Sample Exercises

Hint

(a)

\(\begin{aligned}
&\mathrm{m}=2 \mathrm{~kg} \\
&\mathrm{~s}=10 \mathrm{~m} \\
&\text { Let, acceleration }=a, \text { Initial velocity } u=0 \\
&\mathrm{~S}=\mathrm{ut}+1 / 2 \mathrm{at}^{2} \\
&\Rightarrow 10=1 / 2 a\left(2^{2}\right) \Rightarrow 10=2 a \Rightarrow a=5 \mathrm{~m} / \mathrm{s}^{2} \\
&\text { Force: } F=m a=2 \times 5=10 \mathrm{~N}
\end{aligned}\)

Hint

(c) Given, Initial speed of the car, \(u=40 \mathrm{~km} / \mathrm{hr}\)
\(
=\frac{4000}{3600}=11.11 \mathrm{~m} / \mathrm{s}
\)
Final speed of the car, \(v=0\)
Mass of the car, \(m=2000 \mathrm{~kg}\)
Distance to be travelled by the car before coming to rest, \(s=4 \mathrm{~m}\)
Acceleration,
\(
\begin{aligned}
&a=\frac{v^{2}-u^{2}}{2 s} \\
&\Rightarrow a=\frac{0^{2}-(11.11)^{2}}{2 \times 4}=\frac{-123.43}{8}=-15.42 \mathrm{~m} / \mathrm{s}^{2} \\
&\therefore \text { Average force to be applied to stop the car, } \mathrm{F}=\mathrm{ma} \\
&\Rightarrow \mathrm{F}=2000 \times 15.42 \approx 3.1 \times 10^{4} \mathrm{~N}
\end{aligned}
\)

Hint

(b) Initial velocity \(\quad u=0\) (negligible)
\(v=5 \times 10^{6} \mathrm{~m} / \mathrm{s} .\)
\(\mathrm{s}=1 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m} .\)
acceleration \(a=\frac{v^{2}-u^{2}}{2 \mathrm{~s}}=\frac{\left(5 \times 10^{6}\right)^{2}-0}{2 \times 1 \times 10^{-2}}=\frac{25 \times 10^{12}}{2 \times 10^{-2}}=12.5 \times 10^{14} \mathrm{~ms}^{-2}\)
\(\mathrm{~F}=\mathrm{ma}=9.1 \times 10^{-31} \times 12.5 \times 10^{14}=113.75 \times 10^{-17}=1.1 \times 10^{-15} \mathrm{~N} .\)

Hint

(d) The free body diagram for the problem is shown below

From the free-body diagram of the \(0.3 \mathrm{~kg}\) block,
\(
\begin{aligned}
&\mathrm{T}=0.3 \mathrm{~g} \\
&\Rightarrow \mathrm{T}=0.3 \times 10=3 \mathrm{~N}
\end{aligned}
\)
Now, from the free-body diagram of the \(0.2 \mathrm{~kg}\) block,
\(
\begin{aligned}
&\mathrm{T}_{1}=0.2 \mathrm{~g}+\mathrm{T} \\
&\Rightarrow \mathrm{T}_{1}=0.2 \times 10+3=5 \mathrm{~N}
\end{aligned}
\)
\(\therefore\) The tensions in the two strings are \(5 \mathrm{~N}\) and \(3 \mathrm{~N}\), respectively.

Hint

(a) Let a be the common acceleration of the blocks.
For block-1,
\(F-T=m a \dots (1)\)
For block-2,
\(\mathrm{T}=\mathrm{ma} \quad . .(2)\)
Subtracting equation (2) from (1), we get:
\(
\begin{aligned}
&F-2 T=0 \\
&\Rightarrow T=\frac{F}{2}
\end{aligned}
\)

Hint

(a) Given, Mass of the particle, \(\mathrm{m}=50 \mathrm{~g}=5 \times 10^{-2} \mathrm{~kg}\)
Slope of the v-t graph gives acceleration.
At \(t=2 \mathrm{~s}\),
Slope \(=\frac{15}{3}=5 \mathrm{~m} / \mathrm{s}^{2}\)
So, acceleration, \(a=5 \mathrm{~m} / \mathrm{s}^{2}\)
\(F=m a=5 \times 10^{-2} \times 5\)
\(\Rightarrow \mathrm{F}=0.25 \mathrm{~N}\) along the motion.
At \(t=4 \mathrm{~s}\),
Slope \(=0\)
So, acceleration, \(a=0\)
\(
\Rightarrow F=0
\)

At \(t=6 \mathrm{sec}\),
Slope \(=\frac{-15}{3}=-5 m / \mathrm{s}^{2}\)
So, acceleration, \(a=-5 \mathrm{~m} / \mathrm{s}^{2}\)
\(
\mathrm{F}=\mathrm{ma}=-5 \times 10^{-2} \times 5
\)
\(\Rightarrow F=-0.25 \mathrm{~N}\) along the motion
or, \(F=0.25 \mathrm{~N}\) opposite the motion.

Hint

(b) Let, \(F \rightarrow\) contact force between \(m_{A} \& m_{B}\).
And, \(f \rightarrow\) force exerted by experimenter.
\(\mathrm{F}+\mathrm{m}_{\mathrm{A}} \mathrm{a}-\mathrm{f}=0\) \(\Rightarrow \mathrm{F}=\mathrm{f}-\mathrm{m}_{\mathrm{A}} \mathrm{a} \ldots \ldots \ldots\)….(i)
\(m_{B} a-f=0\)

\(\Rightarrow F=m_{B} a \dots (ii)\)
From eqn (i) and eqn (ii)
\(\Rightarrow f-m_{A} a=m_{B} a \Rightarrow f=m_{B} a+m_{A} a \Rightarrow f=a\left(m_{A}+m_{B}\right)\).
\(\Rightarrow f=\frac{F}{m_{B}}\left(m_{B}+m_{A}\right)=F\left(1+\frac{m_{A}}{m_{B}}\right)\) [because \(a=F / m_{B}\) ]
\(\therefore\) The force exerted by the experimenter is \(F\left(1+\frac{m_{A}}{m_{B}}\right)\)

Hint

(c) \(\mathrm{r}=1 \mathrm{~mm}=10^{-3}\) m \(=4 \mathrm{mg}=4 \times 10^{-6} \mathrm{~kg}\) \(\mathrm{~s}=10^{-3} \mathrm{~m}\) \(\mathrm{v}=0\) \(\mathrm{u}=30 \mathrm{~m} / \mathrm{s} .\) So, \(a=\frac{\mathrm{v}^{2}-\mathrm{u}^{2}}{2 \mathrm{~s}}=\frac{-30 \times 30}{2 \times 10^{-3}}=-4.5 \times 10^{5} \mathrm{~m} / \mathrm{s}^{2}\) (decelerating)
Taking magnitude only deceleration is \(4.5 \times 10^{5} \mathrm{~m} / \mathrm{s}^{2}\)
So, force \(F=4 \times 10^{-6} \times 4.5 \times 10^{5}=1.8 \mathrm{~N}\)

Hint

(d) Displacement of the particle from the mean position, \(x=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
\(
\mathrm{k}=15 \mathrm{~N} / \mathrm{m}
\)
Mass of the particle, \(m=0.3 \mathrm{~kg}\)
\(
\begin{aligned}
&\text { Acceleration, } a=\frac{|F|}{m} \\
&\Rightarrow a=\frac{k x}{m}=\frac{15(0.2)}{0.3}=\frac{3}{0.3}=10 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
So, the initial acceleration when the particle is released from a point \(x=20 \mathrm{~cm}\) is \(10 \mathrm{~m} / \mathrm{s}^{2}\).

Hint

(a) Let the block \(m\) be displaced towards left by displacement \(x\).
\(
\begin{aligned}
&\therefore F_{1}=-k_{1} X \text { (compressed) } \\
&F_{2}=-k_{2} X \text { (expanded) } \\
&m a=F_{1}+F_{2} \\
&\Rightarrow a=\frac{-x\left(k_{1}+k_{2}\right)}{m}
\end{aligned}
\)
i.e. \(\left(k_{1}+k_{2}\right) \frac{x}{m}\) opposite the displacement or towards the mean position.

Hint

(c) Mass of block \(A, m=5 \mathrm{~kg}\)
\(
\begin{aligned}
&\mathrm{F}=m a=10 \mathrm{~N} \\
&\Rightarrow a=\frac{10}{5}=2 m / \mathrm{s}^{2}
\end{aligned}
\)
As there is no friction between A and B, when block A moves, block B remains at rest in its position. The initial velocity of \(A, u=0\)
Distance covered by A to separate out,
\(
\mathrm{s}=0.2 \mathrm{~m}
\)
Using \(s=u t+\frac{1}{2} a t^{2}\)
\(
\begin{aligned}
&0.2=0+\frac{1}{2} \times 2 t^{2} \\
&\Rightarrow t^{2}=0.2
\end{aligned}
\)
\(
\Rightarrow t=0.44 \mathrm{~s} \approx 0.45 \mathrm{~s}
\)

Hint

(b) At any depth let the ropes make angle \(\theta\) with the vertical
From the free body diagram
\(F \cos \theta+F \cos \theta-m g=0\)
\(\Rightarrow 2 \mathrm{~F} \cos \theta=\mathrm{mg} \Rightarrow \mathrm{F}=\frac{\mathrm{mg}}{2 \cos \theta}\)
As the man moves up. \(\theta\) increases i.e. \(\cos \theta\) decreases. Thus \(F\) increases.

When the man is at depth \(\mathrm{h}\)
\(
\begin{aligned}
\cos \theta &=\frac{h}{\sqrt{(d / 2)^{2}+h^{2}}} \\
\text { Force } &=\frac{\mathrm{mg}}{\frac{h}{\sqrt{\frac{d^{2}}{4}+h^{2}}}}=\frac{m g}{4 h} \sqrt{d^{2}+4 h^{2}}
\end{aligned}
\)

Hint

(b) From the free body diagram
\(
\begin{aligned}
&\therefore \mathrm{R}+0.5 \times 2-\mathrm{w}=0 \\
&\Rightarrow \mathrm{R}=\mathrm{w}-0.5 \times 2 \\
&=0.5(10-2)=4 \mathrm{~N}
\end{aligned}
\)
So, the force exerted by the block A on the block B, is \(4 \mathrm{~N}\).

