### Related Formula Conservation of Linear Momentum during embedded collision: mu = (M + m)V$mu = (M + m)V$ Conservation of Mechanical Energy during vertical rise: frac12(M+m)V^2 = (M+m)gh implies h = fracV^22g$\frac{1}{2}(M+m)V^2 = (M+m)gh \implies h = \frac{V^2}{2g}$ ### Core Logic Given data: m = 10^-2mathrm~kg$m = 10^{-2}\mathrm{~kg}$, u = 2 times 10^2mathrm~ms^-1$u = 2 \times 10^2\mathrm{~ms}^{-1}$, M = 1mathrm~kg$M = 1\mathrm{~kg}$, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s}^2$. Apply momentum balance: 10^-2 times (2 times 10^2) = (1 + 0.01)V$10^{-2} \times (2 \times 10^2) = (1 + 0.01)V$ 2 = 1.01V implies V approx 2mathrm~ms^-1\ text(since 1.01 approx 1text)$2 = 1.01V \implies V \approx 2\mathrm{~ms}^{-1}\ \text{(since } 1.01 \approx 1\text{)}$ ### Step 1: Compute Maximum Rise Height Using the work-energy relation for the subsequent upward swing: h = fracV^22g = frac2^22 times 10 = frac420 = 0.20mathrm~m$h = \frac{V^2}{2g} = \frac{2^2}{2 \times 10} = \frac{4}{20} = 0.20\mathrm{~m}$ ### Pattern Recognition Perfectly inelastic collision approximation: because m ll M$m \ll M$, we can simplify M+m approx M$M+m \approx M$ during velocity matching to complete the calculation rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power Class 11 Physics: Laws of Motion

### Related Formula Potential energy at the horizontal release point relative to the lowest point: U_i = mgell$U_i = mg\ell$ If 10\%$10\%$ of energy is dissipated, the remaining 90\%$90\%$ is converted entirely to kinetic energy at the lowest point: E_f = 0.90 times U_i = frac12mv^2$E_f = 0.90 \times U_i = \frac{1}{2}mv^2$ ### Core Logic Let the horizontal position be our reference for potential energy relative to the lowest point. * Initial potential energy: U_i = mgell$U_i = mg\ell$ * Energy remaining after 10\%$10\%$ loss: 0.90(mgell)$0.90(mg\ell)$ Equating this to kinetic energy at the bottom: frac910 mgell = frac12 mv^2$\frac{9}{10} mg\ell = \frac{1}{2} mv^2$ ### Step 1: Calculate the Velocity We can cancel m$m$ from both sides: frac910 gell = frac12 v^2$\frac{9}{10} g\ell = \frac{1}{2} v^2$ Substitute ell = 10text m$\ell = 10\text{ m}$ and g = 10text m s^-2$g = 10\text{ m s}^{-2}$: frac910 (10)(10) = frac12 v^2$\frac{9}{10} (10)(10) = \frac{1}{2} v^2$ 90 = frac12 v^2 implies v^2 = 180$90 = \frac{1}{2} v^2 \implies v^2 = 180$ v = sqrt180 = sqrt36 times 5 = 6sqrt5text m s^-1$v = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}\text{ m s}^{-1}$

### Pattern Recognition Whenever there is a fractional energy loss x$x$, use the formula: v = sqrt2(1-x)gell$v = \sqrt{2(1-x)g\ell}$. Substituting x = 0.1$x = 0.1$ gives v = sqrt1.8gell$v = \sqrt{1.8g\ell}$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula For a body to just complete a vertical loop of radius L$L$: * Speed at the lowest point A$A$: v_A = sqrt5gL$v_A = \sqrt{5gL}$ * Speed at the highest point B$B$: v_B = sqrtgL$v_B = \sqrt{gL}$ ### Core Logic The kinetic energy at any point is given by: textK.E. = frac12mv^2$\text{K.E.} = \frac{1}{2}mv^2$ Thus: (textK.E.)_A = frac12m v_A^2 = frac12m(5gL)$(\text{K.E.})_A = \frac{1}{2}m v_A^2 = \frac{1}{2}m(5gL)$ (textK.E.)_B = frac12m v_B^2 = frac12m(gL)$(\text{K.E.})_B = \frac{1}{2}m v_B^2 = \frac{1}{2}m(gL)$ ### Step 1: Calculate the Ratio Taking the ratio of the kinetic energies at A$A$ and B$B$: frac(textK.E.)_A(textK.E.)_B = fracfrac12m(5gL)frac12m(gL) = frac51 = 5:1$\frac{(\text{K.E.})_A}{(\text{K.E.})_B} = \frac{\frac{1}{2}m(5gL)}{\frac{1}{2}m(gL)} = \frac{5}{1} = 5:1$ ### Pattern Recognition Since textK.E. propto v^2$\text{K.E.} \propto v^2$, the ratio of kinetic energies is simply the ratio of the squares of the critical velocities at the bottom and top of the vertical loop: (sqrt5gL)^2 : (sqrtgL)^2 = 5:1$(\sqrt{5gL})^2 : (\sqrt{gL})^2 = 5:1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula K = fracP^22m implies P = sqrt2mK$K = \frac{P^2}{2m} \implies P = \sqrt{2mK}$ Where P$P$ is linear momentum, m$m$ is mass, and K$K$ is kinetic energy. ### Core Logic Given K_1 = K_2$K_1 = K_2$, the momentum ratio simplifies directly to the square root of their masses: fracP_1P_2 = sqrtfracm_1m_2$\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$ ### Step 1: Calculate the value Substitute m_1 = 4text g$m_1 = 4\text{ g}$ and m_2 = 25text g$m_2 = 25\text{ g}$: fracP_1P_2 = sqrtfrac425 = frac25$\frac{P_{1}}{P_{2}} = \sqrt{\frac{4}{25}} = \frac{2}{5}$ ### Pattern Recognition For constant kinetic energy tracking profiles, momentum maps proportionally to sqrtm$\sqrt{m}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula The force vector vecF$\vec{F}$ is related to the potential energy U$U$ by the negative gradient of potential energy: vecF = -vecnabla U = -left( fracpartial Upartial xhati + fracpartial Upartial yhatj + fracpartial Upartial zhatk right)$\vec{F} = -\vec{\nabla} U = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} \right)$ Hence, the x-component of force is: F_x = -fracpartial Upartial x$F_x = -\frac{\partial U}{\partial x}$ ### Core Logic Given the potential energy function: U = 2x^2 + 3y^3 + 2z$U = 2x^2 + 3y^3 + 2z$ Taking the partial derivative with respect to x$x$ (treating y$y$ and z$z$ as constants): fracpartial Upartial x = fracpartialpartial x(2x^2) = 4x$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2) = 4x$ ### Step 1: Substitute Coordinates The x-component of the force is: F_x = -4x$F_x = -4x$ At point P(1, 2, 3) mathrm~m$P(1, 2, 3) \mathrm{~m}$, we substitute x = 1$x = 1$: F_x = -4(1) = -4 mathrm~N$F_x = -4(1) = -4 \mathrm{~N}$ Magnitude of the x-component of force is: |F_x| = 4 mathrm~N$|F_x| = 4 \mathrm{~N}$ ### Pattern Recognition When asked for a specific component (like x-component), only differentiate partially with respect to that specific variable. The remaining coordinates (y, z$y, z$) act purely as constants and vanish. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power