Past NEET Papers

Hint

(b) Work done by a variable force
Work done, \(W=\int_{y_{i}}^{y_{f}} F d y \Rightarrow \int_{y=0}^{y_{f}=1} F \cdot d y\) where, \(\mathrm{F}=20+10 y\)
\(
\begin{aligned}
&\therefore \quad W=\int_{0}^{1}(20+10 y) d y \\
&=\left[20 y+\frac{10 y^{2}}{2}\right]_{0}^{1}=25 J
\end{aligned}
\)

Hint

(c) The area under the force-displacement curve gives the amount of work done.
From the work-energy theorem,
\(
W=\Delta K E
\)

From \(x=0\) to \(x=8 \mathrm{~m}\)
\(
\begin{aligned}
&\frac{1}{2} m v^{2}=100+30 \\
&\therefore v=\sqrt{520}=23 \mathrm{~m} / \mathrm{s} \\
&\text { From } x=0 \text { to } x=12 \mathrm{~m} \\
&\frac{1}{2} m v^{2}=100+30-47.5+20 \\
&\therefore v=\sqrt{410}=20.6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(c) From work-energy theorem,
\(
\begin{aligned}
&\mathrm{W}_{g}+\mathrm{W}_{a}=\Delta \mathrm{K} . \mathrm{E} . \\
&\text { or, } m g h+\mathrm{W}_{a}=\frac{1}{2} m v^{2}-0 \\
&10^{-3} \times 10 \times 10^{3}+\mathrm{W}_{a}=\frac{1}{2} \times 10^{-3} \times(50)^{2} \\
&\Rightarrow \mathrm{W}_{a}=-8.75 \mathrm{~J}
\end{aligned}
\)
which is the work done due to air resistance
Work done due to gravity \( \mathrm{W}_{g} =\mathrm{mgh}\) \(=10^{-3} \times 10 \times 10^{3}=10 \mathrm{~J}\)

Hint

(b) Case (a): Suppose the two springs are stretched by the same distance \(x\). Then \(\frac{W_{P}}{W_{Q}}=\frac{\frac{1}{2} k_{p} x^{2}}{\frac{1}{2} k_{Q} x^{2}}=\frac{k_{p}}{k_{Q}}\)
As \(\mathrm{k}_{\mathrm{p}}>\mathrm{k}_{\mathrm{Q}}\) so, \(\mathrm{W}_{\mathrm{p}}>\mathrm{W}_{\mathrm{Q}}\)
Case (b): Suppose the two springs stretched by distance \(x_{P}\) and \(x_{Q}\) by the same force \(\mathrm{F}\). Then, \(\quad \mathrm{F}=k_{p}{x}_{p}=k_{Q}{x}_{Q}\)

\(
\frac{W_{p}}{W_{Q}}=\frac{\frac{1}{2} k_{p} x_{p}^{2}}{\frac{1}{2} k_{Q} x_{Q}^{2}}=\frac{k_{p} \cdot x_{p} \cdot x_{p}}{k_{Q} \cdot x_{Q} \cdot x_{Q}}=\frac{F x_{p}}{F x_{Q}}=\frac{k_{Q}}{k_{p}}
\)
As \(k_{P}>k_{Q} \quad \therefore \quad \mathrm{W}_{\mathrm{Q}}>\mathrm{W}_{\mathrm{P}}\)

Hint

(c) Given : \(\overrightarrow{\mathrm{F}}=3 \hat{i}+\hat{j}\)
\(
\begin{aligned}
&\vec{r}_{1}=(2 \hat{i}+\hat{k}), \vec{r}_{2}=(4 \hat{i}+3 \hat{j}-\vec{k}) \\
&\vec{r}=\vec{r}_{2}-\vec{r}_{1}=(4 \hat{i}+3 \hat{j}-\vec{k})-(2 \hat{i}+\hat{k})
\end{aligned}
\)
or \(\vec{r}=2 \hat{i}+3 \hat{j}-2 \hat{k}\)
So work done by the given force, \(w=\vec{F} \cdot \vec{r}\)
\(
=(3 \hat{i}+\hat{j}) \cdot(2 \hat{i}+3 \hat{j}-2 \hat{k})=6+3+0=9 \mathrm{~J}
\)

Hint

(b)

(b) Work done is equal to the area under the curve in the F-d graph.

Work done = Area under F-d graph
\(
\begin{aligned}
&=2 \times(7-3)+\frac{1}{2} \times 2 \times(12-7) \\
&=8+\frac{1}{2} \times 10=8+5=13 \mathrm{~J}
\end{aligned}
\)

Hint

(a) \(W=\) Potential energy stored in the spring +
Loss of potential energy of mass
\(
W=m g(h+d)-\frac{1}{2} k d^{2}
\)
The gravitational potential energy of the ball gets converted into the elastic potential energy of the spring.
\(
m g(h+d)=\frac{1}{2} k d^{2}
\)
Net work done \(=\operatorname{mg}(h+d)-\frac{1}{2} k d^{2}=0\)

Hint

(c) Work done by the force = force x displacement
or \(\quad W=F \times s \dots (i)\)
But from Newton’s 2nd law, we have
Force = mass \(\times\) acceleration
i.e. \(F=m a \dots (ii)\)
Hence, from Eqs. (i) and (ii), we get
\(W=m a s=m\left(\frac{d^{2} s}{d t^{2}}\right) s\left(\because a=\frac{d^{2} s}{d t^{2}}\right) \ldots\) (iii)
Now, we have displacement,
\(s=\frac{1}{3} t^{2}\)
\(\therefore \quad \frac{d^{2} s}{d t^{2}}=\frac{d}{d t}\left[\frac{d}{d t}\left(\frac{1}{3} t^{2}\right)\right]\)
\(=\frac{d}{d t} \times\left(\frac{2}{3} t\right)\)
\(=\frac{2}{3} \frac{d t}{d t}=\frac{2}{3}\)
Hence, Eq. (iii) becomes
\(W=\frac{2}{3} m s=\frac{2}{3} m \times \frac{1}{3} t^{2}=\frac{2}{9} m t^{2}\)
We have, \(\quad m=3 \mathrm{~kg}, \mathrm{t}=2 \mathrm{~s}\)
\(
\therefore \quad W=\frac{2}{9} \times 3 \times(2)^{2}=\frac{8}{3} \mathrm{~J}
\)

Hint

(a) Network done in sliding a body up to a height hon inclined plane
\(=\) Work done against gravitational force
+ Work done against frictional force
\(\Rightarrow \quad \mathrm{W}=\mathrm{W}_{g}+\mathrm{W}_{f} \quad \ldots\) (i)
but \(\quad W=300 \mathrm{~J}\)
\(W_{g}=m g h\)
\(=2 \times 10 \times 10=200 \mathrm{~J}\)
Putting in Eq. (i), we get
\(300=200+W_{f}\)
\(\Rightarrow \quad W_{f}=300-200=100 \mathrm{~J}\)

Hint

(c) Work done in moving the object from \(x=0\) to \(x=6 \mathrm{~m}\), is given by area under the F-x curve.
\(
\begin{aligned}
W &=\text { Area of square }+\text { area of triangle } \\
&=3 \times 3+\frac{1}{2} \times 3 \times 3 \\
&=9+4.5=13.5 \mathrm{~J}
\end{aligned}
\)

Hint

(d) Efficiency \(=\frac{\text { output } \text { work }}{\text { input work }}\)
i.e., Efficiency,
\(
=\frac{75 \mathrm{~g} \times 3}{250 \times 12}=\frac{75 \times 10 \times 3}{250 \times 12}=0.75=75 \%
\)

Hint

(a) The mass is \(30 \mathrm{~g}\) or \(0.03 \mathrm{~kg}\) and the position of the particle is \(\mathrm{x}=3 \mathrm{t}-4 \mathrm{t}^{2}+\mathrm{t}^{3}\).
The velocity is given as,
\(
\frac{\mathrm{dx}}{\mathrm{dt}}=3-8 \mathrm{t}+3 \mathrm{t}^{2}
\)
The acceleration is given as,
\(
\frac{d^{2} x}{d t^{2}}=-8+6 t
\)
The work done is given as,
\(\mathrm{dW}=(\mathrm{ma}) \mathrm{dx}\)
\(\mathrm{dW}=(0.03) \times(-8+6 \mathrm{t})\left(3-8 \mathrm{t}+3 \mathrm{t}^{2}\right) \mathrm{dt}\)
\(\mathrm{W}=(0.03) \int_{0}^{4}\left(18 \mathrm{t}^{3}-72 \mathrm{t}^{2}+82 \mathrm{t}-24\right) \mathrm{dt}\)
\(\mathrm{W}=(0.03)\left(18 \times \frac{\mathrm{t}^{4}}{4}-72 \times \frac{\mathrm{t}^{3}}{3}+82 \times \frac{\mathrm{t}^{2}}{2}-24 \mathrm{t}\right)_{0}^{4}\)
\(\mathrm{W}=(0.03) \times 176\)
\(\mathrm{W}=5.28 \mathrm{~J}\)
Thus, the work done during the first \(4 \mathrm{~s}\) is \(5.28 \mathrm{~J}\).

