Sample Exercises

Hint

(b)

\(\begin{aligned}
&M=m_{c}+m_{b}=90 \mathrm{~kg} \\
&u=6 \mathrm{~km} / \mathrm{h}=1.666 \mathrm{~m} / \mathrm{sec} \\
&v=12 \mathrm{~km} / \mathrm{h}=3.333 \mathrm{~m} / \mathrm{sec} \\
&\text { Increase in } \mathrm{K} . \mathrm{E} .=1 / 2 \mathrm{Mv}^{2}-1 / 2 \mathrm{Mu}^{2} \\
&=1 / 290 \times(3.333)^{2}-1 / 2 \times 90 \times(1.66)^{2}=374.8 \mathrm{~J}
\end{aligned}\)

Hint

(c)

\(\begin{aligned}
&m =2 \mathrm{~kg} . \\
&u=10 \mathrm{~m} / \mathrm{sec} \\
&a=3 \mathrm{~m} / \mathrm{aec}^{2} \\
&t=5 \mathrm{sec} \\
&\mathrm{v}=\mathrm{u}+\mathrm{at}=10+315=25 \mathrm{~m} / \mathrm{sec} . \\
&\therefore \text { Final Kinetic Energy }=1 / 2 \mathrm{mv}^{2}=625 \mathrm{~J} .
\end{aligned}\)

Hint

(a) Resisting force acting on the box, \(F=100 \mathrm{~N}\)
Displacement of the box, \(S=4 \mathrm{~m}\)
Also,
\(
\theta=180^{\circ}
\)
\(\therefore\) Work done by the resisting force,
\(
W=\vec{F} \cdot \vec{S}=100 \times 4 \times \cos 180^{\circ}=-400 J
\)

Hint

(d)
Mass of the block, \(M=5 \mathrm{~kg}\)
Angle of inclination, \(\theta=30^{\circ}\)
Gravitational force acting on the block,
\(F=m g\)
Work done by the force of gravity depends only on the height of the object, not on the path length covered by the object.
Height of the object, \(h=10 \times \sin 30^{\circ}\)
\(
=10 \times \frac{1}{2}=5 m
\)
\(\therefore\) Work done by the force of gravity, w \(=\mathrm{mgh}\)
\(
=5 \times 9.8 \times 5=245 J
\)

Hint

(a) Given:
\(\mathrm{F}=2.50 \mathrm{~N}, \mathrm{~S}=2.5 \mathrm{~m}\) and \(\mathrm{m}=15 \mathrm{~g}=0.015 \mathrm{~kg}\)
Work done by the force,
\(W=F \cdot S \cos 0^{\circ}\) ( acting along the same line )
\(
=2.5 \times 2.5=6.25 \mathrm{~J}
\)
Acceleration of the particle is,
\(
\begin{aligned}
&a=\frac{F}{m}=\frac{2.5}{0.015} \\
&=\frac{500}{3} \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Applying the work-energy principle for finding the final velocity of the particle,
\(
\begin{aligned}
&\frac{1}{2} m v^{2}-0=6.25 \\
&\Rightarrow \nu=\sqrt{\frac{6.25 \times 2}{0.15}}=28.86 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
So, time taken by the particle to cover \(2.5 \mathrm{~m}\) distance,
\(t=\frac{\nu-u}{a}=\frac{(28.86) \times 3}{500}\)
\(
\begin{aligned}
&\therefore \text { Average power }=\frac{W}{t} \\
&=\frac{6.25 \times 500}{(28.86) \times 3}=36.1 W
\end{aligned}
\)

Hint

(d) Initial position vector, \(\overrightarrow{r_{1}}=2 \vec{i}+3 \vec{j}\) Final position vector, \(\vec{r}_{2}=3 \vec{i}+2 \vec{j}\)
So, displacement vector,
\(
\begin{aligned}
&\vec{r}=\vec{r}_{2}-\vec{r}_{1} \\
&=(3 \vec{i}+2 \vec{j})-(2 \vec{i}+\vec{j}) \\
&=\vec{i}-\vec{j}
\end{aligned}
\)
Force acting on the particle,\(\vec{F}=5 \vec{i}+5 \vec{j}\)
So, work done \(=\vec{F} \cdot \vec{S}=5 \times 1+5(-1)=0\)

Hint

(c) Given:
Mass of the block, \(m=2 \mathrm{~kg}\)
Distance covered by the man, \(s=40 \mathrm{~m}\)
Acceleration of the man, \(\mathrm{a}=0.5 \mathrm{~m} / \mathrm{s}^{2}\)
So, force applied by the man on the box,
\(
\begin{aligned}
&F=m a \\
&=2 \times(0.5) \\
&=1 N
\end{aligned}
\)
Work done by the man on the block, \(W=F \cdot S\)
\(
\begin{aligned}
&=1 \times 40 \\
&=40 J
\end{aligned}
\)

Hint

(a) Given that force is a function of displacement, i.e.
\(
F=a+b x \text {, }
\)
where \(a\) and \(b\) are constants.
So, work done by this force during the displacement \(x=0\) to \(x=d\),
\(
\begin{aligned}
&W=\int_{0}^{d} F \mathrm{dx} \\
&W=\int_{0}^{d}(a+b x) d x \\
&W=\left[a x+\frac{b x^{2}}{2}\right]_{0}^{d} \\
&W=a d+\frac{b d^{2}}{2} \\
&\Rightarrow W=\left(a+\frac{b d}{2}\right) d
\end{aligned}
\)

Hint

(b) Given :
\(
\mathrm{m}=250 \mathrm{~g}, \theta=37^{\circ}, \mathrm{d}=1 \mathrm{~m}
\)
Here, \(R\) is the normal reaction of the block.
As the block is moving with uniform speed, \(F=\mu \mathrm{R}=\mathrm{mg} \sin 37^{\circ}\)
So, work done against the force of friction,
\(
\begin{aligned}
&W=\mathrm{Fd} \cos 0^{\circ} \\
&W=\left(\operatorname{mg} \sin 37^{\circ}\right) \times d \\
&W=\left(0.25 \times 9.8 \times \sin 37^{\circ}\right) \times 1.0 \\
&W=1.5 J
\end{aligned}
\)

Hint

(a) 
Given \(a=\frac{F}{2(M+m)} \)
case-a: From the figure below

\(
\begin{aligned}
&\mathrm{ma}=\mu \mathrm{R} \text { and } \mathrm{R}=\mathrm{mg} \\
&\Rightarrow \mu=\frac{\mathrm{ma}}{\mathrm{R}}=\frac{\mathrm{F}}{2(\mathrm{M}+\mathrm{m}) \mathrm{g}}
\end{aligned}
\)
case b: Frictional force acting on the smaller block \(f=\mu R=\frac{F}{2(M+m) g} \times m g=\frac{m \times F}{2(M+m)}\)
case-c: Work done \(\mathrm{w}=\mathrm{fs}\)
\(
s=d
\)
\(
w=\frac{m F}{2(M+m)} \times d=\frac{m F d}{2(M+m)}
\)

Hint

(a) Given Weight \(=2000 \mathrm{~N}, \mathrm{~s}=20 \mathrm{~m}, \mu=0.2\)
The free-body diagram for the box is shown below:
case-a: From the figure,
\(R+P \sin \theta-2000=0 \ldots(i)\) \(P \cos \theta-0.2 R=0\)… (ii)
From the equation (i) and (ii),
\(
\begin{aligned}
&P \cos \theta-0.2(2000-P \sin \theta)=0 \\
&P(\cos \theta+0.2 \sin \theta)=400 \\
&P=\frac{400}{\cos \theta+0.2 \sin \theta} \ldots(\text { iii })
\end{aligned}
\)
So, work done by the person,
\(
\begin{aligned}
&W=P S \cos \theta \\
&=\frac{8000 \cos \theta}{\cos \theta+0.2 \sin \theta} \\
&=\frac{8000}{1+0.2 \tan \theta} \\
&=\frac{40000}{5+\tan \theta} \ldots(i v)
\end{aligned}
\)
(b) For minimum magnitude of force from equation (iii),
\(
\begin{aligned}
&\frac{d}{d k}(\cos \theta+0.2 \sin \theta)=0 \\
&\Rightarrow \tan \theta=0.2
\end{aligned}
\)
Putting the value in equation (iv),
\(
\begin{aligned}
&W=\frac{40000}{(5+\tan \theta)} \\
&=\frac{40000}{5+0.2} \approx 7692 J
\end{aligned}\)

Hint

(c) Given Weight, \(\mathrm{mg}=100 \mathrm{~N}\)
\(\theta=37^{\circ}\) and \(\mathrm{s}=2 \mathrm{~m}\)
Force, \(F=m g \sin 37^{\circ}\)
\(=100 \times \frac{3}{5}=60 N\)
So, work done when the force is parallel to incline,
\(
\begin{aligned}
&W=F S \cos \theta \\
&=60 \times 2 \times \cos 0^{\circ}=120 J
\end{aligned}
\)
In \(\triangle A B C, A B=2 \mathrm{~m}\)
\(
\begin{aligned}
&A C=\mathrm{h} \\
&=s \times \sin 37^{\circ}=2.0 \times \sin 37^{\circ} \\
&=1.2 \mathrm{~m}
\end{aligned}
\)
\(\therefore\) Work done when the force is in horizontal direction,
\(
\begin{aligned}
&\mathrm{W}=\mathrm{mgh} \\
&=100 \times 1.2=120 J
\end{aligned}
\)

Hint

(b) Given Mass of the car, \(\mathrm{m}=500 \mathrm{~kg}\)
Distance covered by the car, \(\mathrm{s}=25 \mathrm{~m}\)
Initial speed of the car, \(u=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}\)
Final speed of the car, \(v=0 \mathrm{~m} / \mathrm{s}\)
Retardation of the car,
\(
\begin{aligned}
&a=\frac{\left(v^{2}-u^{2}\right)}{2 s} \\
&\Rightarrow a=-\frac{400}{50}=-8 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Frictional force, \(\mathrm{F}=\mathrm{ma}=500 \times 8=4000 \mathrm{~N}\)

Hint

(c) The acceleration of the car is given as,
\(
\begin{aligned}
&\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \\
&\left(72 \times \frac{5}{18}\right)^{2}=0+2 \times \mathrm{a} \times 25 \\
&\mathrm{a}=8 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
The force is given as,
\(
\begin{aligned}
&F=m a \\
&=500 \times 8 \\
&=4000 \mathrm{~N}
\end{aligned}
\)
Thus, the average force needed to accelerate the car is \(4000 \mathrm{~N}\).

