\(\quad = 2 m v \sin 30^{\circ}\)       \(\left[\left|v_{1}\right| v_{2} \mid\right]\)

Average force acting on the wall will be given by
\(
\begin{aligned}
&\Rightarrow \quad F x t=2 m v \sin 30^{\circ} \\
&\Rightarrow \quad F=\frac{2 m v \sin 30^{\circ}}{t} \\
&\Rightarrow \quad F=\frac{2 \times 0.5^{2} \times 12 \times 1}{2 \times 0.25}=24 \\
&\quad F=24 \mathrm{~N}
\end{aligned}
\)

Hint

Force imparted \(=\) rate of change of momentum
\(F=\frac{\Delta p}{\Delta t}\) or \(F=\frac{p_{1}-p_{2}}{\Delta t}\) or \(F=\frac{m\left(v_{1}-v_{2}\right)}{\Delta t}\)
Here, mass of body \(m=150 \mathrm{~g}=0.150 \mathrm{~kg}\),
\(
v_{1}=20 \mathrm{~m} / \mathrm{s} \text {, }
\)
\(
v_{2}=0
\)
\(
\begin{aligned}
&\text { Time taken, } \Delta t=0.1 \mathrm{~s} \\
&\therefore \quad F=\frac{0.150 \times(20-0)}{0.1}=30 \mathrm{~N}
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (a) Change in momentum along the wall } \\
&=m v \cos 60^{\circ}-m v \cos 60^{\circ}=0 \\
&\text { Change in momentum perpendicular to the wall } \\
&=m v \sin 60^{\circ}-\left(-m v \sin 60^{\circ}\right)=2 m v \sin 60^{\circ} \\
&\therefore \text { Applied force }=\frac{\text { Change in momentum }}{\text { Time }} \\
&=\frac{2 m v \sin 60^{\circ}}{0.20} \\
&=\frac{2 \times 3 \times 10 \times \sqrt{3}}{2 \times 20}=50 \times 3 \sqrt{3} \\
&=150 \sqrt{3} \text { newton }
\end{aligned}\)

Hint

To calculate impulse first of all calculate the time during which force becomes zero.
We are given,\(F=600-2 \times 10^{5} t\)
When the bullet leaves the barrel, the force on the bullet becomes zero.
\(
\begin{aligned}
600-2 \times 10^{5} t &=0 \Rightarrow t=\frac{600}{2 \times 10^{5}} \\
&=3 \times 10^{-3} \mathrm{~s}
\end{aligned}
\)
Then, average impulse imparted to the bullet
\(
\begin{aligned}
I &=\int_{0}^{t} F d t \\
&=\int_{0}^{3 \times 10^{-3}}\left(600-2 \times 10^{5} t\right) d t \\
&=\left[600 t-\frac{2 \times 10^{5} t^{2}}{2}\right]_{0}^{3 \times 10^{-3}} \\
&=600 \times 3 \times 10^{-3}-10^{5} \times\left(3 \times 10^{-3}\right)^{2} \\
&=1.8-0.9=0.9 \mathrm{~N}-\mathrm{s}
\end{aligned}
\)

Hint

(b) Given : Mass of rocket \((\mathrm{m})=5000 \mathrm{~kg}\)
Exhaust speed \((v)=800 \mathrm{~m} / \mathrm{s}\)
Acceleration of rocket \(, a=20 \mathrm{~m} / \mathrm{s}^{2}\)
Gravitational acceleration, \( g=10 \mathrm{~m} / \mathrm{s}^{2}\)
Thrust, \(\Rightarrow \frac{v d m}{d t}\)
We know that upward force,
\(
F=m(g+a)=5000(10+20)
\)
\(
=5000 \times 30=150000 \mathrm{~N}
\)
This thrust gives upward force, \(F=\frac{v d m}{d t}\)
We also know that amount of gas ejected
\(
\Rightarrow\left(\frac{d m}{d t}\right)=\frac{F}{v}=\frac{150000}{800}=187.5 \mathrm{~kg} / \mathrm{s}
\)

Hint

(b) According to the second law of motion, the magnitude of the force can be calculated by.

Force \(F=m a\)
Here, \(F=10 \mathrm{~N}\)
\(
a=1 \mathrm{~m} / \mathrm{s}^{2} \Rightarrow \therefore m=\frac{F}{a}=\frac{10}{1}=10 \mathrm{~kg}
\)

Hint

(b) Mass \(=150 \mathrm{gm}=\frac{150}{1000} \mathrm{~kg}\)
Force = Mass \(\times\) acceleration
\(
=\frac{150}{1000} \times 20 \mathrm{~N}=3 \mathrm{~N}
\)
Impulsive force \(=F \cdot \Delta t=3 \times 0.1=0.3 \mathrm{~N}\)

Hint

(c) Velocity of the rocket \(u=300 \mathrm{~m} / \mathrm{s}\) and force \(\mathrm{F}=345 \mathrm{~N}\). According to Newton’s second law, thrust (force) = Rate of change of linear momentum.
\(
\text { Rate of combustion of fuel } =\frac{F}{u}=\left(\frac{d m}{d t}\right)=\left(\frac{345}{300}\right)=1.15 \mathrm{~kg} / \mathrm{sec}
\)

Hint

(b) Thrust on the satellite is the force with which the satellite moves upwards in space. It is given by
\(
F=-u \frac{d m}{d t}
\)
Here, initial velocity \(u=v\), rate of change of mass \(\frac{d m}{d t}=\alpha v\)
As we know that,
\(
F=-v \frac{d m}{d t}=-v(\alpha v)=-\alpha v^{2}
\)
Acceleration \(=\frac{F}{M}=-\frac{\alpha v^{2}}{M}\)

Hint

(b) According to Newton’s first law of motion, a body continues to be in a state of rest or of uniform motion, unless it is acted upon by an external force to change the state. Hence, Newton’s first law of motion is related to the physical independence of force.