Hint

(d) Work done by a variable force \(F\) in displacement from \(x=x_{1}\) to \(x=x_{2}\) is given by
\(
W=\int_{x_{1}}^{x_{2}} F(d x)
\)
Given, \(\quad x_{1}=0, x_{2}=5\)
\(
\begin{aligned}
F &=\left(7-2 x+3 x^{2}\right) \mathrm{N} \\
\therefore \quad W &=\int_{0}^{5}\left(7-2 x+3 x^{2}\right) d x \\
&=\left[7 x-x^{2}+x^{3}\right]_{0}^{5} \\
&=\left[7 \times 5-(5)^{2}+(5)^{3}\right] \\
&=[35-25+125]=135 \mathrm{~J}
\end{aligned}
\)

Hint

\(
\begin{array}{l}
\text { (b) } \mathrm{W}=\Delta \mathrm{E}=\frac{1}{2} m\left(v_{1}^{2}-v_{2}^{2}\right) \\
=\frac{1}{2} \times 0.01\left[(1000)^{2}-(500)^{2}\right]=3750 \mathrm{~J} .
\end{array}
\)

Hint

(a) Given: Mass of particle, \(\mathrm{M}=10 \mathrm{~g}=\) \(\frac{10}{1000} \mathrm{~kg}\)
radius of circle \(R=6.4 \mathrm{~cm}\)
Kinetic energy E of particle \(=8 \times 10^{-4} \mathrm{~J}\)
acceleration \(a=\) ?
\(
\begin{array}{l}
\frac{1}{2} \mathrm{mv}^{2}=\mathrm{E} \Rightarrow \frac{1}{2}\left(\frac{10}{1000}\right) \mathrm{v}^{2}=8 \times 10^{-4} \\
\Rightarrow v^{2}=16 \times 10^{-2} \\
\Rightarrow \quad v=4 \times 10^{-1}=0.4 \mathrm{~m} / \mathrm{s} \\
\text { Now using } \\
v^{2}=u^{2}+2 a s \quad(s=4 \pi \mathrm{R}) \\
(0.4)^{2}=0^{2}+2 a\left(4 \times \frac{22}{7} \times \frac{6.4}{100}\right) \\
\Rightarrow \quad a=(0.4)^{2} \times \frac{7 \times 100}{8 \times 22 \times 6.4}=0.1 \mathrm{~m} / \mathrm{s}^{2}
\end{array}
\)

Hint

(d) From the work-energy theorem,
Work done \(=\) Change in \(\mathrm{KE}\)
\(
\begin{aligned}
\Rightarrow \quad W &=K_{f}-K_{i} \\
\Rightarrow \quad K_{f} &=W+K_{i}=\int_{x_{1}}^{x_{2}} F x d x+\frac{1}{2} m v^{2} \\
&=\int_{20}^{30}-0.1 x d x+\frac{1}{2} \times 10 \times 10^{2} \\
&=-0.1\left[\frac{x^{2}}{2}\right]_{20}^{30}+500 \\
&=-0.05\left[30^{2}-20^{2}\right]+500 \\
&=-0.05[900-400]+500 \\
\quad K_{f} &=-25+500=475 \mathrm{~J}
\end{aligned}
\)

Hint

(b) We have to use the Law of momentum conservation:
\(P(\) initial \()=P(\) final \()\)
\(0=n \cdot m \cdot u+(M-n \cdot m) \cdot v\)
where: \(\mathrm{n}=10, \mathrm{~m}=10 \mathrm{~g}=0.01 \mathrm{~kg}, \mathrm{u}=\) \(800 \mathrm{~m} / \mathrm{s}, \mathrm{M}=100 \mathrm{~kg}\).
\(0=10 \cdot 0.01 \mathrm{~kg} \cdot 800 \mathrm{~m} / \mathrm{s}+(100 \mathrm{~kg}-10\) – \(0.01 \mathrm{~kg}) \cdot \mathrm{v}\)
\(\mathrm{v}=-80  \mathrm{kgm} \mathrm{s}^{-1} / 99.9  \mathrm{kgm}  \mathrm{s}^{-1}\)
\(\mathrm{v}= -0.8 \mathrm{~m} / \mathrm{s}\)
\(\mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=\frac{10 \times 0.01 \mathrm{~kg} \times 800 \mathrm{~m} / \mathrm{s}}{5 \mathrm{~s}}=16 \mathrm{~N}\)

Hint

(c) As the particle is moving in a potential energy region.
\(\therefore\) Kinetic energy \(>0\)
And, total energy \(\mathrm{E}=\mathrm{K} . \mathrm{E} .+\) P.E.
\(\Rightarrow \mathrm{U}(x)<\mathrm{E}\)

Hint

(b) When the work is done by the system against any conservative force the potential energy increases.
\(
-\Delta \mathrm{U}=\mathrm{W}_{\text {conservative force }}
\)

Hint

(a) As \(m\) is the mass per unit length, then rate of mass per second \(=\frac{m x}{t}=m v\)
\(\therefore \quad\) Rate of \(K E=\frac{1}{2}(m v) v^{2}=\frac{1}{2} m v^{3}\)

Hint

(a) When the body is thrown upwards. its K.E is converted into P.E.
The loss of energy due to air friction is the difference of K.E and P.E.
\(
\begin{aligned}
\frac{1}{2} m v^{2}-m g h &=\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \\
&=200-180=20 \mathrm{~J}
\end{aligned}
\)

Hint

(c) Kinetic energy is given by
\(
E=\frac{1}{2} m v^{2}=\frac{1}{2 m}(m v)^{2}
\)
but, \(m v=\) momentum of the particle \(=p\) \(\therefore \quad E=\frac{p^{2}}{2 m}\) or \(p=\sqrt{2 m E}\)
Therefore,
\(
\frac{p_{1}}{p_{2}}=\sqrt{\frac{m_{1} E_{1}}{m_{2} E_{2}}}
\)
but it is given that, \(p_{1}=p_{2}\)
\(\therefore \quad m_{1} E_{1}=m_{2} E_{2}\)
or \(\frac{E_{1}}{E_{2}}=\frac{m_{2}}{m_{1}} \dots (i)\)
Now, \(\quad m_{1}>m_{2}\)
or \(\frac{m_{1}}{m_{2}}>1 \dots (ii)\)
Thus, from Eqs. (i) and (ii), we get
\(
\frac{E_{1}}{E_{2}}<1 \text { or } E_{1}<E_{2}
\)

Hint

(d) Since height is the same for both balls, their velocities on reaching the ground will be the same (Velocity of free-falling body does not depend on its mass, it depends on the height from which it has been dropped.)
\(
\therefore \frac{K \cdot E_{1}}{K \cdot E_{2}}=\frac{\frac{1}{2} m_{1} v_{0}^{2}}{\frac{1}{2} m_{2} v_{0}^{2}}=\frac{m_{1}}{m_{2}}=\frac{2}{4}=\frac{1}{2}
\)

