Work, Power & Energy

$\vec {v} _ {0 1} = (\sqrt {3} \hat {i} + \hat {j}) m / s, \vec {v} _ {0 2} = \vec {0}$
$m _ {1} = 2 m _ {2}$
$\mathrm{Aftercollision}, \vec {v} _ {1} = (\hat {i} + \sqrt {3} \hat {j}) m / s$
$vec {v} _ {2} = ?$
Applying conservation of linear momentum,
$m _ {1} \vec {v} _ {0 1} + m _ {2} \vec {v} _ {0 2} = m _ {1} \vec {v} _ {1} + m _ {2} \vec {v} _ {2}$
$2 m _ {2} (\sqrt {3} \hat {i} + \hat {j}) + 0 = 2 m _ {2} (\hat {i} + \sqrt {3} \hat {j}) + m _ {2} \vec {v} _ {2}$
$\vec {v} _ {2} = 2 (\sqrt {3} \hat {i} + \hat {j}) - 2 (\hat {i} + \sqrt {3} \hat {j}) = 2 (\sqrt {3} \hat {i} - \hat {j}) + 2 (\hat {i} - \sqrt {3} \hat {j})$
$\vec {v} _ {2} = 2 (\sqrt {3} - 1) (\hat {i} - \hat {j})$
For angle between $\vec{v}_1$ and $\vec{v}_2$ ,
$\cos \theta = \frac {\vec {v} _ {1} . \vec {v} _ {2}}{| \vec {v} _ {1} | | \vec {v} _ {2} |} = \frac {2 (\sqrt {3} - 1) (1 - \sqrt {3})}{2 \times 2 \sqrt {2} (\sqrt {3} - 1)}$
$\cos \theta = \frac {1 - \sqrt {3}}{2 \sqrt {2}} \Rightarrow \theta = 1 0 5 ^ {\circ}.$

$P.t = \frac{1}{2} mv^2 \Rightarrow v = \left(\sqrt{\frac{2P}{m}}\right)t^{1/2}$ $s = \int_{0}^{t} v dt = \sqrt{\frac{2P}{m}} \int_{0}^{t} t^{1/2} dt = \sqrt{\frac{2P}{m}} \cdot \frac{t^{3/2}}{3/2}; s = \sqrt{\frac{8P}{9m}}. t^{3/2}$ .

$W = \int_{\bar{r}_i}^{\bar{r}_f}\vec{F}.d\bar{r}$ $W = \int_{1}^{0} - x dx + \int_{0}^{1} y dy \Rightarrow W = \frac{-x^2}{2}\bigg|_1^0 +\frac{y^2}{2}\bigg|_0^1$ $= -\left(\frac{0^2}{2} -\frac{1^2}{2}\right) + \left(\frac{1^2}{2} -\frac{0^2}{2}\right)\Rightarrow W = 1J.$

$x = 3t^{2} + 5$
$v = \frac {d x}{d t} = 6 t$
$\text { at } t = 0 \quad v = 0$
$t = 5 \sec v = 3 0 m / s$
$\mathrm{W.D.} = \Delta K E$
$W. D. = \frac {1}{2} m v ^ {2} - 0 = \frac {1}{2} (2) (3 0) ^ {2} = 9 0 0 J.$

$k_{i} = \frac{1}{2} mv_{0}^{2}$
From linear momentum conservation
$m v _ {0} = (2 m + M) v _ {f} \Rightarrow v _ {f} = \frac {m v _ {0}}{2 m + M}$
$\frac {k _ {i}}{k _ {f}} = \frac {\frac {1}{2} m v _ {0} ^ {2}}{\frac {1}{2} (2 m + M) \left(\frac {m v _ {0}}{2 m + M}\right) ^ {2}} = 6$
$\Rightarrow \frac {2 m + M}{m} = 6 \Rightarrow \frac {M}{m} = 4.$

Time taken for the particles to collide,
$a _ {r e l} = 0$
$v _ {r e l} = 1 0 0 m / s$
$t = \frac {d}{V _ {r e l}} = \frac {1 0 0}{1 0 0} = 1 \sec$

$s _ {b} = 1 0 0 \times 1 - \frac {1}{2} \times 1 0 \times 1 = 9 5 m$
Speed of wood just before collision = gt = 10 m/s and speed of bullet just before collision v-gt = 100 - 10 = 90 m/s

Now, conservation of linear momentum just before and after the collision $P_{i} = P_{f}$
$(0.0 2) \times 9 0 -.0 3 \times 1 0 = (.0 5) v$
$1.5 = .0 5 v \Rightarrow v = 3 0 m / s$
Max. height reached by body $h=\frac{v^{2}}{2g}$
$h = \frac {3 0 \times 3 0}{2 \times 1 0} = 4 5 m$
∴ Height above tower = 45-5=40 m.

$N - mg = \frac{mg}{2} \Rightarrow N = \frac{3mg}{2}$
$S = u t + \frac {1}{2} \left(\frac {g}{2}\right) t ^ {2} = \frac {g t ^ {2}}{4} [ \because u = 0 ]$
Now, work done $W = \vec{N}\vec{S} = \left( \frac{3mg}{2} \right) \left( \frac{1}{4} g t^{2} \right) = \frac{3mg^{2}t^{2}}{8}$

Work done = $\vec{F}.\vec{d} = (3\hat{i} - 12\hat{j}).(4\hat{i}) = 12 J.$
work energy theorem
$W _ {n e t} = \Delta K. E., 1 2 = K _ {f} - 3 \Rightarrow K _ {f} = 1 5 J.$

Velocity of 1 kg block just before it collides with $3 \, kg \, block = \sqrt{2gh} = \sqrt{2000} \, m/s$
Applying momentum conversation just before and just after collision.
$1 \times \sqrt {2 0 0 0} = 4 v \Rightarrow v = \frac {\sqrt {2 0 0 0}}{4} m / s = 5 \sqrt {5} m / s$
initial compression of spring
$1.2 5 \times 1 0 ^ {6} x _ {0} = 3 0 \Rightarrow x _ {0} \approx 0$
applying work energy theorem,
$W _ {g} + W _ {s p} = \Delta K E$

$\Rightarrow 4 0 \times x + \frac {1}{2} \times 1. 2 5 \times 1 0 ^ {6} (0 ^ {2} - x ^ {2}) = 0 - \frac {1}{2} \times 4 \times v ^ {2}$
solving $x\approx 2cm$
Aliter : assuming $5\sqrt{5}$ as maximum velocity
So $V_{\mathrm{max}} = \omega x$
$5 \sqrt {5} = \sqrt {\frac {k}{m}} \times x \Rightarrow 5 \sqrt {5} = \sqrt {\frac {1 . 2 5 \times 1 0 ^ {6}}{4}} \times x$
$x = 2 c m.$

Conserving momentum: $mu = -mv_{1} + MV_{2}$
Collision of nuclear particle to be elastic
$e = \frac {v _ {2} + v _ {1}}{u} = 1$
$\therefore v _ {2} = u - v _ {1}$
$v _ {1} = 0.6 u$
$1.6 m = 0.4 M \Rightarrow M = 4 m.$

Applying linear momentum conservation
$m _ {1} v _ {1} \hat {i} + m _ {2} v _ {2} \hat {i} = m _ {1} v _ {3} \hat {i} + m _ {2} v _ {4} \hat {i}$
$m _ {1} v _ {1} + 0. 5 m _ {1} v _ {2} = m _ {1} (0.5 v _ {1}) + 0. 5 m _ {1} v _ {4}$
$0.5 m _ {1} v _ {1} = 0.5 m _ {1} (v _ {4} - v _ {2}) \Rightarrow v _ {1} = v _ {4} - v _ {2}.$

Mass of the hanging part = $\frac{M}{n}$ $h_{COM} = \frac{L}{2n}$

work done $W = mgh_{COM} = \left(\frac{M}{n}\right)g\left(\frac{L}{2n}\right) = \frac{MgL}{2n^2}$

Linear momentum conservation
$m 2 v + 2 m v = m \times 0 + m \frac {v ^ {\prime}}{\sqrt {2}} \times 2 \Rightarrow v ^ {\prime} = 2 \sqrt {2} v.$

Applying Linear momentum conservation
$m v = (m + M) v _ {c} \Rightarrow v _ {c} = \frac {v}{5}$
applying work energy theorem
$- m g h = \frac {1}{2} (m + M) v _ {c} ^ {2} - \frac {1}{2} m v ^ {2}$
solving we get, $h = \frac{2v^2}{5g}$

$M \times 10 \cos 30^{\circ} + 2M \times 5 \cos 45^{\circ}$
$= 2 M \times v _ {1} \cos 3 0 ^ {\circ} + M v _ {2} \cos 4 5 ^ {\circ}$
$5 \sqrt {3} + 5 \sqrt {2} = 2 v _ {1} \frac {\sqrt {3}}{2} + \frac {v _ {2}}{\sqrt {2}}$
$1 0 \times M \sin 3 0 ^ {\circ} - 2 M \times 5 \sin 4 5 ^ {\circ}$
$= M v _ {2} \sin 4 5 ^ {\circ} - 2 M v _ {1} \sin 3 0 ^ {\circ}$
$5 - 5 \sqrt {2} = \frac {v _ {2}}{\sqrt {2}} - v _ {1}$
Solving $v_{1} = \frac{17.5}{2.7} \simeq 6.5m / s$ , $v_{2} \approx 6.3m / s$ .

By conservation of linear momentum
$2 v _ {0} = 2 \left(\frac {v _ {0}}{4}\right) + m v \Rightarrow 2 v _ {0} = \frac {v _ {0}}{2} + m v$
$\Rightarrow \frac {3 v _ {0}}{2} = m v$...(i)
Since collision is elastic →
$V _ {s e p a r a t i o n} = v _ {a p p r o c h}$
$\Rightarrow v - \frac {v _ {0}}{4} = v _ {0} \Rightarrow \frac {5 v _ {0}}{4} = v$...(ii)
equating (ii) and (i)
$\frac {3 v _ {0}}{2} = m \left(\frac {5 v _ {0}}{4}\right) \Rightarrow m = \frac {6}{5} = 1.2 k g$

According to work energy theorem
Work done by force on the particle = change in KE
Work done = Area under F-x graph
= ∫ F.dx
= 2 × 2 + (2 + 3) × 1/2
$W = KE_final - KE_initial = 6.5$
$KE_initial = 0$
$KE_final = 6.5J.$

(bc)

b) speed of particle after collision $= 2V + v_{0}$ change in speed $= (2V + v_{0}) - v_{0}$ after each collision $= 2V$
No. of collision per unit time (frequency) = $\frac{v}{2L}$ change in speed in dt time = 2V × number of collision in dt time
$\Rightarrow d v = 2 V \left(\frac {v}{2 L}\right) \cdot \frac {d L}{V}$
$d v = \frac {v d L}{L}$
Now, $dv = -\frac{vdL}{L}$ {as $L$ decrease}
$\int_ {V _ {0}} ^ {V} \frac {d v}{v} = - \int_ {L _ {0}} ^ {L _ {0} / 2} \frac {d L}{L}$
$\Rightarrow [ 1 n v ] _ {V _ {0}} ^ {V} = [ 1 n L ] _ {L _ {0}} ^ {L _ {0 / 2}}$
$\Rightarrow v = 2 v _ {0}$
$\Rightarrow K E _ {L _ {0 / 2}} = \frac {1}{2} m (2 v _ {0}) ^ {2}$
and $kE_{L_0} = \frac{1}{2} mv_0^2, \frac{kE_{L_0 / 2}}{kE_{L_0}} = 4$ .

To complete a vertical circle, speed at A should be
$v _ {A} = \sqrt {5 g R}$
using energy conservation
$m g h = \frac {1}{2} m v _ {A} ^ {2}$
$h = \frac {1}{2} \frac {v _ {A} ^ {2}}{g} = \frac {1}{2} \frac {5 g}{g} \frac {D}{2} \quad \left(R = \frac {D}{2}\right)$
$h = \frac {5 D}{4}$

$F = -\frac{\partial u}{\partial r} = \frac{K}{r^3}$ Since it is performing circular motion $F = \frac{mv^2}{r} = \frac{K}{r^3}$ $mv^2 = \frac{K}{r^2}$ $\Rightarrow K.E. = \frac{1}{2}mv^2 = \frac{K}{2r^2}$ Total energy = P.E. + K.E. $= -\frac{K}{2r^2} + \frac{K}{2r^2} = Zero$

initial

$\frac {1}{2} m v _ {1} ^ {2} + \frac {1}{2} m v _ {2} ^ {2} = \frac {3}{2} \left(\frac {1}{2} m v _ {0} ^ {2}\right)$
$\Rightarrow v _ {1} ^ {2} + v _ {2} ^ {2} = \frac {3}{2} v _ {0} ^ {2}$...(i)
from momentum conservation
$m v _ {0} = m (v _ {1} + v _ {2})$...(ii)
$(v _ {1} + v _ {2}) ^ {2} = v _ {0} ^ {2} \Rightarrow v _ {1} ^ {2} + v _ {2} ^ {2} + 2 v _ {1} v _ {2} = v _ {0} ^ {2}$
$2 v _ {1} v _ {2} = - \frac {v _ {0} ^ {2}}{2}$
$(v _ {1} - v _ {2}) ^ {2} = v _ {1} ^ {2} + v _ {2} ^ {2} - 2 v _ {1} v _ {2} = \frac {3}{2} v _ {0} ^ {2} + \frac {v _ {0} ^ {2}}{2}$
$v _ {1} - v _ {2} = \sqrt {2} v _ {0}$

Let initial speed of neutron is $v_{0}$ and kinetic energy is $K$ . $1^{\mathrm{st}}$ collision:

By momentum conservation
$m v _ {0} = m v _ {1} + 2 m v _ {2} \Rightarrow v _ {1} + 2 v _ {2} = v _ {0}$
$\text {by} e = 1 \quad v _ {2} - v _ {1} = v _ {0}$
$\Rightarrow \quad v _ {2} = \frac {2 v _ {0}}{3}; v _ {1} = - \frac {v _ {0}}{3}$

$\Rightarrow P _ {d} = \frac {8}{9} \approx . 8 9$
$2^{\mathrm{nd}}$ collision:

by momentum conservation
$m v _ {0} = m v _ {1} + 1 2 m v _ {2}$
$\Rightarrow v _ {1} + 1 2 v _ {2} = v _ {0}$
$\text {by} e = 1$
$v _ {2} - v _ {1} = v _ {0}$
$v _ {2} = \frac {2 v _ {0}}{1 3}; v _ {1} = \frac {- 1 1 v _ {0}}{1 3}$
Now fraction loss of energy

(abd) $\frac{dk}{dt} = \gamma t$ as $k = \frac{1}{2} mv^2$ $\therefore \frac{dk}{dt} = mv\frac{dv}{dt} = \gamma t$
$\therefore m \int_ {0} ^ {v} v d v = \gamma \int_ {0} ^ {t} t d t$
$\frac {m v ^ {2}}{2} = \frac {\gamma t ^ {2}}{2} \Rightarrow v = \sqrt {\frac {\gamma}{m}} t$
... (i)
$a = \frac {d v}{d t} = \sqrt {\frac {\gamma}{m}} = \text { constant }$
Since $F = ma$
$\therefore F = m \sqrt {\frac {\gamma}{m}} = \sqrt {\gamma m} = \mathrm{constant}$

Initially

Finally
Momentum Conservation
$m v = m v _ {A} + \frac {m}{2} v _ {B}$...(i)
$e = \frac {v _ {B} - v _ {A}}{v} = 1$...(ii)
$2 (v _ {B} - v _ {A}) = 2 v _ {A} + v _ {B}$
$2 v _ {B} - v _ {B} = 4 v _ {A}$
$v _ {B} = 4 v _ {A}$
$\frac {\lambda_ {A}}{\lambda_ {B}} = \frac {m _ {B} v _ {B}}{m _ {A} v _ {A}} = \frac {m / 2}{m} \times 4 = 2$