### Related Formula m_1 v_1 = m_2 v_2$m_1 v_1 = m_2 v_2$ ### Core Logic Since there is no external horizontal force acting on the body, linear momentum along the horizontal axis is conserved. Initial mass m_1 = 1000text kg$m_1 = 1000\text{ kg}$, initial velocity v_1 = 6text m/s$v_1 = 6\text{ m/s}$. Final mass m_2 = 1000 + 200 = 1200text kg$m_2 = 1000 + 200 = 1200\text{ kg}$. ### Step 1: Balance conservation equation 1000 times 6 = 1200 times v_2$1000 \times 6 = 1200 \times v_2$ 6000 = 1200 v_2 implies v_2 = frac60001200 = 5text m/s$6000 = 1200 v_2 \implies v_2 = \frac{6000}{1200} = 5\text{ m/s}$ ### Pattern Recognition Inelastic mass addition transitions are classic momentum-balance equations, scaling velocity inversely with total expanded mass profiles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula Kinetic frictional force (f$f$) on a flat surface: f = mu N = mu m g$f = \mu N = \mu m g$ Work done against friction (W_textagainst$W_{\text{against}}$): W_textagainst = f cdot s = mu m g s$W_{\text{against}} = f \cdot s = \mu m g s$ ### Core Logic Given values: m = 100 mathrm~kg, quad s = 10 mathrm~m, quad mu = 0.4$m = 100 \mathrm{~kg}, \quad s = 10 \mathrm{~m}, \quad \mu = 0.4$ Take g = 10 mathrm~m/s^2$g = 10 \mathrm{~m/s^2}$. ### Step 1: Calculate Friction Force The normal force on a horizontal surface is: N = mg = 100 times 10 = 1000 mathrm~N$N = mg = 100 \times 10 = 1000 \mathrm{~N}$ The force of friction is: f = mu N = 0.4 times 1000 = 400 mathrm~N$f = \mu N = 0.4 \times 1000 = 400 \mathrm{~N}$ ### Step 2: Calculate Work Done The work done against friction is: W_textagainst = f cdot s = 400 mathrm~N times 10 mathrm~m = 4000 mathrm~J$W_{\text{against}} = f \cdot s = 400 \mathrm{~N} \times 10 \mathrm{~m} = 4000 \mathrm{~J}$ Therefore, the work done is 4000 mathrm~J$4000 \mathrm{~J}$. ### Pattern Recognition Work done *by* friction is negative (-4000 mathrm~J$-4000 \mathrm{~J}$) because the frictional force opposes displacement. Work done *against* friction is positive (+4000 mathrm~J$+4000 \mathrm{~J}$) because it represents the external energy that must be spent to sustain slide. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula W_f = f_k cdot d$W_f = f_k \cdot d$ f_k = mu N$f_k = \mu N$ N = mg costheta$N = mg \cos\theta$ ### Core Logic The work done against the frictional force is the product of the kinetic friction force and the displacement along the plane. The normal force N$N$ on the block is given by N = mg costheta$N = mg \cos\theta$, where theta = 60^circ$\theta = 60^{\circ}$. ### Step 1: Calculate Normal and Frictional Force Given: mu = 0.1$\mu = 0.1$ m = 1 mathrm~kg$m = 1 \mathrm{~kg}$ theta = 60^circ$\theta = 60^{\circ}$ d = 10 mathrm~m$d = 10 \mathrm{~m}$ Normal force N = 1 times 10 times cos(60^circ) = 10 times frac12 = 5 mathrm~N$N = 1 \times 10 \times \cos(60^{\circ}) = 10 \times \frac{1}{2} = 5 \mathrm{~N}$. Frictional force f_k = mu N = 0.1 times 5 = 0.5 mathrm~N$f_k = \mu N = 0.1 \times 5 = 0.5 \mathrm{~N}$. ### Step 2: Work Done Against Friction Work done against frictional force = f_k times d$= f_k \times d$ W = 0.5 mathrm~N times 10 mathrm~m = 5 mathrm~J$W = 0.5 \mathrm{~N} \times 10 \mathrm{~m} = 5 \mathrm{~J}$ ### Pattern Recognition Whenever asked for "work done against friction", simply compute mu mg costheta times d$\mu mg \cos\theta \times d$. The applied force (10 mathrm~N$10 \mathrm{~N}$) is irrelevant to the friction calculation itself since it is parallel to the plane. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Work, Energy and Power

