Laws of Motion

$\mathrm{m}=800 \mathrm{kg}$
$\mathrm{r}=300 \mathrm{m}$
$\theta =30°$
${\mu }_{\mathrm{s}}=0.2$
${\mathrm{V}}_{\max }=\sqrt{Rg\left[\frac{\tan \theta +\mu }{1-\mu \tan \theta }\right]}$
$=\sqrt{300\times \mathrm{g}\times \left[\frac{\tan 30°+0.2}{1-0.2\times \tan 30°}\right]}$
$=\sqrt{300\times 10\times \left[\frac{0.57+0.2}{1-0.2\times 0.57}\right]}$
${\mathrm{V}}_{\max }=51.4 \mathrm{m}/\mathrm{s}$

$$\begin{gathered} F_1=k{x}_1 \\ 5=k{x}_1 \\ x_1=\frac{5}{k} \\ 7=k{x}_2 \\ x_2=\frac{7}{k} \end{gathered}$$

$$\begin{gathered} x=5x_1-2x_2 \\ x=\frac{25}{k}-\frac{14}{k}\Rightarrow x=\frac{11}{k} \\ F=kx \\ F=k\times \frac{11}{k} \\ F=11N \end{gathered}$$

$$\begin{gathered} F_1=k{x}_1 \\ 5=k{x}_1 \\ x_1=\frac{5}{k} \\ 7=k{x}_2 \\ x_2=\frac{7}{k} \end{gathered}$$

$$\begin{gathered} x=5x_1-2x_2 \\ x=\frac{25}{k}-\frac{14}{k}\Rightarrow x=\frac{11}{k} \\ F=kx \\ F=k\times \frac{11}{k} \\ F=11N \end{gathered}$$

F = ma

$=m\left(\frac{v-u}{t}\right)$

$=\frac{120}{1000}\times \frac{25}{0.1}$

F = 300 N

F = ma

$=m\left(\frac{v-u}{t}\right)$

$=\frac{120}{1000}\times \frac{25}{0.1}$

F = 300 N

$$\begin{gathered} F_1=k{x}_1 \\ 5=k{x}_1 \\ x_1=\frac{5}{k} \\ 7=k{x}_2 \\ x_2=\frac{7}{k} \end{gathered}$$

$$\begin{gathered} x=5x_1-2x_2 \\ x=\frac{25}{k}-\frac{14}{k}\Rightarrow x=\frac{11}{k} \\ F=kx \\ F=k\times \frac{11}{k} \\ F=11N \end{gathered}$$

\(\begin{aligned} & \left|\overrightarrow{p_1}\right|=\left|\overrightarrow{p_2}\right| \\ & K E=\frac{P^2}{2 M} ; p \text { same } \\ & K E \propto \frac{1}{m} \\ & \frac{K E_1}{K E_2}=\frac{p^2 / 2 M_1}{p^2 / 2 M_2}=\frac{M_2}{M_1}\end{aligned}\)

\(\begin{aligned} & \left|\overrightarrow{p_1}\right|=\left|\overrightarrow{p_2}\right| \\ & K E=\frac{P^2}{2 M} ; p \text { same } \\ & K E \propto \frac{1}{m} \\ & \frac{K E_1}{K E_2}=\frac{p^2 / 2 M_1}{p^2 / 2 M_2}=\frac{M_2}{M_1}\end{aligned}\)

$m=0.1 \mathrm{kg},\overrightarrow{F}=(5\hat{i}+10\hat{j})\mathrm{N}$

$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=(50\hat{i}+100\hat{j})\mathrm{m}/{\mathrm{s}}^2$

$\overrightarrow{s}=\frac{1}{2}\overrightarrow{a}{t}^2=\frac{1}{2}(50\hat{i}+100\hat{j})\left(4\right)$

$=(100\hat{i}+200\hat{j})\mathrm{m}$

Now, on comparing with $(2x\hat{i}+5y\hat{j})\mathrm{m}$.

$2x=100\Rightarrow x=50,5y=200\Rightarrow y=40$

$\therefore x:y=5:4$

$$
\begin{aligned}
& \mathrm{F}=\mathrm{M} \times \mathrm{a} \\
& v=4 \sqrt{x} \\
& v^2=16 x \\
& 2 v \frac{d v}{d x}=16 \\
& \frac{\mathrm{vdv}}{\mathrm{dx}}=\frac{16}{2}=8 \\
& \mathrm{~F}=0.5 \times 8=4 \mathrm{~N}
\end{aligned}
$$

Let:

• $M$ be the initial mass of the balloon.
• $T$ be the tension (upward force) acting on the balloon.

Using Newton’s second law for the initial state:

$$T - Mg = Ma$$

$$T = M(g + a)$$

Let $m$ be the mass that is dropped. The new mass of the balloon becomes $(M - m)$. The equation of motion after dropping mass:

$$T - (M - m)g = (M - m) \cdot 3a$$

Substituting $T = M(g + a)$:

$$M(g + a) - (M - m)g = (M - m) \cdot 3a$$

Expanding:

$$Mg + Ma - Mg + mg = 3a(M - m)$$

$$Ma + mg = 3aM - 3am$$

Rearrange terms:

$$Ma - 3aM = - mg - 3am$$

$$Ma - 3aM = -m(g + 3a)$$

$$m = \frac{2aM}{g + 3a}$$​

The mass that should be dropped is $\frac{2aM}{g+3a}$.

Let:

• $M$ be the initial mass of the balloon.
• $T$ be the tension (upward force) acting on the balloon.

Using Newton’s second law for the initial state:

$$T - Mg = Ma$$

$$T = M(g + a)$$

Let $m$ be the mass that is dropped. The new mass of the balloon becomes $(M - m)$. The equation of motion after dropping mass:

$$T - (M - m)g = (M - m) \cdot 3a$$

Substituting $T = M(g + a)$:

$$M(g + a) - (M - m)g = (M - m) \cdot 3a$$

Expanding:

$$Mg + Ma - Mg + mg = 3a(M - m)$$

$$Ma + mg = 3aM - 3am$$

Rearrange terms:

$$Ma - 3aM = - mg - 3am$$

$$Ma - 3aM = -m(g + 3a)$$

$$m = \frac{2aM}{g + 3a}$$​

The mass that should be dropped is $\frac{2aM}{g+3a}$.