Hint

(a) Given that, Mass \(\mathrm{m}=50 \mathrm{~kg}\)
Acceleration \(\mathrm{a}=1.2 \mathrm{~m} / \mathrm{s}^{2}\)
Now, we have
\(\mathrm{T}-\mathrm{mg}=\mathrm{ma}\)
\(\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\)
(a) The acceleration is upward
\(\mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a})\)
\(\mathrm{T}=\frac{50}{100}(9.8+1.2)\)
\(\mathrm{T}=0.55 \mathrm{~N}\)
(b) The acceleration is upward but negative
\(\mathrm{T}=\mathrm{m}(\mathrm{g}-\mathrm{a})\)
\(
\begin{aligned}
\mathrm{T} &=\frac{50}{1000}(9.8-1.2) \\
\mathrm{T} &=0.43 \mathrm{~N}
\end{aligned}
\)

(c) The velocity is constant and therefore acceleration is zero
\(
\begin{aligned}
\mathrm{T} &=\frac{50}{1000} \times 9.8 \\
\mathrm{~T} &=0.49 \mathrm{~N}
\end{aligned}
\)
(d) The acceleration is downward
\(
\begin{aligned}
\mathrm{T} &=\frac{50}{1000}(9.8-1.2) \\
\mathrm{T} &=0.43 \mathrm{~N}
\end{aligned}
\)
(e) The acceleration is downward but negative
\(
\begin{aligned}
\mathrm{T} &=\frac{50}{1000}(9.8+1.2) \\
\mathrm{T} &=0.55 \mathrm{~N}
\end{aligned}
\)
(f) The velocity is constant therefore acceleration is zero
\(
\begin{aligned}
\mathrm{T} &=\frac{50}{1000} \times 9.8 \\
\mathrm{~T} &=0.49 \mathrm{~N}
\end{aligned}
\)
Hence, this is the required solution

Hint

(d) Consider the acceleration of the elevator first upward and then downward in the same magnitude is \(a\) and, \(m\) be the mass of the person standing on the scale.
Then,
\(\mathrm{mg}+\mathrm{ma}=72 \mathrm{~g}\)
and
\(\mathrm{mg}-\mathrm{ma}=60 \mathrm{~g}\)
Now solving
\(
\begin{aligned}
&\mathrm{m}=\frac{132 \mathrm{~g}}{2 \mathrm{~g}} \\
&=66 \mathrm{~kg}
\end{aligned}
\)
and
\(
\begin{aligned}
&2 \mathrm{ma}=12 \mathrm{~g} \\
&\Rightarrow \mathrm{a}=\frac{6 \mathrm{~g}}{\mathrm{~m}}=6 \times \frac{9.9}{66}=0.9 \mathrm{~ms}^{-2}
\end{aligned}
\)

Hint

(c) Let the acceleration of the \(3 \mathrm{~kg}\) mass relative to the elevator is \(\mathrm{a}\) in the downward direction.
As, shown in the free body diagram
\(\mathrm{T}-1.5 \mathrm{~g}-1.5(\mathrm{~g} / 10)-1.5 \mathrm{a}=0 \quad\) from figure (1)
and, \(T-3 g-3(g / 10)+3 a=0 \quad\) from figure \((2)\)
\(\Rightarrow \mathrm{T}=1.5 \mathrm{~g}+1.5(\mathrm{~g} / 10)+1.5 \mathrm{a} \quad \ldots\) (i)
And \(T=3 g+3(g / 10)-3 a \quad\)… (ii)
Equation (i) \(\times 2 \Rightarrow 3 \mathrm{~g}+3(\mathrm{~g} / 10)+3 a=2 \mathrm{~T}\)
Equation (ii) \(\times 1 \Rightarrow 3 \mathrm{~g}+3(\mathrm{~g} / 10)-3 \mathrm{a}=\mathrm{T}\)
Subtracting the above two equations we get, \(T=6 a\)
Subtracting \(T=6 a\) in equation (ii)
\(6 a=3 g+3(g / 10)-3 a\).
\(\Rightarrow 9 a=\frac{33 g}{10} \Rightarrow a=\frac{(9.8) 33}{10}=32.34\)
\(\Rightarrow \mathrm{a}=3.59 \therefore \mathrm{T}=6 \mathrm{a}=6 \times 3.59=21.55\)
\(\mathrm{T}^{1}=2 \mathrm{~T}=2 \times 21.55=43.1 \mathrm{~N}\) cut is \(\mathrm{T}_{1}\) shown in spring.
Mass \(=\frac{\mathrm{wt}}{\mathrm{g}}=\frac{43.1}{9.8}=4.39=4.4 \mathrm{~kg}\)

Hint

(d) For equilibrium,
Spring force = Weight of the block
Case(1) \(\mathrm{k} \mathrm{x}_{1}=\mathrm{m}_{1} \mathrm{~g} \Rightarrow \mathrm{x}_{1}=\frac{\mathrm{m}_{1} \mathrm{~g}}{\mathrm{k}} \Rightarrow \frac{2(10)}{100}=0.2 \mathrm{~m}\)
Case (2) \(x_{2}=\frac{m_{2} g}{k}=\frac{3(10)}{100}=0.3 \mathrm{~m}\)
Difference in the elongation \(=0.3-0.2=0.1 \mathrm{~m}\)

Hint

(c) Let, the air resistance force is \(F\) and the Buoyant force be B.
Given that
\(\mathrm{F}_{\mathrm{a}} \propto \mathrm{v}\), where \(\mathrm{v} \rightarrow\) velocity
\(\Rightarrow F_{a}=k v\), where \(k \rightarrow\) proportionality constant. When the balloon is moving downward,
\(
\begin{aligned}
&\mathrm{B}+\mathrm{kv}=\mathrm{mg} \\
&\Rightarrow \mathrm{M}=\frac{\mathrm{B}+\mathrm{kv}}{\mathrm{g}}
\end{aligned} \dots (i)
\)
For the balloon to rise with a constant velocity \(v\), (upward) let the mass be \(m\)
Here, \(B-(m g+k v)=0 \quad\)..(ii)
\(\Rightarrow B=m g+k v\)
\(\Rightarrow \mathrm{m}=\frac{\mathrm{B}-\mathrm{kw}}{\mathrm{g}}\)
So, the amount of mass that should be removed \(=M-m\).
\(
=\frac{B+k v}{g}-\frac{B-k v}{g}=\frac{B+k v-B+k v}{g}=\frac{2 k v}{g}=\frac{2(M g-B)}{g}=2\{M-(B / g)\}
\)

Hint

(b) Let buoyant force be \(\mathrm{F}\).
Now F – mg \(=\mathrm{mg} / 6\)
Which gives \(F=7 \mathrm{mg} / 6\)
Let \(M\) be the mass of sand to be put to give acceleration \(\mathrm{g} / 6\) down. Writing equation of motion, \((\mathrm{m}+\mathrm{M}) \mathrm{g}-\mathrm{F}=(\mathrm{m}+\mathrm{M}) \mathrm{g} / 6\)
Putting value of \(F\) we get,
\((\mathrm{m}+\mathrm{M}) \mathrm{g}-7 \mathrm{mg} / 6=(\mathrm{m}+\mathrm{M}) \mathrm{g} / 6\)
\(\mathrm{mg}+\mathrm{Mg}-7 \mathrm{mg} / 6=\mathrm{mg} / 6+\mathrm{Mg} / 6\)
\(\operatorname{Mg}(1-1 / 6)=\operatorname{mg}(1 / 6+7 / 6-1)\)
\(\mathrm{Mg} \times 5 / 6=\mathrm{mg} \times 2 / 6\)
\(M=2 \mathrm{~m} / 5\)

Hint

(a) Given that, \(\vec{F}=\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{mg}}\) act on the particle
For the particle to move undeflected with constant velocity, the net force should be zero.
\(
\begin{aligned}
&\therefore(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{A}})+\overrightarrow{\mathrm{m} g}=0 \\
&\therefore(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{A}})=-\overrightarrow{\mathrm{m} g}
\end{aligned}
\)
Because, \((\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{A}})\) is perpendicular to the plane containing \(\overrightarrow{\mathrm{u}}\) and \(\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{u}}\) should be the \(x z-\) plane
Again u \(A \sin \theta=m g\)
\(
\mathrm{u}=\frac{\mathrm{mg}}{\mathrm{A} \sin \theta}
\)
\(u\) will be minimum when \(\sin \theta=1 \Rightarrow \theta=90^{\circ}\)
\(
\therefore \mathrm{U}_{\min }=\frac{\mathrm{mg}}{\mathrm{A}} \text { along } \mathrm{Z} \text { – axis. }
\)

Hint

(c) \(\mathrm{m}_{1}=0.3 \mathrm{~kg}, \mathrm{~m}_{2}=0.6 \mathrm{~kg}\)
\(\mathrm{~T}-\left(\mathrm{m}_{1} \mathrm{~g}+\mathrm{m}_{1} \mathrm{a}\right)=0 \quad \ldots\) (i) \(\Rightarrow \mathrm{T}=\mathrm{m}_{1} \mathrm{~g}+\mathrm{m}_{1} \mathrm{a}\)
\(\mathrm{T}+\mathrm{m}_{2} \mathrm{a}-\mathrm{m}_{2} \mathrm{~g}=0 \dots (ii)\)

\(\Rightarrow \mathrm{T}=\mathrm{m}_{2} \mathrm{~g}-\mathrm{m}_{2} \mathrm{a}\)
From equation (i) and equation (ii)
\(\mathrm{m}_{1} \mathrm{~g}+\mathrm{m}_{1} \mathrm{a}+\mathrm{m}_{2} \mathrm{a}-\mathrm{m}_{2} \mathrm{~g}=0\), from (i)
\(\Rightarrow \mathrm{a}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)=\mathrm{g}\left(\mathrm{m}_{2}-\mathrm{m}_{1}\right)\)
\(\Rightarrow \mathrm{a}=\mathrm{f}\left(\frac{\mathrm{m}_{2}-\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right)=9.8\left(\frac{0.6-0.3}{0.6+0.3}\right)=3.266 \mathrm{~ms}^{-2}\).
a) \(\mathrm{t}=2\) sec acceleration \(=3.266 \mathrm{~ms} \mathrm{~s}^{-2}\)
Initial velocity u \(=0\)
So, distance travelled by the body is,
\(\mathrm{S}=\) ut \(+1 / 2\) at \({ }^{2} \Rightarrow 0+1 / 2(3.266) 2^{2}=6.5 \mathrm{~m}\)
b) From (i) \(\mathrm{T}=\mathrm{m}_{1}(\mathrm{~g}+\mathrm{a})=0.3(9.8+3.26)=3.9 \mathrm{~N}\)
c) The force exerted by the clamp on the pully is given by
\(\mathrm{F}-2 \mathrm{~T}=0\)
\(\mathrm{~F}=2 \mathrm{~T}=2 \times 3.9=7.8 \mathrm{~N}\).

Hint

(d) Mass per unit length \(=\frac{3}{30} \mathrm{~kg} / \mathrm{cm}=0.1 \mathrm{~kg} / \mathrm{cm}\)
Mass of \(10 \mathrm{~cm}\) part \(=\mathrm{m}_{1}=1 \mathrm{~kg}\)
Mass of \(20 \mathrm{~cm}\) part \(=\mathrm{m}_{2}=2 \mathrm{~kg}\)
Let \(F\) be the force of contact between them
From free body diagram,
\(\mathrm{F}-20-\mathrm{a}=0 \dots (1)\)
\(32-F-2 a=0 \dots (2)\)
\(\Rightarrow 32-20-2 a-a=0\)
\(\Rightarrow 12-3 \mathrm{a}=0\)
\(\Rightarrow \mathrm{a}=\frac{12}{3}=4 \mathrm{~ms}^{-2}\)
Contact force F \(=20+\mathrm{a}=20+4=24 \mathrm{~N}\)
\(\therefore\) force exerted \(=24 \mathrm{~N}\)

Hint

(c) Given Mass of each block is \(1 \mathrm{~kg}\)
\(
\begin{aligned}
&\sin \theta_{1}=\frac{4}{5} \\
&\sin \theta_{2}=\frac{3}{5}
\end{aligned}
\)
The free-body diagrams for both the boxes are shown below:
\(\mathrm{mg} \sin \theta_{1}-\mathrm{T}=\mathrm{ma} \quad \ldots\) (i)
\(
\mathrm{T}-\mathrm{mg} \sin \theta_{2}=\mathrm{ma} \ldots \text { (ii) }
\)
Adding equations (i) and (ii), we get:
\(
\mathrm{mg}\left(\sin \theta_{1}-\sin \theta_{2}\right)=2 \mathrm{ma}
\)
\(
\Rightarrow 2 \mathrm{a}=\mathrm{g}\left(\sin \theta_{1}-\sin \theta_{2}\right)
\)

