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)

### Core Logic Potential energy functions are strictly mathematically defined exclusively for conservative force interactions via the relationship F = -fracdUdx$F = -\frac{dU}{dx}$[cite: 727]. Coulomb's force, Gravitational force, and Spring restoring force are completely path-independent conservative fields[cite: 146, 147, 149]. Frictional force is a path-dependent, dissipative non-conservative force[cite: 148, 727, 728]. Consequently, it is impossible to define a scalar potential energy function for mechanical friction[cite: 727, 728]. ### Pattern Recognition Whenever you encounter a potential energy definition requirement, remember that it is a direct marker for conservative fields. Dissipative forces like friction or viscous drag instantly break this condition[cite: 727, 728]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula a = vfracdvdx$$a = v\frac{dv}{dx}$$ F = m cdot a$F = m \cdot a$ [cite: 778] ### Core Logic Given velocity as a function of position x$x$: [cite: 164, 779] v = 4sqrtx implies v^2 = 16x$$v = 4\sqrt{x} \implies v^2 = 16x$$ [cite: 164, 779] Differentiating both sides with respect to position coordinate x$x$: [cite: 780] 2vfracdvdx = 16 implies vfracdvdx = 8$$2v\frac{dv}{dx} = 16 \implies v\frac{dv}{dx} = 8$$ [cite: 780, 781] Thus, the acceleration of the object is a constant value a = 8\ textm/s^2$a = 8\ \text{m/s}^2$[cite: 781]. Converting mass to kilograms (m = 500\ textg = 0.5\ textkg$m = 500\ \text{g} = 0.5\ \text{kg}$) [cite: 164, 788]: F = 0.5 times 8 = 4\ textN$$F = 0.5 \times 8 = 4\ \text{N}$$ [cite: 788] ### Pattern Recognition When velocity depends on position coordinate x$x$ like v = ksqrtx$v = k\sqrt{x}$, squaring instantly reveals that acceleration is constant, since v^2 = k^2 x$v^2 = k^2 x$ matches the third kinematic profile v^2 = 2ax$v^2 = 2ax$ directly[cite: 779]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula The work done W$W$ by a variable force component F(x)$F(x)$ over a displacement step is given by: W = int_x_1^x_2 F(x) dx$$W = \int_{x_{1}}^{x_{2}} F(x) dx$$ ### Core Logic Assuming the object moves from the origin x=0$x=0$ to x=1text m$x=1\text{ m}$ : W = int_0^1 (alpha + beta x^2) dx = 5text J$$W = \int_{0}^{1} (\alpha + \beta x^{2}) dx = 5\text{ J}$$ ### Step 1: Integration and Variable Isolation Perform the integration step : W = left[ alpha x + fracbeta x^33 ight]_0^1 = alpha + fracbeta3 = 5$$W = \left[ \alpha x + \frac{\beta x^{3}}{3} ight]_{0}^{1} = \alpha + \frac{\beta}{3} = 5$$ Given alpha = 1text N$\alpha = 1\text{ N}$ , substitute this into the equation to find beta$\beta$ : 1 + fracbeta3 = 5 implies fracbeta3 = 4 implies beta = 12text N/m^2$$1 + \frac{\beta}{3} = 5 \implies \frac{\beta}{3} = 4 \implies \beta = 12\text{ N/m}^{2}$$ ### Pattern Recognition For polynomial forces, the integration steps always yield fractional coefficients matching their power index (1$1$ for constant, frac13$\frac{1}{3}$ for squared terms). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula * **Average Power** (P_textavg$P_{\text{avg}}$) = fractextTotal Work DonetextTotal Time = fracvecF cdot vecst$\frac{\text{Total Work Done}}{\text{Total Time}} = \frac{\vec{F} \cdot \vec{s}}{t}$ * **Instantaneous Power** (P_textinst$P_{\text{inst}}$) = vecF cdot vecv(t)$\vec{F} \cdot \vec{v}(t)$ ### Core Logic Given parameters: * Force vector: vecF = 2hati + 3hatj$\vec{F} = 2\hat{i} + 3\hat{j}$ * Displacement coordinates: P(3,4) rightarrow Q(6,10) implies vecs = (6-3)hati + (10-4)hatj = 3hati + 6hatj$P(3,4) \rightarrow Q(6,10) \implies \vec{s} = (6-3)\hat{i} + (10-4)\hat{j} = 3\hat{i} + 6\hat{j}$ * Time window, t = 4text s$t = 4\text{ s}$ Calculate Average Power : W = vecF cdot vecs = (2hati + 3hatj) cdot (3hati + 6hatj) = (2 times 3) + (3 times 6) = 6 + 18 = 24 text J$$W = \vec{F} \cdot \vec{s} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j}) = (2 \times 3) + (3 \times 6) = 6 + 18 = 24 \text{ J}$$ P_textavg = fracWt = frac244 = 6 text W$$P_{\text{avg}} = \frac{W}{t} = \frac{24}{4} = 6 \text{ W}$$ Now, analyze the instantaneous dynamics to extract final velocity vecv$\vec{v}$ at t=4text s$t=4\text{ s}$: Acceleration vector : veca = fracvecFm = frac2hati + 3hatj4 = 0.5hati + 0.75hatj$\vec{a} = \frac{\vec{F}}{m} = \frac{2\hat{i} + 3\hat{j}}{4} = 0.5\hat{i} + 0.75\hat{j}$ Assuming the body starts from rest, velocity at t=4text s$t=4\text{ s}$ is : vecv = veca cdot t = (0.5hati + 0.75hatj) times 4 = 2hati + 3hatj$$\vec{v} = \vec{a} \cdot t = (0.5\hat{i} + 0.75\hat{j}) \times 4 = 2\hat{i} + 3\hat{j}$$ Calculate Instantaneous Power at t=4text s$t=4\text{ s}$ : P_textinst = vecF cdot vecv = (2hati + 3hatj) cdot (2hati + 3hatj) = 2^2 + 3^2 = 4 + 9 = 13 text W$$P_{\text{inst}} = \vec{F} \cdot \vec{v} = (2\hat{i} + 3\hat{j}) \cdot (2\hat{i} + 3\hat{j}) = 2^2 + 3^2 = 4 + 9 = 13 \text{ W}$$ Taking the final ratio : fracP_textavgP_textinst = frac613$$\frac{P_{\text{avg}}}{P_{\text{inst}}} = \frac{6}{13}$$ *(Note: There is a minor kinematic inconsistency in the question data layout regarding matching coordinate parameters independently, but the calculations follow the standard intended framework directly).* ### Pattern Recognition For constant force acceleration from rest, average power equals frac12 F a t$\frac{1}{2} F a t$ while instantaneous power scales linearly as F a t$F a t$, meaning the structural ratio simplifies exactly to 1:2$1:2$. The custom displacement vector here alters that baseline baseline ratio as tracked. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula u = sqrt2gh = sqrt2gR(1 - costheta)$$u = \sqrt{2gh} = \sqrt{2gR(1 - \cos\theta)}$$ v_1 = left(fracm_1 - m_2m_1 + m_2right)u + left(frac2m_2m_1 + m_2right)v_2i$$v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u + \left(\frac{2m_2}{m_1 + m_2}\right)v_2i$$ ### Core Logic

Velocity of bob A just prior to collision : u = sqrt2gleft(R - Rcos 60^circright) = sqrt2gfracR2 = sqrtgR$$u = \sqrt{2g\left(R - R\cos 60^{\circ}\right)} = \sqrt{2g\frac{R}{2}} = \sqrt{gR}$$ Using conservation of momentum and coefficient of restitution e=1$e=1$ for elastic interaction [cite: 660, 662]: m_A u = m_A v_1 + m_B v_2 implies m u = m v_1 + fracm2 v_2 implies 2v_1 + v_2 = 2u$$m_A u = m_A v_1 + m_B v_2 \implies m u = m v_1 + \frac{m}{2} v_2 \implies 2v_1 + v_2 = 2u$$ ### Step 1: Apply Restitution Velocity Difference v_2 - v_1 = u$v_2 - v_1 = u$ Subtracting equations yields : 3v_1 = u implies v_1 = fracu3 = frac13sqrtgR$$3v_1 = u \implies v_1 = \frac{u}{3} = \frac{1}{3}\sqrt{gR}$$ ### Pattern Recognition In an elastic head-on collision where one body hits half its mass at rest, it retains exactly one-third of its initial hitting speed[cite: 661, 663]. ### Chapter Mix Class 11 Physics: Work, Energy and Power