A block of mass 1 kg, moving along x with speed v_i=10mathrm~m/s$v_{i}=10\mathrm{~m/s}$ enters a rough region ranging from x=0.1mathrm~m$x=0.1\mathrm{~m}$ to x=1.9mathrm~m$x=1.9\mathrm{~m}$. The retarding force acting on the block in this range is F_r=-kxmathrm~N$F_{r}=-kx\mathrm{~N}$ with k=10mathrm~N/m$k=10\mathrm{~N/m}$. Then the final speed of the block as it crosses rough region is :

### Related Formula V_textBottom = sqrtV_textTop^2 + 4gR$$V_{\text{Bottom}} = \sqrt{V_{\text{Top}}^2 + 4gR}$$ ### Core Logic Given velocity at the top position : V_textTop = sqrtn^2 gR$$V_{\text{Top}} = \sqrt{n^2 gR}$$ By work-energy theorem, mechanical energy conservation between the top and bottom positions gives : V_textBottom = sqrtn^2 gR + 4gR$$V_{\text{Bottom}} = \sqrt{n^2 gR + 4gR}$$ Since KE = frac12m v^2$KE = \frac{1}{2}m v^2$, the ratio of kinetic energy at the bottom to that at the top is textRatio = fracV_textBottom^2V_textTop^2 = fracn^2 gR + 4gRn^2 gR = fracn^2 + 4n^2$$\text{Ratio} = \frac{V_{\text{Bottom}}^2}{V_{\text{Top}}^2} = \frac{n^2 gR + 4gR}{n^2 gR} = \frac{n^2 + 4}{n^2}$$ ### Pattern Recognition Kinetic energy change in vertical circles always gains a fixed additive value of 2mg(2R) = 4mgR$2mg(2R) = 4mgR$ due to gravity work. ### Chapter Mix Class 11 Physics: Work, Energy and Power

### 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