Hint

(a) GIven,
\(v=a \sqrt{x}(\) uniformly accelerated motion )
Displacement, \(s=d-0=d\)
Putting \(x=0\), we get \(v_{1}=0\)
Putting \(\mathrm{x}=\mathrm{d}\), we get \(v_{2}=a \sqrt{d}\)
\(a=\frac{v_{2}^{2}-v_{1}^{2}}{2 s}=\frac{a^{2} d}{2 d}=\frac{a^{2}}{2}\)
Force, \(\mathrm{F}=\mathrm{m} a=\frac{m a^{2}}{2}\)
Work done, W \(=\) Fs \(\cos \theta\)
\(
=\frac{m a^{2}}{2} \times d=\frac{m a^{2} d}{2}
\)

Hint

(d)

(a) \(m=2 \mathrm{~kg}, \quad \theta=37^{\circ}, \quad \mathrm{F}=20 \mathrm{~N}\)
From the free body diagram
\(
\begin{aligned}
&F=(2 g \sin \theta)+m a \Rightarrow a=(20-20 \sin \theta) / s=4 \mathrm{~m} / \mathrm{sec}^{2} \\
&\mathrm{~S}=u \mathrm{t}+1 / 2 a \mathrm{t}^{2} \quad(\mathrm{u}=0, \mathrm{t}=1 \mathrm{~s}, \mathrm{a}=1.66) \\
&=2 \mathrm{~m}
\end{aligned}
\)
So, work, done w \(=F_{s}=20 \times 2=40 \mathrm{~J}\)
(b)  If \(W=40 J\)
\(
S=\frac{W}{F}=\frac{40}{20}=2 \mathrm{~m}
\)
\(
h=2 \sin 37^{\circ}=1.2 \mathrm{~m}
\)
So, the work done
\(
\begin{aligned}
&W=-\text { mgh } \\
&=-20 \times 1.2=-24 J
\end{aligned}
\)
(c) \(v=u+\) at \(=4 \times 10=40 \mathrm{~m} / \mathrm{sec}\)
So, Kinetic Energy \(=1 / 2 \mathrm{mv}^{2}=1 / 2 \times 2 \times 16=16 \mathrm{~J}\)

Hint

(c) \(\begin{aligned}
&\mathrm{m}=2 \mathrm{~kg}, \quad \theta=37^{\circ}, \quad \mathrm{F}=20 \mathrm{~N}, \quad a=10 \mathrm{~m} / \mathrm{sec}^{2} \\
&\text { a) } \mathrm{t}=1 \mathrm{sec} \\
&\text { So, s= ut }+1 / 2 a t^{2}=5 \mathrm{~m}
\end{aligned}\)
Work done by the applied force \(w=F S \cos 0^{\circ}=20 \times 5=100 \mathrm{~J}\)
b) \(B C(h)=5 \sin 37^{\circ}=3 \mathrm{~m}\)
So, work done by the weight \(W=m g h=2 \times 10 \times 3=60 \mathrm{~J}\)
c) So, frictional force \(f=m g \sin \theta\)
work done by the frictional forces \(w=f s \cos 0^{\circ}=(m g \sin \theta) s=20 \times 0.60 \times 5=60 \mathrm{~J}\)

Hint

(a) Given, \(m=250 \mathrm{~g}=0.250 \mathrm{~kg}\),
\(
u=40 \mathrm{~cm} / \mathrm{sec}=0.4 \mathrm{~m} / \mathrm{sec}
\)
\(
\mu=0.1, \quad v=0
\)
Here, \(\mu R=\operatorname{ma}\{\) where, \(a=\) deceleration \(\}\)
\(
\begin{aligned}
&a=\frac{\mu R}{m}=\frac{\mu \mathrm{mg}}{m}=\mu g=0.1 \times 9.8=0.98 \mathrm{~m} / \mathrm{sec}^{2} \\
&\mathrm{~S}=\frac{\mathrm{v}^{2}-\mathrm{u}^{2}}{2 \mathrm{a}}=0.082 \mathrm{~m}=8.2 \mathrm{~cm}
\end{aligned}
\)
Again, work done against friction is given by
\(
\begin{aligned}
&-W=\mu R S \cos \theta \\
&=0.1 \times 2.5 \times 0.082 \times 1\left(\theta=0^{\circ}\right)=0.02 \mathrm{~J} \\
&\Rightarrow W=-0.02 \mathrm{~J}
\end{aligned}
\)

Hint

(d) Given Height, \(h=50 \mathrm{~m}\)
Mass of water falling per hour, \(\mathrm{m}=1.8 \times 10^{5} \mathrm{~kg}\)
Power of a lamp,
\(P=100\) watt
Potential energy of the water ,
\(
\begin{aligned}
&\text { P.E. = mgh } \\
&=1.8 \times 10^{5} \times 9.8 \times 50 \\
&=882 \times 10^{5} \mathrm{~J}
\end{aligned}
\)
As only half the potential energy of water is converted into electrical energy,
Electrical energy \(=\frac{1}{2} P. \) E . \(=441 \times 10^{5} \mathrm{~J} / \mathrm{hr}\)
So, power in watt \((J / \mathrm{sec})=\left(\frac{441 \times 10^{5}}{60 \times 60}\right)\)
Therefore, the number of \(100 \mathrm{~W}\) lamps that can be lit using this energy,
\(
\mathrm{n}=\frac{441 \times 10^{5}}{3600 \times 100}=122.5 \approx 122
\)

Hint

(c) Given Mass of the bucket with the paint, \(m=6 \mathrm{~kg}\)
Height at which the bucket is placed, \(h=2 \mathrm{~m}\)
Potential energy of the bucket with the paint at the given height,
\(P . E .=m g h\)
\(=6 \times(9.8) \times 2\)
\(=117.6 J\)
\(P . E\). on the floor \(=0\)
Loss in potential energy \(=117.6-0\)
\(
=117.6 J \approx 118 J
\)

Hint

(a) Given Height of the cliff, \(h=40 \mathrm{~m}\)
Initial speed of the projectile, \(u=50 \mathrm{~m} / \mathrm{s}\)
Let the projectile hit the ground with velocity v.
Applying the law of conservation of energy,
\(
\begin{aligned}
&\mathrm{mgh}+\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \mathrm{mv}^{2} \\
&\Rightarrow 10 \times 40+\left(\frac{1}{2}\right) \times 2500=\frac{1}{2} \mathrm{v}^{2} \\
&\Rightarrow \mathrm{v}^{2}=3300 \\
&\Rightarrow \mathrm{v}=57.4 \mathrm{~m} / \mathrm{s}\approx 58 \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)
The projectile hits the ground with a speed of \(58 \mathrm{~m} / \mathrm{s}\).

Hint

(c) Time taken to cover \(200 \mathrm{~m}, \mathrm{t}=1 \mathrm{~min} 57.56\) seconds \(=117.56 \mathrm{~s}\)
Power exerted by her,
\(
\begin{aligned}
&\mathrm{P}=460 \mathrm{~W} \\
&P=\frac{W}{t}
\end{aligned}
\)
Work done, \(\mathrm{W}=\mathrm{Pt}=460 \times 117.56 \mathrm{~J}\)
Again, \(W=F S\)
\(
\begin{aligned}
&\Rightarrow F=\frac{W}{S} \\
&=\frac{460 \times 117.56}{200} \\
&=270.3 \mathrm{~N} \approx 270 \mathrm{~N}
\end{aligned}
\)
\(\therefore\) Resistance force offered by the water during the swim is \(270 \mathrm{~N}\).

Hint

\(
\mathrm{s}=100 \mathrm{~m}, \quad \mathrm{t}=10.54 \mathrm{sec}, \mathrm{m}=50 \mathrm{~kg}
\)
The motion can be assumed to be uniform because the time taken for acceleration is minimum.
a) Speed \(v=S / t=9.487 \mathrm{e} / \mathrm{s}\)
So, K.E. \(=1 / 2 \mathrm{mv}^{2}=2250 \mathrm{~J}\)
b) Weight \(=\mathrm{mg}=490 \mathrm{~J}\)
given \(R=m g / 10=49 \mathrm{~J}\)
so, work done against resistance \(W_{F}=-R S=-49 \times 100=-4900 \mathrm{~J}\)
c) To maintain her uniform speed, she has to exert \(4900 \mathrm{J}\) of energy to over come friction
\(
P=\frac{W}{t}=4900 / 10.54=465 \mathrm{~W}
\)

Hint

Height through which water is lifted, \(h=10 \mathrm{~m}\)
Flow rate of water \(=\left(\frac{\mathrm{m}}{\mathrm{t}}\right)\)
\(=30 \mathrm{~kg} / \mathrm{min}=0.5 \mathrm{~kg} / \mathrm{s}\)
Power delivered by the engine,
\(
\begin{aligned}
&P=\frac{\mathrm{mgh}}{\mathrm{t}} \\
&=(0.5) \times 9.8 \times 10 \\
&=49 \mathrm{~W} \\
&1 \mathrm{hp}=746 \mathrm{~W}
\end{aligned}
\)
So, the minimum horse power (hp) that the engine should possess
\(
\begin{aligned}
&=\frac{\mathrm{p}}{746}=\left(\frac{49}{746}\right) \\
&=6.6 \times 10^{-2} \mathrm{hp}
\end{aligned}
\)

Hint

(a) The correct answer is
\(
\begin{aligned}
&\mathrm{m}=200 \mathrm{~g}=0.2 \mathrm{~kg}, \quad \mathrm{~h}=150 \mathrm{~cm}=1.5 \mathrm{~m}, \quad \mathrm{v}=3 \mathrm{~m} / \mathrm{sec} \text {, Total } \\
&\quad \mathrm{t}=1 \mathrm{sec} \\
&\text { work done } \\
&=1 / 2 \mathrm{mv}^{2}+\mathrm{mgh}=(1 / 2) \times(0.2) \times 9+(0.2) \times(9.8) \times(1.5) \\
&=3.84 \mathrm{~J} \\
&\text { h.p. used }=\frac{3.84}{746}=5.14 \times 10^{-3}
\end{aligned}
\)

Hint

(d) Given Mass of the metal,
\(
\mathrm{m}=2000 \mathrm{~kg}
\)
Distance, \(s=12 \mathrm{~m}\)
Time taken, \(\mathrm{t}=1\) minute \(=60 \mathrm{~s}\)
Force applied by the engine to lift the metal,
\(
\mathrm{F}=\mathrm{mg}
\)
So, work done by the engine,
\(
\begin{aligned}
&\mathrm{W}=\mathrm{F} \times \mathrm{s} \times \cos \theta=\mathrm{mgs} \times \cos 0^{\circ}\left[\theta=0^{\circ} \text { for minimum force }\right] \\
&=2000 \times 10 \times 12 \\
&=240000 \mathrm{~J}
\end{aligned}
\)
So, power exerted by the engine,
\(
\begin{aligned}
&P=\frac{W}{t} \\
&=\frac{240000}{60}=4000 \text { watt }
\end{aligned}
\)
Power in hp,
\(
P=\frac{4000}{746}=5.3 \mathrm{hp}
\)