\(\begin{aligned}
&\Longrightarrow \mathrm{a}=\frac{\mathrm{g}}{5} \times \frac{1}{2} \\
&=\frac{\mathrm{g}}{10} \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}\)

Hint

(b) From the free-body diagram of block of mass \(m_{1}\),
\(
\mathrm{m}_{1} \mathrm{a}=\mathrm{T}-\mathrm{F} \quad \ldots \text { (i) }
\)
From the free-body diagram of block of mass \(m_{2}\),
\(
m_{2} a=m_{2} g-T \quad \ldots \text { (ii) }
\)
Adding both the equations, we get:
\(
\begin{aligned}
&a\left(m_{1}+m_{2}\right)=m_{2} g-\frac{m_{2} g}{2} \ldots \ldots\left[\text { because } \mathrm{F}=\frac{m_{2} g}{2}\right] \\
&\Rightarrow a=\frac{m_{2} g}{2\left(m_{1}+m_{2}\right)} \\
&\text { So, the acceleration of mass } m_{1} \\
&a=\frac{m_{2} g}{2\left(m_{1}+m_{2}\right)}, \text { towards the right } .
\end{aligned}
\)

Hint

(a) Let the acceleration of the blocks be \(a\).
The free-body diagrams for both the blocks are shown below:
From the free-body diagram,
\(
\mathrm{m}_{1} \mathrm{a}=\mathrm{m}_{1} \mathrm{~g}+\mathrm{F}-\mathrm{T} \quad \ldots \text { (i) }
\)
Again, from the free-body diagram,
\(
m_{2} a=T-m_{2} g-F \quad \text {…(ii) }
\)

Adding equations (i) and (ii), we have:
\(
\begin{aligned}
&a=g \frac{m_{1}-m_{2}}{m_{1}+m_{2}} \\
&\Rightarrow a=\frac{3 g}{7}=\frac{29.4}{7} \\
&=4.2 m / s^{2}
\end{aligned}
\)
Hence, acceleration of the block is \(4.2 \mathrm{~m} / \mathrm{s}^{2}\).
After the string breaks, \(m_{1}\) moves downward with force \(F\) acting downward. Then,
\(
\begin{aligned}
&m_{1} a=\mathrm{F}+\mathrm{m}_{1} \mathrm{~g} \\
&5 \mathrm{a}=1+5 \mathrm{~g} \\
&\Rightarrow a=\frac{5 g+1}{5} \\
&=g+0.2 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)

Hint

(d) Suppose the block \(m_{1}\) moves upward with acceleration \(a_{1}\) and the blocks \(m_{2}\) and \(m_{3}\) have relative acceleration \(a_{2}\) due to the difference of weight between them.
So, the actual acceleration of the blocks \(m_{1}, m_{2}\) and \(m_{3}\) will be \(a_{1},\left(a_{1}-a_{2}\right)\) and \(\left(a_{1}+a_{2}\right)\), as shown.
From figure 2, \(T-1 g-1 a_{1}=0 \quad\)…(i)
From figure 3, \(\frac{T}{2}-2 g-2\left(a_{1}-a_{2}\right)=0 . .(i i)\)
From figure 4, \(\frac{T}{2}-3 g-3\left(a_{1}+a_{2}\right)=0 . .(i i i)\)
From equations (i) and (ii), eliminating \(T\), we get:
\(
\begin{aligned}
&1 g+1 a_{2}=4 g+4\left(a_{1}+a_{2}\right) \\
&5 a_{2}-4 a_{1}=3 g \quad \text {…(iv) }
\end{aligned}
\)

From equations (ii) and (iii), we get:
\(
\begin{aligned}
&2 g+2\left(a_{1}-a_{2}\right)=3 g-3\left(a_{1}-a_{2}\right) \\
&5 a_{1}+a_{2}=g \quad \ldots(v)
\end{aligned}
\)
Solving equations (iv) and (v), we get:
\(
a_{1}=\frac{2 g}{29}
\)
and \(a_{2}=g-5 a_{1}\)
\(
\begin{aligned}
&\Rightarrow a_{2}=g-\frac{10 g}{29}=\frac{19 g}{29} \\
&\text { So, } a_{1}-a_{2}=\frac{2 g}{29}-\frac{19 g}{29}=-\frac{17 g}{29}
\end{aligned}
\)
So, accelerations of \(\mathrm{m}_{1}, \mathrm{~m}_{2}\) and \(\mathrm{m}_{3}\) are \(\frac{19 g}{29}(u p), \frac{17 g}{29}\) (down ) and \(\frac{21 g}{29}\) (down)

\(\begin{aligned}
&\text { Now, } \mathrm{u}=0, \mathrm{~s}=20 \mathrm{~cm}=0.2 \mathrm{~m} \\
&a_{2}=\frac{19 g}{29} \\
&\therefore s=u t+\frac{1}{2} a t^{2} \\
&\Rightarrow 0.2=\frac{1}{2} \times \frac{19}{29} g t^{2} \\
&\Rightarrow t=0.25 s
\end{aligned}\)

Hint

(c) The free-body diagrams for both the bodies are shown below:
\(\mathrm{T}+\mathrm{ma}=\mathrm{mg} \quad\)…(i)
and \(T=\) ma \(\quad\)…(ii)
From equations (i) and (ii), we get:
\(
\begin{aligned}
&m a+m a=m g \\
&\Rightarrow 2 m a=g \\
&\Rightarrow a=\frac{g}{2}=\frac{10}{5}=5 m / s^{2}
\end{aligned}
\)
From equation (ii),
\(
\mathrm{T}=\mathrm{ma}=5 \mathrm{~N}
\)

Hint

(a) Let the acceleration of mass M be a.
So, the acceleration of mass \(2 \mathrm{M}\) will be \(\frac{a}{2}\)

\(
\begin{aligned}
&\text { (a) } 2 \mathrm{M}(\mathrm{a} / 2)-2 \mathrm{~T}=0 \\
&\Rightarrow \mathrm{Ma}=2 \mathrm{~T} \\
&\mathrm{~T}+\mathrm{Ma}-\mathrm{Mg}=0 \\
&\Rightarrow \frac{M a}{2}+M a=M g \\
&\Rightarrow 3 M a=2 M g \\
&\Rightarrow a=\frac{2 g}{3}
\end{aligned}
\)
(b) Tension,
\(
T=\frac{M a}{2}=\frac{M}{2} \times \frac{2 g}{3}=\frac{M g}{3}
\)

(c) Let \(\mathrm{T}^{\prime}=\) resultant of tensions
\(
\begin{aligned}
&\therefore T^{\prime}=\sqrt{T^{2}+T^{2}}=\sqrt{2} T \\
&\therefore T^{\prime}=\sqrt{2} T=\frac{\sqrt{2} M g}{3} \\
&\text { Again, } \tan \theta=\frac{T}{T}=1 \\
&\Rightarrow \theta=45^{\circ}
\end{aligned}
\)
So, the force exerted by the clamp on the pulley is \(\frac{\sqrt{2} \mathrm{Mg}}{3}\) at an angle of \(45^{\circ}\) with the horizontal.

Hint

(d) Let acceleration of the block of mass \(2 \mathrm{M}\) be \(\mathrm{a}\).
So, the acceleration of the block of mass M will be \(2 a\).
\(
\begin{aligned}
&M(2 a)+M g \sin \theta-T=0 \\
&\Rightarrow T=2 M a+M g \sin \theta \quad \ldots \text { (i) } \\
&2 T+2 M a-2 M g=0
\end{aligned}
\)
From equation (i),
\(
\begin{aligned}
&2(2 \mathrm{Ma}+\mathrm{Mg} \sin \theta)+2 \mathrm{Ma}-2 \mathrm{Mg}=0 \\
&4 \mathrm{Ma}+2 \mathrm{Mg} \sin \theta+2 \mathrm{Ma}-\mathrm{Mg}=0 \\
&6 \mathrm{Ma}+2 \mathrm{Mg} \sin 30^{\circ}+2 \mathrm{Mg}=0 \\
&6 \mathrm{Ma}=\mathrm{Mg} \\
&\Rightarrow a=\frac{g}{6}
\end{aligned}
\)
Hence, the acceleration of mass
\(M=2 a=2 \times \frac{g}{6}=\frac{g}{3}\), up the plane.

Hint

(a) Block m will have the same acceleration as that of \( \mathrm{M}^{\prime} \)  as it does not slip over \( \mathrm{M}^{\prime}.\)

\(\begin{array}{ll}\mathrm{T}+\mathrm{Ma}-\mathrm{Mg}=0  \ldots \text { (i) (From } Fig(a)) \\ \mathrm{T}-\mathrm{M}^{\prime} \mathrm{a}-\mathrm{R} \sin \theta=0  \ldots \text { (ii) (From Fig(b)) } \\ \mathrm{R} \sin \theta-\mathrm{ma}=0  \ldots \text { (iii) (From } Fig(c)) \\ \mathrm{R} \cos \theta-\mathrm{mg}=0  \ldots \text { (iv) (From Fig(d) } \\ \text {Eliminating } \mathrm{T}, \mathrm{R} \text { and a from the above equation, we get }  \\M=\frac{\mathrm{M}^{\prime}+\mathrm{m}}{\cot \theta-1}\end{array}\)

Hint

(b) Given,
\(
\begin{aligned}
&\mathrm{m}_{1}=100 \mathrm{~g}=0.1 \mathrm{~kg} \\
&\mathrm{~m}_{2}=500 \mathrm{~g}=0.5 \mathrm{~kg} \\
&\mathrm{~m}_{3}=50 \mathrm{~g}=0.05 \mathrm{~kg}
\end{aligned}
\)
The free-body diagram for the system is shown below:

From the free-body diagram of the \(500 \mathrm{~g}\) block,
\(
\mathrm{T}+0.5 a-0.5 g=0 \dots (i)
\)
From the free-body diagram of the \(50 \mathrm{~g}\) block,
\(
T_{1}-0.05 g-0.05 a=0 \dots (ii)
\)
From the free-body diagram of the \(100 \mathrm{~g}\) block,
\(
T_{1}+0.1 a-T+0.1 g \sin 30^{\circ}=0 \dots (iii)
\)

Solving equations (i), (i) & (iii)

We get, a = \(\frac{8}{13} g\), downward

Hint

(b) From the free-body diagram,
\(
\begin{aligned}
&\mathrm{T}-[15 \mathrm{~g}+15(\mathrm{a})]=0 \\
&\mathrm{~T}-[15 \mathrm{~g}+15(1)]=0 \\
&\Rightarrow \mathrm{T}=15(10+1) \\
&\Rightarrow \mathrm{T}=15 \times 11=165 \mathrm{~N}
\end{aligned}
\)
The monkey should apply a force of \(165 \mathrm{~N}\) to the rope.
Initial velocity, \(u=0\)
\(
\mathrm{s}=5 \mathrm{~m}
\)
Using, \(s=u t+\frac{1}{2} a t^{2}\), we get:
\(
\begin{aligned}
&5=0+\left(\frac{1}{2}\right) \times 1 \times t^{2} \\
&\Rightarrow t^{2}=5 \times 2 \\
&\Rightarrow t=\sqrt{10} s
\end{aligned}
\)
Hence, the time required to reach the ceiling is \(\sqrt{10} s\)

Hint

(c) Let the acceleration of monkey A upwards be a, so that a maximum tension of \(30 \mathrm{~N}\) is produced in its tail.

