A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be : (Take 'g' as acceleration due to gravity) [cite: 174, 175]

### Related Formula veca = fracvecFm$$\vec{a} = \frac{\vec{F}}{m}$$ vecv = int veca \, dt$$\vec{v} = \int \vec{a} \, dt$$ vecr = int vecv \, dt$$\vec{r} = \int \vec{v} \, dt$$ ### Core Logic Given vecF = 4t^3hati - 3thatj$\vec{F} = 4t^3\hat{i} - 3t\hat{j}$ and mass m = 4text kg$m = 4\text{ kg}$. Acceleration: veca = fracvecFm = frac4t^3hati - 3thatj4 = t^3hati - frac34thatj$$\vec{a} = \frac{\vec{F}}{m} = \frac{4t^3\hat{i} - 3t\hat{j}}{4} = t^3\hat{i} - \frac{3}{4}t\hat{j}$$ ### Step 1: Calculating Velocity Velocity is the integral of acceleration with v(0) = 0$v(0) = 0$: vecv = int left( t^3hati - frac34thatj right) dt = fract^44hati - frac38t^2hatj$$\vec{v} = \int \left( t^3\hat{i} - \frac{3}{4}t\hat{j} \right) dt = \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j}$$ At t = 2text s$t = 2\text{ s}$: vecv(2) = frac(2)^44hati - frac38(2)^2hatj = 4hati - frac32hatj$$\vec{v}(2) = \frac{(2)^4}{4}\hat{i} - \frac{3}{8}(2)^2\hat{j} = 4\hat{i} - \frac{3}{2}\hat{j}$$ ### Step 2: Calculating Position Position is the integral of velocity with vecr(0) = vec0$\vec{r}(0) = \vec{0}$: vecr = int vecv \, dt = int left( fract^44hati - frac38t^2hatj right) dt$$\vec{r} = \int \vec{v} \, dt = \int \left( \frac{t^4}{4}\hat{i} - \frac{3}{8}t^2\hat{j} \right) dt$$ vecr = fract^520hati - frac324t^3hatj = fract^520hati - fract^38hatj$$\vec{r} = \frac{t^5}{20}\hat{i} - \frac{3}{24}t^3\hat{j} = \frac{t^5}{20}\hat{i} - \frac{t^3}{8}\hat{j}$$ At t = 2text s$t = 2\text{ s}$: vecr(2) = frac(2)^520hati - frac(2)^38hatj = frac3220hati - frac88hatj = frac85hati - hatj$$\vec{r}(2) = \frac{(2)^5}{20}\hat{i} - \frac{(2)^3}{8}\hat{j} = \frac{32}{20}\hat{i} - \frac{8}{8}\hat{j} = \frac{8}{5}\hat{i} - \hat{j}$$ ### Pattern Recognition Recognize F to a to v to r$F \to a \to v \to r$ means consecutive integrations. Because the initial state is from rest at origin, we don't have to worry about constants of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

