Work, Power & Energy

Height fallen up to point Q
$h = R \sin 3 0 ^ {\circ} = 4 0 \times \frac {1}{2} = 2 0 \mathrm{m}$
Since, $W_{nc} = \Delta U + \Delta K$ , so we get
$\begin{array}{r l} & W _ {n c} = - m g h + \frac {1}{2} m v ^ {2} \\ \Rightarrow & W _ {n c} = - (1) (1 0) (2 0) + \frac {1}{2} (1) v ^ {2} \\ \text {Given that} W _ {n c} = - 1 5 0 \mathrm{J} \\ \Rightarrow & - 1 5 0 = - 2 0 0 + \frac {v ^ {2}}{2} \\ \Rightarrow & v ^ {2} = 1 0 0 \\ \Rightarrow & v = 1 0 \mathrm{ms} ^ {- 1} \end{array}$
Hence, the correct answer is (B).

At point Q, component of weight along PQ (radially outwards) is $mg\cos60^{\circ}$ i.e., $\frac{mg}{2}$ .
Normal reaction is radially inwards, so we have
$\begin{array}{r l} & N - \frac {m g}{2} = \frac {m v ^ {2}}{R} \\ \Rightarrow & N = \frac {m g}{2} + \frac {m v ^ {2}}{R} \\ \Rightarrow & N = \frac {1 \times 1 0}{2} + \frac {1 \times (1 0) ^ {2}}{4 0} = 7. 5 \mathrm{N} \end{array}$
Hence, the correct answer is (A).

Conservation of linear momentum along x-direction
$m _ {2} v = m _ {1} v _ {x}$
$\frac {m _ {2} v}{m _ {1}} = v _ {x}$... (i)
Along y direction
$m_{2} \times \frac{v}{2} = m_{1} v_{y} \Rightarrow v_{y} = \frac{v_{x}}{2}$ [from (i)]
or $\tan \theta = \frac{1}{2}$ .

At the highest point, $v = 0$ , $h_1 = 2.5m$ $\therefore$ Total energy, $E_1 = mgh_1 + 0 = mgh_1$ At the lowest point, $v = ?$ , $h_2 = 1.5m$ $\therefore$ Total energy, $E_2 = mgh_2 + \frac{1}{2}mv^2$ According to the law of conservation of mechanical energy $E_1 = E_2$ $mgh_1 = \frac{1}{2}mv^2 + mgh_2 \Rightarrow v^2 = 2g(h_1 - h_2)$ $v = \sqrt{2g(h_1 - h_2)} = \sqrt{2 \times 10 \times (2.5 - 1.5)}$ $= \sqrt{2 \times 10 \times 1} = \sqrt{20} ms^{-1} = 2\sqrt{5} ms^{-1}.$

When stone hits the ground momentum $P = m\sqrt{2gh}$ when same stone dropped from 2h(100% of initial) then momentum $P' = m\sqrt{2g(2h)} = \sqrt{2}P$ . Which is changed by 41% of initial.

Total mass of particle = 4 m
when it explodes three fragments (2 m), (m) and (m)
Total released energy = $\frac{1}{2}(2m)v^{2} + \frac{1}{2}mv^{2} + \frac{1}{2}mv^{2}$ $=\frac{1}{2}v^{2}(2m+m+m)=\frac{1}{2}v^{2}4m=2mv^{2}.$

Constant power of car $P_0 = F.v. = ma.v$
$P _ {0} = m \frac {d v}{d t}. v \Rightarrow P _ {0} d t = m v d v \Rightarrow P _ {0}. t = \frac {m v ^ {2}}{2}$
$v = \sqrt {\frac {2 P _ {0} t}{m}} \Rightarrow v \propto \sqrt {t}.$

Kinetic energy of an object, $K = \frac{1}{2} mv^2$
Where $m$ is the mass of the object and $v$ is the speed of the object.
$\because K \propto v ^ {2}$
Therefore, the curve between speed and kinetic energy is parabola.