Hint

(b) Thrust = weight
\(
\begin{aligned}
&\Rightarrow \frac{u d M}{d t}=m g \Rightarrow \frac{d M}{d t}=\frac{m g}{u} \\
&=\frac{600 \times 10}{1000}=6 \mathrm{~kg} \mathrm{~s}^{-1}
\end{aligned}
\)

Hint

(d) Impulse = final momentum – initial momentum \(=m\left(v_{2}-v_{1}\right)\)

Hint

(b) Given : Mass \(m_{1}=4 \mathrm{~kg}\) and \(m_{2}=6 \mathrm{~kg}\).
Acceleration of the system,
\(
\begin{aligned}
&a=\frac{\left(m_{2}-m_{1}\right) g}{\left(m_{2}+m_{1}\right)} \\
&\therefore a=\frac{(6-4) g}{6+4}=\frac{g}{5}
\end{aligned}
\)

Hint

(b) \(k x=\mathrm{mg}+3 \mathrm{mg}\)
Force on string,
\(\Rightarrow k x=4 \mathrm{mg}\)
After cutting the string

For block A: \(3 m a_{A}=4 \mathrm{mg}-3 \mathrm{mg}\)
\(a_{A}=\frac{4 m g-3 m g}{3 m}=\frac{g}{3}\)
\(
a_{A}=\frac{g}{3}
\)
For block B: \(m a_{B}=m g\); \(a_{B}=g\)

Hint

(a) Net force on particle in uniform circular motion is centripetal force \(\left(\frac{m v^{2}}{\ell}\right)\) which is provided by tension in string so the net force will be equal to tension i.e., \(\mathrm{T}\).

Hint

(b) Given, \(m_{A}=4 \mathrm{~kg}\)

\(m_{B}=2 \mathrm{~kg} \Rightarrow m_{C}=1 \mathrm{~kg}\)

So, total mass \(M=4+2+1=7 \mathrm{~kg}\)
Now, \(\quad F=M a \Rightarrow 14=7 a \Rightarrow a=2 \mathrm{~m} / \mathrm{s}^{2}\)
FBD of block A,

\(\begin{aligned}
&F-F^{\prime}=4 a \\
&F^{\prime}=14-4 \times 2 \Rightarrow F^{\prime}=6 \mathrm{~N}
\end{aligned}\)

Hint

(c) First of all, consider the forces on the blocks
For the Ist block, \(\quad\left[\because m_{1}=m_{2}=m_{3}\right]\)
\(
m g-T_{1}=m \times a \dots (i)
\)
Let us consider 2 nd and 3rd block as a system.
So, \(T_{1}-2 \mu m g=2 m \times a \dots (ii)\)
Solving Eqs. (i) and (ii),
\(
\begin{aligned}
&\Rightarrow \quad m g-T_{1}=m \times a \\
&\Rightarrow T_{1}-2 \mu m g=2 m \times a
\end{aligned}
\)
Addiing Eqs. (i) and (ii)
\(
m g(1-2 \mu)=3 m \times a \Rightarrow a=\frac{g}{3}(1-2 \mu)
\)

Hint

(a) Let upthrust of air be \(F_{a}\) then
When, the balloon is descending down with acceleration a from the free body diagram
\(\mathrm{F}_{\mathrm{a}}=\mathrm{mg}-\mathrm{ma} \Rightarrow F a=m(g-a) \dots (i)\)
For upward motion, \( F_{a}-(\mathrm{m}-\Delta \mathrm{m}) g=(\mathrm{m}-\Delta \mathrm{m}) a \dots (ii)\)
From equation (i) and (ii),
Therefore, \(\Delta \mathrm{m}=\frac{2 m a}{(g+a)}\)

Hint

(d) Since all the blocks are moving with constant velocity and we know that, if velocity is constant, acceleration of the body becomes zero. Hence, the net force on all the blocks will be zero.

Hint

(c) Total mass m
\(
\begin{aligned}
&=\text { Mass of lift }+\text { Mass of person } \\
&=940+60=1000 \mathrm{~kg}
\end{aligned}
\)
So, from the free body diagram below

Hence, \(T-1000 \times 10=1000 \times 1\)
\(
T=11000 \mathrm{~N}
\)

Hint

(a) The components of \(1 \mathrm{~N}\) and \(2 \mathrm{~N}\) forces along \(+x\)-axis \(=1 \cos 60^{\circ}+2 \sin 30^{\circ}\) \(=1 \times \frac{1}{2}+2 \times \frac{1}{2}=\frac{1}{2}+1=\frac{3}{2}=1.5 \mathrm{~N}\)
The component of \(4 \mathrm{~N}\) force along \(-x\)-axis \(=4 \sin 30^{\circ}=4 \times \frac{1}{2}=2 \mathrm{~N}\).
Therefore, if a force of \(0.5 \mathrm{~N}\) is applied along \(+x\) axis, the resultant force along \(x\)-axis will become zero and the resultant force will be obtained only along \(y\)-axis.

Hint

(a) Here, the lift is accelerating upward at the rate of a. Hence, the equation of motion is written as
\(
\begin{gathered}
R-m g=m a \\
28000-20000=2000 a \\
\quad\left[\because g=10 \mathrm{~ms}^{-2}\right] \\
\Rightarrow \quad a=\frac{8000}{2000}=4 \mathrm{~ms}^{-2} \text { upwards }
\end{gathered}
\)

Hint

(d) Condition for the max value of the mass of block B so that two blocks do not move is,
\(
\begin{aligned}
&m_{B} g=\mu_{s} m_{A} g \\
&\Rightarrow m_{B}=\mu_{s} m_{A} \\
&\text { or, } m_{B}=0.2 \times 2=0.4 \mathrm{~kg}
\end{aligned}
\)

Hint

(b) According to the condition,

\(\mathrm{N}=m a \sin \theta+m g \cos \theta\) \dots (i)
Also, \(m g \sin \theta=m a \cos \theta\) \dots (ii)
From (i) \& (ii), \(a=g \tan \theta\)
\(
\begin{aligned}
\therefore N &=m g \frac{\sin ^{2} \theta}{\cos \theta}+m g \cos \theta . \\
&=\frac{m g}{\cos \theta}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\frac{m g}{\cos \theta}
\end{aligned}
\)

Note: The condition for the body to be at rest relative to the inclined plane, \(a=g \sin \theta-b \operatorname{cos} \theta=0\) Horizontal acceleration, \(b=g \tan \theta\).

