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 $\vec{v}_2$ ,
$\cos \theta = \frac {\vec {v} _ {1} . \vec {v} _ {2}}{| \vec {v} _ {1} | | \vec {v} _ {2} |} = \frac {2 (\sqrt {3} - 1) (1 - \sqrt {3})}{2 \times 2 \sqrt {2} (\sqrt {3} - 1)}$
$\cos \theta = \frac {1 - \sqrt {3}}{2 \sqrt {2}} \Rightarrow \theta = 1 0 5 ^ {\circ}.$

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

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

After reading the problem, we conclude that
for $0 \leq x \leq 15$ , F = 200 N and
for $15 \leq x < 30$ , $F = 200 - \frac{100}{15}(x - 15)$ ⇒ $F(\text{in N}) = \begin{bmatrix} 200 & , \text{for } 0 \leq x \leq 15 \\ 300 - \frac{100}{15}x & , \text{for } 15 \leq x < 30 \end{bmatrix}$ Since, $W = \int F dx$ ⇒ $W = \int_{0}^{15} 200 dx + \int_{15}^{30} \left(300 - \frac{100}{15}x\right) dx$ ⇒ $W = (200)(15) + (300)(15) - \left(\frac{100}{15}\right)\left(\frac{(30^2 - 15^2)}{2}\right)$ ⇒ $W = 3000 + 4500 - 2250 = 5250 J$ Hence, the correct answer is (B).

Given that, $U = -\frac{A}{r^{6}} + \frac{B}{r^{12}}$
$\Rightarrow F = - \frac {d U}{d r} = - A (- 6 r ^ {- 7}) + B (- 1 2 r ^ {- 1 3})$
$\Rightarrow 0 = \frac {6 A}{r ^ {7}} - \frac {1 2 B}{r ^ {1 3}}$
$\Rightarrow \frac {6 A}{1 2 B} = \frac {1}{r ^ {6}}$
$\Rightarrow r = \left(\frac {2 B}{A}\right) ^ {1 / 6}$
$\text { So }, U _ {\mathrm{eq}} = U \mid_ {r = \left(\frac {2 B}{A}\right) ^ {1 / 6}} = - \frac {A}{(2 B / A)} + \frac {B}{(4 B ^ {2} / A ^ {2})}$
$\Rightarrow U _ {\mathrm{eq}} = - \frac {A ^ {2}}{2 B} + \frac {A ^ {2}}{4 B} = - \frac {A ^ {2}}{4 B}$
Hence, the correct answer is (C).

Since, P = constant. From Work-Energy Theorem
$W = \frac {1}{2} m v ^ {2}$
Also, we know that $W = Pt$
$\Rightarrow K = P t = \frac {1}{2} m v ^ {2}$
$\Rightarrow v = \frac {d s}{d t} = \sqrt {\frac {2 P t}{m}} = \sqrt {\frac {2 P}{m}} t ^ {1 / 2}$
$\Rightarrow s = \int d s = \sqrt {\frac {2 P}{m}} \int_ {0} ^ {t} t ^ {1 / 2} d t = \frac {2}{3} \sqrt {\frac {2 P}{m}} t ^ {3 / 2}$
Hence, the correct answer is (D).

Since the elevator is moving with a constant speed, so we have

$T = 6 8 0 0 + 9 2 0 0 + 6 0 0 0 = 2 2 0 0 0 \mathrm{N}$
So, power delivered by the motor is
$P = T v = 2 2 0 0 0 \times 3 = 6 6 0 0 0 \mathrm{W}$
Hence, the correct answer is (C).

$\begin{array}{l} W = \int_ {\vec {r} _ {i}} ^ {\vec {r} _ {g}} \vec {F} \cdot d \vec {r} = \int_ {1} ^ {0} - x d x + \int_ {0} ^ {1} y d y = \left(- \frac {x ^ {2}}{2}\right) \bigg | _ {1} ^ {0} + \left(\frac {y ^ {2}}{2}\right) \bigg | _ {0} ^ {1} \\ \Rightarrow W = - \left(\frac {0 ^ {2}}{2} - \frac {1 ^ {2}}{2}\right) + \left(\frac {1 ^ {2}}{2} - \frac {0 ^ {2}}{2}\right) = 1 \text {J} \end{array}$
Hence, the correct answer is (B).

At point y, we have
$m g \sin 3 0 ^ {\circ} = \frac {m v ^ {2}}{R}$
$\Rightarrow v ^ {2} = \frac {g R}{2}$

Applying conservation of mechanical Energy, we get
$\begin{array}{r l} & {\frac {1}{2} m v _ {0} ^ {2} = m g h + \frac {1}{2} m v ^ {2}} \\ {\Rightarrow} & {v _ {0} ^ {2} - 2 g h = \frac {g R}{2}} \end{array}$
Along the path xyz, speed is maximum at points x and z due to which maximum centripetal force will be required.
Hence, (A) and (D) are correct.

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

$k_{i} = \frac{1}{2} mv_{0}^{2}$
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
$m 2 v + 2 m v = m \times 0 + m \frac {v ^ {\prime}}{\sqrt {2}} \times 2 \Rightarrow v ^ {\prime} = 2 \sqrt {2} v.$