Hint

(a) The specification given by the company are
\(
\begin{aligned}
&\mathrm{u}=0, \mathrm{~m}=95 \mathrm{~kg}, \mathrm{P}_{\mathrm{m}}=3.5 \mathrm{hp} \\
&\mathrm{v}_{\mathrm{m}}=60 \mathrm{~km} / \mathrm{h}=50 / 3 \mathrm{~m} / \mathrm{sec},  \mathrm{} \mathrm{t}_{\mathrm{m}}=5 \mathrm{sec}
\end{aligned}
\)
So, the maximum acceleration that can be produced is given by, \(a=\frac{(50 / 3)-0}{5}=\frac{10}{3}\)
So, the driving forces is given by
\(
\mathrm{F}=\mathrm{ma}=95 \times \frac{10}{3}=\frac{950}{3} \mathrm{~N}
\)
So, the velocity that can be attained by maximum h.p. While supplying \(\frac{950}{3}\)N Force will be \(\mathrm{v}=\frac{\mathrm{p}}{\mathrm{F}} \Rightarrow \mathrm{v}=\frac{3.5 \times 746 \times 3}{950}=8.2 \mathrm{~m} / \mathrm{sec}\).
Because, the scooter can reach a maximum of \(8 .\mathrm{2} \mathrm{m} / \mathrm{sec}\) while producing a force of \(950 / 3 \mathrm{~N}\), the specifications given are some what over claimed.

Hint

(c) Given ,Mass of the block, \(\mathrm{m}=30 \mathrm{~kg}\)
Speed acquired by the block, \(v=40 \mathrm{~cm} / \mathrm{s}\) \(=0.4 \mathrm{~m} / \mathrm{s}\)
Distance covered by the block, \(\mathrm{s}=2 \mathrm{~m}\)
Let \(a\) be the acceleration of the block in the downward direction.
The Force applied by the chain on the block(draw the free body diagram),
\(
\begin{aligned}
&\mathrm{F}=(\mathrm{ma}-\mathrm{mg}) \\
&=\mathrm{m}(\mathrm{a}-\mathrm{g}) \\
&\mathrm{a}=\frac{v^{2}-\mathrm{u}^{2}}{2 \mathrm{~s}} \\
&=\frac{0.16}{4}=0.04 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Work done by the chain,
\(
\begin{aligned}
&\mathrm{W}=\text { FS } \cos \theta \\
&=\mathrm{m}(\mathrm{a}-\mathrm{g}) \times \mathrm{s} \cos 0^{\circ} \\
&=30(0.04-9.8) \times 2 \\
&=-30 \times(9.76) \times 2 \\
&=-585.6=-586 \mathrm{~J} \\
&\Rightarrow \mathrm{W}=-586 \mathrm{~J}
\end{aligned}
\)

Hint

(a) Given, \(T=19 \mathrm{~N}\)
From the free-body diagrams,

\(\begin{aligned}
2 m g-T &=2 m a \dots (i)\\
T-m g &=m a \dots (ii)
\end{aligned}\)

From equations (i) and (ii),
\(
\begin{aligned}
&\mathrm{T}=4 \mathrm{ma} \\
&\Rightarrow a=\frac{T}{4 \mathrm{~m}}=\frac{4}{\mathrm{~m}} \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)

Now, displacement \(\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\)
Net mass \(=2 \mathrm{~m}-\mathrm{m}=\mathrm{m}\)
Decrease in potential energy,
P. E. \(=\mathrm{mgh}\)
\(
\begin{aligned}
&=\mathrm{m} \times \mathrm{g} \times\left(\frac{2}{\mathrm{~m}}\right) \\
&=9.8 \times 2=19.6 \mathrm{~J}
\end{aligned}
\)

Hint

(b) Given,
\(
\mathrm{m}_{1}=3 \mathrm{~kg}, \mathrm{~m}_{2}=2 \mathrm{~kg}, \mathrm{t}=\text { during } 4^{\text {th }} \text { second }
\)
From the free-body diagram,
\(
\begin{aligned}
&T-3 g+3 a=0 \ldots \text { (i) } \\
&T-2 g-2 a=0 \ldots \text { (ii) }
\end{aligned}
\)
Equating (i) and (ii), we get:
\(
\begin{aligned}
&3 g-3 a=2 g+2 a \\
&\Rightarrow a=\frac{g}{5} m / s^{2}
\end{aligned}
\)
Distance travelled in the \(4^{\text {th }}\) second,
\(
\begin{aligned}
&\mathrm{s}\left(4^{\text {th }}\right)=\frac{a}{2}(2 \mathrm{n}-1) \\
&=\frac{\left(\frac{g}{5}\right)}{2}(2 \times 4-1) \\
&=\frac{7 \mathrm{~g}}{10}=\frac{7 \times 9.8}{10} \mathrm{~m}
\end{aligned}
\)
Net mass m \(=m_{1}-m_{2}=3-2=1 \mathrm{~kg}\)
So, decrease in potential energy,
P.E. = mgh
\(P . E .=1 \times 9.8 \times\left(\frac{7}{10}\right) \times 9.8=67.2 \mathrm{~J}=67 \mathrm{~J}\)
So, work done by gravity during the fourth second \(=\) P.E. \(=67 \mathrm{~J}\)

Hint

(d) From the problem it is given given,
\(
\begin{aligned}
&\mathrm{m}_{1}=4 \mathrm{~kg}, \mathrm{~m}_{2}=1 \mathrm{~kg} \\
&\mathrm{v}_{2}=0.3 \mathrm{~m} / \mathrm{s} \\
&\mathrm{v}_{1}=2 \times 0.3=0.6 \mathrm{~m} / \mathrm{s} \\
&\left(\mathrm{v}_{1}=2 \mathrm{v}_{2} \text { in this system }\right)
\end{aligned}
\)
Height descended by the \(1 \mathrm{~kg}\) block, \(\mathrm{h}=1 \mathrm{~m}\)
Distance travelled by the \(4 \mathrm{~kg}\) block,
\(
\mathrm{s}=2 \times 1=2 \mathrm{~m}
\)
Initially the system is at rest. So, \(u=0\)
Applying work energy theoremwhich says that
change in \(\mathrm{K} . \mathrm{E} .=\) Work done (for the system )
\(
\left(\frac{1}{2}\right) \mathrm{m}_{1} v_{1}^{2}+\left(\frac{1}{2}\right) \mathrm{m}_{2} v_{2}^{2}=(-\mu \mathrm{R}) \mathrm{s}+\mathrm{m}_{2} \mathrm{gh}
\)
\(
\begin{aligned}
&\frac{1}{2} \times 4 \times(0.36)+\frac{1}{2} \times 1 \times(0.09)[\mathrm{As}, \mathrm{R}=4 \mathrm{~g}=40 \mathrm{~N}] \\
&=-\mu 40 \times 2+1 \times 40 \times 1 \\
&\Rightarrow 0.72+0.045=-80 \mu+10 \\
&\Rightarrow \mu=\frac{9.235}{80}=0.12
\end{aligned}
\)
So, the coefficient of kinetic friction between the block and the table is \(0.12\).

Hint

(a) Given ,Mass of the block, \(\mathrm{m}=100 \mathrm{~g}=0.1 \mathrm{~kg}\),
Velocity of the block at the highest point, \(v=5 \mathrm{~m} / \mathrm{s}\)
Radius of the circular tube, \(r=10 \mathrm{~cm}\)
Work done by the block
\(=\) Total energy at the highest point A- Total energy at the lowest point B
\(
\begin{aligned}
&=\left(\frac{1}{2} \mathrm{~m} v^{2}+\mathrm{mgh}-0\right) \\
&\Rightarrow \mathrm{W}=\frac{1}{2} \times(0.1) \times 25+(0.1) \times 10 \times(0.2) \\
&\text { As, } h=2 r=0.2 \mathrm{~m} \\
&\mathrm{~W}=1.25+0.2=1.45 \mathrm{~J}
\end{aligned}
\)
So, the work done by the tube on the body is \(-1.45\) joule.

Hint

(c) Given,Mass of the car, \(\mathrm{m}=1400 \mathrm{~kg}\)
\(\mathrm{h}=10 \mathrm{~m}\)
Since the car is moving when the motor stops, it has kinetic energy . Thus
\(
\begin{aligned}
&\mathrm{v}_{i}=54 \mathrm{~km} / \mathrm{h} \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s} \\
&\mathrm{v}_{f}=0 \\
&\triangle \mathrm{K}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2} \\
&\triangle \mathrm{K}=\frac{1}{2} \times 1400\left(0^{2}-15^{2}\right) \\
&\triangle \mathrm{K}=-157500 \mathrm{~J}
\end{aligned}
\)
Let the gravitational potential energy be zero at the starting point .
Then the potential energy at the terminal is
\(
\begin{aligned}
&U_{\mathrm{i}}=0 \\
&U_{f}=m g h \\
&U_{f}=1400 \times 9.8 \times 10=137200 \mathrm{~J} \\
&\triangle U=U_{f}-U_{i} \\
&\triangle U=137200-0=137200 \mathrm{~J}
\end{aligned}
\)
Let \(W\) be the work done against friction during ascent.
Then ” – W ” is the work done by the frictional force.
\(
\begin{aligned}
&-\mathrm{W}=\Delta \mathrm{K}+\Delta \mathrm{U} \\
&-\mathrm{W}=-157500+137200 \\
&-\mathrm{W}=-20300 \mathrm{~J} \\
&\mathrm{~W}=20300 \mathrm{~J}
\end{aligned}
\)
So, work done against friction is 20,300 joule.

Hint

(b) Given,
Mass of the block, \(\mathrm{m}=200 \mathrm{~g}=0.2 \mathrm{~kg}\)
Length of the incline, \(s=10 \mathrm{~m}\),
Height of the incline, \(h=3.2 \mathrm{~m}\)
Acceleration due to gravity, \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\)
(a) Work done, \(\mathrm{W}=\mathrm{mgh}=0.2 \times 10 \times 3.2=6.4 \mathrm{~J}\)
(b) Work done to slide the block up the incline
\(
\begin{aligned}
&\mathrm{W}=(\mathrm{mg} \sin \theta) \times \mathrm{s} \\
&=(0.2) \times 10 \times(3.2 / 10) \times 10 \\
&=6.4 \mathrm{~J}
\end{aligned}
\)
(c) Let final velocity be \(v\) when the block falls to the ground vertically.
Change in the kinetic energy = Work done
\(
\frac{1}{2} \mathrm{mv}^{2}-0=6.4 \mathrm{~J}
\)
\(
\Rightarrow v=8 \mathrm{~m} / \mathrm{s}
\)
(d) Let \(v\) be the final velocity of the block when it reaches the ground by sliding.
\(
\begin{aligned}
&\frac{1}{2} m v^{2}-0=6.4 \mathrm{~J} \\
&\Rightarrow v=8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&l=10 \mathrm{~m}, \mathrm{~h}=8 \mathrm{~m}, \quad \mathrm{mg}=200 \mathrm{~N} \\
&f=200 \times \frac{3}{10}=60 \mathrm{~N}
\end{aligned}
\)
a) Work done by the ladder on the boy is zero when the boy is going up because the work is done by the boy himself.
b) Work done against frictional force, \(\mathrm{W}=\mu \mathrm{RS}=f \ell=(-60) \times 10=-600 \mathrm{~J}\)
c) Work done by the forces inside the boy is
\(
W_{b}=(m g \sin \theta) \times 10=200 \times \frac{8}{10} \times 10=1600 \mathrm{~J}
\)

Hint

(c) Given,
Height of the starting point of the track, \(\mathrm{H}=1 \mathrm{~m}\)
Height of the ending point of the track, \(h=0.5 \mathrm{~m}\)
Let \(v\) be the velocity of the particle at the end point on the track.
Applying the law of conservation of energy at the starting and ending point of the track, we get
\(
\begin{aligned}
&\mathrm{mgH}=\frac{1}{2} \mathrm{~m} v^{2}+\mathrm{mgh} \\
&\Rightarrow g-\left(\frac{1}{2}\right) v^{2}=0.5 \mathrm{~g} \\
&\Rightarrow v^{2}=2(\mathrm{~g}-0.5 \mathrm{~g})=\mathrm{g} \\
&\Rightarrow v=\sqrt{\mathrm{g}}=3.1 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
After leaving the track, the body exhibits projectile motion for which,
\(
\begin{aligned}
&\theta=0 \\
&y=-0.5
\end{aligned}
\)
Using equation of motion along the horozontal direction,
\(
\begin{aligned}
&-0.5=(\mathrm{u} \sin \theta) t-\left(\frac{1}{2}\right) \mathrm{gt}^{2} \\
&\Rightarrow 0.5=4.9 \times \mathrm{t}^{2} \\
&\Rightarrow \mathrm{t}=0.31 \mathrm{sec} \\
&\text { So, } \mathrm{x}=(v \cos \theta) \mathrm{t} \\
&=3.1 \times 0.31=1 \mathrm{~m}
\end{aligned}
\)
So, the particle will hit the ground at a horizontal distance of \(1 \mathrm{~m}\) from the other end of the track.

Hint

(d)

\(
\mathrm{mg}=10 \mathrm{~N}, \quad \mu=0.2, \quad \mathrm{H}=1 \mathrm{~m}, \mathrm{u}=\mathrm{v}=0
\)
 work done = change in P.E.
Increase in K.E.
\(
\Rightarrow \mathrm{w}=\mathrm{mgh}=10 \times 1=10 \mathrm{~J}
\)
Again, on the horizontal surface the fictional force
\(
\mathrm{F}=\mu \mathrm{R}=\mu \mathrm{mg}=0.2 \times 10=2 \mathrm{~N}
\)
So, the K.E. is used to overcome friction
\(
\Rightarrow S=\frac{W}{F}=\frac{10 \mathrm{~J}}{2 \mathrm{~N}}=5 \mathrm{~m}
\)

Hint

(d) Let ‘ \(d x^{\text {‘ }}\) be the length of an element at a distance \(x\) from the table
mass of the element of \(d x\) length \(=(\mathrm{m} / l) d x\)
Work done to put \(d x\) part back on the table
\(\mathrm{dW}=(\mathrm{m} / \mathrm{l}) g  \mathrm{x}  \mathrm{dx}\)
So, total work done to put \(l / 3\) part back on the table
\(
\mathrm{W}=\int_{0}^{l / 3}(\mathrm{~m} / \ell) \mathrm{gx} \mathrm{dx} \Rightarrow \mathrm{W}=(\mathrm{m} / \ell) g\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{\frac{\ell}{3}}=\frac{\mathrm{mg} \ell^{2}}{18 \ell}=\frac{\mathrm{mg} \ell}{18}
\)

Hint

(a) Let \(x\) length of the chain be on the table at a particular instant.
Consider a small element of length ‘ \(d x\) ‘ and mass ‘ \(d m\) ‘ on the table. \(\mathrm{dm}=\frac{M}{L} \mathrm{dx}\) Work done by the friction on this element is \(\mathrm{dW}=\mu \mathrm{Rx}=\mu\left(\frac{M}{L} \times \mathrm{gx}\right) \mathrm{dx}\)
Total work done by friction on two third part of the chain,
\(W=\int_{2 L / 3}^{0} \mu \frac{M}{L} \mathrm{gx} \mathrm{dx}\)
\(
\therefore W=\mu \frac{M}{L} \mathrm{~g}\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{2 \mathrm{~L} / 3}
\)
\(
\begin{aligned}
&=-\mu \frac{M}{L} \mathrm{~g}\left[\frac{4 L^{2}}{18}\right] \\
&=-\frac{2 \mu \mathrm{MgL}}{9}
\end{aligned}
\)
The total work done by friction during the period the chain slips off the table is \(-\frac{2 \mu \mathrm{MgL}}{9}\)

Hint

(c) Given,
Mass of the block, \(m=1 \mathrm{~kg}\)
Height of point \(\mathrm{A}, \mathrm{H}=1 \mathrm{~m}\)
Height of point \(B, h=0.8 \mathrm{~m}\)
Work done by friction \(=\) Change in potential energy of the body
\(
\begin{aligned}
&\Rightarrow \mathrm{W}_{1}=\mathrm{mgh}-\mathrm{mgH} \\
&=1 \times 10(0.8-1) \\
&=-1 \times 10 \times(0.2)=-2 \mathrm{~J}
\end{aligned}
\)
The work done by the frictional force on the block during its transit from \(A\) to \(B\) is – 2 joule.

Hint

(d)

\(
\begin{aligned}
&\mathrm{m}=5 \mathrm{~kg}, \quad \mathrm{x}=10 \mathrm{~cm}=0.1 \mathrm{~m}, \quad v=2 \mathrm{~m} / \mathrm{sec}, \\
&\mathrm{h}=? \quad \mathrm{g}=10 \mathrm{~m} / \mathrm{sec}^{2} \\
&\mathrm{~s} o, \mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}=\frac{50}{0.1}=500 \mathrm{~N} / \mathrm{m}
\end{aligned}
\)
Total energy just after the blow \(E=1 / 2 m v^{2}+1 / 2 k x^{2} \dots(i)\)
Total energy at height \(h=1 / 2 k(h-x)^{2}+m g h \dots(ii)\)
\(1 / 2 m v^{2}+1 / 2 k x^{2}=1 / 2 k(h-x)^{2}+m g h\)
On, solving we can get,
\(
\mathrm{h}=0.2 \mathrm{~m}=20 \mathrm{~cm}
\)

Hint

(a) Given,Mass of the block, \(\mathrm{m}=250 \mathrm{~g}=0.25 \mathrm{~kg}\),
Spring constant, \(k=100 \mathrm{~N} / \mathrm{m}\)
Compression in the string, \(x=10 \mathrm{~cm}=0.1 \mathrm{~m}\),
Acceleration due to gravity, \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\)
Let the block rises to height \(h\).
Applying law of conservation of energy which says that the total energy should always remain conserved.
\(
\begin{aligned}
&\frac{1}{2} \mathrm{kx}^{2}=\mathrm{mgh} \\
&\Rightarrow h=\frac{1}{2}\left(\frac{\mathrm{kx}^{2}}{\mathrm{mg}}\right) \\
&=\frac{100 \times 0.01}{2 \times(0.250) \times 10} \\
&=0.2 \mathrm{~m}=20 \mathrm{~cm}
\end{aligned}
\)
So, the block rises to \(20 \mathrm{~cm}\).

Hint

(b) Given:Mass of the block, \(\mathrm{m}=2 \mathrm{~kg}\)
Initial distance of the block from the spring, \(\mathrm{S}_{1}=4.8 \mathrm{~m}\),
Comression in the spring, \(x=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Final distance of the block from the spring, \(\mathrm{S}_{2}=1 \mathrm{~m}\)
As \(\theta=37^{\circ}\),
\(\sin 37^{\circ}=0.60=\frac{3}{5}\), \(\cos 37^{\circ}=0.80=\frac{4}{5}\)
Applying the work-energy principle for downward motion of the block,
\(
\begin{aligned}
&0-0=\mathrm{mg} \sin 37^{\circ}(\mathrm{x}+4.8)-\mu \mathrm{R} \times 5-\frac{1}{2} \mathrm{kx}^{2} \\
&\Rightarrow 20 \times(0.06) \times 5-\mu \times 20 \times(0.80) \times 5-\frac{1}{2} \mathrm{k}(0.2)^{2}=0 \\
&\Rightarrow 60-80 \mu-0.02 \mathrm{k}=0 \\
&\Rightarrow 80 \mu+0.02 \mathrm{k}=60 \ldots \text { (i) }
\end{aligned}
\)
Similarly for the upward motion of the body the equation is
\(
\begin{aligned}
&0-0=\left(-\mathrm{mg} \sin 37^{\circ}\right)-\mu \mathrm{R} \times 1+\frac{1}{2} \mathrm{k}(-2)^{2} \\
&\Rightarrow 20 \times(0.06) \times 1-\mu \times 20 \times(0.80) \times 1-\frac{1}{2} \mathrm{k}(0.2)^{2} \\
&\Rightarrow 12-16 \mu+0.02 \mathrm{k}=0 \dots (ii)
\end{aligned}
\)
Adding equations (i) and (ii), we get:
\(
\begin{aligned}
&96 \mu=48 \\
&\Rightarrow \mu=0.5
\end{aligned}
\)
Now putting the value of \(\mu\) in equation (i), we get:
\(
k=1000 \mathrm{~N} / \mathrm{m}
\)

Hint

(b) Let the velocity of the body at \(A\) be \(v\).
So, the velocity of the body at \(B\) is \(\frac{v}{2}\). Energy at point \(A=\) Energy at point \(B\)
\(
\begin{aligned}
&\text { So, } \frac{1}{2} \mathrm{~m} v_{A}^{2}-\left(\frac{1}{2}\right) \mathrm{mv}_{B}^{2}=\frac{1}{2} \mathrm{kx}^{2} \\
&\Rightarrow \frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \mathrm{~m}\left(v_{A}^{2}-v_{B}^{2}\right) \\
&\Rightarrow \mathrm{kx}^{2}=\mathrm{m}\left(v^{2}-\frac{v^{2}}{4}\right) \\
&\Rightarrow \mathrm{kx}^{2}=\mathrm{m} \frac{\left(4 v^{2}-v^{2}\right)}{4} \\
&\Rightarrow \mathrm{k}=\frac{3 \mathrm{m{v^{2}}}}{4 \mathrm{x}^{2}}
\end{aligned}
\)

Hint

(a) Mass of the body \(=m\)
Let the elongation be \(x\)
So, \(1 / 2 k x^{2}=m g x\)
\(
\Rightarrow x=2 \mathrm{mg} / \mathrm{k}
\)

Hint

(a) The body is displaced \(x\) towards right
Let the velocity of the body be \(v\) at its mean position Applying law of conservation of energy
\(
\begin{aligned}
1 / 2 m v^{2}=1 / 2 k_{1} x^{2}+1 / 2 k_{2} x^{2} \\ \Rightarrow m v^{2}=x^{2}\left(k_{1}+k_{2}\right) \\ \Rightarrow v^{2}=\frac{x^{2}\left(k_{1}+k_{2}\right)}{m} \\
\Rightarrow v=x \sqrt{\frac{k_{1}+k_{2}}{m}}
\end{aligned}
\)

Hint

(b) Let the compression be \(x\)
According to the law of conservation of energy
\(1 / 2 m v^{2}=1 / 2 k x^{2} \\ \Rightarrow x^{2}=m v^{2} / k  \\ \Rightarrow x=v \sqrt{(m / k)}\)

b) No. It will be in the opposite direction and the magnitude will be less due to loss in spring.

Hint

(c) Given Mass of the block, \(\mathrm{m}=100 \mathrm{~g}=0.1 \mathrm{~kg}\)
Compression in the spring, \(x=5 \mathrm{~cm}=0.05 \mathrm{~m}\)
Spring constant, \(k=100 \mathrm{~N} / \mathrm{m}\)
Let \(v\) be the velocity of the block when it leaves the spring.
Applying the law of conservation of energy,
Elastic potential energy of the spring = Kinetic energy of the block
\(\frac{1}{2} \mathrm{~m} \nu^{2}=\frac{1}{2} \mathrm{kx}{ }^{2}\)
\(\Rightarrow v=x \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\)
\(=(0.05) \times \sqrt{\frac{(100)}{0.1}}\)
\(=1.58 \mathrm{~m} / \mathrm{s}\)
For the projectile motion,
\(
\begin{aligned}
&\theta=0^{\circ}, \mathrm{y}=-2 \\
&\text { Now, } \mathrm{y}=(\mathrm{u} \cdot \sin \theta) \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2} \\
&-2=\left(-\frac{1}{2}\right) \times(9.8) \times \mathrm{t}^{2} \\
&\Rightarrow \mathrm{t}=0.63 \mathrm{sec}, \\
&\text { So, } \mathrm{x}=(\mathrm{u} cos \theta) \mathrm{t} \\
&=(1.58) \times(0.36)=1 \mathrm{~m}
\end{aligned}
\)
Therefore, the block hits the ground at \(1 \mathrm{~m}\) from the free end of the spring in the horizontal direction.

Hint

(d) Let the velocity of the body at \(A\) is \(v\), for minimum velocity given at \(A\) velocity of the body at point B is zero.
Applying law of conservation of energy at \(A\) & \(B\)
\(
1 / 2 \mathrm{mv}^{2}=\mathrm{mg}(2 \ell) \Rightarrow \mathrm{v}=\sqrt{(4 \mathrm{~g} \ell)}=2 \sqrt{\mathrm{g} \ell}
\)

Hint

(b)

\(
\begin{aligned}
&\mathrm{m}=320 \mathrm{~g}=0.32 \mathrm{~kg} \\
&\mathrm{k}=40 \mathrm{~N} / \mathrm{m} \\
&\mathrm{h}=40 \mathrm{~cm}=0.4 \mathrm{~m} \\
&\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
From the free body diagram,
\(\mathrm{kx} \cos \theta=\mathrm{mg}\)
(when the block breaks off the surface below it then \(R=0\) )
\(\Rightarrow \cos \theta=\mathrm{mg} / \mathrm{kx}\)
So, \(\frac{0.4}{0.4+x}=\frac{3.2}{40 \times x} \Rightarrow 16 x=3.2 x+1.28 \Rightarrow x=0.1 \mathrm{~m}\)
\(
\text { So, } \mathrm{s}=\mathrm{AB}=\sqrt{(\mathrm{h}+\mathrm{x})^{2}-\mathrm{h}^{2}}=\sqrt{(0.5)^{2}-(0.4)^{2}}=0.3 \mathrm{~m}
\)
Let the velocity of the body at \(B\) be \(v\)
Change in K.E. \(=\) work done (for the system)
\(
\begin{aligned}
\left(1 / 2 \mathrm{mv}^{2}+1 / 2 \mathrm{mv}^{2}\right)=-1 / 2 \mathrm{kx}^{2}+\mathrm{mgs} \\
\Rightarrow(0.32) \times \mathrm{v}^{2}=-(1 / 2) \times 40 \times(0.1)^{2}+0.32 \times 10 \times(0.3) \\  \mathrm{v}=1.5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a)

Given \(\theta=37^{\circ}, \mathrm{l}=\) natural length = \(\mathrm{h}\),  Let the velocity be \(v\).
\(
\begin{aligned}
&\cos 37^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}=0.8=\frac{4}{5} \\
&{AC}=(\mathrm{h}+\mathrm{x})=\frac{5 \mathrm{~h}}{4}
\end{aligned}
\)
Applying the law of conservation of energy,
\(
\begin{aligned}
&\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{2} \mathrm{~m} v^{2} \\
&\Rightarrow v=\mathrm{x} \sqrt{\left(\frac{\mathrm{k}}{\mathrm{m}}\right)}=\frac{\mathrm{h}}{4} \sqrt{\left(\frac{\mathrm{k}}{\mathrm{m}}\right)}
\end{aligned}
\)

Hint

(d) The minimum velocity required to cross the height point \(c=\) \(\sqrt{2 \mathrm{gl}}\)
Let the rod be released from a height \(h\).
Total energy at \(A=\) total energy at \(B\)
\(\mathrm{mgh}=1 / 2 \mathrm{mv}^{2} ; \mathrm{mgh}=1 / 2 \mathrm{~m}(2 \mathrm{~g}l)\)
[Because \(v=\) required velocity at \(B\) such that the block makes a complete circle.]
So, \(h=l\).

Hint

(a) Let the velocity at \(B\) be \(v_{2}\).
\(
\begin{aligned}
&\frac{1}{2} \mathrm{~m} v_{1}^{2}=\frac{1}{2} \mathrm{~m} v_{2}^{2}+\mathrm{mg} \mathrm{l} \\
&\Rightarrow \frac{1}{2} \mathrm{~m}(10 \mathrm{~gl})=\frac{1}{2} \mathrm{~m} v_{2}^{2}+\mathrm{mgl} \\
&v_{2}^{2}=8 \mathrm{gl}
\end{aligned}
\)
So, the tension in the string at the horizontal position,
\(
\begin{aligned}
&\mathrm{T}=\frac{\mathrm{m} v^{2}}{\mathrm{R}}=\mathrm{m} \frac{8 \mathrm{gl}}{\mathrm{l}} \\
&=8 \mathrm{mg}
\end{aligned}
\)
b) Let the velocity at \(C\) be \(v_{3}\)
\(
\begin{aligned}
& 1 / 2 \mathrm{mv}_{1}^{2}=1 / 2 \mathrm{mv}_{3}^{2}+\mathrm{mg}(2 \mathrm{l}) \\
\Rightarrow & 1 / 2 \mathrm{~m}(10g l)=1 / 2 \mathrm{mv}_{3}^{2}+2 \mathrm{mgl}\\
\Rightarrow & \mathrm{v}_{3}^{2}=6 \mathrm{mgl}
\end{aligned}
\)
So, the tension in the string is given by
\(
\mathrm{T}_{\mathrm{c}}=\frac{\mathrm{mv}^{2}}{\mathrm{l}}-\mathrm{mg}=\frac{6 \mathrm{glm}}{\mathrm{I}} \mathrm{mg}=5 \mathrm{mg}
\)
c) Let the velocity at point \(D\) be \(v_{4}\)
\(
\begin{aligned}
& \text { Again, } 1 / 2 \mathrm{mv}_{1}^{2}=1 / 2 \mathrm{mv}_{4}{ }^{2}+\mathrm{mgh} \\
& 1 / 2 \times \mathrm{m} \times(10 \mathrm{gl})=1.2 \mathrm{mv}_{4}^{2}+\mathrm{mgl}\left(1+\cos 60^{\circ}\right) \\
\Rightarrow & \mathrm{v}_{4}{ }^{2}=7 \mathrm{gl}
\end{aligned}
\)
So, the tension in the string is
\(
\begin{aligned}
&\mathrm{T}_{\mathrm{D}}=\left(\mathrm{mv}^{2} / \mathrm{l}\right)-\mathrm{mg} \cos 60^{\circ} \\
&=\mathrm{m}(7 \mathrm{gl}) / \mathrm{l}-\mathrm{l}-0.5 \mathrm{mg} \Rightarrow 7 \mathrm{mg}-0.5 \mathrm{mg}=6.5 \mathrm{mg} .
\end{aligned}
\)

Hint

(c) From the figure, \(\cos \theta=A C / A B\) \(\Rightarrow A C=A B \cos \theta \Rightarrow(0.5) \times(0.8)=0.4\).
So, \(C D=(0.5)-(0.4)=(0.1) \mathrm{m}\)
Energy at \(D=\) energy at \(B\)
\(1 / 2 \mathrm{mv}^{2}=\mathrm{mg}(C D)\)
\(v^{2}=2 \times 10 \times(0.1)=2\)
So, the tension is given by,
\(\mathrm{T}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}+\mathrm{mg}=(0.1)\left(\frac{2}{0.5}+10\right)=1.4 \mathrm{~N}\).

Hint

(a) Given, \(N=m g\)
As shown in the figure, \(\mathrm{mv}^{2} / \mathrm{R}=\mathrm{mg}\) \(\Rightarrow v^{2}=g R\)
Total energy at point \(A=\) energy at \(P\)
\(
\begin{aligned}
&1 / 2 \mathrm{kx}^{2}=\frac{\mathrm{mgR}+2 \mathrm{mgR}}{2} \quad \text { [because } v^{2}=g R \text { ] } \\
&\Rightarrow x^{2}=3 \mathrm{mgR} / \mathrm{k} \Rightarrow \mathrm{x}=\sqrt{(3 \mathrm{mgR}) / \mathrm{k}}
\end{aligned}
\)

Hint

(b) Suppose the string becomes slack at point \(P\).
Let the bob rise to a height \(h\).
\(
h=l+l \cos \theta
\)
From the work-energy theorem,
\(
\begin{aligned}
&\frac{1}{2} \mathrm{~m} v^{2}-\frac{1}{2} \mathrm{~m} u^{2}=-\mathrm{mgh} \\
&v^{2}=\mathrm{u}^{2}-2 \mathrm{~g}(\mathrm{l}+\mathrm{l} \cos \theta) \ldots(\mathrm{i})
\end{aligned}
\)
Again, \(\frac{\mathrm{m} v^{2}}{l}=\mathrm{mg} \cos \theta\)
\(
v^{2}=\lg \cos \theta \text {. }
\)
Using equation (i) and (ii) and the value of \(u\), we get,
\(
\begin{aligned}
&g l \cos \theta=3 g l-2 g l-2 g l \cos \theta \\
&3 \cos \theta=1 \\
&\theta=\cos ^{-1}\left(\frac{1}{3}\right)
\end{aligned}
\)

So, angle rotated before the string becomes slack
\(
=180^{\circ}-\cos ^{-1}(1 / 3)=\cos ^{-1}(-1 / 3)
\)

Hint

Given Length of the string, \(L=1.5 \mathrm{~m}\)
Initial speed of the particle, \(\mathrm{u}=\sqrt{57} \mathrm{~m} / \mathrm{s}\)
\text { a) } \begin{aligned}
& m g \cos \theta=m v^{2} / l \\
& v^{2}=\lg \cos \theta \dots (1)\\
& \text { change in K.E. = work done } \\
& 1 / 2 \mathrm{mv}^{2}-1 / 2 \mathrm{mu}^{2}=\mathrm{-mgh} \\
\Rightarrow & \mathrm{v}^{2}-57=-2 \times 1.5 \mathrm{~g}(1+\cos \theta) \ldots(2) \\
\Rightarrow & \mathrm{v}^{2}=57-3 \mathrm{~g}(1+\cos \theta) \\
& \text { Putting the value of } \mathrm{v} \text { from equation (1) } \\
& 15 \cos \theta=57-3 \mathrm{~g}(1+\cos \theta) \Rightarrow 15 \cos \theta=57-30-30 \cos \theta \\
\Rightarrow & 45 \cos \theta=27 \Rightarrow \cos \theta=3 / 5 . \\
\Rightarrow & \theta=\cos ^{-1}(3 / 5)=53^{\circ} \\
\text { b) } & v=\sqrt{57-3 \mathrm{~g}(1+\cos \theta)} \text { from equation (2) } \\
=& \sqrt{9}=3 \mathrm{~m} / \mathrm{sec} .
\end{aligned}[/latex]

Hint

(b)Let the velocity be v when the body leaves the surface.
From the free body diagram,
\(\frac{m v^{2}}{R}=m g \cos \theta\) [normal reaction]
\(\mathrm{v}^{2}=\mathrm{Rg} \cos \theta \quad\)..(1)
Again, from the work-energy principle,
Change in K.E. = work done
\(
\begin{aligned}
&\Rightarrow 1 / 2 m v^{2}-0=m g(R-R \cos \theta) \\
&\Rightarrow v^{2}=2 g R(1-\cos \theta) \quad . .(2)
\end{aligned}
\)
From (1) and (2)
\(R g \cos \theta=2 g R(1-\cos \theta)\)
\(\begin{aligned}
&3 g R \cos \theta=2 g R \\
&\cos \theta=2 / 3 \\
&\theta=\cos ^{-1}(2 / 3)
\end{aligned}\)

Hint

(a) When the particle is released from rest, the centrifugal force is zero.
\(\mathrm{N}\) force \(=\mathrm{mg} \cos \theta=\mathrm{mg} \cos 30^{\circ}\)
\(
=\frac{\sqrt{3}}{2} \mathrm{mg}
\)
b) When the particle leaves contact with the surface (fig-2), \(N=0\).
So, \(\frac{m v^{2}}{R} = m g \cos \theta\)
\(
\Rightarrow v^{2}=R g \cos \theta \quad \text {..(1) }
\)
Again, \(1 / 2 m v^{2}=m g R\left(\cos 30^{\circ}-\cos \theta\right)\)
\(
\Rightarrow v^{2}=2 \operatorname{Rg}\left(\frac{\sqrt{3}}{2}-\cos \theta\right) \quad . .(2)
\)
From equation (1) and equation (2)
\(R g \cos \theta=\sqrt{3} R g-2 R g \cos \theta\)
\(
\begin{aligned}
&\Rightarrow 3 \cos \theta=\sqrt{3} \\
&\Rightarrow \cos \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{aligned}
\)
So, the distance travelled by the particle before leaving contact, \(l=R(\theta-\pi / 6)\left[\right.\) because \(30^{\circ}=\pi / 6\) ]
putting the value of \(\theta\), we get \(l=0.43 R\)

Hint

a) Radius \(=R\)
Horizontal speed \(=v\)
From the above diagram Fig-1:
Normal force,
\(
\mathrm{N}=\mathrm{mg}-\frac{\mathrm{m} v^{2}}{\mathrm{R}}
\)
(b) When the particle is given maximum velocity so that the centrifugal force balances the weight, the particle does not slip on the sphere.
\(
\begin{aligned}
&\text { So }, \frac{\mathrm{m} v^{2}}{R}=\mathrm{mg} \\
&\Rightarrow v=\sqrt{\mathrm{gR}}
\end{aligned}\)
(c) If the body is given velocity \(v_{1}\) at the top such that,
\(
\begin{aligned}
&v_{1}=\frac{\sqrt{g R}}{2} \\
&v_{1}^{2}=\frac{g R}{4}
\end{aligned}
\)
Let the velocity be \(v_{2}\) when it loses contact with the surface, as shown below in Fig-2.
So,
\(
\begin{aligned}
&\frac{\mathrm{m} v_{2}^{2}}{R}=\mathrm{mg} \cos \theta \\
&\Rightarrow v_{2}^{2}=\mathrm{Rg} \cos \theta \ldots(\mathrm{i}) \\
&\text { Again, }\left(\frac{1}{2}\right) \mathrm{m} v_{2}^{2}-\left(\frac{1}{2}\right) m v_{1}^{2} \\
&=\mathrm{mgR}(1-\cos \theta) \\
&\Rightarrow v_{2}^{2}=v_{1}^{2}+2 \mathrm{gR}(1-\cos \theta) \ldots(i i)
\end{aligned}
\)
From equations (i) and (ii),
\(
\begin{aligned}
&R g \cos \theta=\left(\frac{R g}{4}\right)+2 g R(1-\cos \theta) \\
&\Rightarrow \cos \theta=\left(\frac{1}{4}+2-2 \cos \theta\right) \\
&\Rightarrow 3 \cos \theta=\left(\frac{9}{4}\right) \\
&\Rightarrow \theta=\cos ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
\)

Hint

a) Net force on the particle between \(A \& B, F=m g \sin \theta\)
work done to reach \(B, W=F S=m g \sin \theta l\)
Again, work done to reach \(B\) to \(C=m g h=m g R(1-\cos \theta)\)
So, Total workdone \(=\mathrm{mg}[l \sin \theta+\mathrm{R}(1-\cos \theta)]\)
Now, change in K.E. = work done
\(
\begin{aligned}
&\Rightarrow 1 / 2 \mathrm{mv}_{\mathrm{o}}^{2}=\mathrm{mg}[\ell \sin \theta+\mathrm{R}(1-\cos \theta) \\
&\Rightarrow \mathrm{v}_{\mathrm{o}}=\sqrt{2 g(R(1-\cos \theta)+\ell \sin \theta)}
\end{aligned}
\)
b) When the block is projected at a speed \(2 \mathrm{v}_{0}\).
Let the velocity at \(C\) will be \(V_{c}\).
Applying energy principle,
\(
\begin{aligned}
&1 / 2 m v_{c}^{2}-1 / 2 m\left(2 v_{o}\right)^{2}=-m g[l \sin \theta+R(1-\cos \theta)] \\
&\Rightarrow v_{c}^{2}=4 v_{o}-2 g[e \sin \theta+R(1-\cos \theta)]
\end{aligned}
\)
\(4.2 g[l \sin \theta+R(1-\cos \theta)]-2 g[l \sin \theta+R(1-\cos \theta)\)
\(=6 g[\ell \sin \theta+R(1-\cos \theta)]\)
So, force acting on the body,
\(
\Rightarrow N=\frac{{m v_{c}}^{2}}{R}=6 m g[(l / R) \sin \theta+1-\cos \theta]
\)
c) Let the block loose contact after making an angle \(\theta\), \(\frac{m v^{2}}{R}=m g \cos \theta \Rightarrow v^{2}=R g \cos \theta\)..(1)
Again, \(1 / 2 \mathrm{mv}^{2}=m g(R-R \cos \theta) \Rightarrow \mathrm{v}^{2}=2 g R(1-\cos \theta) \dots(2)\)
From (1) and (2) \(\cos \theta=2 / 3 \Rightarrow \theta=\cos ^{-1}(2 / 3)\)

Hint

(d) Suppose the sphere moves to the left with acceleration a
Let \(m\) be the mass of the particle.
The particle \(m\) will also experience inertia due to acceleration  \(a\) as it is in the sphere. It will also experience the tangential inertia force
\(\left[\mathrm{m}\left(\frac{d v}{d t}\right)\right]\) and centrifugal force \(\left(\frac{m v^{2}}{R}\right)\)
From the diagram,
\(
\begin{aligned}
&m \frac{d v}{d t}=\mathrm{ma} \cos \theta+\mathrm{mg} \sin \theta \\
&\Rightarrow \mathrm{m} v \frac{d v}{d t}=\mathrm{ma} \cdot \cos \theta\left(R \frac{d \theta}{d t}\right)+\mathrm{mg} \sin \theta\left(R \frac{d \theta}{d t}\right)\left(\text { because, } v=R \frac{d \theta}{d t}\right) \\
&\Rightarrow v \mathrm{d} v=a \mathrm{R} \cos \theta \mathrm{d} \theta+\mathrm{gR} \sin \theta \mathrm{d} \theta \\
&\text { Integrating both sides, we get: } \\
&\frac{v^{2}}{2}=\mathrm{aR} \sin \theta-\mathrm{gR} \cos \theta+C \\
&\text { Given: } \\
&\theta=0, v=0 \\
&\text { So, } \\
&C=\mathrm{gR} \\
&\Rightarrow \frac{v^{2}}{2}=\mathrm{aR} \sin \theta-\mathrm{gR} \cos \theta+\mathrm{gR} \\
&\Rightarrow v^{2}=2 \mathrm{R}(a \sin \theta+g-g \cos \theta) \\
&\Rightarrow v=[2 \mathrm{R}(a \sin \theta+g-g \cos \theta)]^{1 / 2}
\end{aligned}
\)

Hint

(c) Mass of the trolley, \(M=200 \mathrm{~kg}\)
Speed of the trolley, \(v=36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s}\)
Mass of the boy, \(m=20 \mathrm{~kg}\)
Initial momentum of the system of the boy and the trolley
\(
\begin{aligned}
&=(M+m) v \\
&=(200+20) \times 10 \\
&=2200 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Let \(v^{\prime}\) be the final velocity of the trolley with respect to the ground.
Final velocity of the boy with respect to the ground \(=v^{\prime}-4\)
Final momentum \(=M v^{\prime}+m\left(v^{\prime}-4\right)\)
\(
\begin{aligned}
&=200 v^{\prime}+20 v^{\prime}-80 \\
&=220 v^{\prime}-80
\end{aligned}
\)
As per the law of conservation of momentum:
Initial momentum \(=\) Final momentum
\(
\begin{aligned}
&2200=220 v^{\prime}-80 \\
&\therefore v^{\prime}=2280 / 220=10.36 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Length of the trolley, \(l=10 \mathrm{~m}\)
Speed of the boy, \(v^{\prime \prime}=4 \mathrm{~m} / \mathrm{s}\)
Time taken by the boy to run, \(\mathrm{t}=10 / 4=2.5 \mathrm{~s}\)
\(\therefore\) Distance moved by the trolley \(=v^{\prime} \times t=10.36 \times 2.5=25.9 \mathrm{~m}\)

Hint

(b) Mass of the bolt, \(m=0.3 \mathrm{~kg}\)
Speed of the elevator \(=7 \mathrm{~m} / \mathrm{s}\)
Height, \(h=3 \mathrm{~m}\)
Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.
Heat produced = Loss of potential energy
\(=\mathrm{mgh}=0.3 \times 9.8 \times 3\)
\(=8.82 \mathrm{~J}\)
The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

Hint

(d) From the above figure
\(
\begin{aligned}
&R=m g \cos \theta \\
&F=\mu R=\mu m g \cos \theta
\end{aligned}
\)
Net Force on the block down the incline
\(
\begin{aligned}
&=m g \sin \theta-F \\
&=m g \sin \theta-\mu m g \cos \theta \\
&=m g(\sin \theta-\mu \cos \theta)
\end{aligned}
\)
Here distance moved \(x=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
In equilibrium
work done = Potential energy of stretched spring
\(m g(\sin \theta-\mu \cos \theta) x=\frac{1}{2} k x^{2}\)
or \(2 m g(\sin \theta-\mu \cos \theta)=k x\)
or \(2 \times 1 k g \times 10 m s^{-2}\left(\sin 37^{\circ}-\mu \cos 37^{\circ}\right)=100 \times 0.1 m\)
or \(20(0.601-\mu \times 0.798)=10\)
or \(0.601-0.798 \mu=\frac{10}{20}=0.5\)
or \(-0.798 \mu=0.5-0.601=-0.101\)
or \(\mu=-\frac{0.101}{-0.798}=\frac{101}{798}=0.126\)
Hence \(\mu=0.126\)

Hint

(a)

\(A B\) and \(A C\) are two smooth planes inclined to the horizontal at \(\angle \theta_{1}\) and \(\angle \theta_{2}\) respectively.
The height of both the planes is the same, therefore, both the stones will reach the bottom with same speed.
As P.E. at \(O=K . E\). at \(A=K . E\). at \(B\)
Therefore,
\(
\begin{gathered}
m g h=1 / 2 m v_{1}^{2}=1 / 2 m v_{2}^{2} \\
v_{1}=v_{2}
\end{gathered}
\)
As it is clear from fig. above, acceleration of the two blocks are
\(
a_{1}=g \sin \theta_{1},
\)
\(
a_{2}=g \sin \theta_{2}
\)
As \(\theta_{2}>\theta_{1}\)
\(
\therefore \quad a_{2}>a_{1}
\)
From \(v=u+a t\)
\(
\begin{aligned}
&=0+a t \\
\Rightarrow \quad t &=v / a
\end{aligned}
\)
As \(t \propto 1 / a\), and \(a_{2}>a_{1}\)
\(
\therefore \quad t_{2}<t_{1}
\)
That is, the second stone will take lesser time and reach the bottom earlier than the first stone.

Hint

(c) Mass of the bullet, \(\mathrm{m}=0.012 \mathrm{~kg}\)
Initial speed of the bullet, \(\mathrm{u}_{\mathrm{b}}=70 \mathrm{~m} / \mathrm{s}\)
Mass of the wooden block, \(M=0.4 \mathrm{~kg}\)
Initial speed of the wooden block, \(\mathrm{u}_{\mathrm{w}}=\mathrm{0}\)
Final speed of the system of the bullet and the block \(=\mathrm{v}   \mathrm{m} / \mathrm{s}\)
Applying the law of conservation of momentum:
\(
\begin{aligned}
&\mathrm{mu}_{\mathrm{b}}+\mathrm{Mu}_{\mathrm{w}}=(\mathrm{m}+\mathrm{M}) \mathrm{v} \\
&0.012 \times 70+0.4 \times 0=(0.012+0.4) \mathrm{v} \\
&\mathrm{v}=0.84 / 0.412 \\
&=2.04 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
For the system of the bullet and the wooden block:
Mass of the system, \(\mathrm{m}^{\prime}=0.412 \mathrm{~kg}\)
Velocity of the system \(=2.04 \mathrm{~m} / \mathrm{s}\)
Height up to which the system rises \(=h\)
Applying the law of conservation of energy to this system:
Potential energy at the highest point = Kinetic energy at the lowest point
\(\begin{aligned}
&\mathrm{m}^{\prime} g \mathrm{~h}=(1 / 2) \mathrm{m}^{\prime} \mathrm{v}^{2} \\
&\mathrm{~h}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}} \\
&=\frac{(2.04)^{2}}{2 \times 9.8} \\
&=0.2123 \mathrm{~m} \\
&\text { The wooden block will rise to a height of } 0.2123 \mathrm{~m} . \\
&\text { The heat produced = Kinetic energy of the bullet }-\text { Kinetic energy of the system } \\
&=(1 / 2) \mathrm{mu}{ }^{2}-(1 / 2) \mathrm{m}^{\prime} \mathrm{v}^{2} \\
&=(1 / 2) \times 0.012 \times(70)^{2}-(1 / 2) \times 0.412 \times(2.04)^{2} \\
&=29.4-0.857=28.54 \mathrm{~J}
\end{aligned}\)

Hint

(a)
Power used by the family, \(P=8 \mathrm{~kW}\)
\(=8 \times 10^{3} \mathrm{~W}\)
Solar energy received per square metre \(=200 \mathrm{~W}\)
Efficiency of conversion from solar to electricity energy \(=20 \%\)
Area required to generate the desired electricity \(=A\)
We have
\(8 \times 10^{3}=20 \% \times(A \times 200)\)
\(=(20 / 100) \times A \times 200\)
\(\therefore \quad A=8 \times 10^{3} / 40=200 \mathrm{~m}^{2}\)
(b)
The area of a solar plate required to generate \(8 \mathrm{~kW}\) of electricity is almost equivalent to the area of the roof of a building having dimensions \(14 \mathrm{~m}\) \(\times 14 \mathrm{~m} .(\approx \sqrt{200})\).

Hint

(a) is the right option. Given,
(a) Mass of the weight, \(m=10 \mathrm{~kg}\)
Height to which the person lifts the weight, \(h=0.5 \mathrm{~m}\)
Number of times the weight is lifted, \(n=1000\)
Therefore, Work done against gravitational force,
\(
\begin{aligned}
W &=n(m g h) \\
&=1000 \times 10 \times 9.8 \times 0.5 \\
&=49 \mathrm{~kJ}
\end{aligned}
\)
(b) Energy equivalent of \(1 \mathrm{~kg}\) of fat \(=3.8 \times 10^{7} \mathrm{~J}\)
Efficiency rate \(=20 \%\)
Mechanical energy supplied by the person’s body,
\(
\begin{aligned}
E &=(20 / 100) \times 3.8 \times 10^{7} \mathrm{~J} \\
&=(1 / 5) \times 3.8 \times 10^{7} \mathrm{~J}
\end{aligned}
\)
Equivalent mass of fat lost by the dieter,
\(
\begin{aligned}
m &=\left[1 /(1 / 5) \times 3.8 \times 10^{7}\right] \times 49 \times 10^{3} \\
&=(245 / 3.8) \times 10^{-4} \\
&=6.45 \times 10^{-3} \mathrm{~kg}
\end{aligned}
\)

Hint

(c) Area of the circle swept by the windmill \(=A\)
The velocity of the wind \(=v\)
Density of air \(=\rho\)
(a) Volume of the wind flowing through the windmill per sec \(=A v\)
Mass of the wind flowing through the windmill per \(\sec =\rho A v\)
Mass \(m\), of the wind flowing through the windmill in time \(t=\rho A v t\)
(b) Kinetic energy of air \(=(1 / 2) m v^{2}\)
\(
=(1 / 2)(\rho A v t) v^{2}=(1 / 2) \rho A v^{3} t
\)
(c) Area of the circle swept by the windmill \(=A=30 \mathrm{~m}^{2}\)
Velocity of the wind \(=v=36 \mathrm{~km} / \mathrm{h}\)
Density of air, \(\rho=1.2 \mathrm{~kg} \mathrm{~m}^{-3}\)
Electric energy produced \(=25 \%\) of the wind energy
\(=(25 / 100) \times\) Kinetic energy of air
\(
=(1 / 8) \rho A v^{3} t
\)
Electrical power \(=\) Electrical energy \(/\) Time
\(
\begin{aligned}
&=(1 / 8) \rho A v^{3} t / t \\
&=(1 / 8) \rho A v^{3} \\
&=(1 / 8) \times 1.2 \times 30 \times(10)^{3} \\
&=4.5 \mathrm{~kW}
\end{aligned}
\)