Hint

(a) Reading of the scale = Apparent wt. of the man \(=m(g+a)\)
\(
=80(10+5)=1200 \mathrm{~N}
\)

Hint

(a) Maximum bearable tension in the rope \(T=25 \times 10=250 \mathrm{~N}\)
From the figure below

\(T-m g=m a \quad\) or acceleration \(a=\frac{T-m g}{m}\)
Given, mass \(m=20 \mathrm{~kg}\),
\(
g=10 \mathrm{~m} / \mathrm{s}^{2} \text {, }
\)

Taking, \(T=250 \mathrm{~N}\)
Hence, \(a=\frac{250-20 \times 10}{20}=\frac{50}{20}\) \(=2.5 \mathrm{~m} / \mathrm{s}^{2}\)

Hint

(b) When lift move upwards with the same acceleration, then according to free body diagram of the left \(T-m g=m a\) or \(T=m(g+a)\)
Given,
\(
m=1000 \mathrm{~kg}, a=1 \mathrm{~m} / \mathrm{s}^{2} \text {, }
\)
\(
g=9.8 \mathrm{~m} / \mathrm{s}^{2}
\)
Thus, \(T=1000(9.8+1)=1000 \times 10.8\)
\(
=10800 \mathrm{~N}
\)

Hint

\(\text { (a) Let } \mathrm{T} \text { be the tension in the string. }\)

\(
\begin{aligned}
&\therefore 10 \mathrm{~g}-\mathrm{T}=10 a \dots (i)\\
&T-5 g=5 a \dots (ii)
\end{aligned}
\)
Adding (i) and (ii),
\(
5 g=15 a \Rightarrow a=\frac{g}{3} \mathrm{~m} / \mathrm{s}^{2}
\)
Note: When the pulley have a finite mass \(m\) then acceleration,
\(
a=\frac{m_{1}-m_{2}}{m_{1}+m_{2}+\frac{m}{2}}
\)

Hint

(a)
(i) When, mass is lifted upwards with an acceleration a, then according to free body diagram

\(\begin{aligned}
T_{1}-m g &=m a \Rightarrow T_{1}=m g+m a \\
T_{1} &=m(g+a)
\end{aligned}\)
Substituting the values, we obtain \(\therefore \quad T_{1}=(1)(9.8+4.9)=14.7 \mathrm{~N}\)
(ii) When, mass is lowered downwards with an acceleration a, then
\(
\begin{gathered}
m g-T_{2}=m a \\
\Rightarrow T_{2}=m g-m a=m(g-a)
\end{gathered}
\)
Substituting the values, we have
Then, ratio of tensions
\(
\frac{T_{1}}{T_{2}}=\frac{14.7}{4.9}=\frac{3}{1} \Rightarrow T_{1}: T_{2}=3: 1
\)

Hint

(c) Let \(\mathrm{T}\) be the tension in the branch of a tree when monkey is descending with acceleration a. Then \(\mathrm{mg}-T=m a\); and \(T=75 \%\) of weight of monkey,
\(
\therefore m a=m g-\left(\frac{75}{100}\right) m g=\left(\frac{1}{4}\right) m g
\)
or \(a=\frac{g}{4}\).

Hint

We can draw the situation in the following diagram

The frictional force, \(f=\mu N=\mu \mathrm{mg}\)
\(
[\because N=m g]
\)
From the Free body diagram(FBD), the resultant force is
\(
\begin{aligned}
|\mathbf{F}| &=\sqrt{N^{2}+f^{2}} \\
&=\sqrt{(m g)^{2}+(\mu m g)^{2}} \\
&=m g \sqrt{1+\mu^{2}}
\end{aligned}
\)
This is the minimum force required to move the object. But as the body is not moving
\(
\therefore \quad|\mathbf{F}| \leq m g \sqrt{1+\mu^{2}}
\)

Hint

(d) Coefficient of friction or sliding friction has no dimension.
\(
f=\mu_{\mathrm{s}} \mathrm{N} \Rightarrow \mu_{\mathrm{s}}=\frac{f}{N}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]
\)

Hint

Since the wedge is accelerating towards the right with a, thus a pseudo force acts in the left direction in order to keep the block stationary. As the system is in equilibrium.

\(\begin{aligned}
& \therefore & \Sigma F_{x} &=0 \\
\text { or } & & \Sigma F_{y} &=0 \\
\Rightarrow & R \sin \theta &=m a \\
\text { or } & m g \sin \theta &=m a \dots (i) \\
\text { Similarly, } & R \cos \theta &=m g
\end{aligned}\)
or \(m g \cos \theta=m g \dots (ii)\)
Dividing Eq. (i) by Eq (ii), we get
\(
\begin{aligned}
\frac{m g \sin \theta}{m g \cos \theta} &=\frac{m a}{m g} \\
\Rightarrow \quad \tan \theta &=\frac{a}{g}
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (b) For the motion of both the blocks }\\
&m_{1} \mathrm{a}=\mathrm{T}-\mu_{\mathrm{k}} \mathrm{m}_{1} \mathrm{~g}\\
&m_{2} g-T=m_{2} a\\
&\mathrm{a}=\frac{\mathrm{m}_{2} \mathrm{~g}-\mu_{\mathrm{k}} \mathrm{m}_{1} \mathrm{~g}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\\
&\mathrm{m}_{2} \mathrm{~g}-\mathrm{T}=\left(\mathrm{m}_{2}\right)\left(\frac{\mathrm{m}_{2} \mathrm{~g}-\mu_{\mathrm{k}} \mathrm{m}_{1} \mathrm{~g}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right)\\
&\text { solving we get tension in the string }\\
&\mathrm{T}=\frac{\mathrm{m}_{1} \mathrm{~m}_{2} \mathrm{~g}\left(1+\mu_{\mathrm{k}}\right) \mathrm{g}}{\mathrm{m}_{1}+\mathrm{m}_{2}}
\end{aligned}\)

Hint

(a) Coefficient of static friction, $\mu_{s}=\tan \theta$
\(
\Rightarrow \mu_{s}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=0.577 \cong 0.6
\)

So, distance covered by plank,
\(
\mathrm{S}=u t+\frac{1}{2} \mathrm{at}^{2}
\)
where, \(u=0\) and [/latex]a=g \sin \theta-\mu_{k}(g) \cos \theta[/latex]
\(
\begin{aligned}
&4=\frac{1}{2} a(4)^{2} \Rightarrow a=\frac{1}{2}=0.5 \\
&{\left[\because \theta=30^{\circ}, s=4 \mathrm{~m} \text { and } t=4 \mathrm{~s} \text { given }\right]} \\
&\Rightarrow \mu_{k}=\frac{0.9}{\sqrt{3}}=0.5
\end{aligned}
\)