\(
\begin{aligned}
&T-5 g-30-5 a=0 \quad \text {…(i) } \\
&30-2 g-2 a=0 \quad \text {…(ii) }
\end{aligned}
\)
From equations (i) and (ii), we have:
\(
T=105 \mathrm{~N} \text { (max.) }
\)
and \(a=5 \mathrm{~m} / \mathrm{s}^{2}\)
So, A can apply a maximum force of \(105 \mathrm{~N}\) on the rope to carry monkey B with it.
For minimum force, there is no acceleration of A and B.
\(\mathrm{T}_{1}=\) weight of monkey \(\mathrm{B}\)
\(
\Rightarrow \mathrm{T}_{1}=20 \mathrm{~N}
\)
Rewriting equation (i) for monkey A, we get:
\(
\begin{aligned}
&T-5 \mathrm{~g}-20=0 \\
&\Rightarrow T=70 \mathrm{~N}
\end{aligned}
\)
Therefore, to carry monkey B with it, monkey A should apply a force of magnitude between \(70 \mathrm{~N}\) and \(105 \mathrm{~N}\).

Hint

(d)

(i) Given, Mass of man \(=60 \mathrm{~kg}\).
Let \(\mathrm{R}=\) apparent weight of man in this case.
Now, \(R+T-60 \mathrm{~g}=0\) [From Free body diagram of the man] \(\Rightarrow T=60 g-R \quad\)…(i)
\(T-R-30 g=0 \quad\)…(ii) [ From Free body diagram of the box] \(\Rightarrow 60 \mathrm{~g}-\mathrm{R}-\mathrm{R}-30 \mathrm{~g}=0\) [ From (i)]
\(\Rightarrow R^{\prime}=15 \mathrm{~g}\) The weight shown by the machine is \(15 \mathrm{~kg}\).

(ii) From the Free Body Diagram of the man
\(
\begin{aligned}
&\mathrm{T}+R-60 \mathrm{~g}-60 \mathrm{a}=0 \\
&\Rightarrow \mathrm{T}-60 \mathrm{a}=0[\therefore \mathrm{R}=60 \mathrm{~g}] \\
&\Rightarrow \mathrm{T}=60 \mathrm{a} \dots (i)
\end{aligned}
\)

From the Free Body Diagram of the box
\(
\begin{aligned}
&\mathrm{T}-\mathrm{R}-30 \mathrm{~g}-30 \mathrm{a}=0 \\
&\Rightarrow \mathrm{T}-60 \mathrm{~g}-30 \mathrm{~g}-30 \mathrm{a}=0 \\
&\Rightarrow \mathrm{T}-30 \mathrm{a}=90 \mathrm{~g}=900 \\
&\Rightarrow \mathrm{T}=30 \mathrm{a}-900 \dots (ii)
\end{aligned}
\)

From eqn (i) and eqn (ii) we get \(\mathrm{T}=2 \mathrm{T}-1800 \Rightarrow \mathrm{T}=1800 \mathrm{~N}\).
So, he should exert \(1800 \mathrm{~N}\) force on the rope to get the correct reading.

Hint

(b) The force on the block which makes the body move down the plane is the component of its weight parallel to the inclined surface.
\(
\mathrm{F}=\mathrm{mg} \sin \theta
\)

So, acceleration \( a=g \sin \theta\)
Initial velocity of block \(u=0\).
\(
s=\ell, a=g \sin \theta
\)
Now, \(S=u t+1 / 2\) at \({ }^{2}\)
\(
\Rightarrow \ell=0+1 / 2(g \sin \theta) t^{2} \Rightarrow g^{2}=\frac{2 \ell}{g \sin \theta} \Rightarrow t=\sqrt{\frac{2 \ell}{g \sin \theta}}
\)
Time taken is \(\sqrt{\frac{2 \ell}{g \sin \theta}}\)

Hint

(a) Let the pendulum (formed by the ball and the string) make angle \(\theta\) with the vertical. (i) From the free-body diagram,
\(
\begin{aligned}
&\mathrm{T} \cos \theta-\mathrm{mg}=0 \\
&\mathrm{~T} \cos \theta=\mathrm{mg} \\
&\Rightarrow T=\frac{m g}{\cos \theta} \ldots(i) \\
&\text { And, ma-T } \sin \theta=0 \\
&\Rightarrow \mathrm{ma}=\mathrm{T} \sin \theta \\
&\Rightarrow T=\frac{m a}{\sin \theta} \ldots(i i) \\
&\Rightarrow \tan \theta=\frac{a}{g} \\
&\Rightarrow \theta=\tan ^{-1} \frac{a}{g}
\end{aligned}
\)
So, the angle formed by the ball with the vertical is \(\tan ^{-1}\left(\frac{a}{g}\right)\)

(ii) Let the angle of the incline be \(\theta\).
From the diagram,
\(\Rightarrow m a \cos \theta=m g \sin \theta\)
\(
\frac{\sin \theta}{\cos \theta}=\frac{a}{g}
\)
\(\tan \theta=\frac{a}{g} \theta=\tan ^{-1}\left(\frac{a}{g}\right)\)
So, the angle of incline is \(\tan ^{-1}\left(\frac{a}{g}\right)\)

Hint

(d) The two bodies are separated because the elevator is moving downward with an acceleration of \(12 \mathrm{~m} / \mathrm{s}^{2}\) (>g) and the body moves with acceleration, \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) [Freely falling body]
Now, for the block:
\(
g=10 \mathrm{~m} / \mathrm{s}^{2}, u=0, t=0.2 \mathrm{~s}
\)
So, the distance travelled by the block is given by
\(
\begin{aligned}
&s=u t+\frac{1}{2} a t^{2} \\
&=0+\frac{1}{2} 10 \times(0.2)^{2}=5 \times 0.04 \\
&=0.2 \mathrm{~m}=20 \mathrm{~cm}
\end{aligned}
\)
The displacement of the body is \(20 \mathrm{~cm}\) during the first \(0.2 \mathrm{~s}\).

Hint

(d) Here \(m=20 \mathrm{~kg}, F=-50 \mathrm{~N}\) (retardation force)
As \(F=m a\)
\(
\Rightarrow a=\frac{F}{m}=\frac{-50}{20}=-2.5 m s^{-2}
\)
Using equation \(v=u+ at\)
Given \(u=15 m s^{-1}, v=0\)
Now, \(0=15+(-2.5) t\)
\(
\text { or } t=6 s
\)

Hint

(a) Mass of the body, \(m=3 \mathrm{~kg}\)
Initial speed of the body, \(u=2 \mathrm{~m} / \mathrm{s}\)
Final speed of the body, \(v=3.5 \mathrm{~m} / \mathrm{s}\)
Time, \(t=25 \mathrm{~s}\)
Using the first equation of motion, the acceleration a produced in the body can be calculated as:
\(
\begin{aligned}
&\mathrm{v}=\mathrm{u}+\mathrm{at} \\
&\therefore a=\frac{v-u}{t} \\
&=\frac{3.5-2}{25}=\frac{1.5}{25}=0.06 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
As per Newton’s second law of motion, force is given as:
\(
\begin{aligned}
&F=m a \\
&=3 \times 0.06=0.18 \mathrm{~N}
\end{aligned}
\)
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

Hint

(b) Here \(m=5 \mathrm{~kg}\)
\(F_{1}=8 N\) and \(F_{2}=6 N\)
The resultant force on the body
\(
\begin{aligned}
&F=\sqrt{F_{1}^{2}+F_{2}^{2}}=\sqrt{8^{2}+6^{2} N} \\
&=>F=\sqrt{64+36} N=10 \mathrm{~N}
\end{aligned}
\)
The acceleration \(a=\frac{F}{m}\)
\(\Rightarrow a=\frac{10}{5}=2 m s^{-2}\) in the same direction as the resultant force
The direction of acceleration
\(
\tan \beta=\frac{6}{8}=\frac{3}{4}=0.75
\)
or \(\beta=\tan ^{-1}(0.75)\)
\(=37^{\circ}\) with \(8 \mathrm{~N}\) force

Hint

(d) Here mass of three-wheeler \(m_{1}=400 \mathrm{~kg}\), mass of driver \(=m_{2}=65 \mathrm{~kg}\), initial speed of auto, \(u=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}\) \(=10 \mathrm{~ms}^{-1}\), final speed, \(v=0\) and \(t=4 \mathrm{~s}\)
As acceleration, \(\quad a=\frac{v-u}{t}=\frac{0-10}{4}=-2.5 \mathrm{~ms}^{-2}\)
Now
\(
\begin{aligned}
F &=\left(m_{1}+m_{2}\right) a=(400+65) \times(-2.5) \\
&=-1162.5 \mathrm{~N}=-1.2 \times 10^{3} \mathrm{~N}
\end{aligned}
\)
The – ve sign shows that the force is retarding force.

Hint

(c) Given \(\mathrm{m}=20000 \mathrm{~kg}\)
\(\mathrm{a}=5 \mathrm{~ms}^{-2}\) (against gravity) since the rocket has to move upwards against gravity the total initial thrust of the blast is given by
\(
\begin{aligned}
&F=m a+m g \\
&=m(a+g)=20000(5+9.8) \\
&=2.96 \times 10^{5} \mathrm{~N}
\end{aligned}
\)

Hint

(a) Initial velocity of the truck, \(u=0\)
Acceleration, \(a=2 \mathrm{~m} / \mathrm{s}^{2}\)
Time, \(t=10 \mathrm{~s}\)
As per the first equation of motion, the final velocity is given as:
\(
\begin{aligned}
&v=u+a t \\
&=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
The final velocity of the truck and hence, of the stone is \(20 \mathrm{~m} / \mathrm{s}\).
At \(t=11 \mathrm{~s}\), the horizontal component \(\left(v_{x}\right)\) of velocity, in the absence of air resistance, remains unchanged, i.e.,
\(
v_{x}=20 \mathrm{~m} / \mathrm{s}
\)
The vertical component \(\left(v_{y}\right)\) of velocity of the stone is given by the first equation of motion as:
\(
v_{y}=u+a_{y} \Delta t
\)

Where, \(\Delta t=11-10=1 \mathrm{~s}\) and \(a_{y}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\)
\(
\therefore v_{y}=0+10 \times 1=10 \mathrm{~m} / \mathrm{s}
\)
The resultant velocity \((v)\) of the stone is given as:
\(
\begin{aligned}
\mathrm{v} &=\sqrt{v_{x}^{2}+v_{y}^{2}} \\
&=\sqrt{20^{2}+10^{2}}=\sqrt{400+100} \\
&=\sqrt{500}=22.36 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Let \(\theta\) be the angle made by the resultant velocity with the horizontal component of velocity, \(v_{x}\)

\(
\begin{aligned}
&\therefore \tan \theta=\frac{v_{y}}{v_{x}} \\
&\theta=\frac{\tan ^{-1}(10)}{20} \\
&=\tan ^{-1}(0.5) \\
&=26.57^{\circ}
\end{aligned}
\)
(b) When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is \(10 \mathrm{~m} / \mathrm{s}^{2}\) and it acts vertically downward.

