### Related Formula By conservation of linear momentum, the common velocity v_textcm$v_{\text{cm}}$ of the combined mass system is: m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_textcm$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{\text{cm}}$ By conservation of energy, the loss in Kinetic Energy during maximum compression converted to the potential energy of the spring: Delta K = frac12 k x^2 Rightarrow frac12 m_1 v^2 - frac12 (m_1 + m_2) v_textcm^2 = frac12 k x^2$\Delta K = \frac{1}{2} k x^2 \Rightarrow \frac{1}{2} m_1 v^2 - \frac{1}{2} (m_1 + m_2) v_{\text{cm}}^2 = \frac{1}{2} k x^2$ ### Core Logic Given parameters: - m_1 = 10mathrm~kg$m_1 = 10\mathrm{~kg}$, m_2 = 5mathrm~kg$m_2 = 5\mathrm{~kg}$ - Initial velocity of A: v = 3mathrm~m/s$v = 3\mathrm{~m/s}$ - Initial velocity of B: v_2 = 0$v_2 = 0$ - Spring constant k = 3000mathrm~N/m$k = 3000\mathrm{~N/m}$ ### Step 1: Calculate the common center of mass velocity (v_textcm$v_{\text{cm}}$) v_textcm = frac10 times 3 + 5 times 010 + 5 = frac3015 = 2mathrm~m/s$v_{\text{cm}} = \frac{10 \times 3 + 5 \times 0}{10 + 5} = \frac{30}{15} = 2\mathrm{~m/s}$ ### Step 2: Apply Energy Conservation to find spring compression ($ ### Step 2: Apply Energy Conservation to find spring compression ($x) $) $frac12 k x^2 = K_i - K_f$\frac{1}{2} k x^2 = K_i - K_f$ $ $frac12 (3000) x^2 = left[ frac12 (10) (3^2) right] - left[ frac12 (10 + 5) (2^2) right]$\frac{1}{2} (3000) x^2 = \left[ \frac{1}{2} (10) (3^2) \right] - \left[ \frac{1}{2} (10 + 5) (2^2) \right]$ $ $1500 x^2 = frac12(90) - frac12(15)(4)$1500 x^2 = \frac{1}{2}(90) - \frac{1}{2}(15)(4)$ $ $1500 x^2 = 45 - 30 = 15$1500 x^2 = 45 - 30 = 15$ $ $x^2 = frac151500 = frac1100$x^2 = \frac{15}{1500} = \frac{1}{100}$ $ $x = frac110mathrm~m = 0.1mathrm~m$x = \frac{1}{10}\mathrm{~m} = 0.1\mathrm{~m}$ ### Pattern Recognition For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy. $ ### Pattern Recognition For maximum compression in block-spring-block collisions, the relative kinetic energy gets fully transformed into spring potential energy. $frac12 mu v_textrel^2 = frac12 k x^2$\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{1}{2} k x^2$ where $ where $\mu = \frac{m_1 m_2}{m_1 + m_2} is the reduced mass. Here: $ is the reduced mass. Here: $mu = frac10 times 515 = frac103mathrm~kg$\mu = \frac{10 \times 5}{15} = \frac{10}{3}\mathrm{~kg}$ $ $frac12 left(frac103right) (3)^2 = frac12 (3000) x^2 Rightarrow 15 = 1500 x^2 Rightarrow x = 0.1mathrm~m$\frac{1}{2} \left(\frac{10}{3}\right) (3)^2 = \frac{1}{2} (3000) x^2 \Rightarrow 15 = 1500 x^2 \Rightarrow x = 0.1\mathrm{~m}$$ Using reduced mass simplifies center-of-mass collision problems instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula The electrical input power P_textin$P_{\text{in}}$ delivered to the motor is: P_textin = V cdot I$P_{\text{in}} = V \cdot I$ The fraction of power lost is determined by the efficiency eta$\eta$: P_textlost = P_textin - P_textout = P_textin (1 - eta)$P_{\text{lost}} = P_{\text{in}} - P_{\text{out}} = P_{\text{in}} (1 - \eta)$ To convert watts (mathrmJ/s$\mathrm{J/s}$) to calories per second: P_textlost (mathrmcal/s) = fracP_textlost (mathrmW)4.2$P_{\text{lost}} (\mathrm{cal/s}) = \frac{P_{\text{lost}} (\mathrm{W})}{4.2}$ ### Core Logic Given parameters: - Potential $ ### Core Logic Given parameters: - Potential $V = 100\mathrm{~V} - Current $ - Current $I = 1\mathrm{~A} - Efficiency $ - Efficiency $\eta = 91.6\% = 0.916 ### Step 1: Calculate Input Power ($ ### Step 1: Calculate Input Power ($P_{\text{in}}) $) $P_textin = 100 times 1 = 100mathrm~W$P_{\text{in}} = 100 \times 1 = 100\mathrm{~W}$ ### Step 2: Calculate Power Loss in Watts $ ### Step 2: Calculate Power Loss in Watts $P_textlost = 100 times (1 - 0.916) = 100 times 0.084 = 8.4mathrm~W$P_{\text{lost}} = 100 \times (1 - 0.916) = 100 \times 0.084 = 8.4\mathrm{~W}$ ### Step 3: Convert Power Loss to cal/s $ ### Step 3: Convert Power Loss to cal/s $P_textlost = frac8.44.2 = 2mathrm~cal/s$P_{\text{lost}} = \frac{8.4}{4.2} = 2\mathrm{~cal/s}$ ### Pattern Recognition Efficiency calculations often feature clean conversion rates. Recognizing that $ ### Pattern Recognition Efficiency calculations often feature clean conversion rates. Recognizing that $8.4 is exactly twice $ is exactly twice $4.2$ (the conversion factor for Joules to calories) simplifies the final step. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula By the Work-Energy Theorem, the work done by the retarding force equals the change in kinetic energy: W = Delta K = K_f - K_i$W = \Delta K = K_f - K_i$ W = int_x_i^x_f F_r(x) dx = frac12 m v_f^2 - frac12 m v_i^2$W = \int_{x_i}^{x_f} F_r(x) dx = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$ ### Core Logic Given parameters: - Mass m = 1mathrm~kg$m = 1\mathrm{~kg}$ - Initial velocity v_i = 10mathrm~m/s$v_i = 10\mathrm{~m/s}$ - Region bounds: x_i = 0.1mathrm~m$x_i = 0.1\mathrm{~m}$, x_f = 1.9mathrm~m$x_f = 1.9\mathrm{~m}$ - Retarding force F_r = -kx = -10xmathrm~N$F_r = -kx = -10x\mathrm{~N}$ ### Step 1: Calculate Work Done by the Retarding Force (W$W$) W = int_0.1^1.9 (-10x) dx = -10 left[ fracx^22 right]_0.1^1.9 = -5 left[ (1.9)^2 - (0.1)^2 right]$W = \int_{0.1}^{1.9} (-10x) dx = -10 \left[ \frac{x^2}{2} \right]_{0.1}^{1.9} = -5 \left[ (1.9)^2 - (0.1)^2 \right]$ Using the algebraic identity a^2 - b^2 = (a-b)(a+b)$a^2 - b^2 = (a-b)(a+b)$: (1.9)^2 - (0.1)^2 = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6$(1.9)^2 - (0.1)^2 = (1.9 - 0.1)(1.9 + 0.1) = (1.8)(2.0) = 3.6$ W = -5 times 3.6 = -18mathrm~J$W = -5 \times 3.6 = -18\mathrm{~J}$ ### Step 2: Solve for final velocity (v_f$v_f$) Apply the Work-Energy Theorem: -18 = frac12 (1) v_f^2 - frac12 (1) (10^2)$-18 = \frac{1}{2} (1) v_f^2 - \frac{1}{2} (1) (10^2)$ -18 = 0.5 v_f^2 - 50$-18 = 0.5 v_f^2 - 50$ 0.5 v_f^2 = 50 - 18 = 32$0.5 v_f^2 = 50 - 18 = 32$ v_f^2 = 64 Rightarrow v_f = 8mathrm~m/s$v_f^2 = 64 \Rightarrow v_f = 8\mathrm{~m/s}$ ### Pattern Recognition Notice that integrating a linear force $ ### Pattern Recognition Notice that integrating a linear force $F = -kx yields a potential-energy-like term $ yields a potential-energy-like term $\frac{1}{2}k(x_f^2 - x_i^2). Combining this with the Work-Energy theorem gives $. Combining this with the Work-Energy theorem gives $\frac{1}{2} m v_f^2 + \frac{1}{2} k x_f^2 = \frac{1}{2} m v_i^2 + \frac{1}{2} k x_i^2$, which is identical to conservation of mechanical energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula Instantaneous power P$P$ generated by a force is given by: P = vecF cdot vecv$P = \vec{F} \cdot \vec{v}$ Newton's second law: veca = fracvecFm = fracmathrmdvecvmathrmdt implies vecv = int veca \,mathrmdt$\vec{a} = \frac{\vec{F}}{m} = \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \implies \vec{v} = \int \vec{a} \,\mathrm{d}t$ ### Core Logic Convert mass to SI units: m = 1000 mathrm~g = 1 mathrm~kg$m = 1000 \mathrm{~g} = 1 \mathrm{~kg}$ Calculate acceleration: veca = frac2thatmathbfi + 3t^2hatmathbfj1 = 2thatmathbfi + 3t^2hatmathbfj$\vec{a} = \frac{2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}}{1} = 