Hint

(b) Net work done by the block in going from top to bottom of the inclined plane, must be equal to the work done by frictional force. The block may be stationary, when \(m g \sin \theta \cdot L=\mu m g \cos \theta \frac{L}{2}\)
or
\(
\begin{aligned}
\mu &=\frac{m g \sin \theta \cdot L}{m g \cos \theta \cdot \frac{L}{2}} \\
&=2 \frac{\sin \theta}{\cos \theta}=2 \tan \theta \\
\mu &=2 \tan \theta
\end{aligned}
\)

Hint

(c) Frictional force on the box \(f=\mu \mathrm{mg}\)
\(\therefore\) Acceleration in the box, \(a=\frac{f}{m}=\frac{\mu m g}{m}\)
\(
\begin{aligned}
& \mathrm{a}=\mu \mathrm{g}=5 \mathrm{~ms}^{-2} \\
& \mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \\
\Rightarrow \quad & 0=2^{2}+2 \times(5) \mathrm{s} \\
\Rightarrow & \mathrm{s}=-\frac{2}{5} \text { w.r.t. belt } \\
\Rightarrow & \text { distance }=0.4 \mathrm{~m}
\end{aligned}
\)

Hint

(c) Forces acting on the block are as shown in the figure below. Normal reaction \(\mathrm{N}\) is provided by the force ma due to acceleration \(\alpha\)

For the block not to fall, frictional force,
\(
\begin{aligned}
&f_{r} \geq \mathrm{mg} \\
&\Rightarrow \quad \mu \mathrm{N} \geq \mathrm{mg} \\
&\Rightarrow \quad \mu \mathrm{m\alpha} \geq \mathrm{mg} \\
&\Rightarrow \quad \alpha \geq \mathrm{g} / \mu
\end{aligned}
\)

Hint

(d) Block B will come to rest, if the force applied to it will vanish due to frictional force acting between block B and the surface, i.e. frictional force = force applied

\(
\begin{array}{ll}
\text { i.e. } & \mu m g=m a \\
\text { or } & \mu m g=m\left(\frac{v}{t}\right) \text { or } t=\frac{v}{\mu g}
\end{array}
\)

Hint

(b) \(a=\frac{F-\mu R}{m}=\frac{100-0.5 \times(10 \times 10)}{10}=5 \mathrm{~ms}^{-2}\)

Hint

(a) Maximum friction force \(=\mu \mathrm{mg}\) \(=.6 \times 1 \times 9.8=5.88 \mathrm{~N}\)
But here required friction force \(=m a=1 \times 5=5 \mathrm{~N}\)

Hint

(b) As there is only a gravitational field that works. We know it is a conservative field and depends only on the endpoints. So, \(V_{M}=V_{B}\)

Hint

(d) The angle of repose or angle of sliding is defined as the minimum angle of inclination of a plane with the horizontal, such that a body placed on the plane just begins to slide down.

\(A B\) is an inclined plane such that a body placed on it just begins to slide down
\(\angle B A C=\theta=\) angle of repose
In equilibrium,
\(
\begin{aligned}
& f =m g \sin \theta \\
\text { and } & R =m g \cos \theta \\
\therefore & \frac{f}{R} =\frac{m g \sin \theta}{m g \cos \theta}=\tan \theta \\
\text { i.e. } & \mu =\tan \theta
\end{aligned}
\)

Hint

(b) Here \(\mathrm{u}=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0\); \(a=-\mu g=-0.5 \times 10=-5 \mathrm{~m} / \mathrm{s}^{2}\)
As \(v^{2}=u^{2}+2 a s\),
\(
\therefore \quad s=\frac{\left(v^{2}-u^{2}\right)}{2 a}=\frac{\left(0-(20)^{2}\right.}{2 \times(-5)}=40 \mathrm{~m}
\)

Hint

(a) The force of friction should balance the weight of the chain hanging. If M is the mass of the whole chain of length L and x is the length of the chain hanging to balance, and \(\mu\) is the coefficient of friction then
\(
\begin{array}{ll}
& \mu \frac{M}{L}(L-x) g=\frac{M}{L} x g \\
\text { or } & \mu(L-x)=x \\
\text { or } & x=\frac{\mu L}{\mu+1}=\frac{0.25 L}{1.25} \quad(A s, \mu=0.25) \\
\therefore & x=\frac{L}{5} \text { or } \frac{x}{L}=\frac{1}{5}=\frac{1}{5} \times 100=20 \%
\end{array}
\)

Hint

(c) Given the mass of the block, \(\mathrm{m}=10 \mathrm{~kg}\); radius of a cylindrical drum, \(\mathrm{r}-1 \mathrm{~m}\); coefficient of friction between the block and the inner wall of the cylinder \(\mu=0.1\);
Minimum angular velocity \(\omega_{\text {min }}\)

From the below figure, it can be concluded that the block will be stationary when the limiting friction \(\left(f_{L}\right)\) is equal to or greater than the downward force or weight of the block, i.e.

\(
\begin{aligned}
&\mathrm{f}_{\mathrm{L}} \geq \mathrm{mg} \\
&\Rightarrow \mu \mathrm{N} \geq \mathrm{mg} \\
&\Rightarrow \mu \mathrm{r} \omega^{2} \geq \mathrm{mg}
\end{aligned}
\)
Hrere, \(\mathrm{N}=\mathrm{mr} \omega^{2}\)
or, \(\mathrm{m} \geq \sqrt{\frac{\mathrm{g}}{\mathrm{r} \mu}}\)
\(
\begin{aligned}
&\text { or, } \omega_{\min }=\sqrt{\frac{\mathrm{g}}{\mathrm{r} \mu}} \\
&\therefore \omega_{\min }=\sqrt{\frac{10}{0.1 \times 1}}=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)

Hint

When, a plane is inclined to the horizontal at an angle \(\theta\), which is greater than the angle of repose, the body placed on the inclined plane slides down with an acceleration a.