Hint

(a) Vertically downward
At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.
(b) Parabolic path
At the mean position, the velocity of the bob is \(1 \mathrm{~m} / \mathrm{s}\). The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Hint

(a) Accleration \(=\frac{600 N}{10 k g+20 k g}=20 m s^{-2}\)
(i) When force is applied on \(10 \mathrm{~kg}\) mass \(600-T=10 \times 20\) or
\(
\mathrm{T}=400 \mathrm{~N}
\)
(ii) When force is applied on \(20 \mathrm{~kg}\) mass \(600-T=20 \times 20\) or
\(
\mathrm{T}=200 \mathrm{~N}
\)

Hint

(b) For block \(m_{2} \rightarrow m_{2} g-T=m_{2} a\)
and for block \(\mathrm{m}_{1} \rightarrow \mathrm{T}-\mathrm{m}_{1} \mathrm{~g}=\mathrm{m}_{1} \mathrm{a}\)…(ii)
Adding \(\mathrm{i}\) and ii we obtain
\(
\begin{aligned}
&\left(m_{2}-m_{1}\right) g=\left(m_{2}+m_{1}\right) a \\
&\text { or } a=\left(\frac{m_{2}-m_{1}}{m_{2}+m_{1}}\right) g \\
&=\frac{12-8}{12+8} \times 10 \\
&=\frac{4 \times 10}{20}=2 m s^{-2}
\end{aligned}
\)
Substituting value of a in equation ii we obtain
\(
\begin{aligned}
&T=m_{1}(g+a) \\
&=8 \times(10+2) \\
&=8 \times 12=96 \mathrm{~N}
\end{aligned}
\)

Hint

(c) Initial momentum of each ball before the collision
\(
=0.05 \times 6 \mathrm{~kg} \mathrm{~ms}-1=0.3 \mathrm{~kg} \mathrm{~ms}^{-1}
\)
Final momentum of each ball after collision
\(=-0.05 \times 6 \mathrm{~kg} \mathrm{~ms}^{-1}=-0.3 \mathrm{~kg} \mathrm{~ms}^{-1}\)

Impulse imparted to each ball due to the other
= final momentum – initial momentum \(=-0.3 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}-0.3 \mathrm{~kg} \mathrm{~ms}^{-1}\)
\(=-0.6 \mathrm{~kg} \mathrm{~ms}^{-1}=0.6 \mathrm{~kg} \mathrm{~ms}^{-1}\) (in magnitude)
The two impulses are opposite in direction.

Hint

(d)
\(\begin{aligned}
&\mathrm{m}=0.02 \mathrm{~kg}, \mathrm{M}=100 \mathrm{~kg}, \mathrm{v}=80 \mathrm{~ms}^{-1}, \mathrm{~V}=? \\
&V=-\frac{m v}{M}=-\frac{0.020 \mathrm{~kg} \times 80 \mathrm{~ms}^{-1}}{100 \mathrm{~kg}} \\
&=-0.016 \mathrm{~ms}^{-1}
\end{aligned}
\)
The negative sign indicates that the gun moves in a direction opposite to the direction of motion of the bullet.

Hint

(a) Given: Mass of the ball is \(0.15 \mathrm{~kg}\), its initial speed is \(54 \mathrm{kmh}^{-1}\) and the ball is deflected at an angle of \(45^{\circ}\). The situation is shown in the figure below.

Here, \(A O\) is the original path of the ball, \(O B\) is the path followed by the ball after deflection. Angle \(A O B\) is the angle between the incident and deflected paths of the ball.
From the figure,
\(\angle \mathrm{AOP}=\angle \mathrm{BOP} \angle \mathrm{AOP}+\angle \mathrm{BOP}=45^{\circ} \angle \mathrm{AOP}=22.5^{\circ}\)
The horizontal components of velocities suffer no change. The vertical components of velocities are in opposite directions.
Impulse imparted to the ball is given as,
\(I=m v \cos \theta-(-m v \cos \theta)=2 m v \cos \theta\)
Where, \(m\) is the mass of the ball and \(v\) is the velocity of the ball.
By substituting the given values in the above expression, we get \(\mathrm{I}=2 \times 0.15 \times 54 \times 10003600 \times \cos 22.5=4.16 \mathrm{kgms}^{-1}\)
Thus, the impulse imparted to the ball is \(4.16 \mathrm{kgms}^{-1}\).

Hint

(b) Mass of the stone, \(\mathrm{m}=0.25 \mathrm{~kg}\)
Radius of the circle, \(\mathrm{r}=1.5 \mathrm{~m}\)
Number of revolution per second, \(n=40 / 60=2 / 3 \mathrm{rps}\)
Angular velocity, \(\omega=2 \pi n\)
The centripetal force for the stone is provided by the tension \(T\), in the string, i.e.,
\(
\begin{aligned}
\mathrm{T} &=m \omega^{2} \mathrm{r} \\
&=0.25 \times 1.5 \times(2 \times 3.14 \times(2 / 3))^{2} \\
&=6.57 \mathrm{~N}
\end{aligned}
\)
Maximum tension in the string, \(\mathrm{T}_{\max }=200 \mathrm{~N}\)
\(
\begin{aligned}
&\mathrm{T}_{\max }=\frac{\mathrm{mv}_{\max }^{2}}{\mathrm{r}} \\
&\mathrm{v}_{\max }=\left(\frac{\mathrm{T}_{\max } \mathrm{r}}{\mathrm{m}}\right)^{1 / 2} \\
&=(200 \times 1.5 / 0.25)^{1 / 2} \\
&=(1200)^{1 / 2}=34.64 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Therefore, the maximum speed of the stone is \(34.64 \mathrm{~m} / \mathrm{s}\).

Hint

(d) Here, \(v=15 \mathrm{~m} \mathrm{~s}^{-1}\)
Area of cross section, \(A=10^{-2} \mathrm{~m}^{2}\)
Density of water, \(\rho=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\)
Mass of water hitting the wall per second
\(
\begin{aligned}
&=\rho \times A \times v \\
&=10^{3} \mathrm{~kg} \mathrm{~m}^{-3} \times 10^{-2} \mathrm{~m}^{2} \times 15 \mathrm{~ms}^{-1}=150 \mathrm{~kg} \mathrm{~s}^{-1}
\end{aligned}
\)
Force exerted on the wall = Momentum loss of water per second
\(
=150 \mathrm{~kg} \mathrm{~s}^{-1} \times 15 \mathrm{~ms}^{-1}=2250 \mathrm{~N}=2.25 \times 10^{3} \mathrm{~N}
\)

Hint

(b) Here \(v=720 \mathrm{~km} / \mathrm{h}=720 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=200 \mathrm{~m} / \mathrm{s}\) and angle of banking \(\theta=15^{\circ}\)
From the relation
\(
\begin{aligned}
&\tan \theta=\frac{v^{2}}{r g} \text { we have } \\
&r=\frac{v^{2}}{g \tan \theta}=\frac{200 \times 200}{10 \times \tan 15^{\circ}}=\frac{200 \times 200}{10 \times 0.2679} \\
&\Rightarrow r=14931 \mathrm{~m}=14.9 \mathrm{~km}
\end{aligned}
\)

Hint

(a) Mass of the man, \(\mathrm{m}=70 \mathrm{~kg}\)
Radius of the drum, \(r=3 \mathrm{~m}\)
Coefficient of friction, \(\mu=0.15\)
Frequency of rotation, \(\omega=200 \mathrm{rev} / \mathrm{min}=200 / 60=10 / 3 \mathrm{rev} / \mathrm{s}\)
The necessary centripetal force required for the rotation of the man is provided by the normal force \(\left(F_{N}\right)\).
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force ( \(\left.f=F_{N}\right)\) acting upward.
Hence, the man will not fall until:
\(\mathrm{mg}<\mathrm{F}_{\mathrm{N}}=\mu \mathrm{mr} \omega^{2}\)
\(
\begin{aligned}
&\omega>\sqrt{\mathrm{g} / \mathrm{r} \mu} \\
&=\left(\frac{10}{(0.15 \times 3)}\right)^{1 / 2}=4.71 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)

Hint

(a)

(i) mg downward by the earth (gravity),
(ii) \(N\) normal force by the incline and
(iii) \(f\) up the plane, (friction) by the incline.
Taking components parallel to the incline and writing Newton’s second law,
\(
\begin{array}{r}
m g \sin 30^{\circ}-f=m g / 4 \\
f=m g / 4 .
\end{array}
\)

There is no acceleration perpendicular to the incline. Hence,
\(
=N=m g \cos 30^{\circ}=m g \cdot \frac{\sqrt{3}}{2} .
\)
As the block is slipping on the incline, friction is \(f=\mu_{k} N\).
So, \(\quad \mu_{k}=\frac{f}{N}=\frac{m g}{4 m g \sqrt{3 / 2}}=\frac{1}{2 \sqrt{3}}\).

Hint

(b) For vertical equilibrium
\(
F \cos \theta+m g={N} .
\)
Eliminating \({N}\) from these equations,
\(
F \sin \theta=\mu F \cos \theta+\mu m g
\)
or,
\(
F=\frac{\mu m g}{\sin \theta-\mu \cos \theta} .
\)

Hint

(c) Equilibrium of the block \(m\) gives
\(T=\mu  N_{1}\) and \(N_{1}=m g\)
which gives
\(T=\mu m g \dots (i)\)
Next, consider the equilibrium of the block \(M\). Taking components parallel to the incline
\(
T+\mu N_{2}=M g \sin \theta .
\)
Taking components normal to the incline

\(N_{2}=M g \cos \theta\)
These give \(\quad T=M g(\sin \theta-\mu \cos \theta) \dots (ii).\)
From (i) and (ii), \(\mu m g=M g(\sin \theta-\mu \cos \theta)\)

\( M / m=\frac{\mu}{\sin \theta-\mu \cos \theta}\)

Hint

(d) From the free body diagram,
\(
\mathrm{R}-\mathrm{mg}=0
\)
(where \(R\) is the normal reaction force and \(g\) is the acceleration due to gravity)
\(
\Rightarrow \mathrm{R}=\mathrm{mg}
\)
Again \(\mathrm{ma}-\mu_{\mathrm{k}} \mathrm{R}=0\)
(where \(\mu_{\mathrm{k}}\) is the coefficient of kinetic friction and a is deceleration)
or ma \(=\mu_{k} R\)
\(
\begin{aligned}
&\mathrm{ma}=\mu_{\mathrm{k}} \mathrm{mg} \\
&\Rightarrow \mathrm{a}=\mu_{\mathrm{k}} \mathrm{g} \\
&\Rightarrow 4=\mu_{\mathrm{k}} \mathrm{g} \\
&\Rightarrow \mu_{\mathrm{k}}=\frac{4}{g}=\frac{4}{10}=0.4
\end{aligned}
\)
Hence, the coefficient of the kinetic friction between the block and the plane is \(0.4\).

Hint

(c) The friction force acting on the block will decelerate it. Let the deceleration be ‘a’.
\(\mathrm{R}-\mathrm{mg}=0\)
(where \(R\) is the normal reaction force)
\(\Rightarrow \mathrm{R}=\mathrm{mg} \dots (i)\)
Again, ma \(-\mu_{k} R=0\)
(where \(\mu_{\mathrm{k}}\) is the coefficient of kinetic friction)
From Equation (i),

\(
\begin{aligned}
\Rightarrow m a &=\mu_{k} \mathrm{mg} \\
\Rightarrow a &=\mu_{k} g=0.1 \times 10 \\
&=1 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Given the initial velocity, \(u=10 \mathrm{~m} / \mathrm{s}\)
final velocity, \(v=0 \mathrm{~m} / \mathrm{s}\) (block comes to rest)
\(
a=-1 \mathrm{~m} / \mathrm{s}^{2} \text { (deceleration) }
\)
Using equation of motion \(v^{2}-u^{2}=2 a s\)
(where \(s\) is the distance travelled before coming to rest)
\(
s=\frac{v^{2}-u^{2}}{2 a}
\)
On substituting the respective values, we get
\(
\begin{aligned}
&=0-\frac{10^{2}}{2}(-1) \\
&=\frac{100}{2}=50 \mathrm{~m}
\end{aligned}
\)
Therefore, the block will travel \(50 \mathrm{~m}\) before coming to rest.

