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 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

### Related Formula v_1' = v_2 quad textand quad v_2' = v_1$$v_1' = v_2 \quad \text{and} \quad v_2' = v_1$$ for perfectly elastic collision (e=1$e=1$) when masses are identical (m_1 = m_2$m_1 = m_2$). ### Core Logic Reason (R) states that identical masses exchange their velocities during elastic collision, which is mathematically correct. Let us trace the sequence of collisions chronologically: 1. Since v_A = 5mathrm~m/s$v_A = 5\mathrm{~m/s}$ and v_B = 2mathrm~m/s$v_B = 2\mathrm{~m/s}$, sphere A collides with sphere B. After this collision, they swap velocities: v_A' = 2mathrm~m/s, quad v_B' = 5mathrm~m/s$$v_A' = 2\mathrm{~m/s}, \quad v_B' = 5\mathrm{~m/s}$$ 2. Now sphere B has velocity v_B' = 5mathrm~m/s$v_B' = 5\mathrm{~m/s}$ and sphere C has v_C = 4mathrm~m/s$v_C = 4\mathrm{~m/s}$. Sphere B will collide with C. After swapping: v_B'' = 4mathrm~m/s, quad v_C' = 5mathrm~m/s$$v_B'' = 4\mathrm{~m/s}, \quad v_C' = 5\mathrm{~m/s}$$ 3. Looking at the values now: v_A' = 2mathrm~m/s$v_A' = 2\mathrm{~m/s}$ and v_B'' = 4mathrm~m/s$v_B'' = 4\mathrm{~m/s}$. No more collisions occur. Therefore, the final velocities are v_A = 2mathrm~m/s, v_B = 4mathrm~m/s, v_C = 5mathrm~m/s$v_A = 2\mathrm{~m/s}, v_B = 4\mathrm{~m/s}, v_C = 5\mathrm{~m/s}$. The values given in Assertion (A) are wrong. Hence, (A) is false but (R) is true.

### Pattern Recognition Velocity exchange happens pairwise in sequential order. Do not try to solve simultaneous conservation laws across all three blocks at once; handle each collision step-by-step from left to right. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Core Logic Let's apply the comprehensive Work-Energy theorem framework across key layout tracking nodes:

mathrmW_textall = Delta mathrmK$$\mathrm{W}_{\text{all}} = \Delta \mathrm{K}$$ mathrmM g (mathrmR + mathrmR cos 60^circ) + frac12 mathrmk (mathrmR^2 - 0^2) = frac12 mathrmm v^2$$\mathrm{M g} (\mathrm{R} + \mathrm{R} \cos 60^{\circ}) + \frac{1}{2} \mathrm{k} (\mathrm{R}^2 - 0^2) = \frac{1}{2} \mathrm{m v}^2$$ Simplifying the gravitational shift and potential expansions: mathrmM g frac3mathrmR2 + fracmathrmkmathrmR^22 = frac12 mathrmm v^2$$\mathrm{M g} \frac{3\mathrm{R}}{2} + \frac{\mathrm{k}\mathrm{R}^2}{2} = \frac{1}{2} \mathrm{m v}^2$$ ### Step 1: Final Kinematic Value mathrmv = sqrt3mathrmgR + fracmathrmkmathrmR^2mathrmm$$\mathrm{v} = \sqrt{3\mathrm{gR} + \frac{\mathrm{k}\mathrm{R}^2}{\mathrm{m}}}$$ Matches parameters specified by option (4). ### Pattern Recognition Isolate spring metrics at node points: Initial extension equals 2mathrmR - mathrmR = mathrmR$2\mathrm{R} - \mathrm{R} = \mathrm{R}$. Final extension is 0$0$ since spring length matching mathrmR$\mathrm{R}$ satisfies unextended conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Core Logic Assertion A: Central force configurations depend solely on the positional distance parameter mathrmr$\mathrm{r}$. They are strictly conservative fields, making path metrics completely irrelevant for total work evaluation. (True) Reason R: Non-conservative profiles like friction or drag dissipate thermal energy paths and do not possess any state potential energy function. (True) Since statement R provides general information about non-conservative forces rather than stating why central fields are path independent, it fails as a direct explanatory bridge. ### Step 1: Final Conclusion Both assertions are factually accurate, but R is not the appropriate explanation for A. This selects option (2). ### Pattern Recognition Conservative fields allow defining potential profiles (F = -nabla mathrmU$F = -\nabla \mathrm{U}$); non-conservative structures completely break this relation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula Conservation of Mechanical Energy: E_texttotal = K + U = textconstant$$E_{\text{total}} = K + U = \text{constant}$$ At the initial height S$S$ (velocity v = 0$v = 0$): E_texttotal = mgS$$E_{\text{total}} = mgS$$ At any height x$x$ above the ground: U = mgx quad textand quad K = frac12mv^2$$U = mgx \quad \text{and} \quad K = \frac{1}{2}mv^2$$ ### Core Logic Let the height of the particle at that instant be x$x$. We are given: K = 3U$K = 3U$ Substitute the energy terms: frac12mv^2 = 3mgx$$\frac{1}{2}mv^2 = 3mgx$$ By energy conservation: K + U = E_texttotal$$K + U = E_{\text{total}}$$ 3U + U = mgS implies 4U = mgS$$3U + U = mgS \implies 4U = mgS$$ 4(mgx) = mgS implies x = fracS4$$4(mgx) = mgS \implies x = \frac{S}{4}$$ ### Step 1: Calculating the Speed Now find the speed v$v$ at this height x = S/4$x = S/4$. Since K = 3U$K = 3U$: frac12mv^2 = 3mgx$$\frac{1}{2}mv^2 = 3mgx$$ frac12mv^2 = 3mgleft(fracS4right)$$\frac{1}{2}mv^2 = 3mg\left(\frac{S}{4}\right)$$ v^2 = frac6gS4 = frac3gS2$$v^2 = \frac{6gS}{4} = \frac{3gS}{2}$$ v = sqrtfrac3gS2$$v = \sqrt{\frac{3gS}{2}}$$ ### Pattern Recognition Standard ratio trick: If K = n U$K = n U$, then by energy conservation (n+1)U = E_texttotal$(n+1)U = E_{\text{total}}$. This immediately yields: x = fracSn+1$$x = \frac{S}{n+1}$$ Here n = 3$n = 3$, so x = S/4$x = S/4$. This rapid shortcut lets you find the height in a split second! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power