As it is clear from the figure
\(R=m g \cos \theta \dots (i)\)
Net force on the body down the inclined plane which means it is sliding downwards
\(F=m g \sin \theta-f \dots (ii)\)

i.e. \(\quad F=m a=m g \sin \theta-\mu R \quad(f=\mu R) \\\) \(\therefore \quad m a=m g \sin \theta-\mu m g \cos \theta\) \(=m g(\sin \theta-\mu \cos \theta)\) Hence, \(a=g(\sin \theta-\mu \cos \theta)\) \(\therefore\) Time taken by body to slide down the plane \(\quad t_{1}=\sqrt{\frac{2 s}{a}}=\sqrt{g(\sin \theta-\mu \cos \theta)}\)

When friction is absent, then time taken to slide down the plane
\(t_{2}=\sqrt{\frac{2 s}{g \sin \theta}} \Rightarrow \because t_{1}=2 t_{2} \quad\) (given)
\(\therefore \quad t_{1}^{2}=4 t_{2}^{2}\)
or \(\frac{2 s}{g(\sin \theta-\mu \cos \theta)}=\frac{2 s \times 4}{g \sin \theta}\)
or \(\sin \theta=4 \sin \theta-4 \mu \cos \theta\)
or \(\mu=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 45^{\circ}\)
\(\therefore \quad \mu=\frac{3}{4}=0.75\)

Hint

\(\begin{aligned}
&\text { (c) } \mathrm{T}-\mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}[\text { centripetal force }]=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \\
&\Rightarrow \quad \mathrm{T}=\mathrm{mg}+\frac{\mathrm{mv}^{2}}{\mathrm{r}}
\end{aligned}\)

As the velocity is maximum at the lowest point so tension is maximum at the lowest position of the mass. so, the chance of breaking is maximum.

Hint

(b) According to the question, a car is negotiating a curved road of radius \(R\). The road is banked at angle \(\theta\) and the coefficient of friction between the tires of the car and the road is \(\mu_{s}\). So, this given situation can be drawn as shown in the figure below.

Considering the case of vertical equilibrium
\(\quad N \cos \theta=m g+f_{l} \sin \theta\) \(\Rightarrow \quad m g=N \cos \theta-f_{l} \sin \theta \quad \ldots\) (i) Considering the case of horizontal equilibrium, \(\quad N \sin \theta+f_{l} \cos \theta=\frac{m v^{2}}{R} \quad\)…(ii)
Divide eqs. (i) and (ii), we get
\(
\begin{aligned}
\frac{v^{2}}{R g} &=\frac{\sin \theta+\mu_{s} \cos \theta}{\cos \theta-\mu_{s} \sin \theta} \\
\Rightarrow \quad v &=\sqrt{R g\left(\frac{\sin \theta+\mu_{s} \cos \theta}{\cos \theta-\mu_{s} \sin \theta}\right)} \\
\Rightarrow \quad v &=\sqrt{R g\left(\frac{\tan \theta+\mu_{s}}{1-\mu_{s} \tan \theta}\right)}
\end{aligned}
\)

Hint

(d) According to the question, we have
Let the tension at point \(A\) be \(T_{A}\). So, from Newton’s second law
\(
T_{A}-m g=\frac{m v_{c}^{2}}{R}
\)
Energy at point \(A=\frac{1}{2} m v_{0}^{2}\)
Energy at point \(C\) is
\(
\frac{1}{2} m v_{c}^{2}+m g \times 2 R
\)

Applying Newton’s 2nd law at point \(C\)
\(
T_{c}+m g=\frac{m v_{c}^{2}}{R}
\)
To complete the loop \(T_{c} \geq 0\)
So, \(m g=\frac{m v_{c}^{2}}{R}\)
\(\Rightarrow v_{c}=\sqrt{g R}\)
From Eqs. (i) and (ii)by conservation of energy
\(
\begin{aligned}
& \frac{1}{2} m v_{0}^{2} &=\frac{1}{2} m v_{c}^{2}+2 m g R \\
\Rightarrow \quad & \frac{1}{2} m v_{0}^{2} &=\frac{1}{2} m g R+2 m g R \times 2 \\
\Rightarrow \quad & v_{0}^{2} &=g R+4 g R \\
\Rightarrow \quad & v_{0} &=\sqrt{5 g R}
\end{aligned}
\)

Hint

(d) According to question, two stones experience same centripetal force
i.e., \(\mathrm{F}_{\mathrm{C}_{1}}=\mathrm{F}_{\mathrm{C}_{2}}\)
or, \(\frac{\mathrm{mv}_{1}^{2}}{\mathrm{r}}=\frac{2 \mathrm{mv}_{2}^{2}}{(\mathrm{r} / 2)} \quad\) or, \(\mathrm{V}_{1}^{2}=4 \mathrm{~V}_{2}^{2}\)
So, \(\mathrm{V}_{1}=2 \mathrm{~V}_{2}\) i.e., \(n=2\)

Hint

(d) Given; speed \(=10 \mathrm{~m} / \mathrm{s}\); radius \(r=10 \mathrm{~m}\)
Angle made by the wire with the vertical is \(\theta\).
Then, \(\mathrm{T} \cos \theta=\mathrm{mg}\)
\(
\begin{aligned}
&\mathrm{T} \sin \theta=\frac{m v^{2}}{r} \\
&\text { So, } \\
&\tan \theta=\frac{v^{2}}{r g}=\frac{10^{2}}{10 \times 10}=1 \quad \frac{m v^{2}}{r} \\
&\Rightarrow \theta=45^{\circ}=\frac{\pi}{4}
\end{aligned}
\)

Hint

(b) The angle of banking
\(
\tan \theta=\frac{v^{2}}{r g}
\)
Given, \(\quad \theta=45^{\circ}\)
Radius of banked curve road
\(
\begin{array}{rl}
r=90 & \mathrm{~m} \text { and } g=10 \mathrm{~m} / \mathrm{s}^{2} \\
\Rightarrow \quad \tan 45^{\circ} & =\frac{v^{2}}{90 \times 10} \\
v & =\sqrt{90 \times 10 \times \tan 45^{\circ}} \\
& =\sqrt{90 \times 10 \times 1}=30 \mathrm{~m} / \mathrm{s}
\end{array}
\)