Hint

(a) Body is kept on the horizontal table. If no force is applied, no frictional force will be there. That means the frictional force is zero.

Hint

(d) From the free body diagram,
\(
\mathrm{R}-\mathrm{mg} \cos \theta=0 \Rightarrow \mathrm{R}=\mathrm{mg} \cos \theta \dots (i)\)
For the block
\(u=0, \quad s=8 m, t=2 \mathrm{sec}\)
\(\therefore s=u t+1 / 2 a t^{2} \Rightarrow 8=0+1 / 2 (a 2^{2}) \Rightarrow a=4 \mathrm{~m} / \mathrm{s}^{2}\)
Again, \(\mu R+m a-m g \sin \theta=0\)
\(\Rightarrow \mu \mathrm{mg} \cos \theta+\mathrm{ma}-\mathrm{mg} \sin \theta=0\)
From equation (i)
\(\Rightarrow \mathrm{m}(\mu \mathrm{g} \cos \theta+\mathrm{a}-\mathrm{g} \sin \theta)=0\)
\(\Rightarrow \mu \times 10 \times \cos 30^{\circ}=g \sin 30^{\circ}-a\)
\(\Rightarrow \mu \times 10 \times \sqrt{(3 / 3)}=10 \times(1 / 2)-4\)
\(\Rightarrow(5 / \sqrt{3}) \mu=1 \Rightarrow \mu=1 /(5 / \sqrt{3})=0.11\)

Hint

(b) From the free body diagram
\(
\begin{aligned}
&4-4 a-\mu R+4 g \sin 30^{\circ}=0  \dots (1)\\
&R-4 g \cos 30^{\circ}=0 \dots (2)\\
&\Rightarrow R=4 g \cos 30^{\circ}
\end{aligned}
\)
Putting the values of \(R\) in eq. (1)
\(
4-4 a-0.11 \times 4 g \cos 30^{\circ}+4 g \sin 30^{\circ}=0
\)
\(\Rightarrow 4-4 a-0.11 \times 4 \times 10 \times(\sqrt{3} / 2)+4 \times 10 \times(1 / 2)=0\)
\(
\Rightarrow 4-4 a-3.81+20=0 \Rightarrow a \approx 5 \mathrm{~m} / \mathrm{s}^{2}
\)
For the block \(u=0, t=2 \mathrm{sec}, \quad a=5 \mathrm{~m} / \mathrm{s}^{2}\)
Distance \(s=u t+1 / 2\) at \({ }^{2} \Rightarrow s=0+(1 / 2) 5 \times 2^{2}=10 \mathrm{~m}\) The block will move \(10 \mathrm{~m}\).

Hint

(d)

(case-a) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.
Applied force \(=\mu \mathrm{R}+2 \mathrm{~g} \sin 30^{\circ}\) \dots(1)
(where \(\mu\) is the coefficient of static friction)
\(R=m g \cos 30^{\circ}\)
Substituting the respective values in Equation (1), we get
\(
\begin{aligned}
&=0.2 \times(9.8) \sqrt{3}+2 \times 9.8 \times\left(\frac{1}{2}\right) \\
&3.39+9.8 \approx 13 \mathrm{~N}
\end{aligned}
\)

With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.

(case-b) The net force acting down the incline is given by
\(F=2 g \sin 30^{\circ}-\mu R\)
\(
\begin{aligned}
&=2 \times 9.8 \times \frac{1}{2}-3.99 \\
&=6.41 \mathrm{~N}
\end{aligned}
\)
Because \(F=6.41 \mathrm{~N}\), the body will move down the incline with acceleration, hence the force required is zero.

Hint

(d) From the free body diagram
\(
\begin{aligned}
&g=10 \mathrm{~m} / \mathrm{s}^{2}, \quad \quad \mathrm{~m}=2 \mathrm{~kg}, \quad \theta=30^{\circ}, \quad \mu=0.2 \\
&R-m g \cos \theta-F \sin \theta=0 \\
&\Rightarrow R=m g \cos \theta+F \sin \theta \quad \ldots(1) \\
&\text { And } m g \sin \theta+\mu R-F \cos \theta=0 \\
&\Rightarrow m g \sin \theta+\mu(\mathrm{mg} \cos \theta+F \sin \theta)-F \cos \theta=0 \\
&\Rightarrow m g \sin \theta+\mu \mathrm{mg} \cos \theta+\mu F \sin \theta-F \cos \theta=0 \\
&\Rightarrow F=\frac{(m g \sin \theta-\mu \mathrm{mg} \cos \theta)}{(\mu \sin \theta-\cos \theta)} \\
&\Rightarrow F=\frac{2 \times 10 \times(1 / 2)+0.2 \times 2 \times 10 \times(\sqrt{3} / 2)}{0.2 \times(1 / 2)-(\sqrt{3} / 2)}\approx17.5N
\end{aligned}
\)

Hint

(a) From the free body diagram:
\(
\begin{aligned}
&R-m g \cos 45^{\circ}=0 \\
&R=m g \cos 45^{\circ}=\frac{m g}{\sqrt{2}}
\end{aligned}
\)
Net force acting on the boy, making him slide down \(=\mathrm{mg} \sin 45^{\circ}-\mu \mathrm{R}\)
\(
\begin{aligned}
&=m g \sin 45^{\circ}-\mu \mathrm{mg} \cos 45^{\circ} \\
&=m \times 10 \times\left(\frac{1}{\sqrt{2}}\right)-0.6 \times m \times 10 \times\left(\frac{1}{\sqrt{2}}\right) \\
&=m(5 \sqrt{2}-3 \sqrt{2})=m \times 2 \times \sqrt{2}
\end{aligned}
\)
The acceleration of the boy \(=\frac{\text { Force }}{\text { Mass }}\)
\(
\begin{aligned}
&=\frac{m(2 \sqrt{2})}{m} \\
&=2 \sqrt{2} \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)

Hint

(c) Let its acceleration be a. For the first half-meter-
\(
\begin{aligned}
&\mathrm{u}=0 \quad \mathrm{t}=0.5 \mathrm{~s} \\
&\mathrm{~s}=\frac{1}{2} \mathrm{at}^{2} \\
&\Rightarrow 0.5=\frac{1}{2} \mathrm{a} \times 0.5^{2} \\
&\Rightarrow \mathrm{a}=4 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Now let at time \(\mathrm{t}\) it completes another half meter and hence total distance is \(1 \mathrm{~m}\). This also includes the time taken to cover the first half meter.

\(\mathrm{s}=\frac{1}{2} a \mathrm{t}^{2}\)
\(
\Rightarrow 1=\frac{1}{2} 4 t^{2}
\)
\(
\Rightarrow \mathrm{t}=\frac{1}{\sqrt{2}}=0.707 \mathrm{~s}
\)
Hence time to cover next half meter only \(=0.707 \mathrm{~s}-0.5 \mathrm{~s}=0.21 \mathrm{~s}\)(approx.)

Hint

(b) From the free body diagram
\(
\begin{aligned}
&\mu_{1} \mathrm{R}+1-16=0 \\
&\Rightarrow \mu_{1}(2 g)+(-15)=0 \\
&\Rightarrow \mu_{1}=15 / 20=0.75 \\
&\mu_{2} R+4 \times 0.5+16-4 g \sin 30^{\circ}=0 \\
&\Rightarrow \mu_{2}(20 \sqrt{3})+2+16-20=0 \\
&\Rightarrow \mu_{2}=\frac{2}{20 \sqrt{3}}=\frac{1}{17.32}=0.057 \approx 0.06
\end{aligned}
\)
\(\therefore\) Co-efficient of friction \(\mu_{1}=0.75 \& \mu_{2}=0.06\)

Hint

(d) From the Free body diagram of blocks ( 15 kg block assumed to move downwards) and writing Newton’s 2nd law of motion, \(F=m a\)
for the 15 kg block : \(15 \mathrm{~g}-\mathrm{T}_{1}=15 \mathrm{a} \ldots\) (1)
for the 5 kg block : \(\mathrm{T}_{2}-5 \mathrm{~g}=5 \mathrm{a} \ldots\) (2)
for \(5 \mathrm{~kg}\) block on the horizontal surface, we use Newton’s second law in two direction (i.e X – axis and Y- axis)

Balancing force along Y-direction: \(N=5 \mathrm{~g}\), and using concept of friction \(\mathrm{f}=\) \(\mu \mathrm{N}\), we get \(\mathrm{f}=0.2 \times 5 \times \mathrm{g}=\mathrm{g}\)
Now along X-direction : \(T_{1}-T_{2}-f=5 a\)….(3)
using value of \(f\) and adding eq-1 , eq-2 and eq-3 we get
\(
\begin{array}{ll}
\Rightarrow & 15 \mathrm{~g}-5 \mathrm{~g}-\mathrm{f}=(15+5+5) \times \mathrm{a} \\
\Rightarrow & 9 \mathrm{~g}=25 \mathrm{a} \\
\Rightarrow & \mathrm{a}=\frac{9 \mathrm{~g}}{25}=\frac{9 \times 10}{25}=\frac{18}{5} \frac{\mathrm{m}}{\mathrm{s}^{2}}(\text { since } \mathrm{g}=10)
\end{array}
\)
now using eq \(1: T_{1}=150-\frac{15 \times 18}{5}=150-54=96 \mathrm{~N}\)
using eq \(2: \mathrm{T}_{2}=50+\frac{5 \times 18}{5}=68 \mathrm{~N}\)
Hence values of \(\mathrm{T}_{1}=96 \mathrm{~N}\) and \(\mathrm{T}_{2}=68 \mathrm{~N}\)

Hint

(a) Given,
initial velocity of the vehicle, \(u=36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s}\)
final velocity of the vehicle, \(v=0\)
\(\mathrm{s}=5 \mathrm{~m}, \mu=\frac{4}{3}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\)
Let the maximum angle of incline be \(\theta\).
Using the equation of motion
\(
\begin{aligned}
&a=\frac{v^{2}-u^{2}}{2 s}=\frac{0-10}{2 \times 5} \\
&=-10 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
From the free-body diagram

\(R-m g \cos \theta=0\)
\(\Rightarrow R=m g \cos \theta \dots (1)\)
Again, \(
\begin{aligned}
&\mathrm{ma}+\mathrm{mg} \sin \theta-\mu \mathrm{R}=0 \\
&\Rightarrow \mathrm{ma}+\mathrm{mg} \sin \theta-\mu \mathrm{mg} \cos \theta=0 \\
&\Rightarrow \mathrm{a}+\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta=0 \\
&\Rightarrow 10+10 \sin \theta-\left(\frac{4}{3}\right) \times 10 \cos \theta=0 \\
&\Rightarrow 30+30 \sin \theta-40 \cos \theta=0 \\
&\Rightarrow 3+3 \sin \theta-4 \cos \theta=0 \\
&\Rightarrow 4 \cos \theta-3 \sin \theta=3 \\
&\Rightarrow 4 \sqrt{1-\sin ^{2} \theta}=3+3 \sin \theta
\end{aligned}
\)

On squaring, we get
\(16\left(1-\sin ^{2} \theta\right)=9+9 \sin ^{2} \theta+18 \sin \theta\)

\(25 \sin ^{2} \theta+18 \sin \theta-7=0\) \(\Rightarrow \sin \theta=\frac{18+\sqrt{18^{2}-4(25)(-7)}}{2 x 25}\)

\(=\frac{-18+32}{50}=\frac{14}{50}=0.28\)

\(\Rightarrow \theta=\sin ^{-1}(0.28)=16^{\circ}\)
Therefore, the maximum incline of the road, \(\theta=16^{\circ}\).