For the same force, $F = k_{1}x_{1} = k_{2}x_{2}$
... (1)
Work done on spring $S_{1}$
$W _ {1} = \frac {1}{2} k _ {1} x _ {1} ^ {2} = \frac {(k _ {1} x _ {1}) ^ {2}}{2 k _ {1}} = \frac {F ^ {2}}{2 k _ {1}}$
{using (1)}
Work done on spring $S_{2}$ is
$W _ {2} = \frac {1}{2} k _ {2} x _ {2} ^ {2} = \frac {(k _ {2} x _ {2}) ^ {2}}{2 k _ {2}} = \frac {F ^ {2}}{2 k _ {2}}$
{using (1)}
$\Rightarrow \frac {W _ {1}}{W _ {2}} = \frac {k _ {2}}{k _ {1}}$
Since, $W_{1} > W_{2}$
$\Rightarrow k _ {2} > k _ {1}$
$\Rightarrow k _ {1} < k _ {2}$
Statement 2 is true.
For the same extension, i.e. for $x_{1} = x_{2} = x$ ... (2)
Work done on spring $S_{1}$ is $W_{1} = \frac{1}{2} k_{1}x_{1}^{2} = \frac{1}{2} k_{1}x^{2}$
Work done on spring $S_{2}$ is $W_{2} = \frac{1}{2} k_{2}x_{2}^{2} = \frac{1}{2} k_{2}x^{2}$
$\Rightarrow \frac {W _ {1}}{W _ {2}} = \frac {k _ {1}}{k _ {2}}$
Since, $k_{1} < k_{2}$
$\Rightarrow W _ {1} < W _ {2}$
Statement 1 is false.
Hence, the correct answer is (D).

By Work Energy Theorem
$\begin{array}{r l} & W = \vec {F} \cdot \Delta \vec {r} = \Delta K \\ \Rightarrow & \Delta K = (7 \hat {i} + 4 \hat {j} + 3 \hat {k}) \cdot (2 \hat {i} + 3 \hat {j} + 4 \hat {k}) \\ \Rightarrow & \Delta K = 1 4 + 1 2 + 1 2 = 3 8 \mathrm{J} \end{array}$
Hence, the correct answer is (A).

According to work-energy theorem
W=Change in kinetic energy
$F S \cos \theta = \frac {1}{2} m v ^ {2} - \frac {1}{2} m u ^ {2}$
Substituting the given values, we get
$20 \times 4 \times \cos \theta = 4 0 - 0$
$[ \because u = 0 ]$
$\cos \theta = \frac {4 0}{8 0} = \frac {1}{2}$
$\theta = \cos^ {- 1} \left(\frac {1}{2}\right) = 6 0 ^ {\circ}.$

$P = \vec{F} \cdot \vec{v} = F v \cos \theta$ Just before hitting $\theta$ is zero and both F, v are maximum.

$U_{1} = mgh_{1}$ and $U_{2} = mgh_{2}$
$U _ {1} = m g h _ {1} \text { and } U _ {2} = m g h _ {2}$
$\% \text { energy lost } = \frac {U _ {1} - U _ {2}}{U _ {1}} \times 1 0 0$
$= \frac {m g h _ {1} - m g h _ {2}}{m g h _ {1}} 1 0 0 = \left(\frac {h _ {1} - h _ {2}}{h _ {1}}\right) \times 1 0 0$
$ = \frac {2 - 1 . 5}{2} \times 1 0 0 = 2 5 \%.$

Before collision

After collision
Here, $m = 0.25kg$ , $u_{1} = 3ms^{-1}$ , $u_{2} = -1ms^{-1}$
It is an inelastic collision.
According to conservation of momentum
$m u _ {1} + m u _ {2} = (m + m) v$
$\Rightarrow v = \frac {m u _ {1} + m u _ {2}}{2 m} = \frac {u _ {1} + u _ {2}}{2} = \frac {3 - 1}{2} = 1 m s ^ {- 1}.$

$U = \frac{a}{x^{12}} - \frac{b}{x^6}$ $F = -\frac{dU}{dx} = +12\frac{a}{x^{13}} - \frac{6b}{x^7} = 0 \Rightarrow x = \left(\frac{2a}{b}\right)^{1/6}$ $U(x = \infty) = 0$ $U_{\text{equilibrium}} = \frac{a}{\left(\frac{2a}{b}\right)^2} - \frac{b}{\left(\frac{2a}{b}\right)} = -\frac{b^2}{4a}$ $\therefore U(x = \infty) - U_{\text{equilibrium}} = 0 - \left(-\frac{b^2}{4a}\right) = \frac{b^2}{4a}$ .

$\because$ Speed is constant
∴ Work done by forces =0
∴ Power = $\frac{Work}{Time}$ = 0.

Power $= F v = v \left( \frac{m}{t} \right) v = v^{2}(\rho A v)$ $= \rho A v^{3} = (100)(2)^{3} = 800 W$

By conservation of linear momentum:
$2m = mv_1 + 2mv_2$
$\Rightarrow v_1 + 2v_2 = 2$
By definition of coefficient of restitution, $e$:
$e = \dfrac{1}{2} = \dfrac{v_2-v_1}{2-0}$
$\Rightarrow v_2-v_1=1$
$\Rightarrow v_1=0$ and $v_2=1,\text{m s}^{-1}$.

In an inelastic collision, only momentum is conserved whereas in elastic collision both momentum and kinetic energy are conserved.