Hint

(d) For smooth driving maximum speed of car \(v\) then,
\(
\begin{aligned}
&\frac{m v^{2}}{R}=\mu_{s} m g \\
&v=\sqrt{\mu_{s} R g}
\end{aligned}
\)

Hint

(c) Centripetal force \(=\frac{m v^{2}}{r}=\frac{500 \times(10)^{2}}{50}\)
\(
=1000 \mathrm{~N} \quad[\because 36 \mathrm{~km} / \mathrm{hr}=10 \mathrm{~m} / \mathrm{s}]
\)

Hint

(a) Given : Mass \(m=0.4 \mathrm{~kg}\)
Its frequency \(n=2 \mathrm{rev} / \mathrm{sec}\)
Radius \(r=1.2 \mathrm{~m}\). We know that linear velocity of the body \((v)=\omega r=(2 \pi n) r\)
\(
=2 \times 3.14 \times 2 \times 1.2=15.08 \mathrm{~m} / \mathrm{s} \text {. }
\)
Therefore, tension in the string when the body is at the top of the circle (T)
\(
\begin{aligned}
&=\frac{\mathrm{mv}^{2}}{\mathrm{r}}-\mathrm{mg}=\frac{0.4 \times(15.08)^{2}}{2}-(0.4 \times 9.8) \\
&=45.78-3.92=41.56 \mathrm{~N}
\end{aligned}
\)

Hint

(a) \(r=30 \mathrm{~m}\) and \(\mu=0.4\).
\(
v_{\max }=\sqrt{\mu r g}=\sqrt{0.4 \times 30 \times 9.8}=10.84 \mathrm{~m} / \mathrm{s}
\)

Hint

(c) On the diametrically opposite points, the velocities have the same magnitude but in opposite directions. Therefore, the change in momentum is MV \(-(-M V)=2 \mathrm{MV}\)

Hint

(b) Cream gets separated from churned milk due to centrifugal force.

Note: Centrifugal force is a pseudo force that is equal and opposite to the centripetal force. Centripetal force can be a mechanical, electrical, or magnetic force.

Hint

(b) Given, the mass of the ball dropped from the height, \(m=0.15 \mathrm{~kg}\)
The height from the ball dropped,
\(h=10 \mathrm{~m}\)
We know that,
\(\mid\) mpulse \(|=m| \Delta \bar{v} \mid\)
where, \(\quad \Delta \bar{v}=v_{2}-v_{1}\)
Here, \(v_{2}\) is velocity reaches to the same height,
\(v_{1}\) is velocity just before striking to the ground.

For case (1), ball dropped from the \(10 \mathrm{~m}\) height and strikes to the ground. Now, the velocity of the ball just before striking to the ground is
\(
\begin{aligned}
&\quad v_{1}=-\sqrt{2 g h} \\
&\Rightarrow \quad v_{1}=-\sqrt{2(10)(10)} \\
&\Rightarrow \quad v_{1}=-10 \sqrt{2} \mathrm{~m} / \mathrm{s} \\
&\text { For case }(2) \text { ball rebounds to the same } \\
&\text { height. } \\
&\text { The velocity with which the ball just } \\
&\text { reaches to the same height, } \\
&\qquad v_{2}=\sqrt{2 g h} \\
&\Rightarrow \quad v_{2}=\sqrt{2(10)(10)} \\
&\Rightarrow \quad \quad v_{2}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Now, magnitude of the impulse imparted to the ball,
\(
\begin{aligned}
\text { |lmpulse } \mid &=m|\Delta \bar{v}| \\
&=0.15|10 \sqrt{2}-(-10 \sqrt{2})| \\
&=0.15|2 \times 10 \sqrt{2}| \\
&=4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}
\end{aligned}
\)

Hint

(b) As the truck move to the right, so the bob will move to the left due to inertia of rest with acceleration a. From the diagram below  as the string moves by an angle of \(\theta\) with the vertical then the tangent angle is
\(
\tan \theta=\frac{m a}{m g} \Rightarrow \theta=\tan ^{-1}\left(\frac{a}{g}\right)
\)

Hint

(a) As we know, the impulse is imparted due to changes in perpendicular components of the momentum of the ball.
\(
\begin{aligned}
&J=\Delta p=m v_{f}-m v_{i} \\
&=m v \cos 60^{\circ}-\left(-m v \cos 60^{\circ}\right) \\
&=2 m v \cos 60^{\circ}=2 m v \times \frac{1}{2}=m v
\end{aligned}
\)

Hint

(c) According to the law of conservation of momentum, \(p_{i}=p_{f}\)
\(
\Rightarrow \quad(0.01) \times 400+0=2 v+(0.01) v^{\prime} \ldots \text { (i) }
\)
Also, velocity \(v\) of the block just after the collision is
\(
\begin{aligned}
&\quad v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 0.1}=\sqrt{2} \quad \text {…(ii) } \\
&\Rightarrow \text { From Eqs. (i) and (ii), we have } \\
&\quad v^{\prime} \approx 120 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(b) Momentum is conserved before and after the collision.
We have, \(\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3}=0 \quad[\because p=m v]\)
\(
\begin{array}{lrlr}
\therefore & & 1 \times 12 \mathbf{i}+2 \times 8 \mathbf{j}+\mathbf{p}_{3}=0 \\
\Rightarrow & & 12 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}+\mathbf{p}_{3}=0 \\
\Rightarrow & & \mathbf{p}_{3}  =-(12 \mathbf{i}+16 \mathbf{j}) \\
& \therefore & \mathbf{p}_{3}  =\sqrt{(12)^{2}+(16)^{2}} \\
& & =\sqrt{144+256} \\
& & =20 \mathrm{~kg}-\mathrm{m} / \mathrm{s} \\
& \text { Now, } & \mathbf{p}_{3}  =m_{3} v_{3} \\
\Rightarrow & & m_{3}  =\frac{\mathbf{p}_{3}}{v_{3}}=\frac{20}{4}=5 \mathrm{~kg}
\end{array}
\)

Hint

(c) Here, \(\mathbf{p}_{1}=m_{2} v \mathbf{i}+m_{1} \times 0\)
\(
\mathbf{p}_{f}=m_{2} \frac{v}{2} \hat{\mathbf{j}}+m_{1} \times \mathbf{v}_{1}
\)
Law of conservation of momentum

\(\begin{aligned}
\mathbf{p}_{i} &=\mathbf{p}_{f} \\
m_{2} v \hat{\mathbf{i}} &=m_{2} \frac{v}{2} \hat{\mathbf{j}}+m_{1} \times \mathbf{v}_{1} \\
\mathbf{v}_{1} &=\frac{m_{2}}{m_{1}} v \hat{\mathbf{i}}+\frac{m_{2}}{m_{1}} \frac{v}{2} \hat{\mathbf{j}}
\end{aligned}\)

From this equation, we can find \(\tan \theta=\frac{y}{x}=\frac{1}{2}, \theta=\tan ^{-1}\left(\frac{1}{2}\right)\) to the \(x\)-axis.

Hint

(b) As, \(m r=\) constant
\(m_{1} r_{1}=m_{2} r_{2}\)
\(\Rightarrow \quad r_{2}=\frac{m_{1} r_{1}}{m_{2}}=\frac{0.5 \times 10}{50}=0.1\)
So, distance travelled by the man will be \(10+0.1=10.1 \mathrm{~m}\)

Hint

(b) According to Newton’s 2nd law, force applied on an object is equal to the rate of change of momentum.
\(
\begin{array}{ll}
\text { i.e. } \quad & \mathbf{F}=\frac{d \mathbf{p}}{d t} \\
\text { or } & \mathbf{F}=m \frac{d v}{d t} \dots (i)
\end{array}
\)
Given, \(m=3 \mathrm{~kg}, t=3 \mathrm{~s}, \mathbf{F}=\left(6 \mathrm{t}^{2} \hat{\mathbf{i}}+4 \mathrm{t} \hat{\mathbf{j}}\right) \mathrm{N}\)
Substituting these values in Eq. (i), we get
\(
\left(6 t^{2} \hat{\mathbf{i}}+4 t \hat{\mathbf{j}}\right)=3 \frac{d \mathbf{v}}{d t}
\)
or \(\quad d \mathbf{v}=\frac{1}{3}\left(6 t^{2} \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\right) d \mathrm{t}\)
Now, taking integration of both sides, we get
\(
\begin{aligned}
\int d \mathbf{v} &=\int_{0}^{t} \frac{1}{3}\left(6 t^{2} \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\right) d t \\
\mathbf{v} &=\frac{1}{3} \int_{0}^{t}\left(6 t^{2} \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\right) d t
\end{aligned}
\)

\(\begin{aligned}
&\text { but } t=3 s \quad \text { (given) }\\
&\therefore \quad \mathbf{v}=\frac{1}{3} \int_{0}^{3}\left(6 t^{2} \hat{\mathbf{i}}+4 \hat{\mathrm{j}}\right) d t\\
&\mathbf{v}=\frac{1}{3}\left[\frac{6 t^{3}}{3} \hat{\mathbf{i}}+\frac{4 t^{2}}{2} \hat{\mathbf{j}}\right]_{0}^{3}\\
&\mathbf{v}=\frac{1}{3}\left[2(3)^{3} \hat{\mathbf{i}}+2(3)^{2} \hat{\mathbf{j}}\right]\\
&\mathbf{v}=\frac{1}{3}[54 \hat{\mathbf{i}}+18 \hat{\mathbf{j}}]\\
&\text { or } \mathbf{v}=18 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}
\end{aligned}\)

Hint

(b) Apply conservation of momentum with direction.
Let \(u\) be the velocity and \(\theta\) the direction of the third piece as shown.

The square of momentum of third piece is equal to sum of squares of momentum first and second pieces.
\(
\begin{aligned}
p_{3}^{2} &=p_{1}^{2}+p_{2}^{2} \\
\text { or } \quad p_{3} &=\sqrt{p_{1}^{2}+p_{2}^{2}} \\
& \text { or } \quad 3 m v_{3} &=\sqrt{(m \times 30)^{2}+(m \times 30)^{2}} \\
\text { or } & v_{3} &=\frac{30 \sqrt{2}}{3}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a) According to 1 st equation of motion, velocity of particle after \(5 \mathrm{~s}\)
\(
\begin{aligned}
v &=u-g t \\
v &=100-10 \times 5 \\
&=100-50=50 \mathrm{~m} / \mathrm{s} \quad \text { (upwards) }
\end{aligned}
\)
Applying conservation of linear momentum gives
\(
M v=m_{1} v_{1}+m_{2} v_{2} \dots (i)
\)
Taking upward direction positive, the velocity \(v_{1}\) will be negative.
\(
\therefore \quad v_{1}=-25 \mathrm{~m} / \mathrm{s}, \quad v=50 \mathrm{~m} / \mathrm{s}
\)
Also, \(M=1 \mathrm{~kg}, \mathrm{~m}_{1}=400 \mathrm{~g}=0.4 \mathrm{~kg}\)
and \(m_{2}=\left(M-m_{1}\right)=1-0.4=0.6 \mathrm{~kg}\)
Thus, Eq. (i) becomes,
\(
\begin{aligned}
1 \times 50 &=0.4 \times(-25)+0.6 v_{2} \\
\text { or } \quad 50 &=-10+0.6 v_{2}
\end{aligned}
\)

or \(0.6 v_{2}=60\) or \(v_{2}=\frac{60}{0.6}=100 \mathrm{~m} / \mathrm{s}\)
As \(v_{2}\) is positive, therefore the other part will move upwards with a velocity of \(100 \mathrm{~m} / \mathrm{s}\).

Hint

(a) Whenever there is change in the mass w.r.t. time, apply \(F=-v \frac{d m}{d t}\)
Thrust force on the rocket
\(F_{t}=v_{r}\left(-\frac{d m}{d t}\right)\)  (upwards)
Rate of combustion of fuel
\(
\begin{aligned}
&\text { Rate of combustion of fuel } \\
&\qquad \frac{d m}{d t}=\frac{F_{t}}{v_{r}} \\
&\text { Given, } \quad \begin{aligned}
F_{t} &=210 \mathrm{~N} \\
v_{r} &=300 \mathrm{~m} / \mathrm{s} \\
-\frac{d m}{d t} &=\frac{210}{300}=0.7 \mathrm{~kg} / \mathrm{s}
\end{aligned}
\end{aligned}
\)

Hint

(b) Thrust on the satellite is the force with which the satellite moves upwards in space. It is given by \(F=-u \frac{d m}{d t}\)
Here, initial velocity \(u=v\), rate of change of mass \(\frac{d m}{d t}=\alpha v\)
As we know,
\(
F=-v \frac{d m}{d t}=-v(\alpha v)=-\alpha v^{2}
\)
Acceleration \(=\frac{F}{M}=-\frac{\alpha v^{2}}{M}\)

Hint

(a) The given situation is shown in the following diagram.