Hint

(b) to reach in minimum time, he has to move with maximum possible acceleration.
Let, the maximum acceleration is \(a\)
\(\therefore \mathrm{ma}-\mu \mathrm{R}=0 \Rightarrow \mathrm{ma}=\mu \mathrm{mg}\)
\(\Rightarrow a=\mu g=0.9 \times 10=9 \mathrm{~m} / \mathrm{s}^{2}\)
a) Initial velocity \(u=0, t=\)?
\(a=9 \mathrm{~m} / \mathrm{s}^{2}, \mathrm{~s}=50 \mathrm{~m}\)
\(s=u t+1 / 2 a t^{2} \Rightarrow 50=0+(1 / 2) 9 t^{2} \Rightarrow t=\sqrt{\frac{100}{9}}=\frac{10}{3}\) sec.
b) After covering \(50 \mathrm{~m}\), velocity of the athelete is
\(
V=u+a t=0+9 \times(10 / 3)=30 \mathrm{~m} / \mathrm{s}
\)
He has to stop in minimum time. So deceleration, \( -a=-9 \mathrm{~m} / \mathrm{s}^{2}\) (max)

\(\begin{aligned}
&\mathrm{R}=\mathrm{mg} \\
&\mathrm{ma}=\mu \mathrm{R} \\
&\mathrm{ma}=\mu \mathrm{mg} \\
&\Rightarrow \mathrm{a}=\mu \mathrm{g} \\
&\qquad=9 \mathrm{~m} / \mathrm{s}^{2} \text { (deceleration) } \\
&\begin{aligned}
\mathrm{u}_{1}=& 30 \mathrm{~m} / \mathrm{s}, \mathrm{v}=0 \\
\Rightarrow t=\frac{v_{1}-u_{1}}{a} \\
=\frac{0-30}{-a} \\
=\frac{-30}{-a}=\frac{10}{3} \mathrm{~s}
\end{aligned}
\end{aligned}\)

Hint

(d) Hardest break means the maximum force of friction is developed between cars tire and the road.
Max frictional force \(=\mu \mathrm{R}\)
From the free-body diagram
\(
\begin{aligned}
&\mathrm{R}-\mathrm{mg} \cos \theta=0 \\
&\Rightarrow \mathrm{R}=\mathrm{mg} \cos \theta \ldots \text { (i) } \\
&\text { and } \mu \mathrm{R}+\mathrm{ma}-\mathrm{mg} \sin )=0 \ldots \text { (ii) } \\
&\Rightarrow \mu \mathrm{mg} \cos \theta+\mathrm{ma}-\mathrm{mg} \sin \theta=0 \\
&\Rightarrow \mu g \cos \theta+\mathrm{a}-10 \times(1 / 2)=0 \\
&\Rightarrow \mathrm{a}=5-\{1-(2 \sqrt{3})\} \times 10(\sqrt{3} / 2)=2.5 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
When, hardest brake is applied the car move with acceleration \(2.5 \mathrm{~m} / \mathrm{s}^{2}\)

Hint

(b) Let, a maximum acceleration produced in the car.
\(\therefore \mathrm{ma}=\mu \mathrm{R}\) [For more acceleration, the tyres will slip]
\(\Rightarrow \mathrm{ma}=\mu \mathrm{mg} \Rightarrow a=\mu \mathrm{g}=1 \times 10=10 \mathrm{~m} / \mathrm{s}^{2}\)
For crossing the bridge in minimum time, it has to travel with maximum acceleration
\(
\begin{aligned}
&u=0, \quad s=500 m, \quad a=10 \mathrm{~m} / \mathrm{s}^{2} \\
&\mathrm{~s}=u \mathrm{t}+1 / 2 \text { at }^{2} \\
&\Rightarrow 500=0+(1 / 2) 10 \mathrm{t}^{2} \Rightarrow \mathrm{t}=10 \mathrm{sec} .
\end{aligned}
\)
If acceleration is less than \(10 \mathrm{~m} / \mathrm{s}^{2}\), time will be more than \(10 \mathrm{sec}\). So one can’t drive through the bridge in less than \(10 \mathrm{sec}\).

Hint

(a) Since, \(\mu_{1}<\mu_{2}\), acceleration of \(2-\mathrm{kg}\) block down the plane will be more than the acceleration of \(4 \mathrm{~kg}\) block, if allowed to move separately.

In this case, both blocks are treated as a system of mass \((4+2)=6 \mathrm{~kg}\) and will move down with the same acceleration. Net force down the plane is
\(
\begin{aligned}
F =\left(m_{1}+m_{2}\right) g \sin \theta-\mu_{1} m_{1} g \cos \theta-\mu_{2} m_{2} g \cos \theta \\
=(4+2) g \sin 30^{\circ}-(0.2)(2) g \cos 30^{\circ}-(0.3)(4) g \cos 30^{\circ} \\
=(6)(10)\left(\frac{1}{2}\right)-(0.4)(10)\left(\frac{\sqrt{3}}{2}\right)-(1.2)(10)\left(\frac{\sqrt{3}}{2}\right) \\
\text { Therefore, acceleration } =30-13.76=16.24 \mathrm{~N}
\end{aligned}
\)
Therefore, acceleration of both the blocks down the plane will be \(a=\frac{F}{m_{1}+m_{2}}=\frac{16.24}{4+2}=2.7 \mathrm{~m} \mathrm{~s}^{-2}\)

(b) can be solved. In this case, the \(4 \mathrm{~kg}\) block will travel with more acceleration because, coefficient of friction is less than that of \(2 \mathrm{~kg}\). So, they will move separately. Drawing the free body diagram of \(2 \mathrm{~kg}\) mass only, it can be found that, \(a=2.4 \mathrm{~m} / \mathrm{s}^{2}\).

Hint

(c)

\(\begin{aligned}
&R_{1}=M_{1} g \cos \theta \quad \ldots \text { (i) } \\
&R_{2}=M_{2} g \cos \theta \quad \ldots \text { (ii) } \\
&T+M_{1} g \sin \theta-m_{1} a-\mu R_{1}=0 \quad \ldots \text { (iii) }\\
&T-M_{2}-M_{2} a+\mu R_{2}=0 \quad \ldots \text { (iv) }\\
&\text {Eq (iii)} \Rightarrow T+M_{1} g \sin \theta-M_{1} a-\mu M_{1} g \cos \theta=0 \\
&\text {Eq (iv)} \Rightarrow T-M_{2} g \sin \theta+M_{2} a+\mu M_{2} g \cos \theta=0 \dots (v) \\
&\text { Eq (iv) \& (v) }, \sin \theta\left(M_{1}+M_{2}\right)-a\left(M_{1}+M_{2}\right)-\mu g \cos \theta\left(M_{1}+M_{2}\right)=0 \\
& a\left(M_{1}+M_{2}\right)=g \sin \theta\left(M_{1}+M_{2}\right)-\mu g \cos \theta\left(M_{1}+M_{2}\right) \\
&\Rightarrow a=g(\sin \theta-\mu \cos \theta) \\
&\therefore \text { The blocks (system has acceleration } g(\sin \theta-\mu \cos \theta) \\
&\text {The force exerted by the rod on one of the blocks is tension.} \\
&\text { Tension } T=-M_{1} g \sin \theta+M_{1} a+\mu M_{1} g \sin \theta \\
&T=-M_{1} g \sin \theta+M_{1}(g \sin \theta-\mu g \cos \theta)+\mu M_{1} g \cos \theta \\
&\Rightarrow T=0
\end{aligned}\)

Hint

(d) Let \({p}\) be the force applied to at an angle \(\theta\)
From the free body diagram
\(R+P \sin \theta-m g=0\) \(\Rightarrow R=-P \sin \theta+m g \quad \dots (i);\mu R-P \cos \theta \quad \quad \ldots\) (ii) Eq. (i) is \(\mu(\mathrm{mg}-P \sin \theta)-P \cos \theta=0\) \(\Rightarrow \mu m g=\mu \rho \sin \theta-P \cos \theta \Rightarrow \rho=\frac{\mu m g}{\mu \sin \theta+\cos \theta}\)
Applied force \(P\) should be minimum, when \(\mu \sin \theta+\cos \theta\) is maximum.
Again, \(\mu \sin \theta+\cos \theta\) is maximum when its derivative is zero.
\(\therefore \mathrm{d} / \mathrm{d} \theta(\mu \sin \theta+\cos \theta)=0\)
\(\Rightarrow \mu \cos \theta-\sin \theta=0 \Rightarrow \theta=\tan ^{-1} \mu\)
So, \(P=\frac{\mu \mathrm{mg}}{\mu \sin \theta+\cos \theta}=\frac{\mu \mathrm{mg} / \cos \theta}{\frac{\mu \sin \theta}{\cos \theta}+\frac{\cos \theta}{\cos \theta}}=\frac{\mu \mathrm{mg} \sec \theta}{1+\mu \tan \theta}=\frac{\mu \mathrm{mgsec} \theta}{1+\tan ^{2} \theta}\)
\(
=\frac{\mu \mathrm{mg}}{\sec \theta}=\frac{\mu \mathrm{mg}}{\sqrt{\left(1+\tan ^{2} \theta\right.}}=\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}
\)
Minimum force is \(\frac{\mu \mathrm{mg}}{\sqrt{1+\mu^{2}}}\) at an angle \(\theta=\tan ^{-1} \mu\).

Hint

(b) Let, the max force exerted by the man is T.
From the free-body diagram
\(
\begin{aligned}
&\mathrm{R}+\mathrm{T}-\mathrm{Mg}=0 \\
&\Rightarrow \mathrm{R}=\mathrm{Mg}-\mathrm{T} \dots (i)\\
&\mathrm{R}_{1}-\mathrm{R}-\mathrm{mg}=0 \\
&\Rightarrow \mathrm{R}_{1}=\mathrm{R}+\mathrm{mg} \dots (ii)
\end{aligned}
\)
And \(T-\mu R_{1}=0\)

\(
\begin{aligned}
&\Rightarrow T-\mu(R+m g)=0 \quad \text { [From equn. (ii)] } \\
&\Rightarrow T-\mu R-\mu \mathrm{mg}=0 \\
&\Rightarrow T-\mu(\mathrm{Mg}+\mathrm{T})-\mu \mathrm{mg}=0 \quad \text { [from (i)] } \\
&\Rightarrow T(1+\mu)=\mu \mathrm{Mg}+\mu \mathrm{mg} \\
&\Rightarrow T=\frac{\mu(\mathrm{M}+\mathrm{m}) \mathrm{g}}{1+\mu}
\end{aligned}
\)
Maximum force exerted by man is \(\frac{\mu(M+m) g}{1+\mu}\)

Hint

(a)

\(\begin{aligned}
&\mathrm{R}_{1}-2 \mathrm{~g}=0 \\
&\Rightarrow \mathrm{R}_{1}=2 \times 10=20 \\
&2 \mathrm{a}+0.2 \mathrm{R}_{1}-12=0 \\
&\Rightarrow 2 \mathrm{a}+0.2(20)=12 \\
&\Rightarrow 2 \mathrm{a}=12-4=8 \\
&\Rightarrow a=4 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}\)

