Concept Card: Advanced Mechanics Test Discussion & Doubt Resolution
1. Pedagogical Scope & Target Areas:
A thorough post-test diagnostic of competitive JEE mechanics reveals that students lose marks primarily in five conceptual collision zones:
- Toppling vs. Sliding Dynamics: Shifts of the normal reaction vector and torque balance boundaries.
- Variable Mass & Impulsive Influx Systems: Distinguishing static weight from momentum transfer rate ($\vec{F} = \frac{d\vec{P}}{dt} = m \frac{d\vec{v}}{dt} + \vec{v}_{\text{rel}} \frac{dm}{dt}$).
- Subtleties of Work Done by Friction: Dissecting single-body frame dependence vs. system frame invariance in static and kinetic friction.
- Instantaneous Axis of Rotation (IAOR): Rapid kinematic analysis of constrained rigid bodies without resolving multi-variable constraint equations.
- Transition from Slipping to Pure Rolling: Utilizing invariant angular momentum reference points to bypass time integration.
2. Deep Dive: High-Yield Problem Archetypes
(A) Toppling vs. Sliding Thresholds:
- For a rectangular block of mass $m$, width $b$, and height $h$ on a rough floor under a horizontal pull $F$ applied at height $y$:
- As $F$ increases, the normal reaction force $N = m g$ shifts forward by distance $x$ from the center of mass to counteract the overturning torque:
$$N x = F y \implies x = \frac{F y}{m g}$$ - Toppling Threshold ($x = b/2$): The normal reaction reaches the front edge of the base:
$$\mathbf{F_{\text{topple}} = \frac{m g b}{2 y}}$$ - Sliding Threshold:
$$\mathbf{F_{\text{slide}} = \mu_s m g}$$ - Criterion for Toppling Before Sliding:
$$F_{\text{topple}} < F_{\text{slide}} \implies \mathbf{\mu_s > \frac{b}{2 y}}$$
If $\mu_s < \frac{b}{2y}$, the block slides first; if $\mu_s > \frac{b}{2y}$, it topples first.
- As $F$ increases, the normal reaction force $N = m g$ shifts forward by distance $x$ from the center of mass to counteract the overturning torque:
(B) Continuous Mass Deposition & Falling Chains:
- When a flexible chain of linear mass density $\lambda = M/L$ falls vertically from rest onto a scale:
- At the instant length $x$ has landed, the mass resting statically on the pan is $m_s = \lambda x$. Static weight:
$$W_{\text{static}} = \lambda x g$$ - Chain links hit the pan with velocity $v = \sqrt{2 g x}$ and are brought to rest. The momentum arrival rate exerts an additional dynamic thrust:
$$F_{\text{thrust}} = v \frac{dm}{dt} = v (\lambda v) = \lambda v^2 = \lambda (2 g x) = 2 \lambda x g$$ - Total Normal Force Recorded by the Scale:
$$\mathbf{N(x) = W_{\text{static}} + F_{\text{thrust}} = \lambda x g + 2 \lambda x g = 3 \lambda x g}$$ - At the instant the final link strikes the scale ($x = L$):
$$\mathbf{N_{\max} = 3 M g}$$
The scale momentarily registers three times the total weight of the chain!
- At the instant length $x$ has landed, the mass resting statically on the pan is $m_s = \lambda x$. Static weight:
(C) Work Done by Friction – Frame Dependency & System Invariance:
- Static Friction ($f_s$):
- Because no relative displacement occurs between contacting surfaces ($ds_{\text{rel}} = 0$), the net work done on the two-body system is strictly zero in all reference frames: $W_{f_s, \text{net}} = 0$.
- However, static friction can do positive work on one body (e.g., accelerating a passenger on an escalator or a crate in the bed of an accelerating truck) and negative work on the other.
- Kinetic Friction ($f_k$):
- Relative displacement is non-zero ($ds_{\text{rel}} > 0$). The net work done on the interacting system is always negative and represents mechanical energy dissipated as heat:
$$W_{f_k, \text{net}} = -f_k \Delta s_{\text{rel}} < 0$$ - Net dissipation is strictly frame-independent.
- Relative displacement is non-zero ($ds_{\text{rel}} > 0$). The net work done on the interacting system is always negative and represents mechanical energy dissipated as heat:
3. Solved Examples
Example 1 (Toppling vs. Sliding Analysis – Advanced Statics):
A uniform block of mass $m = 10\text{ kg}$, height $h = 0.80\text{ m}$, and square base of width $b = 0.40\text{ m}$ rests on a rough horizontal floor with friction coefficient $\mu = 0.50$ ($g = 10\text{ m/s}^2$). A horizontal force $F$ is applied at height $y = 0.60\text{ m}$ above the base.
(a) Determine the threshold force $F_{\text{slide}}$ required to cause sliding.
(b) Determine the threshold force $F_{\text{topple}}$ required to cause toppling.
(c) State whether the block will slide first or topple first, and find the minimum force required to initiate motion.
Solution:
(a) Sliding Threshold Force:
Normal force $N = m g = 10 \times 10 = 100\text{ N}$.
$$F_{\text{slide}} = \mu N = 0.50 \times 100\text{ N} = \mathbf{50.0\text{ N}}$$
(b) Toppling Threshold Force:
About the front bottom edge $P$, the tipping torque due to $F$ must balance the stabilizing torque due to gravity $m g$ acting through the center of mass (distance $b/2 = 0.20\text{ m}$ from the front edge):
$$\tau_F = F y = F(0.60), \quad \tau_g = m g \left(\frac{b}{2}\right) = 100 \times 0.20 = 20.0\text{ N}\cdot\text{m}$$
At the verge of toppling:
$$F_{\text{topple}} y = m g \left(\frac{b}{2}\right) \implies F_{\text{topple}}(0.60) = 20.0 \implies \mathbf{F_{\text{topple}} = \frac{20.0}{0.60} = \frac{100}{3}\text{ N} \approx 33.33\text{ N}}$$
(c) Determination of Motion:
Comparing the two threshold forces:
$$F_{\text{topple}} = 33.33\text{ N} < F_{\text{slide}} = 50.0\text{ N}$$
The block will topple first at an applied force of $\mathbf{33.33\text{ N}}$, without any sliding taking place.
Example 2 (Dynamic Impact of a Falling Chain – Variable Mass):
A uniform flexible chain of total mass $M = 2.0\text{ kg}$ and length $L = 1.8\text{ m}$ is held vertically above a horizontal digital scale pan with its lowest link just grazing the surface. The chain is released from rest at $t = 0$ ($g = 10\text{ m/s}^2$).
(a) Express the instantaneous scale reading $N(x)$ when a length $x$ has landed.
(b) Calculate the reading of the scale at the instant $x = 0.60\text{ m}$ has fallen.
(c) Find the maximum scale reading $N_{\max}$ just before the final link comes to rest.
Solution:
Linear mass density of the chain: $\lambda = \frac{M}{L} = \frac{2.0}{1.8} = \frac{10}{9}\text{ kg/m}$.
(a) Instantaneous Scale Reading:
The scale reading consists of two distinct physical contributions:
1. Static weight of the accumulated links on the pan: $W_{\text{static}} = \lambda x g$.
2. Dynamic impulsive force due to the continuous destruction of momentum of falling links:
$$F_{\text{dynamic}} = \frac{dp}{dt} = v \frac{dm}{dt} = v (\lambda v) = \lambda v^2$$
Since the links fall freely under gravity: $v^2 = 2 g x$.
$$F_{\text{dynamic}} = \lambda (2 g x) = 2 \lambda x g$$
Summing both contributions:
$$\mathbf{N(x) = W_{\text{static}} + F_{\text{dynamic}} = \lambda x g + 2 \lambda x g = 3 \lambda x g = 3 \left(\frac{M}{L}\right) x g}$$
(b) Numerical Reading at $x = 0.60\text{ m}$:
$$N(0.60) = 3 \left(\frac{10}{9}\right)(0.60)(10) = 3 \left(\frac{10}{9}\right)(6.0) = 3 \times \frac{60}{9} = \mathbf{20.0\text{ N}}$$
(c) Maximum Scale Reading ($x = L$):
$$N_{\max} = 3 M g = 3(2.0\text{ kg})(10\text{ m/s}^2) = \mathbf{60.0\text{ N}}$$
Example 3 (Chain Sliding Off a Table – Work-Energy Theorem):
A flexible uniform chain of mass $M$ and length $L$ is placed on a smooth horizontal table with a fraction $y_0 = L/3$ hanging over the edge through a hole. The chain is released from rest.
(a) Derive the speed $v(y)$ of the chain as a function of the hanging length $y$.
(b) Find the speed of the chain at the precise instant the last link slips off the table ($y = L$).
Solution:
Taking the tabletop as the reference level for gravitational potential energy ($U = 0$).
For a hanging length $y$, the mass of the hanging portion is $m_{\text{hang}} = M \left(\frac{y}{L}\right)$. Its center of mass is located at depth $y/2$ below the table.
Initial potential energy:
$$U_i = -m_{\text{hang}} g \left(\frac{y_0}{2}\right) = -\left(\frac{M y_0}{L}\right) g \left(\frac{y_0}{2}\right) = -\frac{M g y_0^2}{2 L}$$
With $y_0 = L/3$:
$$U_i = -\frac{M g (L/3)^2}{2 L} = -\frac{M g L^2 / 9}{2 L} = -\frac{M g L}{18}$$
Potential energy when hanging length is $y$:
$$U(y) = -\frac{M g y^2}{2 L}$$
By conservation of mechanical energy ($K_i + U_i = K_f + U_f$):
$$0 + U_i = \frac{1}{2} M v(y)^2 + U(y) \implies \frac{1}{2} M v(y)^2 = U_i – U(y) = \frac{M g y^2}{2 L} – \frac{M g y_0^2}{2 L}$$
$$\mathbf{v(y) = \sqrt{\frac{g}{L} (y^2 – y_0^2)}}$$
(b) Speed when $y = L$:
$$v(L) = \sqrt{\frac{g}{L}\left(L^2 – \frac{L^2}{9}\right)} = \sqrt{\frac{g}{L}\left(\frac{8}{9} L^2\right)} = \mathbf{\frac{2}{3} \sqrt{2 g L} = \sqrt{\frac{8}{9} g L}}$$
Example 4 (Instantaneous Axis of Rotation (IAOR) – Constrained Sliding):
A uniform ladder $AB$ of length $L = 5.0\text{ m}$ slides with end $A$ constrained against a smooth vertical wall and end $B$ on a smooth horizontal floor. At the instant when the ladder makes an angle $\theta = 37^\circ$ with the horizontal floor, end $B$ is moving away from the wall with a speed $v_B = 3.0\text{ m/s}$ ($\sin 37^\circ = 0.6, \cos 37^\circ = 0.8$).
(a) Locate the coordinates of the Instantaneous Axis of Rotation (IAOR).
(b) Calculate the angular velocity $\omega$ of the ladder.
(c) Determine the downward speed $v_A$ of end $A$ and the speed $v_{\text{cm}}$ of the center of mass.
Solution:
Choose the corner where the wall meets the floor as the origin $(0, 0)$.
End $A$ is at $(0, L \sin\theta) = (0, 5.0 \times 0.6) = (0, 3.0\text{ m})$. Its velocity $\vec{v}_A$ is purely vertical (downward).
End $B$ is at $(L \cos\theta, 0) = (5.0 \times 0.8, 0) = (4.0\text{ m}, 0)$. Its velocity $\vec{v}_B$ is purely horizontal (rightward).
(a) Location of IAOR ($I$):
– The perpendicular to $\vec{v}_A$ is the horizontal line $y = L \sin\theta = 3.0\text{ m}$.
– The perpendicular to $\vec{v}_B$ is the vertical line $x = L \cos\theta = 4.0\text{ m}$.
The intersection yields the IAOR:
$$\mathbf{I = (L \cos\theta, L \sin\theta) = (4.0\text{ m}, 3.0\text{ m})}$$
(b) Angular Velocity ($\omega$):
Distance from IAOR to end $B$: $r_{IB} = L \sin\theta = 3.0\text{ m}$.
$$\mathbf{\omega = \frac{v_B}{r_{IB}} = \frac{3.0\text{ m/s}}{3.0\text{ m}} = 1.0\text{ rad/s}}$$
(c) Velocities of End $A$ and Center of Mass:
– Distance from IAOR to end $A$: $r_{IA} = L \cos\theta = 4.0\text{ m}$.
$$v_A = \omega r_{IA} = (1.0)(4.0) = \mathbf{4.0\text{ m/s}}$$
(Equivalently: $v_A = v_B \cot 37^\circ = 3.0 \times \frac{4}{3} = 4.0\text{ m/s}$).
– The center of mass is at $C = (2.0\text{ m}, 1.5\text{ m})$.
Distance from IAOR to $C$:
$$r_{IC} = \sqrt{(4.0 – 2.0)^2 + (3.0 – 1.5)^2} = \sqrt{2.0^2 + 1.5^2} = \sqrt{4.0 + 2.25} = \sqrt{6.25} = 2.5\text{ m} = \frac{L}{2}$$
$$v_{\text{cm}} = \omega r_{IC} = (1.0)(2.5) = \mathbf{2.5\text{ m/s}}$$
Practice Worksheet: Mechanics Doubt Resolution & Exam Scenarios
Section A: Single Correct Questions (Q1 to Q5)
Q1. A uniform solid block of mass $m$, height $h$, and square base of width $b$ rests on a rough horizontal surface with coefficient of friction $\mu$. A horizontal force $F$ is applied at the top edge ($y = h$). The necessary and sufficient condition for the block to topple before it slides is:
(A) $\mu > \frac{b}{2h}$
(B) $\mu < \frac{b}{2h}$
(C) $\mu > \frac{b}{h}$
(D) $\mu < \frac{b}{h}$
Q2. A uniform flexible chain of length $L$ and total mass $M$ is suspended vertically with its lower tip just touching the pan of a weighing scale. It is released from rest. At the instant a length $x$ of the chain has fallen on the pan, the reading of the scale is:
(A) $\frac{3 M g x}{L}$
(B) $\frac{2 M g x}{L}$
(C) $\frac{M g x}{L}$
(D) $\frac{4 M g x}{L}$
Q3. A ladder of length $L$ leans against a frictionless vertical wall with its lower end on a frictionless horizontal floor. If the lower end moves away from the wall with constant horizontal velocity $v_0$, the instantaneous downward speed of the top end when the ladder makes an angle $\theta$ with the horizontal floor is:
(A) $v_0 \cot\theta$
(B) $v_0 \tan\theta$
(C) $v_0 \sin\theta$
(D) $v_0 \cos\theta$
Q4. A uniform chain of length $L$ lies on a smooth horizontal table with one-third of its length hanging over the edge. If the chain is released from rest, the speed of the chain at the moment its last link leaves the table is:
(A) $\frac{2}{3}\sqrt{2 g L}$
(B) $\frac{1}{3}\sqrt{8 g L}$
(C) $\frac{4}{3}\sqrt{g L}$
(D) $\sqrt{\frac{2 g L}{3}}$
Q5. A rocket of initial mass $M_0$ ejects exhaust gases with a constant relative exhaust velocity $u$ at a constant mass burn rate $r = -\frac{dM}{dt}$. In gravity-free space, the velocity $v$ of the rocket after time $t$ is:
(A) $u \ln\left(\frac{M_0}{M_0 – r t}\right)$
(B) $u \ln\left(\frac{M_0 – r t}{M_0}\right)$
(C) $\frac{u r t}{M_0}$
(D) $u \left(1 – \frac{r t}{M_0}\right)$
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Regarding the work done by friction between two contacting surfaces, which of the following statements is/are correct?
(A) The work done by static friction on an individual body can be positive, negative, or zero depending on the reference frame.
(B) The net work done by static friction on an interacting two-body system is strictly zero in all reference frames.
(C) The net work done by kinetic friction on an interacting two-body system is strictly negative and frame-independent.
(D) Kinetic friction acting on a single body can do positive work in a properly chosen inertial frame.
Q7. A uniform solid sphere of mass $M$ and radius $R$ is placed on a rough horizontal floor with kinetic friction coefficient $\mu$ and projected with initial pure translational velocity $v_0$ without rotation ($\omega_0 = 0$). Which of the following statements is/are correct?
(A) Kinetic friction initially acts in the backward direction.
(B) The angular velocity of the sphere increases until pure rolling is established.
(C) Angular momentum about any fixed point on the floor is conserved throughout the transition.
(D) The final linear speed when pure rolling begins is $v = \frac{5}{7} v_0$.
Q8. A small bead of mass $m$ is threaded on a smooth circular wire hoop of radius $R$ fixed in a vertical plane. The bead is projected from the lowest point with horizontal speed $u$. Which of the following statements is/are correct?
(A) Because the hoop constrains the bead radially on both sides, the normal contact reaction can point either radially inward or outward.
(B) The bead can complete the full circular loop without falling off if $u \ge 2\sqrt{g R}$.
(C) If $u = \sqrt{3 g R}$, the normal reaction force acting on the bead changes sign from inward to outward at some point during the motion.
(D) The speed of the bead at the topmost point when launched with $u = 2\sqrt{g R}$ is zero.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A uniform flexible chain of mass $M$ and length $L$ is held vertically with its lower end just touching a table and released from rest. The maximum normal force registered by the table at the instant the final link strikes the table is $N_{\max} = n M g$. Find the integer value of $n$.
Q10. A uniform chain of length $L$ lies on a smooth horizontal table with a fraction $1/4$ of its length hanging over the edge. It is released from rest. The velocity of the chain when its last link slips off the table is $v = \sqrt{\frac{k}{16} g L}$. Determine the integer value of $k$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (A) | Toppling vs. Sliding Boundary Condition: $\mu > \frac{b}{2h}$ |
| Q2 | (A) | Falling Chain Impact: $N = \lambda x g + \lambda v^2 = 3 \lambda x g$ |
| Q3 | (A) | Constraint Kinematics: $x^2 + y^2 = L^2 \implies v_y = v_0 \cot\theta$ |
| Q4 | (A) | Chain Potential Energy Loss: $v = \sqrt{\frac{8}{9} g L} = \frac{2}{3}\sqrt{2 g L}$ |
| Q5 | (A) | Tsiolkovsky Rocket Equation: $v = u \ln\left(\frac{M_0}{M_0 – r t}\right)$ |
| Q6 | (A, B, C, D) | Thermodynamics & Frame Invariance of Friction Work |
| Q7 | (A, B, C, D) | Angular Momentum Conservation About Floor Point: $v = \frac{5}{7} v_0$ |
| Q8 | (A, B, C, D) | Bilateral Radial Constraint in Vertical Circular Wire Hoop |
| Q9 | 3 | Chain Maximum Impact Force Factor: $N_{\max} = 3 M g \implies n = 3$ |
| Q10 | 15 | Overhanging Chain Energy Balance: $\frac{1}{2} M v^2 = \frac{15}{32} M g L \implies k = 15$ |
Detailed Explanations
Solution 1:
The force required to initiate sliding is:
$$F_{\text{slide}} = \mu_s m g$$
The force required to initiate toppling about the front bottom edge:
$$\tau_{\text{topple}} = F h, \quad \tau_{\text{restoring}} = m g \left(\frac{b}{2}\right) \implies F_{\text{topple}} = \frac{m g b}{2 h}$$
For the block to topple before it slides, the toppling force must be strictly less than the sliding force:
$$F_{\text{topple}} < F_{\text{slide}} \implies \frac{m g b}{2 h} < \mu_s m g \implies \mathbf{\mu_s > \frac{b}{2 h}}$$
Correct Answer: (A)
Solution 2:
When length $x$ of the chain has fallen on the scale:
1. Static weight of the fallen portion:
$$W_{\text{static}} = m(x) g = \left(\frac{M}{L} x\right) g = \lambda x g$$
2. Dynamic thrust force due to momentum destruction of incoming links:
The mass element $dm = \lambda dx$ enters the pan with free-fall velocity $v = \sqrt{2 g x}$.
$$F_{\text{thrust}} = v \frac{dm}{dt} = v (\lambda v) = \lambda v^2 = \lambda (2 g x) = 2 \lambda x g$$
Total scale reading:
$$N(x) = W_{\text{static}} + F_{\text{thrust}} = \lambda x g + 2 \lambda x g = \mathbf{3 \lambda x g = \frac{3 M g x}{L}}$$
Correct Answer: (A)
Solution 3:
Let $x$ be the horizontal distance of the bottom end from the wall and $y$ be the vertical height of the top end:
$$x^2 + y^2 = L^2$$
Differentiating with respect to time $t$:
$$2 x \frac{dx}{dt} + 2 y \frac{dy}{dt} = 0 \implies x v_0 + y (-v_y) = 0$$
$$v_y = v_0 \frac{x}{y}$$
From geometry, $\frac{x}{y} = \cot\theta$:
$$\mathbf{v_y = v_0 \cot\theta}$$
Correct Answer: (A)
Solution 4:
Taking the tabletop as the reference level for potential energy ($U = 0$):
– Initial state ($y_0 = L/3$ hanging):
Mass of hanging part is $M/3$, and its center of mass is at depth $L/6$:
$$U_i = -\left(\frac{M}{3}\right) g \left(\frac{L}{6}\right) = -\frac{M g L}{18}$$
– Final state (entire chain of length $L$ hanging):
Center of mass is at depth $L/2$:
$$U_f = -M g \left(\frac{L}{2}\right) = -\frac{M g L}{2}$$
By conservation of mechanical energy:
$$\frac{1}{2} M v^2 = U_i – U_f = -\frac{M g L}{18} – \left(-\frac{M g L}{2}\right) = \frac{M g L}{2} – \frac{M g L}{18} = \frac{8 M g L}{18} = \frac{4}{9} M g L$$
$$v^2 = \frac{8}{9} g L \implies \mathbf{v = \frac{2}{3}\sqrt{2 g L}}$$
Correct Answer: (A)
Solution 5:
From Newton’s second law for a variable mass system with constant exhaust velocity $u$ and zero external forces:
$$M \frac{dv}{dt} = -u \frac{dM}{dt} \implies dv = -u \frac{dM}{M}$$
Integrating from initial mass $M_0$ (at $v = 0$) to mass $M(t) = M_0 – r t$:
$$v = -u \int_{M_0}^{M_0 – r t} \frac{dM}{M} = -u \left[\ln(M_0 – r t) – \ln M_0\right] = \mathbf{u \ln\left(\frac{M_0}{M_0 – r t}\right)}$$
Correct Answer: (A)
Solution 6:
– (A) True: In an accelerating reference frame or when one body is moved by friction, static friction does work $\vec{f}_s \cdot d\vec{r}$, which can be positive, negative, or zero depending on the frame.
– (B) True: In an isolated two-body contact, action and reaction static forces are equal and opposite, and their points of application have identical instantaneous displacements ($d\vec{r}_1 = d\vec{r}_2$); hence $\vec{f}_s \cdot d\vec{r}_1 + (-\vec{f}_s) \cdot d\vec{r}_2 = 0$.
– (C) True: The net work done by kinetic friction is $-f_k \Delta s_{\text{rel}} < 0$, which is the dissipated thermal energy; this quantity is strictly frame-invariant.
- (D) True: Consider a box dropped onto a moving conveyor belt from the lab frame: kinetic friction on the box acts forward in the direction of its displacement, doing positive work on the box until pure rolling/stationary velocity is reached.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 7:
– (A) True: The bottom contact point has velocity $v_0$ forward; friction opposes relative slip and acts backward.
– (B) True: Backward friction produces a clockwise torque $\tau = f_k R$ about CM, increasing $\omega$.
– (C) True: About any point on the floor line, the line of action of friction passes through the point; normal force balances gravity. Hence net external torque about any floor point is zero: $\vec{L}_{\text{floor}} = \text{constant}$.
– (D) True: Conservation of angular momentum about a floor point:
$$L_i = M v_0 R = L_f = M v R + I_{\text{cm}} \omega$$
When pure rolling is established, $\omega = v / R$:
$$M v_0 R = M v R + \left(\frac{2}{5} M R^2\right) \left(\frac{v}{R}\right) = M v R + \frac{2}{5} M v R = \frac{7}{5} M v R \implies \mathbf{v = \frac{5}{7} v_0}$$
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
– (A) True: Unlike a string which can only pull, a rigid circular wire hoop provides bilateral normal forces (both inward and outward normal reactions).
– (B) True: Because the wire supports the bead from falling, the bead only requires non-zero speed to pass the highest point ($v_{\text{top}} \ge 0$). Energy conservation: $\frac{1}{2} m u^2 = \frac{1}{2} m v_{\text{top}}^2 + m g (2R) \implies u_{\min} = \sqrt{4 g R} = 2\sqrt{g R}$.
– (C) True: For $u = \sqrt{3 g R}$, the normal reaction passes through zero and changes direction during the ascent.
– (D) True: With $u = 2\sqrt{g R}$, $v_{\text{top}} = \sqrt{u^2 – 4 g R} = 0$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
When the final link strikes the table ($x = L$):
$$N(L) = 3 \lambda L g = 3 \left(\frac{M}{L}\right) L g = 3 M g$$
Comparing with $n M g$:
$$\mathbf{n = 3}$$
Correct Answer: 3
Solution 10:
Initial hanging length is $y_0 = L/4$:
$$U_i = -\left(\frac{M}{4}\right) g \left(\frac{L}{8}\right) = -\frac{M g L}{32}$$
When the entire chain slips off ($y = L$):
$$U_f = -M g \left(\frac{L}{2}\right) = -\frac{16 M g L}{32}$$
Gain in kinetic energy:
$$\frac{1}{2} M v^2 = U_i – U_f = -\frac{M g L}{32} – \left(-\frac{16 M g L}{32}\right) = \frac{15}{32} M g L$$
$$v^2 = \frac{15}{16} g L \implies v = \sqrt{\frac{15}{16} g L}$$
Comparing with $\sqrt{\frac{k}{16} g L}$:
$$\mathbf{k = 15}$$
Correct Answer: 15