If  \(a\) be the acceleration of the system then, equation of motion of \(10 \mathrm{~kg}\) trolly,
\(
\begin{gathered}
T-\mu R=10 a \\
\Rightarrow \quad T-0.05 \times 10 g=10 a \\
{[\because \mu=0.05, R=10 g]} \\
\Rightarrow \quad T-0.05 \times 10 \times 10=10 a \\
T – 5 =10 a \dots (i)\\
\text { Equation of motion of } 2 \mathrm{~kg} \text { block, } \\
2 g-T=2 a \\
2 \times 10-T=2 a \\
20-T=2 a  \dots (ii)\\
\text { Adding Eqs. (i) and (ii), we have } \\
20-5=12 a \\
15=12 a \\
\Rightarrow \quad a=\frac{15}{12}=\frac{5}{4}=1.25 \mathrm{~ms}^{-2}
\end{gathered}
\)

Hint

(c) Free body diagram of the block is

From Newton’s second law along \(X\)-axis
i.e. \(\quad F-f=m a\)
or \(\quad F-\mu m g=m a\)
or \(\quad a=\frac{F-\mu m g}{m}\)
Given, \(F=100 \mathrm{~N}, \mu=0.5, m=10 \mathrm{~kg}\),
\(
g=10 \mathrm{~m} / \mathrm{s}^{2}
\)
Substituting, the values in the above relation for the acceleration of block,
\(
a=\frac{(100)-(0.5)(10)(10)}{(10)}=5 \mathrm{~m} / \mathrm{s}^{2}
\)

Hint

(b) When static friction is present, then the acceleration of a body is given by \(a=-\mu \mathrm{g}\) Here, initial velocity
\(
\mathrm{u}=72 \mathrm{~km} \mathrm{~h}^{-1}=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s}
\)
Final velocity \(v=0\)
\(
\therefore \quad a=-\mu g=-0.5 \times 10=-5 \mathrm{~m} / \mathrm{s}^{2}
\)
Now, from third equation of motion,
i.e. \(v^{2}=u^{2}+2\) as
\(
\mathrm{s}=\frac{v^{2}-u^{2}}{2 a}=\frac{0-(20)^{2}}{2 \times(-5)}=40 \mathrm{~m}
\)

Hint

(d) According to given question, a uniform circular disc of radius \(50 \mathrm{~cm}\) at rest is free to turn about an axis having perpendicular to its plane and passes through its centre.

\(\therefore\) Angular acceleration, \(\alpha=2\) rad s \(^{-2}\) (given)
Angular speed, \(\omega=\alpha t=4 \mathrm{rad} \mathrm{s}^{-1}\)
because Centripetal acceleration, \(a_{c}=\omega^{2} r\) \(=(4)^{2} \times 0.5\)
\(=16 \times 0.5\)
\(a_{c}=8 \mathrm{~m} / \mathrm{s}^{2}\)
because Linear acceleration at the end of \(2 \mathrm{~s}\),
\(
a_{t}=\alpha r=2 \times 0.5 \Rightarrow a_{t}=1 \mathrm{~m} / \mathrm{s}^{2}
\)
Therefore, the net acceleration at the end of \(2.0 \mathrm{~s}\) is given by

\(\begin{gathered}
a=\sqrt{a_{c}^{2}+a_{t}^{2}} \\
a=\sqrt{(8)^{2}+(1)^{2}}=\sqrt{65} \Rightarrow a \approx 8 \mathrm{~m} / \mathrm{s}^{2}
\end{gathered}\)

Hint

(c) When the disc spins, the frictional force between the gramophone record and coin is \(\mu \mathrm{mg}\). The coin will revolve with the record, if

\(F_{\text {frictional }} \geq F_{\text {centripetal }}\)
i.e. \(\mu m g \geq m \omega^{2} r\)
\(
\frac{\mu g}{r} \geq \omega^{2}
\)

Hint

(a) For a ball to move in a horizontal circle, the ball should satisfy the condition
Tension in the string = Centripetal force
\(
\begin{aligned}
&\Rightarrow \quad T_{\max }=\frac{M v_{\max }^{2}}{R} \\
&\Rightarrow \quad v_{\max }=\sqrt{\frac{T_{\max } \cdot R}{M}}
\end{aligned}
\)
Making substitution, we obtain
\(
v_{\max }=\sqrt{\frac{25 \times 1.96}{0.25}}=\sqrt{196}=14 \mathrm{~m} / \mathrm{s}
\)
In a vertical circle, the tension at the highest point is zero and at the lowest point is maximum.

Hint

(d) As both cars take the same time to complete the circle and as \(\omega=\frac{2 \pi}{t}\), therefore the ratio of angular speeds of the cars will be \(1: 1\).

Hint

\(
\begin{aligned}
& S_{n^{t h}}=u+\frac{1}{2} a(2 n-1) \\
& S_{1^{s t}}=\frac{1}{2} g(2 \times 1-1)=\frac{g}{2} \\
& S_{2^{n d}}=\frac{1}{2} g(2 \times 2-1)=3\left(\frac{1}{2} g\right)
\end{aligned}
\)
\(
\begin{aligned}
& S_{3^{\text {rd }}}=\frac{1}{2} g(2 \times 3-1)=5 \times\left(\frac{1}{2} g\right) \\
& S_{4^{\text {th }}}=\frac{1}{2} g(2 \times 4-1)=7 \times\left(\frac{1}{2} g\right) \\
& S_{1^{s t}}: S_{2^{\text {nd }}}: S_{3^{\text {rd }}}: S_{4^{\text {th }}} \\
& =1: 3: 5: 7
\end{aligned}
\)

Hint

\(
\begin{aligned}
& m a_{\max }=\mu_s m g \\
& a_{\max }=\mu_s g=0.15 \times 10=1.5 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)