\(\begin{aligned}
&4 a-\mu R_{1}=0 \\
&\Rightarrow 4 a=\mu R_{1}=0.2(20) \\
&\Rightarrow 4 a=4 \\
&\Rightarrow a=1 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}\)

\(2 \mathrm{~kg} \text { block has acceleration } 4 \mathrm{~m} / \mathrm{s}^{2},\text { that of } 4 \mathrm{~kg} \text { is } 1 \mathrm{~m} / \mathrm{s}^{2}\)

Similarly for case (b) we can show that \(a=2 \mathrm{~m} / \mathrm{s}^{2}\)

Hint

(c)

(a) When the \(10 \mathrm{~N}\) force applied on \(2 \mathrm{~kg}\) block, it experiences maximum frictional force \(\mu R_{1}=\mu \times 2 \mathrm{~kg}=(0.2) \times 20=4 \mathrm{~N}\) from the \(3 \mathrm{~kg}\) block.
So, the \(2 \mathrm{~kg}\) block experiences a net force of \(10-4=6 \mathrm{~N}\)
So, \(a_{1}=6 / 2=3 \mathrm{~m} / \mathrm{s}^{2}\)
But for the \(3 \mathrm{~kg}\) block, (Fig-3) the frictional force from \(2 \mathrm{~kg}\) block (4N) becomes the driving force and the maximum frictional force between \(3 \mathrm{~kg}\) and \(7 \mathrm{~kg}\) block is
\(
\mu_{2} \mathrm{R}_{2}=(0.3) \times 5 \mathrm{~kg}=15 \mathrm{~N}
\)
So, the \(3 \mathrm{~kg}\) block cannot move relative to the \(7 \mathrm{~kg}\) block. The \(3 \mathrm{~kg}\) block and \(7 \mathrm{~kg}\) block both will have same acceleration \(\left(a_{2}=a_{3}\right)\) which will be due to the \(4 \mathrm{~N}\) force because there is no friction from the floor. \(\therefore a_{2}=a_{3}=4 / 10=0.4 \mathrm{~m} / \mathrm{s}^{2}\)

(b) When the \(10 \mathrm{~N}\) force is applied to the \(3 \mathrm{~kg}\) block, it can experience maximum frictional force of \(15+4\) \(=19 \mathrm{~N}\) from the \(2 \mathrm{~kg}\) block and \(7 \mathrm{~kg}\) block.
So, it can not move with respect to them. As the floor is frictionless, all the three bodies will move together
\(
\therefore a_{1}=a_{2}=a_{3}=10 / 12=(5 / 6) \mathrm{m} / \mathrm{s}^{2}
\)
(c) Similarly, it can be proved that when the \(10 \mathrm{~N}\) force is applied to the \(7 \mathrm{~kg}\) block, all the three blocks will move together.
Again \(a_{1}=a_{2}=a_{3}=(5 / 6) \mathrm{m} / \mathrm{s}^{2}\)

Hint

(c)
\(\begin{aligned}
&\text { Net force on the block }=\sqrt{20^{2}+15^{2}}-(0.5) \times 40 \\
&=25-20=5 \mathrm{~N} \\
&\therefore \tan \theta=\frac{20}{15}=\frac{4}{3} \\
&\Rightarrow \theta=\tan ^{-1}\left(\frac{4}{3}\right)=53^{\circ}
\end{aligned}
\)
Therefore, the block will move at \(53^{\circ}\) angle with the \(15 \mathrm{~N}\) force.

Hint

(d) Mass of man \(=50 \mathrm{~kg} . \quad \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\)
Frictional force developed between hands, legs, and backside with the wall the wt of man. So he remains in equilibrium. He gives the equal force on both the walls so gets equal reaction R from both the walls. If he applies unequal forces R should be different he can’t rest between the walls. Frictional force \(2 \mu \mathrm{R}\) balance his weight.
From the free-body diagram
\(
\mu \mathrm{R}+\mu \mathrm{R}=40 \mathrm{~g} \Rightarrow 2 \mu \mathrm{R}=40 \times 10 \Rightarrow \mathrm{R}=\frac{40 \times 10}{2 \times 0.8}=250 \mathrm{~N}
\)
(a) The normal force is \(250 \mathrm{~N}\).

Hint

(b) Let \(a_{1}\) and \(a_{2}\) be the accelerations of m and M respectively. Here, \(a_{1}>a_{2}\) so that m moves on M
Suppose, after time t m separate from M.
In this time, \(m\) covers \(S_{m}=v t+1 / 2 a_{1} t^{2}\) and \(S_{M}=v t+1 / 2 a_{2} t^{2}\)
For m to separate from M. \(vt +1 / 2 a_{1} t^{2}=v t+1 / 2 a_{2} t^{2}+\ell \dots (1)\)

Again from free body diagram
\(\mathrm{ma}_{1}+\mu / 2 \mathrm{R}=0\)
\(
\Rightarrow \mathrm{ma}_{1}=-(\mu / 2) \mathrm{mg}=-(\mu / 2) \mathrm{m} \times 10 \Rightarrow \mathrm{a}_{1}=-5 \mu
\)
Again,
\(
\begin{aligned}
&\mathrm{Ma}_{2}+\mu(\mathrm{M}+\mathrm{m}) \mathrm{g}-(\mu / 2) \mathrm{mg}=0 \\
&\Rightarrow 2 \mathrm{Ma}_{2}+2 \mu(\mathrm{M}+\mathrm{m}) \mathrm{g}-\mu \mathrm{mg}=0 \\
&\Rightarrow 2 \mathrm{M} \mathrm{a}_{2}=\mu \mathrm{mg}-2 \mu \mathrm{Mg}-2 \mu \mathrm{mg} \\
&\Rightarrow \mathrm{a}_{2} \frac{-\mu \mathrm{mg}-2 \mu \mathrm{Mg}}{2 \mathrm{M}}
\end{aligned}
\)
Putting values of \(a_{1}\) and \(a_{2}\) in equation (1) we can find that
\(
T=\sqrt{\left(\frac{4 m l}{(M+m) \mu g}\right)}
\)

Hint

(a) Let the acceleration of block M is a towards right. So, the block m must go down with an acceleration 2a. As the block m is in contact with the block M, it will also have acceleration a towards right. So, it will experience two inertia forces as shown in the free body diagram Fig-1. From the free body diagram Fig-1

\(\begin{aligned}
&\mathrm{R}_{1}-\mathrm{ma}=0 \Rightarrow \mathrm{R}_{1}=\mathrm{ma} \dots (i)\\
&\text { Again, } 2 m a+T-m g+\mu_{1} R_{1}=0 \\
&\begin{aligned}
&\Rightarrow T=m g-\left(2-\mu_{1}\right) m a \dots (ii)\\
&\text { From free body diagram Fig-2 }
\end{aligned}\\
&\mathrm{T}+\mu_{1} \mathrm{R}_{1}+\mathrm{mg}-\mathrm{R}_{2}=0\\
&\Rightarrow \mathrm{R}_{2}=\mathrm{T}+\mu_{1} \mathrm{ma}+\mathrm{Mg} \quad \text { [Put } \mathrm{R}_{1} \text { from (i)] }\\
&=\left(\mathrm{mg}-2 \mathrm{ma}-\mu_{1} \mathrm{ma}\right)+\mu_{1} \mathrm{ma}+\mathrm{Mg} \quad \text { [Put } \mathrm{T} \text { from (ii)] }\\
&\therefore R_{2}=M g+m g-2 m a \dots (iii) \\
&\text { Again, form the free body diagram }Fig-2\\
&\mathrm{T}+\mathrm{T}-\mathrm{R}-\mathrm{Ma}-\mu_{2} \mathrm{R}_{2}=0\\
&\Rightarrow 2 \mathrm{~T}-\mathrm{MA}-\mathrm{mA}-\mu_{2}(\mathrm{Mg}+\mathrm{mg}-2 \mathrm{ma})=0 \quad \text { [Put } \mathrm{R}_{1} \text { and } \mathrm{R}_{2} \text { from (i) and (iii)] }\\
&\Rightarrow 2 \mathrm{~T}=(\mathrm{M}+\mathrm{m})+\mu_{2}(\mathrm{Mg}+\mathrm{mg}-2 \mathrm{ma}) \quad \ldots \text { (iv) }\\
&\text { From equation (ii) and (iv) }\\
&2 \mathrm{~T}=2 \mathrm{mg}-2\left(2+\mu_{1}\right) \mathrm{mg}=(\mathrm{M}+\mathrm{m}) \mathrm{a}+\mu_{2}(\mathrm{Mg}+\mathrm{mg}-2 \mathrm{ma})\\
&\Rightarrow 2 \mathrm{mg}-\mu_{2}(\mathrm{M}+\mathrm{m}) \mathrm{g}=\mathrm{a}\left(\mathrm{M}+\mathrm{m}-2 \mu_{2} \mathrm{~m}+4 \mathrm{~m}+2 \mu_{1} \mathrm{~m}\right)\\
&\Rightarrow a=\frac{\left[2 m-\mu_{2}(M+m)\right] g}{M+m\left[5+2\left(\mu_{1}-\mu_{2}\right)\right]}
\end{aligned}\)

Hint

(d) Because the block slips on the table, maximum frictional force acts on it. From the free body diagram
\(
\mathrm{R}=\mathrm{mg}
\)
\(\therefore F-\mu \mathrm{R}=0 \Rightarrow \mathrm{F}=\mu \mathrm{R}=\mu \mathrm{mg}\)
But the table is at rest. So, the frictional force at the legs of the table is not \(\mu R_{1}\). Let be \(\mathrm{f}\), so form the free body diagram.
\(
f-\mu \mathrm{R}=0 \Rightarrow f=\mu \mathrm{R}=\mu \mathrm{mg} \text {. }
\)
Total frictional force on table by floor is \(\mu \mathrm{mg}\).

Hint

(a)

\(\begin{aligned}
&R_{1}=m g \quad \ldots \text { (i) } \\
&F-\mu R_{1}-T=0 \Rightarrow F-\mu m g-T=0 \quad \ldots \text { (ii) }
\end{aligned}\)

\(\begin{aligned}
&\mathrm{T}-\mu \mathrm{R}_{1}=0 \\
&\Rightarrow \mathrm{T}=\mu \mathrm{mg}
\end{aligned}\)

\(\therefore F=\mu \mathrm{mg}+\mu \mathrm{mg}=2 \mu \mathrm{mg}\)
[putting \(\mathrm{T}=\mu \mathrm{mg}\) ]

Hint

(b)

\(\begin{aligned}
&R_{1}+Q E-m g=0 \\
&R_{1}=m g-Q E \dots (i)\\
&F-T-\mu R_{1}=0 \\
&\Rightarrow F-T \mu(\mathrm{mg}-Q E)=0 \\
&\Rightarrow F-T-\mu \mathrm{mg}+\mu Q E=0 \ldots(ii) \\
&T-\mu R_{1}=0 \\
&\Rightarrow T=\mu R_{1}=\mu(\mathrm{mg}-Q E)=\mu \mathrm{mg}-\mu Q E \\
&\text { Now equation (ii) is } F-m g+\mu Q E-\mu \mathrm{mg}+\mu \mathrm{QE}=0 \\
&\Rightarrow F-2 \mu \mathrm{mg}+2 \mu \mathrm{QE}=0 \\
&\Rightarrow F=2 \mu \mathrm{mg}-2 \mu \mathrm{QE} \\
&\Rightarrow F=2 \mu(\mathrm{mg}-\mathrm{QE})
\end{aligned}
\)
The maximum horizontal force that can be applied is \(2 \mu(\mathrm{mg}-Q E)\)