Consider two blocks A and B of masses m_1=10mathrm~kg$m_{1}=10\mathrm{~kg}$ and m_2=5mathrm~kg$m_{2}=5\mathrm{~kg}$ that are placed on a frictionless table. The block A moves with a constant speed v=3mathrm~m/s$v=3\mathrm{~m/s}$ towards the block B kept at rest. A spring with spring constant k=3000mathrm~N/m$k=3000\mathrm{~N/m}$ is attached with the block B as shown in the figure.

After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)

### Related Formula F = -fracdVdx$$F = -\frac{dV}{dx}$$ ### Core Logic The magnitude of the force is given by the absolute value of the slope of the potential energy versus position curve: |F| = |textslope of V-x graph|$|F| = |\text{slope of V-x graph}|$. Comparing the slopes of different regions from the graph: - Region BC has the steepest slope (largest magnitude). - Region AB has a moderate positive slope. - Region DE has a moderate negative slope, but less steep than AB. - Region CD is perfectly flat (slope = 0). ### Step 1: Ranking Forces Therefore, magnitudes of the force follow the order: F_BC > F_AB > F_DE > F_CD$$F_{BC} > F_{AB} > F_{DE} > F_{CD}$$ ### Pattern Recognition In a V-x$V-x$ graph, steeper slopes always mean a stronger restoring or repulsive force. A flat line means zero force (equilibrium). ### 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