Hint

(a) If \(k\) be the spring constant, then
\(
\begin{array}{l}
U=\frac{1}{2} \times k \times(2)^{2}=2 k \\
U_{\text {final }}=\frac{1}{2} \times k \times(10)^{2}=50 k \\
\Rightarrow \frac{U}{U_{\text {final }}}=\frac{2 k}{50 k}=\frac{1}{25} \\
\Rightarrow \mathrm{U}_{\text {final }}=25 \mathrm{U}
\end{array}
\)

Hint

(d) Kinetic energy
\(
K=\frac{1}{2 m}\left(p^{2}\right) \text { or } p=\sqrt{2 m K}
\)
If kinetic energy of a body is increased by \(300 \%\), let its momentum becomes \(p^{\prime}\).
New kinetic energy
\(
K^{\prime}=K+\frac{300}{100} K=4 K\left(\begin{array}{l}
\text { initial } K E=K \\
\text { final } K E=K^{\prime}
\end{array}\right)
\)
Therefore, momentum is given by
\(
\begin{array}{l}
p^{\prime}=\sqrt{2 m \times 4 K} \\
=2 \sqrt{2 m K}=2 p
\end{array}
\)
Hence, percentage change (increase) in momentum
\(
\begin{aligned}
\frac{\Delta p}{p} \times 100 &=\frac{p^{\prime}-p}{p} \times 100 \\
&=\left(\frac{p^{\prime}}{p}-1\right) \times 100 \\
&=\left(\frac{2 p}{p}-1\right) \times 100 \\
&=100 \%
\end{aligned}
\)

Hint

(d) If \(l\) is length of pendulum and \(\theta\) be angular amplitude then height
\(
\mathrm{h}=\mathrm{AB}-\mathrm{AC}=l-l \cos \theta=l(1-\cos \theta)
\)
At extreme position, potential energy is maximum and kinetic energy is zero; At mean (equilibrium) position potential energy is zero and kinetic energy is maximum, so from principle of conservation of energy. \((\mathrm{KE}+\mathrm{PE})\) at \(\mathrm{P}=(\mathrm{KE}+\mathrm{PE})\) at \(\mathrm{B}\)
\(
\begin{array}{l}
0+m g h=\frac{1}{2} m v^{2}+0 \\
\Rightarrow v=\sqrt{2 g h}=\sqrt{2 g l(1-\cos \theta)}
\end{array}
\)

Hint

(c) As we know that, relation between kinetic energy and momentum is given by
\(
K E=\frac{p^{2}}{2 m}
\)
If \(p_{1}=p_{2}\) for two bodies
So, \(\quad K E_{1} \propto \frac{1}{m_{1}}\)
and \(\quad K E_{2} \propto \frac{1}{m_{2}}\)
Therefore, ratio of two masses is given by
\(
\frac{m_{1}}{m_{2}}=\frac{K E_{2}}{K E_{1}}=\frac{1}{4} \quad\left[\because \frac{K E_{1}}{K E_{2}}=\frac{4}{1}\right]
\)

Hint

(b) Potential energy = Kinetic energy
i.e. \(m g h=\frac{1}{2} m v^{2} \Rightarrow v=\sqrt{2 g h}\)
If \(h_{1}\) and \(h_{2}\) are the initial and final heights, then
\(
v_{1}=\sqrt{2 g h_{1}}, v_{2}=\sqrt{2 g h_{2}}
\)
Loss in velocity
\(
\Delta v=v_{1}-v_{2}=\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}
\)
\(\therefore\) Fractional loss in velocity
\(
=\frac{\Delta v}{v_{1}}=\frac{\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}}{\sqrt{2 g h_{1}}}=1-\sqrt{\frac{h_{2}}{h_{1}}}
\)
Substituting the values, we have
\(
\begin{aligned}
\therefore \quad \frac{\Delta v}{v_{1}} &=1-\sqrt{\frac{1.8}{5}}=1-\sqrt{0.36}=1-0.6 \\
&=0.4=\frac{2}{5}
\end{aligned}
\)

Hint

\(
\begin{array}{l}
\text { (c) } m_{1}=m, m_{2}=4 \mathrm{~m} \\
\mathrm{~K} . \mathrm{E}_{1}=\mathrm{K} . \mathrm{E}_{2} \\
\frac{1}{2} m_{1} v_{1}^{2}=\frac{1}{2} m_{2} v_{2}^{2} ; \frac{1}{2} m v_{1}^{2}=\frac{1}{2} 4 m v_{2}^{2} \\
\frac{v_{1}}{v_{2}}=2 \Rightarrow v_{1}=2 v_{2} \\
\frac{\text { Linear momentum of first body }}{\text { Linear momentum of second body }} \\
=\frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m \cdot 2 v_{2}}{4 m v_{2}}=\frac{1}{2}
\end{array}
\)

Hint

(d) K.E. \(=\frac{1}{2} m v^{2}\)
Further, \(v^{2}=u^{2}+2 a s=0+2 a d=2 a d\) \(=2(\mathrm{~F} / \mathrm{m}) d\)
Hence, K.E. \(=\frac{1}{2} m \times 2(F / m) d=F d\)
or, K.E. acquired \(=\) Work done \(=\mathrm{F} \times d=\) constant.
i.e., it is independent of mass \(\mathrm{m}\).

Hint

(d0 Let \(p_{1}\) be the initial momentum and \(p_{2}\) be the inversed momentum
So, \(\quad p_{2}=\frac{150}{100} p_{1}\)
i.e. \(m v_{2}=\frac{15}{10} m v_{1} \quad\left(\begin{array}{l}p_{1}=m v_{1} \\ p_{2}=m v_{2}\end{array}\right)\)
or \(\quad v_{2}=\frac{15}{10} v_{1}\)
Now, \(\frac{E_{2}}{E_{1}}=\frac{\frac{1}{2} m v_{2}^{2}}{\frac{1}{2} m v_{1}^{2}}=\left(\frac{v_{2}}{v_{1}}\right)^{2}=\left(\frac{15}{10}\right)^{2}=\frac{225}{100}\)
Clearly, \(E_{2}>E_{1}\)
So, percentage increase in \(K E\)
\(
\begin{array}{l}
=\frac{\left(E_{2}-E_{1}\right)}{E_{1}} \times 100 \\
=\left(\frac{225}{100}-1\right) \times 100=125 \%
\end{array}
\)

Hint

(d) Force due to friction = kinetic energy
\(\mu m g s=\frac{1}{2} m v^{2}\)
[Here, \(v=72 \mathrm{~km} / \mathrm{h}=\frac{72000}{60 \times 60}=20 \mathrm{~m} / \mathrm{s}\) ]
or, \(s=\frac{v^{2}}{2 \mu g}=\frac{20 \times 20}{2 \times 0.5 \times 10}=40 \mathrm{~m}\)
If a body of mass \(m\) moves with velocity \(u\) on a rough surface and stops after travelling a distance \(S\) due to friction. Then, frictional force, \(\mathrm{F}=\mathrm{ma}=\) \(\mu \mathrm{R} \Rightarrow \mathrm{ma}=\mu \mathrm{mg} \Rightarrow a=\mu \mathrm{g}\)
Using \(v^{2}=u^{2}-2 a s\)
\(\Rightarrow 0=u^{2}-2 \mu g S\)
\(\Rightarrow S=\frac{u^{2}}{2 \mu g}\)

Hint

(a) The relation between \(\mathrm{KE}\) and pis given by
\(
\begin{array}{ll}
 K E=\frac{p^{2}}{2 m} \\
\Rightarrow \quad  p^{2}=2 m K E \\
\Rightarrow \quad \quad p=\sqrt{2 m K E} \\
\text { If KE of two bodies are equal. } \\
\text { So, } \quad \quad p_{1} \propto \sqrt{m_{1}} \\
\text { and } \quad p_{2} \propto \sqrt{m_{2}} \\
\Rightarrow \quad \frac{p_{1}}{p_{2}}=\sqrt{\frac{m_{1}}{m_{2}}}=\sqrt{\frac{1}{9}}=\frac{1}{3}
\end{array}
\)

Hint

(c) \(E=\frac{1}{2} m v^{2}\). Hence, \(m v=(2 \mathrm{mE})^{1 / 2}\).
For same KE, momentum \(\propto \sqrt{m}\).
Hence, the ratio is \(2: 1\).

Hint

(d) Given force \(\vec{F}=2 t \hat{i}+3 t^{2} \hat{j}\)
According to Newton’s second law of motion,
\(
\begin{array}{l}
m \frac{d \vec{v}}{d t}=2 t \hat{i}+3 t^{2} \hat{j}(\mathrm{~m}=1 \mathrm{~kg}) \\
\Rightarrow \quad \int_{0}^{\vec{v}} d \vec{v}=\int_{0}^{t}\left(2 t \hat{i}+3 t^{2} \hat{j}\right) d t \\
\Rightarrow \quad \vec{v}=t^{2} \hat{i}+t^{3} \hat{j}
\end{array}
\)
Power P \(=\vec{F} \cdot \vec{v}\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \cdot\left(t^{2} \hat{i}+t^{3} \hat{j}\right)\) \(=\left(2 t^{3}+3 t^{5}\right) \mathrm{W}\)

Hint

(d) As we know power \(\mathrm{P}=\frac{d w}{d t}\)
\(
\Rightarrow w=P t=\frac{1}{2} \mathrm{mV}^{2}
\)
So, \(v=\sqrt{\frac{2 \mathrm{P}t}{\mathrm{m}}}\)
Hence, acceleration \(a=\frac{d V}{d t}=\sqrt{\frac{2 P}{m}} \cdot \frac{1}{2 \sqrt{t}}\) Therefore, force on the particle at time ‘ \(t\) ‘
\(
=m a=\sqrt{\frac{2 K m^{2}}{m}} \cdot \frac{1}{2 \sqrt{t}}=\sqrt{\frac{K m}{2 t}}=\sqrt{\frac{m K}{2}} t^{-1 / 2}
\)

Hint

(d) Given, pressure \(=150 \mathrm{~mm}\) of \(\mathrm{Hg}\)
Pumping rate of heart of a man
\(
=\frac{d V}{d t}=\frac{5 \times 10^{-3}}{60} \mathrm{~m}^{3} / \mathrm{s}
\)
Power of heart \(=P \cdot \frac{d V}{d t}=\rho g h \cdot \frac{d V}{d t}\)
\(
\begin{aligned}
\frac{\left(13.6 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)\left(10 \times 0.15 \times 5 \times 10^{-3}\right)}{60} = 1.70 \mathrm{~W}
\end{aligned}
\)

Hint

(a) Power \(P=\frac{W}{t}\)
\(
\Rightarrow \frac{P_{1}}{P_{2}}=\frac{t_{2}}{t_{1}}=\frac{30 s}{1 \text { minute }}=\frac{30 s}{60 s}=\frac{1}{2}
\)
\(\left(t_{1}=1\right.\) minute; \(t_{2}=30\) second given)

Hint

(b) Constant power of car \(\mathrm{P}_{0}=\mathrm{F} \cdot \mathrm{V}=\mathrm{ma} . \mathrm{V}\)
\(
\begin{array}{c}
P_{0}=m \frac{d v}{d t} \cdot v \\
\mathrm{P}_{0} d t=m v d v \text { Integrating } \\
P_{0} \cdot t=\frac{m v^{2}}{2} \\
\quad v=\sqrt{\frac{2 P_{0} t}{m}} \\
\because \mathrm{P}_{0}, \mathrm{~m} \text { and } 2 \text { are constant } \\
\therefore \quad v \propto \sqrt{t}
\end{array}
\)
Note: At constant power, \(P=\) constant
\(\mathbf{P}=F v=\frac{m d v}{d t} v=P \quad\left(\because F=\frac{m d v}{d t}\right)\)
\(\Rightarrow v d v=\frac{P}{m} d t\)
Integrating both sides, we get
\(\int v d v=\int \frac{P}{m} t\)
\(\Rightarrow \frac{v^{2}}{2}=\frac{P}{m} t+C_{1}\)
At \(t=0, v=0\)
\(\therefore C_{1}=0\)
From equation (i) we get
\(
\mathbf{V}^{2}=\frac{2 P t}{m} \Rightarrow \mathbf{V}=\left(\frac{2 P t}{m}\right)^{1 / 2}
\)

Hint

(b) We know that, Power, \(P=\mathbf{F} \cdot \mathbf{V}=F V \cos \theta\) So, just before hitting, \(\theta\) is zero, power will be maximum.

Hint

(a) \(
\text { Power }=\text { K.E. of water flowing per second }
\)

\(\begin{aligned}
P &=\frac{d k}{d t} \\
P &=\frac{d}{d t}\left(\frac{1}{2} m v^{2}\right) \\
&=\frac{1}{2} v^{2} \frac{d m}{d t} \\
&=\frac{1}{2} v^{2} \frac{d m}{d l} \frac{d l}{d t} \\
&=\frac{1}{2} v^{3} \frac{d m}{d l}=\frac{1}{2} \times 8 \times 100 \mathrm{w} = 400 \mathrm{w} \\
\end{aligned}\)

Hint

(a)

\(\begin{aligned}
P_{\text {generated }} &=P_{\text {input }} \times \frac{90}{100}=\frac{m g h}{t} \times \frac{90}{100} \\
&=\frac{15 \times 10 \times 60}{1} \times \frac{90}{100}=8.1 \mathrm{~kW}
\end{aligned}\)

Hint

(c) Power of a body is defined as the rate at which the body can do the work, i.e.
\(
\text { Power }=\frac{\text { work }}{\text { time }}=\frac{W}{t}
\)
Given, power, \(P=2 \mathrm{~kW}=2000 \mathrm{~W}\)
\(
\begin{array}{l}
\begin{aligned}
W=M g h &=M \times 10 \times 10 \quad\left[\because g=10 \mathrm{~m} / \mathrm{s}^{2}\right] \\
&=100 \mathrm{M}
\end{aligned} \\
\text { Time, } t=60 \mathrm{~s} \\
\therefore \quad 2000=\frac{100 M}{60} \\
\therefore \quad M=1200 \mathrm{~kg} \text { and } V=1200 \mathrm{~L}
\end{array}
\)
Its volume \(=1200\) litre as 1 litre of water contains \(1 \mathrm{~kg}\) of its mass.

Hint

(b) Fractional loss of K.E. of colliding body
\(
\begin{array}{l}
\frac{\Delta K E}{K E}=\frac{4\left(m_{1} m_{2}\right)}{\left(m_{1}+m_{2}\right)^{2}} \\
=\frac{4(4 m) 2 m}{(4 m+2 m)^{2}} \\
=\frac{32 m^{2}}{36 m^{2}}=\frac{8}{9}
\end{array}
\)

Hint

\(
\begin{array}{l}
\text { (a) } \vec{P}_{i}=\vec{P}_{f} \\
m \vec{v}_{i}=\left(\frac{m}{6} \vec{v}_{1}+\frac{5 m}{6} \vec{v}_{2}\right) \\
\vec{v}_{i}=\left(\frac{\vec{v}_{1}}{6}+\frac{5}{6} \vec{v}_{2}\right) \\
20 \hat{i}+25 \hat{j}-12 \hat{k}=\frac{(100 \hat{i}+35 \hat{j}+8 \hat{k})}{6}+\frac{5 \vec{v}_{2}}{6} \\
\vec{v}_{2}=\frac{20 \hat{i}+115 \hat{j}-80 \hat{k}}{5} \\
\therefore \quad \vec{v}_{2}=4 \hat{i}+23 \hat{j}-16 \hat{k}
\end{array}
\)

Hint

(b) Bullets collide at time \(t\)
\(
\begin{array}{l}
x_{1}+x_{2}=100 \mathrm{~m} \\
25 t+25 t=100
\end{array}
\)
\(\begin{array}{l}
t=2 s \\
y=\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times 2^{2}=20 \mathrm{~m} \\
h=200-20=180 \mathrm{~m}
\end{array}\)

Hint

(a) The particle of mass \(5 \mathrm{~m}\) breaks in three fragments of mass \(m, m\) and \(3 \mathrm{~m}\) respectively. Two fragments of mass \(m\) each, move in the perpendicular direction with velocity v, and the left fragment will move in a direction with velocity \(\mathbf{v}^{\prime}\) such that the total momentum of the system must remain conserved.
By law of conservation of momentum,
\(
\begin{aligned}
& 5 m \times 0=m v \hat{\mathbf{i}}+m \hat{\mathbf{j}}+3 m \mathbf{v}^{\prime} \\
\Rightarrow \quad \mathbf{v}^{\prime} &=-\frac{v}{3} \hat{\mathbf{i}}-\frac{v}{3} \hat{\mathbf{j}} \\
\therefore &\left|\mathbf{v}^{\prime}\right|=\sqrt{\left(-\frac{v}{3}\right)^{2}+\left(-\frac{v}{3}\right)^{2}}=\frac{v \sqrt{2}}{3}
\end{aligned}
\)
\(\therefore\) Energy released
\(
\begin{aligned}
E &=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2} \times 3 m\left(\frac{v \sqrt{2}}{3}\right)^{2} \\
&=m v^{2}+\frac{m v^{2}}{3}=\frac{4}{3} m v^{2}
\end{aligned}
\)

Hint

(b) Since, the collision mentioned is an elastic head-on collision. Thus, according to the law of conservation of linear momentum, we get
\(
m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}
\)
where, \(m_{1}\) and \(m_{2}\) are the masses of the two blocks, respectively and \(u_{1}\) and \(u_{2}\) are their initial velocities and \(v_{1}\) and \(v_{2}\) are their final velocities, respectively.
Here, \(m_{1}=m, m_{2}=4 m\)
\(
\begin{array}{c}
u_{1}=v, u_{2}=0 \text { and } v_{1}=0 \\
m v+4 m \times 0=0+4 m v_{2} \\
\Rightarrow \quad m v=4 m v_{2} \text { or } v_{2}=\frac{v}{4} \quad \ldots \text { (i) }
\end{array}
\)
As, the coefficient of restitution is given as
\(
\begin{aligned}
e &=\frac{\text { relative velocity of separation after collision }}{\text { relative velocity of approach }} \\
&=\frac{v_{2}-v_{1}}{u_{2}-u_{1}}=\frac{\frac{v}{4}-0}{0-v} \quad \text { [from Eq. (i)] } \\
&=\frac{1}{4} \\
\therefore e &=0.25
\end{aligned}
\)

Hint

(d) For the collision the final position of both particles should be equal.
hence, \(\overrightarrow{r_{1}}+\overrightarrow{v_{1}} t=\overrightarrow{r_{2}}+\overrightarrow{v_{2}} t\)
\(
\begin{array}{l}
\text { or } \overrightarrow{r_{1}}-\overrightarrow{r_{2}}=\left(\overrightarrow{\mathrm{v}}_{2}-\overrightarrow{\mathrm{v}}_{1}\right) t \\
\text { or } \overrightarrow{\mathrm{r}_{1}}-\overrightarrow{\mathrm{r}_{2}}=\left(\overrightarrow{\mathrm{v}_{2}}-\overrightarrow{\mathrm{v}_{1}}\right) \frac{\left|\overrightarrow{\vec{r}_{1}}-\overrightarrow{\mathrm{r}_{2}}\right|}{\left|\overrightarrow{\mathrm{v}}_{2}-\overrightarrow{\mathrm{v}_{1}}\right|} \\
\text { or } \frac{\overrightarrow{\mathrm{r}_{1}}-\overrightarrow{\mathrm{r}_{2}}}{\left|\overrightarrow{\mathrm{r}_{1}}-\overrightarrow{\mathrm{r}_{2}}\right|}=\frac{\overrightarrow{\mathrm{v}_{2}}-\overrightarrow{\mathrm{v}_{1}}}{\left|\overrightarrow{\mathrm{v}_{2}}-\overrightarrow{\mathrm{v}_{1}}\right|}
\end{array}
\)

Hint

(a) According to law of conservation of kinetic energy, we have
\(
\begin{aligned}
& \frac{1}{2} M v^{2}+0=\frac{1}{2} M\left(\frac{v}{3}\right)^{2}+\frac{1}{2} M v_{2}^{2} \\
\Rightarrow & v^{2}=\frac{v^{2}}{9}+v_{2}^{2} \\
\Rightarrow & v^{2}-\frac{v^{2}}{9}=v_{2}^{2} \Rightarrow \frac{8 v^{2}}{9}
\end{aligned}
\)
Velocity of second block after collision
\(
v_{2}=\frac{2 \sqrt{2}}{3} v
\)

Hint

(a)
Suppose a ball rebounds with speed \(v_{\text {, }}\)
\(
\begin{aligned}
v &=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20} \\
&=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Energy of a ball just after rebound,
\(
E=\frac{1}{2} m v^{2}=200 \mathrm{~m}
\)
As, \(50 \%\) of energy loses in collision means just before collision energy is \(400 \mathrm{~m}\).

According to law of conservation of energy, we have
\(
\begin{aligned}
& \frac{1}{2} m v_{0}^{2}+m g h=400 \mathrm{~m} \\
\Rightarrow & \frac{1}{2} m v_{0}^{2}+m \times 10 \times 20=400 \mathrm{~m} \\
\Rightarrow & v_{0}=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(b) By conservation of linear momentum Magnitude of the momentum of heavier piece of mass \((2 \mathrm{~m})=\) Magnitude of the vector sum of momentum of each piece of mass (m)
\(
\begin{aligned}
&(2 \mathrm{~m}) v_{1}=\sqrt{(m v)^{2}+(m v)^{2}} \\
\Rightarrow & 2 m v_{1}=\sqrt{2} \mathrm{mv} \Rightarrow \mathrm{v}_{1}=\frac{\mathrm{v}}{\sqrt{2}}
\end{aligned}
\)
As two masses of each of mass m move perpendicular to each other.
Total Kinetic Energy generated
\(
\begin{array}{l}
=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}+\frac{1}{2}(2 m) v_{1}^{2} \\
=m v^{2}+\frac{m v^{2}}{2}=\frac{3}{2} m v^{2}
\end{array}
\)

Hint

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

Hint

(c) We know that for the rolling body, the initial kinetic energy is equal to potential energy of the spring
\(
\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}
\)
Potential energy of the spring \(=\frac{k \Delta x^{2}}{2}\)
Here, \(\Delta \mathrm{x}=\) Maximum compression of the spring
Kinetic energy of rolling body \(=\) Translational Kinetic energy +
Rotational kinetic energy
\(
\begin{array}{l}
=\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2} \mathrm{I} \omega^{2} \\
=\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2}\left(\frac{\mathrm{MR}}{2}\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2} \\
=\frac{3}{4} \mathrm{mV}^{2}
\end{array}
\)
Now
\(
\begin{array}{l}
\frac{\mathrm{K} \Delta \mathrm{x}^{2}}{2}=\frac{3}{4} \mathrm{mV}^{2} \\
\Delta \mathrm{x}=\sqrt{\frac{3 \mathrm{mV}^{2}}{2 \mathrm{~K}}} \\
=\sqrt{\frac{3 \times 3 \times 4^{2}}{2 \times 200}} \\
=0.6 \mathrm{~m}
\end{array}
\)

Hint

(d) According to law of conservation of linear momentum along \(x\)-axis, we get \(\Rightarrow m_{1}(0)+m_{2} v=m_{1} v^{\prime} \cos \theta\) \(\Rightarrow \quad \cos \theta=\frac{m_{2} v}{m_{1} v^{\prime}} \quad\)…(i)
Similarly, the law of conservation of linear momentum along the \(y\)-axis, we get
\(\Rightarrow m_{1}(0)+m_{2}(0)=m_{1} v^{\prime} \sin \theta+m_{2}\left(\frac{v}{2}\right)\) \(\Rightarrow \quad \sin \theta= -\frac{m_{2} v}{2 m_{1} v^{\prime}} \quad \ldots\) (ii) From equations, (i) and (ii),
\(\tan \theta=-\left(\frac{m_{2} v}{2 m v^{\prime}}\right) \times\left(\frac{m_{1} v^{\prime}}{m_{2} v}\right)=-\frac{1}{2}\)
\(\theta=\tan ^{-1}\left(\frac{-1}{2}\right)\) to the \(x\)-axis.

Hint

(a) As the two masses stick together after the collision, hence it is an inelastic collision. Therefore, only momentum is conserved.
\(\begin{aligned}
\therefore & m v \hat{i}+3 m(2 v) \hat{j}=(m + 3m) \vec{v} \\
\vec{v} &=\frac{v}{4} \hat{i}+\frac{6}{4} v \hat{j} \\
&=\frac{v}{4} \hat{i}+\frac{3}{2} v \hat{j}
\end{aligned}\)

Hint

(a) If two bodies collide head on with coefficient of restitution
\(
e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}
\)
From, the law of conservation of linear momentum
\(
\begin{array}{r}
m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} \\
\Rightarrow v_{1}=\left[\frac{m_{1}-e m_{2}}{m_{1}+m_{2}}\right] u_{1}+\left[\frac{(1+e) m_{2}}{m_{1}+m_{2}}\right] u_{2}
\end{array}
\)
Substituting \(u_{1}=2 \mathrm{~ms}^{-1}, u_{2}=0, m_{1}=m\)
and \(m_{2}=2 m,  e=0.5\)
we get, \(\quad v_{1}=\left[\frac{m-m}{m+2 m}\right] \times 2\)
\(\Rightarrow \quad v_{1}=0\)
Similarly,
\(
\begin{aligned}
v_{2} &=\left[\frac{(1+e) m_{1}}{m_{1}+m_{2}}\right] u_{1}+\left[\frac{m_{2}-e m_{1}}{m_{1}+m_{2}}\right] u_{2} \\
&=\left[\frac{1.5 \times m}{3 m}\right] \times 2=1 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

(a) Let the initial velocity of the shell be \(v\), then by the conservation of momentum \(m v=\mathrm{M} v^{\prime}\) where \(v^{\prime}=\) velocity of gun.
\(
\therefore \quad \mathrm{v}^{\prime}=\left(\frac{\mathrm{m}}{\mathrm{M}}\right) \mathrm{v}
\)
Now, total K.E. \(=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \frac{1}{2} \mathrm{Mv}^{\prime 2}\)
\(
\begin{array}{l}
=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{M}\left(\frac{\mathrm{m}}{\mathrm{M}}\right)^{2} \mathrm{v}^{2} \\
=\frac{1}{2} \mathrm{mv}^{2}\left[1+\frac{\mathrm{m}}{\mathrm{M}}\right] \\
=\left(\frac{1}{2} \times 0.2\right)\left(1+\frac{0.2}{4}\right) v^{2}=(0.1 \times 1.05) \mathrm{v}^{2}
\end{array}
\)
But total K.E. \(=1.05 \mathrm{~kJ}=1.05 \times 10^{3} \mathrm{~J}\) \(\therefore 1.05 \times 10^{3}=0.1 \times 1.05 \times v^{2}\)
\(
\begin{array}{l}
\Rightarrow v^{2}=\frac{1.05 \times 10^{3}}{0.1 \times 1.05}=10^{4} \\
\therefore v=10^{2}=100 \mathrm{~ms}^{-1}
\end{array}
\)

Hint

(b) From conservation of linear momentum
\(
\begin{array}{l}
m_{1} v_{1}+m_{2} v_{2}=0 \\
v_{2}=\left(\frac{-m_{1}}{m_{2}}\right) v_{1}=\left(\frac{-18}{12}\right) 6=-9 \mathrm{~ms}^{-1} \\
\text { K.E. }=\frac{1}{2} m_{2} v_{2}^{2}=\frac{1}{2} \times 12 \times 9^{2}=486 \mathrm{~J}
\end{array}
\)

Hint

(b) Speed of moving mass \(\mathrm{v}=1.5 \mathrm{~m} / \mathrm{s}\)
Given : \(k=50 \mathrm{~N} / \mathrm{m}\) and \(\mathrm{m}=0.5 \mathrm{~kg}\)
The kinetic energy of the moving mass will be converted into potential energy in the spring when the spring is at maximum compression \(\mathrm{x}_{0}\).
\(
\therefore \frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{kx}_{\mathrm{o}}^{2}
\)
or \(x_{0}=\sqrt{\frac{m}{k}} v\)
\(
\Rightarrow \mathrm{x}_{\mathrm{o}}=\sqrt{\frac{0.5}{50}} \times 1.5=0.15 \mathrm{~m}
\)

Hint

(c) From the conservation of linear momentum,
Initial momentum \(p_{\text {initial }}=\) Final
momentum \(p_{\text {final }}\)
\(
0=m_{1} v_{1}-m_{2} v_{2}
\)
or \(\quad m_{1} v_{1}=m_{2} v_{2}\)
or \(\quad \frac{v_{1}}{v_{2}}=\frac{m_{2}}{m_{1}}\)
Thus, ratio of kinetic energies,
\(
\frac{K_{1}}{K_{2}}=\frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{2}^{2}}=\frac{m_{1}}{m_{2}} \times\left(\frac{m_{2}}{m_{1}}\right)^{2}=\frac{m_{2}}{m_{1}}
\)

Hint

(b) Speed of bomb after 5 second, \(v=u-g t=100-10 \times 5=50 \mathrm{~m} / \mathrm{s}\)
Momentum of \(400 \mathrm{~g}\) fragment \(=\frac{400}{1000} \times(-25)\) [downward]
Momentum of \(600 \mathrm{~g}\) fragment \(=\frac{600}{1000} v\)
Momentum of bomb \(=1 \times 50=50\)
From conservation of momentum
Total momentum before splitting \(=\) total momentum after splitting.
\(\Rightarrow 50=-\frac{400}{1000} \times 25+\frac{600}{1000} v\)
\(\Rightarrow v=100 \mathrm{~m} / \mathrm{s} \quad\) [upward]

Hint

(d) Given,
\(u_{1}=3 \mathrm{~m} / \mathrm{s}, u_{2}=-5 \mathrm{~m} / \mathrm{s}, m_{1}=m_{2}=m A c c\)
According to principle of conservation of linear momentum,
\(
\begin{aligned}
m_{1} u_{1}+m_{2} u_{2} &=m_{1} v_{1}+m_{2} v_{2} \\
m \times 3-m \times 5 &=m v_{1}+m v_{2} \\
\text { or } \quad v_{1}+v_{2} &=-2
\end{aligned}
\)
In an elastic collision,
\(
\begin{aligned}
e &=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \\
\Rightarrow \quad v_{2}-v_{1} &=e\left(u_{1}-u_{2}\right)
\end{aligned}
\)
\(
\Rightarrow v_{2}-v_{1}=(1)(3+5) \quad(\because e=1)
\)
\(
\Rightarrow v_{1}-v_{2}=-8
\)
Adding Eqs. (i) and (ii), we obtain
\(2 v_{1}=-10\)
\(\Rightarrow \quad v_{1}=-5 \mathrm{~m} / \mathrm{s}\)
From Eq. (i),
\(
v_{2}=-2-v_{1}=-2+5=3 \mathrm{~m} / \mathrm{s}
\)
Thus, \(v_{1}=-5 \mathrm{~m} / \mathrm{s}, v_{2}=+3 \mathrm{~m} / \mathrm{s}\)
If two bodies collide elastically, then their velocities are interchanged. Since, it is an elastic collision hence, velocities after collision will be \(-5 \mathrm{~m} / \mathrm{s}\) and \(3 \mathrm{~m} / \mathrm{s}\).