Hint

(d) Mass of the body, \(m=0.5 \mathrm{~kg}\)
Velocity of the body is governed by the equation, \(v=a x^{3 / 2}\) and \(a=5 \mathrm{~m}^{1 / 2} \mathrm{~s}^{-1}\)
Initial velocity, \(u(\) at \(x=0)=0\)
Final velocity \(v(\) at \(x=2 \mathrm{~m})=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
Work done, \(W=\) Change in kinetic energy
\(
\begin{aligned}
&=(1 / 2) m\left(v^{2}-u^{2}\right) \\
&=(1 / 2) \times 0.5\left[(10 \sqrt{2})^{2}-0^{2}\right] \\
&=(1 / 2) \times 0.5 \times 10 \times 10 \times 2 \\
&=50 \mathrm{~J}
\end{aligned}
\)

Hint

(d) Since the hole is on the floor, that means the sand is falling vertically with respect to the trolley. Therefore there is no force in the horizontal direction hence in the horizontal direction momentum is conserved.
Let \(\mathrm{M}=\) mass of trolley
\(\mathrm{m}=\) mass of sandbag
\(\mathrm{v}_{1}=\) initial velocity
\(\mathrm{v}_{2}=\) final velocity ( to be found)
Then \(P_{1}=(M+m) V_{1}\) when the sand bag is empty the momentum is
\(
\mathrm{P}_{2}=(\mathrm{M}+0) \mathrm{v}_{2}
\)
Momentum is conserved in horizontal direction so
\(
\begin{aligned}
&P_{1}=P_{2} \\
&\Rightarrow=v_{2}=\frac{(M+m)}{M} v_{1}=\frac{300+25}{300} \times 27 \times \frac{5}{18}=8.215 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a) The bob A will not rise because when two bodies of the same mass undergo an elastic collision, their velocities are interchanged. After the collision, ball A will come to rest and bob B would move with the velocity of A. Thus bob A will not rise after the collision.

Hint

(b) Let \(m\) be the mass of bob and \(I\) be the length of the string. If the potential energy of bob is taken zero at \(B\), then the potential energy of bob at A will be mgl. When bob reaches the point \(B, 5 \%\) of the potential energy is dissipated against air resistance, and the rest of \(95 \%\) is converted into kinetic energy.
\(\therefore \quad \frac{1}{2} \mathrm{mv}^{2}=0.95 \mathrm{mgl}\)
or
\(=\sqrt{2 \times 0.95 \times 9.8 \times 1.5}\)
\(=5.28 \mathrm{~m} / \mathrm{s}\)

Hint

(d) Here, Volume of water \(=30 \mathrm{~m}^{3}\)
\(\mathrm{t}=15 \mathrm{~min}=15 \times 60 \mathrm{~s}=900 \mathrm{~s} ;\) Height, \(h=40 \mathrm{~m}\)
Efficiency, \(\eta=30 \%\)
Density of water \(=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\)
\(\therefore\) Mass of water pumped=Volume \(\times\) Density
\(=\left(30 \mathrm{~m}^{3}\right)\left(10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\right)=3 \times 10^{4} \mathrm{~kg}\)
\(P_{\text {output }}=\frac{W}{t}=\frac{\mathrm{mgh}}{\mathrm{t}}=\frac{\left(3 \times 10^{4} \mathrm{~kg}\right)\left(10 \mathrm{~ms}^{-2}\right)(40 \mathrm{~m})}{900 \mathrm{~s}}\)
\(=\frac{4}{3} \times 10^{4} \mathrm{~W}\)
Efficiency, \(\eta=\frac{P_{\text {output }}}{P_{\text {input }}}\)
\(P_{\text {input }}=\frac{P_{\text {output }}}{\eta}=\frac{4 \times 10^{4}}{3 \times \frac{30}{100}}=\frac{4}{9} \times 10^{5}\)
\(
=44.4 \times 10^{3} \mathrm{~W}=44.4 \mathrm{~kW} \text {. }
\)

Hint

(a) Since the molecule approaches the wall of the container at a particular speed and then rebounds from the container at the same speed. Since there is no loss in the velocity due to any factor and the mass is the same. So, the momentum of the system remains conserved.

The molecule approach and rebound with the same speed. So, the kinetic energy of approach and rebound remains the same. Therefore it will be an elastic collision.

Hint

(c) Given,Radius of the rain drop, \(r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Volume of the rain drop, \(V=\frac{4}{3} \pi r^{3}\)
\(
=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \mathrm{~m}^{-3}
\)
Density of water, \(\rho=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\)
Mass of the rain drop, \(m=\rho V\)
\(
=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \mathrm{~kg}
\)
Gravitational force, \(F=m g\)
\(
=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 \mathrm{~N}
\)
The work done by the gravitational force on the drop in the first half of its journey,
\(
\begin{aligned}
W_{1} &=F . s \\
&=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 \times 250 \\
&=0.082 \mathrm{~J}
\end{aligned}
\)
This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey.
\(
\text { i.e., } W_{1}=0.082 \mathrm{~J}
\)
According to the law of conservation of energy,
If no resistive force is present, then the total energy of the rain drop will remain the same.
\(\therefore\) Total energy at the top,
\(
\begin{aligned}
E_{\mathrm{T}} &=m g h+0 \\
&=(4 / 3) \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 \times 500 \times 10^{-5} \\
&=0.164 \mathrm{~J}
\end{aligned}
\)
Due to the presence of a resistive force, the drop hits the ground with a velocity of \(10 \mathrm{~m} / \mathrm{s}\).
\(\therefore\) Total energy at the ground is,
\(
\begin{aligned}
E_{G} &=(1 / 2) \mathrm{mv}^{2}+0 \\
&=(1 / 2) \times(4 / 3) \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 \times(10)^{2} \\
&=1.675 \times 10^{-3} \mathrm{~J}
\end{aligned}
\)
\(\therefore\) Resistive force \(=E_{G}-E_{T}=-0.1623 \mathrm{~J}\)

Hint

(a) Here \(K_{e}=10 \mathrm{KeV}\) and \(K_{p}=100 \mathrm{keV}\)
\(m_{e}=9.11 \times 10^{-31} \mathrm{~kg}\) and \(m_{p}=1.67 \times 10^{-27} \mathrm{~kg}\)
As \(\mathrm{K}=\frac{1}{2} m v^{2}\) or \(\mathrm{v}=\operatorname{sqrt}\left(\frac{2 K}{m}\right)\)
Hence \(\frac{v_{e}}{v_{p}}=\sqrt{\frac{K_{e}}{K_{p}} \times \frac{m_{p}}{m_{e}}}=\sqrt{\frac{10 k e V}{100} k e V} \times \frac{1.67 \times 10^{-27} \mathrm{~kg}}{9.11 \times 10^{-31} \mathrm{~kg}}\) \(=13.54\)
\(
\Rightarrow v_{e}=13.54 v_{p}
\)
Thus electron is travelling faster

Hint

(c) Force exerted on the body, \(\mathrm{F}=-\hat{\imath}+2 \hat{\jmath}+3 \hat{k}  \mathrm{~N}\)
Displacement, \(\mathrm{s}=4 \mathrm{~km}\)
Work done, W = F.S
\(
\begin{aligned}
&=(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \cdot(4 \hat{k}) \\
&=0+0+3 \times 4 \\
&=12 \mathrm{~J}
\end{aligned}
\)
Hence, \(12 \mathrm{~J}\) of work is done by the force on the body?

Hint

(c) Power is given by the relation:
\(P=F v\) \(=\operatorname{mav}-\mathrm{mv}(\mathrm{dv} / \mathrm{dt})=\) Constant \((\) Say, \(\mathrm{k})\) \(\therefore v d v=\frac{k}{m} d t\) Intergarit both sides \(\frac{v^{2}}{2}=\frac{k}{m} t\), \(v=\sqrt{\frac{2 k t}{m}}\)
For displacement x of the body, we have
\(
\begin{aligned}
&v=\frac{d x}{d t}=\sqrt{\frac{2 \mathrm{k}}{m}} t^{\frac{1}{2}} \\
&d x=k \prime t^{\frac{1}{2}} d t
\end{aligned}
\)
where \(k \prime=\sqrt{\frac{2 k}{3}}=\) New Constant On integrating both sides, we get:
\(
\begin{aligned}
&x=\frac{2}{3} k \prime t^{\frac{3}{2}} \\
&\therefore x \propto t^{\frac{3}{2}}
\end{aligned}
\)

Hint

(b) Let \(u\) be the initial velocity, a be the constant acceleration, \(P\) be the power, \(F\) be the force, \(v\) be the final velocity, \(t\) be the time, and \(m\) be the mass.
As the body is initially at rest, thus \(\mathrm{u}=0\)
According to the equation of motion, \(\mathrm{v}=\mathrm{u}+\) at \(=0+\) at \(=\) at
Force is the product of mass and acceleration \(\Rightarrow \mathrm{F}=\mathrm{ma}\)
Power is given by the product of force and velocity.
\(
\begin{aligned}
&\Rightarrow \mathrm{P}=\mathrm{Fv} \\
&\therefore \mathrm{P}=\mathrm{ma} \times \text { at }=\mathrm{ma}^{2} \mathrm{t}
\end{aligned}
\)
If mass \(\mathrm{m}\) and acceleration a is constant then, \(\mathrm{P} \alpha \mathrm{t}\)
The power delivered to it is proportional to \(t\).

Hint

(d) (a) False. In an elastic collision of two bodies, the momentum and energy of the system is conserved.
(b) False. Some energy is lost in heat, so it is not conserved.
(c) False. It is true only for conservative forces. For non-conservative forces, work done over one cycle is not zero.
(d) True. Some energy is lost in internal friction and heat.

Hint

(c) Total energy of the particle, \(\mathrm{E}=1 \mathrm{~J}\)
Force constant, \(\mathrm{k}=0.5 \mathrm{~N} / \mathrm{m}\)
Kinetic energy of the particle,
\(
\mathrm{K}=(1 / 2) \mathrm{mv}^{2}
\)
According to the conservation law: \(\mathrm{E}=\mathrm{V}+\mathrm{K} =(1 / 2) \mathrm{kx}^{2}+(1 / 2) \mathrm{mv}^{2}\)
At the moment of turn back, velocity (and hence \(\mathrm{K}\) ) becomes zero.
\(1=(1 / 2) \mathrm{kx}^{2}\)
\((1 / 2) 0.5 \mathrm{x}^{2}=1\)
\(x=\pm 2 m\)
Hence, the particle turns back when it reaches \(x=\pm 2 m\)