2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}$ ### Step 1: Determine Velocity Vector Assuming the object starts from rest at t = 0$t = 0$: vecv = int_0^t (2thatmathbfi + 3t^2hatmathbfj) \,mathrmdt = t^2hatmathbfi + t^3hatmathbfj$\vec{v} = \int_{0}^{t} (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \,\mathrm{d}t = t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}}$ ### Step 2: Calculate Power Compute the dot product of force and velocity: P = vecF cdot vecv = (2thatmathbfi + 3t^2hatmathbfj) cdot (t^2hatmathbfi + t^3hatmathbfj)$P = \vec{F} \cdot \vec{v} = (2t\hat{\mathbf{i}} + 3t^2\hat{\mathbf{j}}) \cdot (t^2\hat{\mathbf{i}} + t^3\hat{\mathbf{j}})$ P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 mathrm~W$P = (2t)(t^2) + (3t^2)(t^3) = 2t^3 + 3t^5 \mathrm{~W}$ ### Pattern Recognition Sees: Time-dependent force $ ### Pattern Recognition Sees: Time-dependent force $\vec{F} \propto t^n on a 1 kg mass. Shortcut: For $ on a 1 kg mass. Shortcut: For $m=1 kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $ kg, velocity is the integral of the force components. Power is the dot product of the force vector and its integral. Since $\int at^n \mathrm{d}t = \frac{a}{n+1} t^{n+1}, power component becomes $, power component becomes $\frac{a^2}{n+1} t^{2n+1}. Here, $. Here, $2^2/2 t^3 + 3^2/3 t^5 = 2t^3 + 3t^5 \mathrm{~W}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula E = K_i + U_i = K_f + U_f$E = K_i + U_i = K_f + U_f$ U = frac12 k y^2$U = \frac{1}{2} k y^2$ where, E$E$ = total mechanical energy K = frac12 m v^2$K = \frac{1}{2} m v^2$ = kinetic energy U$U$ = potential energy of spring with deformation y$y$ ### Core Logic Given parameters: - Mass, m = 2mathrm~kg$m = 2\mathrm{~kg}$ - Natural length of spring, L_0 = 2mathrm~m$L_0 = 2\mathrm{~m}$ - Spring constant, k = 200mathrm~N/m$k = 200\mathrm{~N/m}$ Initial State (when block is pushed): - Length of spring is 1mathrm~m$1\mathrm{~m}$. - Deformation (compression), y_i = L_0 - 1 = 2 - 1 = 1mathrm~m$y_i = L_0 - 1 = 2 - 1 = 1\mathrm{~m}$. - Released from rest: v_i = 0 implies K_i = 0$v_i = 0 \implies K_i = 0$. Final State (at distance x$x$ from the wall): - Since the spring is attached to the wall, its length is xmathrm~m$x\mathrm{~m}$. - Deformation (compression) at this position, y_f = L_0 - x = (2 - x)mathrm~m$y_f = L_0 - x = (2 - x)\mathrm{~m}$. - Kinetic energy K_f = frac12 m v^2 = frac12 (2) v^2 = v^2$K_f = \frac{1}{2} m v^2 = \frac{1}{2} (2) v^2 = v^2$. ### Step 1: Conservation of Energy Equation Equate initial and final energies: K_i + U_i = K_f + U_f$K_i + U_i = K_f + U_f$ 0 + frac12 k y_i^2 = frac12 m v^2 + frac12 k y_f^2$0 + \frac{1}{2} k y_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} k y_f^2$ Substitute the parameters: frac12 (200) (1)^2 = v^2 + frac12 (200) (2 - x)^2$\frac{1}{2} (200) (1)^2 = v^2 + \frac{1}{2} (200) (2 - x)^2$ 100 = v^2 + 100 (2 - x)^2$100 = v^2 + 100 (2 - x)^2$ v^2 = 100 left[ 1 - (2 - x)^2 right]$v^2 = 100 \left[ 1 - (2 - x)^2 \right]$ v = 10 left[ 1 - (2 - x)^2 right]^1/2mathrm~m/s$v = 10 \left[ 1 - (2 - x)^2 \right]^{1/2}\mathrm{~m/s}$ ### Pattern Recognition Sees: Horizontal spring-mass energy conservation. Trap: The deformation is not x$x$; it is the difference from natural length, i.e., (L_0 - x) = (2 - x)$(L_0 - x) = (2 - x)$. Shortcut: Writing out energy conservation directly allows mass to cancel beautifully, simplifying the algebra immediately. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power Class 11 Physics: Oscillations