Hint

(a) Initial momentum \(=m v\)
Final momentum \(=m(-v)\)
change in momentum
\(=m v-m(-v)=2 m v\)

Hint

(c) Initial momentum = Final momentum
\(
\therefore \quad m_{1} v_{1}+m_{2} v_{2}=\left(m_{1}+m_{2}\right) v
\)
Given, \(v_{1}=36 \mathrm{~km} / \mathrm{h}\)
\(
\begin{aligned}
&=36 \times \frac{5}{18}=10 \mathrm{~m} / \mathrm{s}, \\
v_{2} &=0 \\
m_{1} &=2 \mathrm{~kg}, m_{2}=3 \mathrm{~kg} \\
\therefore \quad v &=\frac{m_{1} v_{1}+m_{2} v_{2}}{m_{1}+m_{2}} \\
&=\frac{2 \times 10+3 \times 0}{2+3} \\
\text { or } \quad v &=\frac{20}{5}=4 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Loss in kinetic energy
\(
\begin{array}{l}
=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}-\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2} \\
=\frac{1}{2} \times 2 \times(10)^{2}+0-\frac{1}{2}(2+3) \times(4)^{2} \\
=100-40=60 \mathrm{~J}
\end{array}
\)

Hint

(a) Let \(v\) be the velocity of combined mass after the collision.
Applying law of conservation of linear momentum

Initial momentum = Final momentum
\(
\begin{aligned}
m \times 3+2 m \times 0 &=(m+2 m) v \\
\text { or } \quad 3 m &=3 m v \text { or } v=1 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

(c) When the shell explodes in mid-air its chemical energy is partly converted into mechanical energy, hence K.E. increases.

Hint

(b) When the identical balls collide head-on, their velocities are exchanged.

Hint

(a) Momentum of Ist part \(=\mathrm{m} \times 21=21 \mathrm{~m}\)
Momentum of 2 nd part \(=m \times 21=21 \mathrm{~m}\)
Resultant momentum, \(=\sqrt{(\mathrm{m} 21)^{2}+(\mathrm{m} 21)^{2}}\)
Resultant momentum \(=\) momentum of \(3 \mathrm{rd}\) part
\(\Rightarrow \quad 21 \sqrt{2} \mathrm{~m}=3 \mathrm{mv}\)
\(\Rightarrow \quad v=7 \sqrt{2} \mathrm{~m} / \mathrm{sec}\)
Note: Masses of the pieces are \(1,1,3 \mathrm{~kg}\). Hence
\((1 \times 21)^{2}+(1 \times 21)^{2}=(3 \times \mathrm{v})^{2}\)
That is, \(v=7 \sqrt{2} \mathrm{~m} / \mathrm{s}\)

Hint

(d) \(e=\left|v_{1}-v_{2}\right| /\left|u_{1}-u_{2}\right|\) which is 1 for a perfectly elastic collision.

Hint

(d) Let the particle at height \(S\) from the surface of the Earth is as shown below. Given, at height \(x\), kinetic energy
\(=3\) (potential energy)
\(\Rightarrow \quad \mathrm{KE}=3 \mathrm{mgx}\)
At height \(S\), total energy, \(T E=P E+K E\)
\(
\mathrm{TE}=m g S+0=m g S
\)
At height \(x\), total energy,
\(\begin{array}{r}
\mathrm{TE}=\mathrm{PE}+\mathrm{KE} \\
\mathrm{TE}=\mathrm{mgx}+3 \mathrm{mgx} \\
\mathrm{mgS}=4 \mathrm{mgx} \\
x=\frac{\mathrm{S}}{4} \dots (i)
\end{array}\)
Now, we shall determine the speed of the particle at this height.
As, \(K E=3 \times m g x\)
\(\begin{aligned} K E &=3 \times m g \frac{S}{4} \quad \text { [from Eq. (i)] } \\ \frac{1}{2} m v^{2} &=\frac{3}{4} m g S \\ \Rightarrow \quad v &=\sqrt{\frac{3}{2} g S} \end{aligned}\)

Hint

(b) Given, \(E=10^{-20} \mathrm{~J}\)
In terms of eV, we get
\(
E=\frac{10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=0.06 \mathrm{eV}
\)

Hint

(d) Position vectors of the particles are \(\mathbf{r}_{1}=-2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}\) and \(\mathbf{r}_{2}=4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
\(\therefore\) Displacement of the particle,
\(\Delta \mathbf{s}=\mathbf{r}_{2}-\mathbf{r}_{1}\)
\(
=4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}-(-2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}})=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\)
Force on the particle, \(\mathbf{F}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \mathrm{N}\)
\(\therefore\) Work done, \(W=\mathbf{F} \cdot \Delta \mathbf{s}\)
\(
=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=8-3=5 \mathrm{~J}
\)

Hint

(c) Given, \(K_{p}>K_{0}\)
In case (a), the elongation is same
i.e. \(x_{1}=x_{2}=x\)
So, \(W_{p}=\frac{1}{2} K_{p} x^{2}\) and \(W_{0}=\frac{1}{2} K_{0} x^{2}\)
\(\therefore \quad \frac{W_{p}}{W_{0}}=\frac{K_{p}}{K_{0}}>1 \Rightarrow W_{p}>W_{0}\)
In case (b), the spring force is same i.e. \(\quad F_{1}=F_{2}=F\)
\(\begin{array}{l}
\text { So, } \quad x_{1}=\frac{F}{K_{p}}, x_{2}=\frac{F}{K_{0}}\\
\therefore \quad W_{p}=\frac{1}{2} K_{p} x_{1}^{2}=\frac{1}{2} K_{p} \frac{F^{2}}{K_{p}^{2}}=\frac{1}{2} \frac{F^{2}}{K_{p}}\\
\text { and } W_{0}=\frac{1}{2} K_{0} x_{2}^{2}=\frac{1}{2} K_{0} \cdot \frac{F^{2}}{K_{0}^{2}}=\frac{1}{2} \frac{F^{2}}{K_{0}}\\
\therefore \quad \frac{W_{p}}{W_{0}}=\frac{K_{0}}{K_{p}}<1\\
\Rightarrow \quad W_{p}<W_{0}
\end{array}\)

Hint

(c) Conserving angular momentum
\(
\begin{aligned}
L_{i} &=L_{f} \\
\Rightarrow \quad m v_{0} R_{0} &=m v^{\prime}\left(\frac{R_{0}}{2}\right) \Rightarrow v^{\prime}=2 v_{0}
\end{aligned}
\)
So, final kinetic energy of the particle is
\(
\begin{aligned}
K_{f} &=\frac{1}{2} m v^{\prime 2}=\frac{1}{2} m\left(2 v_{0}\right)^{2} \\
&=4 \frac{1}{2} m v_{0}^{2}=2 m v_{0}^{2}
\end{aligned}
\)

Hint

(b) Given, the potential energy of a particle in a force field, \(U=\frac{A}{r^{2}}-\frac{B}{r^{1}}\)
For stable equilibrium, \(F=-\frac{d U}{d r}=0\)
\(
=\frac{d U}{d r}=-2 A r^{-3}+B r^{-2}
\)
\(
0=-\frac{2 A}{r^{3}}+\frac{B}{r^{2}} \quad\left(\text { As } \frac{-d U}{d r}=0\right)
\)
or \(\frac{2 A}{r}=B\)
The distance of the particle from the centre of the field
\(
r=\frac{2 A}{B}
\)

Hint

(b) Let \(x\) be the extension in the spring.
Applying conservation of energy
\(
\begin{aligned}
\quad M g x-\frac{1}{2} k x^{2} &=0-0 \\
\Rightarrow \quad x &=\frac{2 M g}{k}
\end{aligned}
\)

Hint

(a)
At the highest point
\(
\begin{aligned}
v_{x} &=u \cos \theta \\
v_{y} &=0 \\
K_{H} &=\frac{1}{2} m v_{x}^{2} \\
K_{H} &=\frac{1}{2} m u^{2} \cos ^{2} \theta
\end{aligned}
\)
Initial kinetic energy is
\(
K=\frac{1}{2} m u^{2}
\)
From Eqs. (i) and (ii), we get
\(
\begin{aligned}
K_{H} &=K \cos ^{2} \theta \\
&=K \cos ^{2} 45^{\circ} \\
&=K \times\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{K}{2}
\end{aligned}
\)

Hint

(a) From energy conservation
\(
\frac{1}{2} m v_{\max }^{2}=m g\left(H_{2}-H_{1}\right)
\)
Here, \(H_{1}=\) minimum height of swing from the earth’s surface
\(
\begin{aligned}
&=0.75 \mathrm{~m} \\
H_{2} &=\text { maximum height of swing from } \\
&=2 \mathrm{~m} \quad \text { earth’s surface }
\end{aligned}
\)
\(
\therefore \quad \frac{1}{2} m v_{\max }^{2}=m g(2-0.75)
\)
or \(v_{\max }=\sqrt{2 \times 10 \times 1.25}\)
\(
=\sqrt{25}=5 \mathrm{~m} / \mathrm{s}
\)

Hint

(b) Given, the flow rate of the water,
\(
\frac{m}{t}=15 \mathrm{~kg} / \mathrm{s}
\)
The height of the water fall, \(h=60 \mathrm{~m}\)
Loss due to frictional force \(=10 \%\)
The power used in the turbine
\(
=(100-10) \%=90 \%
\)
The acceleration due to gravity,
\(
g=10 \mathrm{~m} / \mathrm{s}^{2}
\)
We know that, power generated by the turbine
\(=\) change in potential energy \(=0.90 \frac{\mathrm{mgh}}{\mathrm{t}}\)
\(
=0.90 \times 15 \times 10 \times 60=8100 \mathrm{~W}=8.1 \mathrm{~kW}
\)

Hint

(c) Velocity of point mass in vertical circle at lowest point, \(V_{1}=\sqrt{7 g r}\)
\(
\therefore \quad V_{1}=\sqrt{7 g r}>\sqrt{5 g r}
\)
Hence, point mass will have completed the vertical circular path.
We know that,
\(\begin{aligned}
& T_{\text {bottom }}-m g=\frac{m v^{2}}{r}=\frac{m}{r}(\sqrt{7 g r})^{2} \\
\Rightarrow \quad & T_{\text {bottom }}-m g=7 \mathrm{mg} \\
\Rightarrow \quad \quad T_{\text {bottom }} &=8 \mathrm{mg}
\end{aligned}\)

Hint

(b) Let the mass m which is attached to a thin wire and is whirled in a vertical circle is shown in the figure below.
The tension in the string at any point \(P\) be T.

According to Newton’s law of motion, In equilibrium, net force towards the centre = centripetal force \(\Rightarrow T-m g \cos \theta=\frac{m v^{2}}{l}\)
Here, \(l=\) length of wire and \(v=\) linear velocity of the particle whirling in a circle.
\(
\Rightarrow T=m g \cos \theta+\frac{m v^{2}}{l}
\)
At \(A, \theta=0^{\circ} \Rightarrow T_{A}=m g+\frac{m v^{2}}{l}\)
At \(B, \theta=90^{\circ} \Rightarrow T_{B}=\frac{m v^{2}}{l}\)
At C, \(\theta=180^{\circ} \Rightarrow T_{C}=-m g+\frac{m v^{2}}{l}\)
At \(D_{1} \theta=90^{\circ} \Rightarrow T_{D}=T_{B}=\frac{m v^{2}}{l}\)
So, from the above analysis, it can be concluded that the tension is maximum at \(A\) i.e. the lowest point of circle, So chance of breaking is maximum.

Hint

(d) If a body is moving on a frictionless surface, then its total mechanical energy remains conserved. According to the conservation of mechanical energy,
\(
\begin{aligned}
&(T E)_{\text {inital }}=(T E)_{\text {final }} \\
\Rightarrow \quad(K E)_{i}+(P E)_{i} &=(K E)_{f}+(\mathrm{PE})_{f} \\
& 0+m g h=\frac{1}{2} m v_{A}^{2}+0 \\
\Rightarrow \quad & g h=\frac{v_{A}^{2}}{2} \text { or } h=\frac{v_{A}^{2}}{2 g} \dots (i)
\end{aligned}
\)
In order to complete the vertical circle, the velocity of the body at point \(A\)
should be
\(
v_{A}=v_{\text {min }}=\sqrt{5 g R}
\)
where, \(R\) is the radius of the body.
Here, \(\quad R=\frac{A B}{2}=\frac{D}{2}\)
\(
\Rightarrow \quad v_{\min }=v_{A}=\sqrt{\frac{5}{2} g D}
\)
Substituting the value of \(v_{A}\) in Eq. (i), we get
\(
\begin{aligned}
h &=\frac{\left(\sqrt{\left(\frac{5}{2} g D\right)}\right)^{2}}{2 g} \\
&=\frac{5 g D}{2 \times 2 g}=\frac{5}{4} D
\end{aligned}
\)

Hint

(c) According to the question, we have
Let the tension at point \(A\) be \(T_{A}\). So, from Newton’s second law
\(
T_{A}-m g=\frac{m v_{c}^{2}}{R}
\)
Energy at point \(A=\frac{1}{2} m v_{0}^{2} \dots (i)\)
Energy at point \(C\) is
\(
\frac{1}{2} m v_{c}^{2}+m g \times 2 R \dots (ii)
\)
Applying Newton’s 2nd law at point \(C\)
\(
T_{c}+m g=\frac{m v_{c}^{2}}{R}
\)
To complete the \(\operatorname{loop} T_{c} \geq 0\)
So, \(m g=\frac{m v_{c}^{2}}{R}\)
\(\Rightarrow v_{c}=\sqrt{g R} \dots (iii)\)
From Eqs. (i) and (ii) by conservation of energy
\(
\begin{aligned}
& \Rightarrow \quad \frac{1}{2} m v_{0}^{2} =\frac{1}{2} m v_{c}^{2}+2 m g R \\
\Rightarrow \quad & \frac{1}{2} m v_{0}^{2} =\frac{1}{2} m g R+2 m g R \\
\Rightarrow & v_{0}^{2} =g R+4 g R \\
\Rightarrow & v_{0} =\sqrt{5 g R}
\end{aligned}
\)

Hint

(b) When string makes an angle \(\theta\) with the vertical in a vertical circle, then balancing the force we get
\(
T-m g \cos \theta=\frac{m v^{2}}{l}
\)
or \(T=m g \cos \theta+\frac{m v^{2}}{l}\)
Tension is maximum when \(\cos \theta=+1\)
i.e. \(\theta=0\)
Thus, \(\theta\) is zero at lowest point B. At this point tension is maximum. So, string will break at point \(B\).

Hint

\(
\begin{aligned}
& \text { Energy }=\text { Power } \times \text { time } \\
& \begin{aligned}
E & =100 \times 10^3 \times 3600 \\
& =36 \times 10^7 \mathrm{~J}
\end{aligned}
\end{aligned}
\)

Hint

Amount of energy required \(=[S \times \Delta A] \times 2\)
\(
\begin{aligned}
\Rightarrow \text { Energy required } & =\left[0.03 \times 4 \times \pi \times 4 \times 10^{-4}\right] \times 2 \\
& =3.015 \times 10^{-4} \mathrm{~J}
\end{aligned}
\)

Hint

Potential energy stored in spring \(U=\frac{1}{2} K x^2\)
\(
\begin{aligned}
& U=\frac{1}{2} K(2)^2 \text { where } x=2 \mathrm{~cm} \\
& U=\frac{1}{2}(K) \cdot(4) \\
& U=2 K \dots(i) \\
& U^{\prime}=\frac{1}{2} K(8)^2 \\
& U^{\prime}=\frac{1}{2} K \times 64=32 K \dots(ii)
\end{aligned}
\)
On dividing (i) by (ii)
\(
\begin{aligned}
& \frac{U}{U}=\frac{2 K}{32 K}-\frac{1}{16} \\
& U=16 U
\end{aligned}
\)