### Related Formula For a system in static equilibrium: sum F_x = 0 quad textand quad sum F_y = 0$$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$$ ### Core Logic Let's resolve the tension forces vecT_1$\vec{T}_1$ and vecT_2$\vec{T}_2$ into horizontal and vertical components: - T_1$T_1$ makes 60^circ$60^\circ$ with the horizontal. - T_2$T_2$ makes 30^circ$30^\circ$ with the horizontal. - Downward gravitational force: W = m g = 1 times 10 = 10 \ mathrmN$W = m g = 1 \times 10 = 10 \ \mathrm{N}$. 1. **Horizontal Equilibrium (sum F_x = 0$\sum F_x = 0$):** T_1 cos 60^circ = T_2 cos 30^circ$$T_1 \cos 60^\circ = T_2 \cos 30^\circ$$ T_1 cdot frac12 = T_2 cdot fracsqrt32 implies T_1 = T_2 sqrt3$$T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{3}}{2} \implies T_1 = T_2 \sqrt{3}$$ 2. **Vertical Equilibrium (sum F_y = 0$\sum F_y = 0$):** T_1 sin 60^circ + T_2 sin 30^circ = m g = 10$$T_1 \sin 60^\circ + T_2 \sin 30^\circ = m g = 10$$ T_1 cdot fracsqrt32 + T_2 cdot frac12 = 10$$T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 10$$ ### Step 1: Solve for Tensions Substitute T_1 = T_2 sqrt3$T_1 = T_2 \sqrt{3}$ into the vertical equilibrium equation: (T_2 sqrt3) fracsqrt32 + fracT_22 = 10$$(T_2 \sqrt{3}) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10$$ frac3 T_22 + fracT_22 = 10 implies 2 T_2 = 10 implies T_2 = 5 \ mathrmN$$\frac{3 T_2}{2} + \frac{T_2}{2} = 10 \implies 2 T_2 = 10 \implies T_2 = 5 \ \mathrm{N}$$ Substitute T_2$T_2$ back to obtain T_1$T_1$: T_1 = 5 sqrt3 \ mathrmN$$T_1 = 5 \sqrt{3} \ \mathrm{N}$$ Thus, the tension magnitudes are T_1 = 5sqrt3 \ mathrmN$T_1 = 5\sqrt{3} \ \mathrm{N}$ and T_2 = 5 \ mathrmN$T_2 = 5 \ \mathrm{N}$. ### Pattern Recognition Sees: Suspending particle static equilibrium with asymmetric strings. Trap: Associating components with incorrect trigonometry axes or swapping T_1$T_1$ and T_2$T_2$ in options. Shortcut: Since the incline of T_1$T_1$ (60^circ$60^\circ$) is steeper than that of T_2$T_2$ (30^circ$30^\circ$), T_1$T_1$ must carry a larger portion of the load, meaning T_1 > T_2$T_1 > T_2$. From the choices, only (2) satisfies this hierarchy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula For a block sliding down an inclined plane of inclination theta$\theta$: mg sintheta - f_k = ma$$mg \sin\theta - f_k = ma$$ Where normal reaction is N = mg costheta$N = mg \cos\theta$ and kinetic friction is: f_k = mu_k N = mu_k mg costheta$$f_k = \mu_k N = \mu_k mg \cos\theta$$ ### Core Logic Substitute f_k$f_k$ into the equation of motion: mg sintheta - mu_k mg costheta = ma$$mg \sin\theta - \mu_k mg \cos\theta = ma$$ Divide by m$m$: g sintheta - mu_k g costheta = a$$g \sin\theta - \mu_k g \cos\theta = a$$ ### Step 1: Substitute Values Given a = fracg2$a = \frac{g}{2}$ and theta = 60^circ$\theta = 60^\circ$: g sin 60^circ - mu_k g cos 60^circ = fracg2$$g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2}$$ fracsqrt32 - fracmu_k2 = frac12$$\frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2}$$ sqrt3 - mu_k = 1 implies mu_k = sqrt3 - 1$$\sqrt{3} - \mu_k = 1 \implies \mu_k = \sqrt{3} - 1$$ ### Pattern Recognition Sees: Block sliding down with acceleration on an incline. Shortcut: The acceleration on an incline is a = g(sintheta - mu_k costheta)$a = g(\sin\theta - \mu_k \cos\theta)$. For theta = 60^circ$\theta = 60^\circ$, this is a = gleft(fracsqrt32 - fracmu_k2right)$a = g\left(\frac{\sqrt{3}}{2} - \frac{\mu_k}{2}\right)$. Equating to g/2$g/2$ gives mu_k = sqrt3 - 1$\mu_k = sqrt{3} - 1$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula vecF = m veca$$\vec{F} = m \vec{a}$$ vecv = vecu + vecat$$\vec{v} = \vec{u} + \vec{a}t$$ where, vecF$\vec{F}$ = constant force vector m$m$ = mass of body veca$\vec{a}$ = acceleration vector vecu$\vec{u}$ = initial velocity vector vecv$\vec{v}$ = final velocity vector ### Core Logic Given parameters: - Mass, m = 2mathrm~kg$m = 2\mathrm{~kg}$ - Initial velocity, vecu = 3hati + 4hatjmathrm~m/s$\vec{u} = 3\hat{i} + 4\hat{j}\mathrm{~m/s}$ - Force, vecF = 6hatkmathrm~N$\vec{F} = 6\hat{k}\mathrm{~N}$ (directed along positive z-axis) - Time interval, t = frac53mathrm~s$t = \frac{5}{3}\mathrm{~s}$ Calculate the acceleration vector veca$\vec{a}$: veca = fracvecFm = frac6hatk2 = 3hatkmathrm~m/s^2$$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k}\mathrm{~m/s}^{2}$$ ### Step 1: Compute Final Velocity Using the kinematic equation of motion: vecv = vecu + vecat$$\vec{v} = \vec{u} + \vec{a}t$$ vecv = (3hati + 4hatj) + (3hatk) left(frac53right)$$\vec{v} = (3\hat{i} + 4\hat{j}) + (3\hat{k}) \left(\frac{5}{3}\right)$$ vecv = 3hati + 4hatj + 5hatkmathrm~m/s$$\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$$ Thus, the emerging velocity of the body is 3hati + 4hatj + 5hatkmathrm~m/s$3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$. ### Pattern Recognition Sees: Orthogonal initial velocity and force field direction. Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged (3hati + 4hatj$3\hat{i} + 4\hat{j}$). Simply compute the z-component change: v_z = a_z t = left(frac62right) left(frac53right) = 5$v_z = a_z t = \left(\frac{6}{2}\right) \left(\frac{5}{3}\right) = 5$. Result: 3hati + 4hatj + 5hatk$3\hat{i} + 4\hat{j} + 5\hat{k}$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

### Related Formula F_textthrust = left(fracdmdtright)v$$F_{\text{thrust}} = \left(\frac{dm}{dt}\right)v$$ P = F cdot v$P = F \cdot v$ ### Core Logic Given the sand dropping rate rule: fracdmdt = C sqrtv quad (textwhere C text is a constant)$$\frac{dm}{dt} = C \sqrt{v} \quad (\text{where } C \text{ is a constant})$$ To maintain a constant velocity v$v$, the continuous force applied by the conveyor system must balance the rate of gain of momentum of the dropped sand: F = left(fracdmdtright) v = (C sqrtv) cdot v = C v^3/2$$F = \left(\frac{dm}{dt}\right) v = (C \sqrt{v}) \cdot v = C v^{3/2}$$ Power delivered is the product of force and speed: P = F cdot v = (C v^3/2) cdot v = C v^5/2$$P = F \cdot v = (C v^{3/2}) \cdot v = C v^{5/2}$$ Squaring both sides of the expression: P^2 propto v^5$P^2 \propto v^5$ ### Pattern Recognition In variable mass problems involving dropping dust/sand at rest onto a moving frame, the thrust force always simplifies to v cdot fracdmdt$v \cdot \frac{dm}{dt}$, which makes power scale as v^2 cdot fracdmdt$v^2 \cdot \frac{dm}{dt}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion