Work Power and Energy

Elevator moving with constant speed hence
Net force = T = 6800 + 9200 + 6000
T = 22000 N
Required power for motor
= T.V = 22000 × 3 = 66000 W .

Particles will collide after time $t_{0}=\frac{h}{\sqrt{2gh}}$
$v _ {A} = g t _ {0}, v _ {B} = u _ {B} - g t _ {0}$
$\Rightarrow v _ {A} = - v _ {B}$
$\overrightarrow {P _ {i}} = \overrightarrow {P _ {f}}$
$m \overrightarrow {v _ {A}} + m \overrightarrow {v _ {B}} = 2 m \overrightarrow {v _ {f}}$
$\overrightarrow {v _ {f}} = 0$

and height from ground $= h - \frac{1}{2} gt_0^2 = h - \frac{h}{4} = \frac{3h}{4}$ Before collision After collision


Time taken by combined mass to reach the ground
time = $\sqrt{\frac{2 \times 3h/4}{g}} = \sqrt{\frac{3h}{2g}}$ .

$W = \int_{\vec{r}_{i}}^{\vec{r}_{f}} \vec{F}. d\vec{r}$
$W = \int_ {1} ^ {0} - x d x + \int_ {0} ^ {1} y d y \Rightarrow W = \frac {- x ^ {2}}{2} \Bigg | _ {1} ^ {0} + \frac {y ^ {2}}{2} \Bigg | _ {0} ^ {1}$
$= - \left(\frac {0 ^ {2}}{2} - \frac {1 ^ {2}}{2}\right) + \left(\frac {1 ^ {2}}{2} - \frac {0 ^ {2}}{2}\right) \Rightarrow W = 1 J.$

By momentum conservation.
$\frac {m u}{2} + m u = 2 m v ^ {\prime} \Rightarrow v ^ {\prime} = \frac {3 u}{4}$
Range after collision = $\frac{3u}{4}\sqrt{\frac{2H}{g}}$
$= \frac {3 u}{4} \sqrt {\frac {2 . u ^ {2} \sin^ {2} 6 0 ^ {\circ}}{2 g ^ {2}}} \Rightarrow \frac {3}{4} \frac {\sqrt {3}}{2}. \frac {u ^ {2}}{g} = \frac {3 \sqrt {3} u ^ {2}}{8 g}.$

From momentum conservation
$mu\hat{i} + 0 = mv\hat{j} + 3mv'\Rightarrow \vec{v'} = \frac{u}{3}\hat{i} -\frac{v}{3}\hat{j}$
From kinetic energy conservation
$\frac {1}{2} m u ^ {2} = \frac {1}{2} m v ^ {2} + \frac {1}{2} (3 m) \left(\left(\frac {u}{3}\right) ^ {2} + \left(\frac {v}{3}\right) ^ {2}\right)$
Solving, $v = \frac{u}{\sqrt{2}}$ .

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

Conservation of linear momentum $0.1 \times 20 = (0.1 + 1.9) \times v \quad m = 0.1 \, \text{kg}, v_b = 20 \, \text{m/s}$
$v = 1 m / s$
Using work energy theorem
$W _ {g} = \Delta k$

$2 \times g \times 1 = k - \frac {1}{2} \times 2 \times 1 ^ {2}$
$\therefore k = 2 1 J$

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So let the final velocity be $v'$ , so on applying momentum conservation $mv = 16m v' \Rightarrow v' = v / 16$
Now, initial energy $E_{i} = \frac{1}{2} mv^{2}$
Final energy: $E_{f} = \frac{1}{2}\times 16m\times \left(\frac{v}{16}\right)^{2} = \frac{1}{2} m\frac{v^{2}}{16}$
Energy loss: $E_{i} - E_{f} = \frac{1}{2} mv^{2}\left[\frac{15}{16}\right]$
$\% p = \frac {\text {Energy loss}}{\text {Original energy}} \times 100 = 93.75 \%$
$\Rightarrow$ Value of $P$ is close to 94.

$W = \text { area } = (2 0 0 \times 1 5) + \frac {1}{2} (1 0 0 + 2 0 0) \times 1 5$
$W = 3 0 0 0 + 2 2 5 0 = 5 2 5 0 J$

$\vec{v}_{01} = (\sqrt{3}\hat{i} +\hat{j})m / s,\vec{v}_{02} = \vec{0}, m_1 = 2m_2$
After collision, $\vec{v}_1 = (\hat{i} +\sqrt{3}\hat{j})m / s$
$\vec {v} _ {2} = ?$
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 $\vec{v}_2$ ,
$\cos \theta = \frac {\vec {v} _ {1} . \vec {v} _ {2}}{| \vec {v} _ {1} | | \vec {v} _ {2} |} = \frac {2 (\sqrt {3} - 1) (1 - \sqrt {3})}{2 \times 2 \sqrt {2} (\sqrt {3} - 1)}$
$\cos \theta = \frac {1 - \sqrt {3}}{2 \sqrt {2}} \Rightarrow \theta = 1 0 5 ^ {\circ}.$

$x = 3t^{2} + 5$ $v = \frac{dx}{dt}$ $v = 6t + 0$ at t = 0 $v = 0$ at t = 5 sec $v = 30 m/s$ W.D. = $\Delta KE$ W.D. = $\frac{1}{2}mv^{2} - 0 = \frac{1}{2}(2)(30)^{2} = 900 J$ .

$k_{i} = \frac{1}{2} mv_{0}^{2}$ From linear momentum conservation $mv_{0} = (2m + M)v_{f}$ $\Rightarrow v_{f} = \frac{mv_{0}}{2m + M}$ $\frac{k_i}{k_f} = \frac{\frac{1}{2}mv_0^2}{\frac{1}{2}(2m + M)\left(\frac{mv_0}{2m + M}\right)^2} = 6$ $\Rightarrow \frac{2m + M}{m} = 6\Rightarrow \frac{M}{m} = 4.$

Time taken for the particles to collide,

Time taken for the particles to collide, $a_{rel}=0$ $v_{rel}=100m/s$ $t=\frac{d}{V_{rel}}=\frac{100}{100}=1sec$ $s_b=100\times1-\frac{1}{2}\times10\times1=95m$ 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 $P_{i}=P_{f}$ $(0.02)\times90-0.03\times10=(0.05)v$ 1.5=0.05v
v=30m/s
Max. height reached by body $h=\frac{v^{2}}{2g}$ $h=\frac{30\times30}{2\times10}=45m$ ∴ Height above tower = 45 - 5 = 40 m.

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

Work done = $\vec{F}.\vec{d} = (3\hat{i} - 12\hat{j}).(4\hat{i})$ =12 J.
work energy theorem $W_{net} = \Delta K.E.$ 12 = $K_{f} - 3$ $K_{f} = 15J.$

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. $1 \times \sqrt{2000} = 4v \Rightarrow v = \frac{\sqrt{2000}}{4}m/s = 5\sqrt{5}m/s$ initial compression of spring $1.25 \times 10^{6}x_{0} = 30 \Rightarrow x_{0} \approx 0$ applying work energy theorem, $W_{g} + W_{sp} = \Delta KE$ $\Rightarrow 40 \times x + \frac{1}{2} \times 1.25 \times 10^{6}(0^{2} - x^{2})$ $= 0 - \frac{1}{2} \times 4 \times v^{2}$ solving $x \approx 2cm$ .
Aliter : assuming $5\sqrt{5}$ as maximum velocity
So $V_{\max} = \omega x$ $5\sqrt{5} = \sqrt{\frac{k}{m}} \times x \Rightarrow 5\sqrt{5} = \sqrt{\frac{1.25 \times 10^{6}}{4}} \times x$ $x = 2cm.$

$m \xrightarrow{\alpha} u M; \quad \xleftarrow{\alpha} v_1 M \xrightarrow{} v_2$ Conserving momentum: $mu = -mv_1 + MV_2$ ... (i)
Collision of nuclear particle to be elastic $e = \frac{v_2 + v_1}{u} = 1$ $\therefore v_2 = u - v_1$ ... (ii) $v_1 = 0.6u$ $1.6m = 0.4M \Rightarrow M = 4m.$

According to work energy theorem
Work done by force on the particle = change in KE
Work done = Area under F-x graph $=\int F.dx=2\times2+\frac{(2+3)\times1}{2}$ $W=KE_{final}-KE_{initial}=6.5$ $KE_{initial}=0$ $KE_{final}=6.5J.$

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

Mass of the hanging part = $\frac{M}{n}$
$h _ {C O M} = \frac {L}{2 n}$
work done $W = mgh_{COM} = \left(\frac{M}{n}\right)g\left(\frac{L}{2n}\right) = \frac{MgL}{2n^2}$

$\bullet \longrightarrow 2 v$
$\bullet \longrightarrow v$
$m 2 m$

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

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

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

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

$v _ {1} = \frac {4 m - 2 m}{4 m + 2 m} u = \frac {2 m u}{6 m} = \frac {u}{3}$
$\begin{array}{r l} \text { Fraction of energy lost } & = \frac {\frac {1}{2} (4 m) u ^ {2} - \frac {1}{2} (4 m) \left(\frac {u}{3}\right) ^ {2}}{\frac {1}{2} (4 m) u ^ {2}} \\ & = 1 - \frac {1}{9} = \frac {8}{9}. \end{array}$