### Related Formula tan theta = mu$\tan \theta = \mu$ fracmathrmdymathrmdx = tan theta$\frac{\mathrm{d}y}{\mathrm{d}x} = \tan \theta$ ### Core Logic For a block to remain stationary on a rough surface without slipping, the maximum slope of the surface it can rest on is determined by the angle of repose. tan theta le mu$\tan \theta \le \mu$ The slope of the given parabolic curve at any point (x,y)$(x,y)$ is fracmathrmdymathrmdx$\frac{\mathrm{d}y}{\mathrm{d}x}$. ### Step 1: Evaluate the Slope Given curve equation: y = fracx^24$y = \frac{x^2}{4}$ Differentiating with respect to x$x$: fracmathrmdymathrmdx = frac2x4 = fracx2$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x}{4} = \frac{x}{2}$ ### Step 2: Find Maximum Valid Position At the critical point of slipping, slope equals mu$\mu$: fracx2 = mu$\frac{x}{2} = \mu$ Given mu = 0.5 = frac12$\mu = 0.5 = \frac{1}{2}$, we have: fracx2 = frac12 implies x = 1$\frac{x}{2} = \frac{1}{2} \implies x = 1$ ### Step 3: Calculate Maximum Height Substitute x = 1$x = 1$ back into the curve equation to find the maximum height y$y$: y = frac(1)^24 = frac14 mathrm~m$y = \frac{(1)^2}{4} = \frac{1}{4} \mathrm{~m}$ ### Pattern Recognition When asked for maximum height on a curve y=f(x)$y=f(x)$ without slipping, set fracmathrmdymathrmdx = mu$\frac{\mathrm{d}y}{\mathrm{d}x} = \mu$, solve for x$x$, and plug it back into the original equation to find y$y$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula a = fracF_textnetm_texttotal$a = \frac{F_{\text{net}}}{m_{\text{total}}}$ T = m_textbehind times a$T = m_{\text{behind}} \times a$ ### Core Logic Since the surface is smooth, the entire system of three blocks moves with a common acceleration a$a$. a_mathrmA = a_mathrmB = a_mathrmC = fracFm_mathrmA + m_mathrmB + m_mathrmC$a_{\mathrm{A}} = a_{\mathrm{B}} = a_{\mathrm{C}} = \frac{F}{m_{\mathrm{A}} + m_{\mathrm{B}} + m_{\mathrm{C}}}$ ### Step 1: Calculate Common Acceleration a = frac805 + 3 + 2 = frac8010 = 8 mathrm~m/s^2$a = \frac{80}{5 + 3 + 2} = \frac{80}{10} = 8 \mathrm{~m/s}^2$ ### Step 2: Calculate Tensions

mathrmT_1$\mathrm{T}_1$ pulls only block mathrmA$\mathrm{A}$ (5 mathrm~kg$5 \mathrm{~kg}$): mathrmT_1 = m_mathrmA times a = 5 times 8 = 40 mathrm~N$\mathrm{T}_1 = m_{\mathrm{A}} \times a = 5 \times 8 = 40 \mathrm{~N}$

mathrmT_2$\mathrm{T}_2$ pulls both blocks mathrmA$\mathrm{A}$ and mathrmB$\mathrm{B}$ (5 mathrm~kg + 3 mathrm~kg = 8 mathrm~kg$5 \mathrm{~kg} + 3 \mathrm{~kg} = 8 \mathrm{~kg}$): From free body diagram of block B: mathrmT_2 - mathrmT_1 = m_mathrmB times a$\mathrm{T}_2 - \mathrm{T}_1 = m_{\mathrm{B}} \times a$ mathrmT_2 - 40 = 3 times 8 = 24 implies mathrmT_2 = 64 mathrm~N$\mathrm{T}_2 - 40 = 3 \times 8 = 24 \implies \mathrm{T}_2 = 64 \mathrm{~N}$ ### Pattern Recognition Tension at any point in a train of connected accelerating bodies (without friction) is directly proportional to the total mass being pulled *behind* that point. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion