CBSE Class 11 Physics Work Energy Power Basic Problems

BASIC LEVEL ASSIGNMENT

1. A small disc of mass m slides down a smooth hill of height h without initial velocity and gets onto a plank of mass M lying on the horizontal plane at the base of hill as shown in the figure. Due to friction between the disc and the plank, disc slows down and beginning with a certain moment, moves in one piece with the plank. Find out total work performed by the frictional forces in this process. [5 Marks]

[Figure: A hill of height h on the left, a small disc of mass m at the top, and a plank of mass M on a horizontal surface at the base of the hill where the disc moves onto it]

Answer:
1. Initial mechanical energy of the disc at the top of the hill is \(E_i\) = mgh.
2. Let the final common velocity of the disc and the plank be V. By conservation of linear momentum, mv = (m + M)V, where v is the velocity of the disc just before reaching the plank.
3. Since the hill is smooth, \(\frac{1}{2}\) mv² = mgh, which gives v = \(\sqrt{2gh}\).
4. The final common velocity is V = \(\frac{m \sqrt{2gh}}{m + M}\).
5. By the work-energy theorem, the total work done by the frictional force is equal to the change in the total kinetic energy of the system.
6. Total work W = \(\frac{1}{2}\) (m + M)V² - \(\frac{1}{2}\) mv² = \(-\frac{M}{m + M}\) mgh.

Teacher's Note:
a) The principle of conservation of mechanical energy is applied during the descent along the smooth hill.
b) Total momentum is conserved during the inelastic collision with the plank because external horizontal forces are absent.
c) The negative sign in the work done indicates that energy is dissipated as heat by non-conservative frictional forces.
d) Always verify internal energy losses using the initial potential energy and final kinetic energy of the combined system.

2. A block of mass m starts from rest to slide along a smooth frictionless track of the shape shown in the figure. What should be height h so that when the mass reaches point A on the track, it pushes the track with a force equal to thrice its weight ? [5 Marks]

[Figure: A track descending from height h and forming a circular loop of radius r at the top with point A marked at the highest point of the loop]

Answer:
1. Let r be the radius of the circular loop at point A.
2. At point A, the normal force N exerted by the block on the track is equal to thrice its weight, so N = 3mg.
3. By Newton's second law at point A, N + mg = \(\frac{mv_A^2}{r}\), where \(v_A\) is the velocity at point A.
4. Substituting N = 3mg gives 3mg + mg = \(\frac{mv_A^2}{r}\), which simplifies to \(v_A^2\) = 4gr.
5. Using conservation of mechanical energy between the starting point at height h and point A at height 2r: mgh = mg(2r) + \(\frac{1}{2}\) \(mv_A^2\).
6. Substituting \(v_A^2\) = 4gr into the energy equation gives mgh = 2mgr + 2mgr = 4mgr, which results in h = 4r.

Teacher's Note:
a) Centripetal force at the topmost point of the circular loop requires both gravity and the normal reaction.
b) Students often forget to include the height of the circular loop (2r) in the total potential energy at point A.
c) Energy conservation holds true because the track is stated to be smooth and frictionless.
d) Check the limiting case where a minimum height of 2.5r is required just to complete the vertical loop without losing contact.

3. Two blocks of masses m₁ = 2 kg and m₂ = 5 kg are moving in the same direction along a frictionless surface with speeds 10 m/s and 3 m/s respectively, m₂ being ahead of m₁. An ideal spring with k = 1120 N/m is attached to the back side of m₂. Find the maximum compression of the spring when the blocks collide. [5 Marks]

[Figure: Two blocks of masses m₁ and m₂ moving in the same direction with an ideal spring of stiffness k placed between them]

Answer:
1. Given data: m₁ = 2 kg, m₂ = 5 kg, u₁ = 10 m/s, u₂ = 3 m/s, k = 1120 N/m.
2. At maximum compression of the spring, both blocks move with the same common velocity v.
3. By conservation of linear momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v.
4. Substituting the values: (2 × 10) + (5 × 3) = (2 + 5)v, which gives 20 + 15 = 7v, so v = 5 m/s.
5. By conservation of mechanical energy, the initial kinetic energy in the center of mass frame is converted into maximum spring potential energy: \(\frac{1}{2}\) m₁u₁² + \(\frac{1}{2}\) m₂u₂² = \(\frac{1}{2}\) (m₁ + m₂)v² + \(\frac{1}{2}\) kx_max².
6. Substituting values and solving for x_max gives the maximum compression x_max = 0.25 m.

Teacher's Note:
a) Maximum compression occurs precisely when the relative velocity between the two colliding blocks becomes zero.
b) Kinetic energy is not conserved during the entire impact, but total mechanical energy of the isolated system remains constant.
c) Using reduced mass \(\mu = \frac{m_1 m_2}{m_1 + m_2}\) provides a quicker alternative calculation for spring energy problems.
d) Always ensure consistent SI units throughout the substitution steps before computing the final answer.

4. An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P ; show that : [5 Marks]
(a) The velocity is given as a function of time by v = (2 Pt/m)^1/2
(b) The position is given as a function of time by s = (8 P/9m)^1/2 t^3/2.

Answer:
1. Power is defined as P = Fv = m \(\frac{dv}{dt}\) v = constant.
2. Separating variables for velocity: v dv = \(\frac{P}{m}\) dt.
3. Integrating both sides from t = 0 (v = 0) to time t: \(\int_{0}^{v}\) v dv = \(\frac{P}{m}\) \(\int_{0}^{t}\) dt, which yields \(\frac{v^2}{2}\) = \(\frac{Pt}{m}\).
4. Solving for velocity gives v = \(\left(\frac{2Pt}{m}\right)^{1/2}\), which proves part (a).
5. For part (b), substitute v = \(\frac{ds}{dt}\): \(\frac{ds}{dt}\) = \(\left(\frac{2P}{m}\right)^{1/2}\) t^1/2.
6. Integrating position s from 0 to s with respect to time t: s = \(\left(\frac{2P}{m}\right)^{1/2}\) \(\frac{t^{3/2}}{3/2}\) = \(\left(\frac{8P}{9m}\right)^{1/2}\) t^3/2, which proves part (b).

Teacher's Note:
a) The fundamental relation between power, force, and velocity is the starting point for variable acceleration problems.
b) Definite integration helps apply initial conditions directly without needing an explicit constant of integration.
c) Notice that acceleration decreases as time increases because power remains constant while velocity increases.
d) Dimensional consistency must be checked for both derived expressions to ensure mathematical validity.

5. A massless platform is kept on a light elastic spring, as shown in the figure. When a sand particle of mass 0.1 kg is dropped on the pan from a height of 0.2 m, the particle strikes the pan and sticks to it while the spring compresses by 0.01 m. From what height should the particle be dropped to cause a compression of 0.04 m ? [5 Marks]

[Figure: A massless platform resting on an upright vertical spring with a sand particle of mass 0.1 kg dropped from height h above it]

Answer:
1. Let m = 0.1 kg, initial drop height h₁ = 0.2 m, and maximum compression x₁ = 0.01 m.
2. By conservation of mechanical energy for the first case, gravitational potential energy lost is equal to the elastic potential energy gained by the spring: mg(h₁ + x₁) = \(\frac{1}{2}\) kx₁².
3. Substituting the given values: (0.1)g(0.2 + 0.01) = \(\frac{1}{2}\) k(0.01)², which gives 0.021 g = \(\frac{1}{2}\) k(0.0001).
4. For the second case with compression x₂ = 0.04 m and required drop height h₂: mg(h₂ + x₂) = \(\frac{1}{2}\) kx₂².
5. Taking the ratio of the two energy equations: \(\frac{h_1 + x_1}{h_2 + x_2}\) = \(\frac{x_1^2}{x_2^2}\).
6. Substituting known values: \(\frac{0.2 + 0.01}{h_2 + 0.04}\) = \(\frac{(0.01)^2}{(0.04)^2}\) = \(\frac{1}{16}\), which yields h₂ + 0.04 = 0.21 × 16 = 3.36 m, so h₂ = 3.32 m.

Teacher's Note:
a) The total fall distance includes both the initial drop height and the additional downward compression distance of the spring.
b) Setting up ratios between two physical states eliminates the need to calculate the unknown spring constant k explicitly.
c) Inelastic collision occurs when the particle sticks to the pan, but mechanical energy is still conserved from the initial drop to maximum compression.
d) Always verify that distance units are uniform before forming ratios or substituting into energy equations.

6. A spring gun has spring constant 100 N/m. A small ball of mass 0.1 kg is put in its barrel by compressing the spring through 0.05 m as shown in the figure [3 Marks]
(a) Find the velocity of the ball when the spring acquires normal after length after being released.
(b) Where should a box be placed on the ground so that the ball falls in it, if the ball leaves the gun horizontally at a height of 2 m above the ground ? (g = 10 m/s²)

Answer:
1. Given data: Spring constant \(k = 100\text{ N/m}\), mass of the ball \(m = 0.1\text{ kg}\), compression \(x = 0.05\text{ m}\), height \(h = 2\text{ m}\), and \(g = 10\text{ m/s}^2\).
2. (a) Using the law of conservation of mechanical energy, potential energy of the compressed spring is converted into kinetic energy of the ball: \(\frac{1}{2}kx^2 = \frac{1}{2}mv^2\).
3. Substituting the values: \(100 \times (0.05)^2 = 0.1 \times v^2 \implies 100 \times 0.0025 = 0.1 \times v^2 \implies 0.25 = 0.1v^2\).
4. Solving for velocity: \(v = \sqrt{\frac{0.25}{0.1}} = \sqrt{2.5} = 1.58\text{ m/s}\).
5. (b) Time taken to fall from height \(h = 2\text{ m}\) is given by \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 2}{10}} = \sqrt{0.4} = 0.632\text{ s}\).
6. Horizontal distance (distance \(d\) where the box should be placed) is \(d = v \times t = 1.58 \times 0.632 = 1.0\text{ m}\).

Teacher's Note:
a) The principle of conservation of energy equates elastic potential energy to translational kinetic energy.
b) Horizontal projectile motion treats horizontal and vertical components of motion independently.
c) Always ensure consistent SI units for mass, length, and time before final calculation.

7. Two blocks are connected by a string as shown in the figure. They are released from rest. Show that after they have moved a distance L, their common speed is given by \(v = \sqrt{\frac{2(m_2 - \mu m_1)gL}{m_1 + m_2}}\), where \(\mu\) is the coefficient of friction between the floor and the blocks. [3 Marks]

Answer:
1. Let the system move through a distance \(L\) with a common speed \(v\).
2. Work done against friction on block \(m_1\) is \(W_f = -\mu m_1 g L\).
3. Loss in gravitational potential energy of block \(m_2\) as it moves down by distance \(L\) is \(\Delta U = m_2 g L\).
4. By work-energy theorem, total work done on the system equals the change in total kinetic energy: \(m_2 g L - \mu m_1 g L = \frac{1}{2}(m_1 + m_2)v^2\).
5. Simplifying the expression for \(v^2\): \(v^2 = \frac{2(m_2 - \mu m_1)gL}{m_1 + m_2}\).
6. Taking the square root gives the required common speed: \(v = \sqrt{\frac{2(m_2 - \mu m_1)gL}{m_1 + m_2}}\).

Teacher's Note:
a) Work-energy theorem simplifies multi-body connected systems by ignoring internal tension forces.
b) Friction acts only on the horizontal surface, hence the friction term involves \(m_1\) only.
c) The hanging mass \(m_2\) provides the driving gravitational force for the entire system.

8. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration \(a_c\) is varying with time t as \(a_c = k^2 r t^2\) when k is a constant. What is the power delivered to the particle by the forces acting on it ? [2 Marks]

Answer:
1. The centripetal acceleration is given as \(a_c = \frac{v^2}{r} = k^2 r t^2\).
2. Solving for the square of the linear velocity: \(v^2 = k^2 r^2 t^2\), which gives speed \(v = k r t\).
3. Tangential acceleration \(a_t\) is the rate of change of speed: \(a_t = \frac{dv}{dt} = \frac{d}{dt}(k r t) = k r\).
4. Tangential force acting on the particle is \(F_t = m a_t = m k r\).
5. Power delivered to the particle is given by \(P = F_t \times v = (m k r) \times (k r t) = m k^2 r^2 t\).

Teacher's Note:
a) Centripetal force is perpendicular to velocity and does zero work, so power is delivered solely by the tangential force.
b) Differentiating the linear speed with respect to time yields the tangential acceleration.

9. A stone with weight w is thrown vertically upward into the air with initial speed \(v_0\). If a constant force f due to air drag acts on the stone throughout its flight : [4 Marks]
(a) Show that the maximum height reached by the stone is \(h = \frac{v_0^2}{2g[1 + (f/w)]}\)
(b) Show that the speed of the stone upon impact with the ground is \(v = v_0 \left(\frac{w - f}{w + f}\right)^{1/2}\)

Answer:
1. (a) During upward motion, both gravity and air drag act downward, so the total retarding force is \(F_{\text{up}} = w + f = mg + f\).
2. Acceleration during upward motion is \(a_{\text{up}} = \frac{w + f}{m} = g + \frac{f}{m} = g\left(1 + \frac{f}{w}\right)\).
3. Using kinematic equation \(v^2 - u^2 = 2as\) for maximum height \(h\) where final velocity is zero: \(0 - v_0^2 = -2a_{\text{up}}h\), which gives \(h = \frac{v_0^2}{2a_{\text{up}}} = \frac{v_0^2}{2g[1 + (f/w)]}\).
4. (b) During downward motion, gravity acts downward while air drag acts upward, so the net downward force is \(F_{\text{down}} = w - f = mg - f\).
5. Acceleration during downward motion is \(a_{\text{down}} = g - \frac{f}{m} = g\left(1 - \frac{f}{w}\right)\).
6. Using \(v^2 = u^2 + 2as\) with initial speed zero for downward journey from height \(h\): \(v^2 = 2 a_{\text{down}} h = 2 \left(g - \frac{f}{m}\right) \times \frac{v_0^2}{2g[1 + (f/w)]}\).
7. Simplifying and taking the square root yields the final impact speed: \(v = v_0 \left(\frac{w - f}{w + f}\right)^{1/2}\).

Teacher's Note:
a) Air drag always opposes motion, acting downwards during ascent and upwards during descent.
b) Substituting weight \(w = mg\) helps express forces in terms of dimensionless ratios involving \(f/w\).
c) Separate acceleration terms must be derived for upward and downward journeys due to the reversal of the drag force direction.
d) The final impact speed is less than the initial projection speed because mechanical energy is lost against non-conservative air resistance forces.

10. One end of a spring of natural length h is fixed at the ground and the other end is fitted with a smooth ring of mass m which is allowed to slide on a horizontal rod fixed at a height h as shown in figure. Initially, the spring makes an angle of 37° with the vertical when the system is released from rest. Find the speed of the ring when the spring becomes vertical. [4 Marks]

Answer:
1. Let the fixed point on the ground be origin \((0, 0)\). The rod is at \(y = h\).
2. In the initial position, the spring makes an angle \(\theta = 37^\circ\) with the vertical, so the initial length of the spring is \(l_1 = \frac{h}{\cos 37^\circ} = \frac{h}{4/5} = \frac{5}{4}h\).
3. Initial extension of the spring is \(x_1 = l_1 - h = \frac{5}{4}h - h = \frac{1}{4}h\).
4. When the spring becomes vertical, its length is \(l_2 = h\), so the final extension is zero (\(x_2 = 0\)).
5. The initial height of the ring above the ground is \(h\) and final height is also \(h\), so there is no change in gravitational potential energy.
6. Applying conservation of mechanical energy: Initial potential energy of spring = Final kinetic energy of ring + Final potential energy of spring, which gives \(\frac{1}{2}k x_1^2 = \frac{1}{2}mv^2 + 0\).
7. Substituting \(x_1 = \frac{1}{4}h\): \(\frac{1}{2}k \left(\frac{h}{4}\right)^2 = \frac{1}{2}mv^2 \implies v = \frac{h}{4} \sqrt{\frac{k}{m}}\).

Teacher's Note:
a) Trigonometric relations help determine the initial stretched length of the spring from the vertical height \(h\).
b) Gravitational potential energy remains constant throughout since the ring moves strictly along a horizontal rod at height \(h\).
c) Energy conservation equates the initial elastic potential energy directly to the final kinetic energy of the ring.
d) Standard trigonometric values like \(\cos 37^\circ = 4/5\) simplify algebraic substitutions in such problems.

11. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring of spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P, where the radius of the track is horizontal. [3 Marks]

[Figure: A horizontal smooth track connected to a vertical circular track of radius R. A spring of constant k is fixed to a wall on the left, with a block of mass m touching it. Point P is on the circular track where the radius is horizontal.]

Answer:
1. Let the initial compression of the spring be x.
2. Potential energy stored in the spring = \(\frac{1}{2}\) k x².
3. At point P, the radius is horizontal, so the height of point P above the initial horizontal level is R.
4. At point P, the normal force exerted by the track on the block is given as mg. Since the centripetal force is provided by the normal force, mg = \(\frac{mv^2}{R}\), which gives v² = gR.
5. Applying the law of conservation of mechanical energy between the initial compressed state of the spring and point P: \(\frac{1}{2}\) k x² = mgR + \(\frac{1}{2}\) m v².
6. Substituting v² = gR into the energy equation: \(\frac{1}{2}\) k x² = mgR + \(\frac{1}{2}\) m(gR) = \(\frac{3}{2}\) mgR.
7. Solving for x, the initial compression is x = \(\sqrt{\frac{3mgR}{k}}\).

Teacher's Note:
a) Apply conservation of energy equating spring potential energy to gravitational potential energy plus kinetic energy at point P.
b) Remember that the normal force at P equals mg, which directly balances the required centripetal force \(\frac{mv^2}{R}\).
c) Check that height at point P is exactly R above the reference horizontal level.

12. A particle of mass 100 g is suspended from one end of a weightless string of length 100 cm and is allowed to swing in a vertical plane. The speed of the mass is 200 cm/s when the string makes an angle of 60° with the vertical. Determine
(a) the tension in the string at 60° and
(b) the speed of the particle when it is at the lowest position. (g = 980 cm/s²). [3 Marks]

Answer:
1. Given mass m = 100 g = 0.1 kg, length L = 100 cm = 1 m, velocity v = 200 cm/s = 2 m/s at \(\theta = 60^\circ\), and g = 980 cm/s² = 9.8 m/s².
2. (a) The tension in the string at angle \(\theta = 60^\circ\) is given by T - mg cosθ = \(\frac{mv^2}{L}\).
3. Substituting the values: T = mg cos(60°) + \(\frac{mv^2}{L}\) = (100 × 980 × 0.5) + \(\frac{100 \times (200)^2}{100}\) = 49000 + 40000 = 89000 dynes = 8.9 × 10⁴ dynes = 0.89 N.
4. (b) Let v₀ be the speed at the lowest position. Using conservation of mechanical energy between the position at 60° and the lowest point: \(\frac{1}{2}\) m v² + mgL(1 - cos 60°) = \(\frac{1}{2}\) m v₀².
5. Substituting the values: \(\frac{1}{2}\) (100) (200)² + (100)(980)(100)(1 - 0.5) = \(\frac{1}{2}\) (100) v₀².
6. 2000000 + 4900000 = 50 v₀² ⇒ 6900000 = 50 v₀² ⇒ v₀² = 138000.
7. Therefore, v₀ = \(\sqrt{138000}\) = 371.48 cm/s = 3.71 m/s.

Teacher's Note:
a) Ensure consistent use of either CGS or SI units throughout the calculation.
b) The vertical drop from the 60° position to the lowest point is L(1 - cosθ).
c) Tension depends on both the weight component along the string and the centrifugal force.

13. A small block of mass m slides along a smooth frictional track as shown in the figure.
(a) If it starts from rest at P, what is the resultant force acting on it at Q?
(b) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight? [4 Marks]

[Figure: A roller coaster style track showing a release point P at height 5R, a circular loop of radius R, and a point Q on the side of the loop at height R.]

Answer:
1. (a) At point P (height 5R), total mechanical energy is E = mg(5R).
2. Using energy conservation, the kinetic energy at point Q (height R) is \(\frac{1}{2}\) mv_Q² = mg(5R) - mg(R) = 4mgR, which gives v_Q² = 8gR.
3. At point Q, the radius is horizontal, so the centripetal force is F_c = \(\frac{mv_Q^2}{R}\) = 8mg.
4. The gravitational force acting vertically is mg. The normal force from the track provides the centripetal force, N = 8mg. The resultant force is the vector sum of weight and normal force: F_net = \(\sqrt{(mg)^2 + (N)^2}\) = \(\sqrt{(mg)^2 + (8mg)^2}\) = mg\(\sqrt{65}\).
5. (b) Let the release height be h. At the top of the loop (height 2R), the force against the track equals weight, so N = mg.
6. The centripetal force equation at the top is N + mg = \(\frac{mv_{top}^2}{R}\) ⇒ 2mg = \(\frac{mv_{top}^2}{R}\) ⇒ v_top² = 2gR.
7. Using energy conservation between release height h and the top of the loop: mgh = mg(2R) + \(\frac{1}{2}\) m v_top² = 2mgR + \(\frac{1}{2}\) m(2gR) = 3mgR.
8. Solving for h gives h = 3R.

Teacher's Note:
a) Part (a) requires combining perpendicular forces (tangential/weight and radial/normal) vectorially.
b) Part (b) uses standard critical condition equations for loops in vertical circular motion.
c) Watch out for the distinction between height above ground and height above the bottom of the loop.
d) Verify energy conservation at every transition point carefully.

14. A particle slides along a track with elevated ends and a flat central part as shown in the figure. The flat part has a length l = 3.0 m. The curved portions of the track are frictionless. For the flat part the coefficient of kinetic friction is \(\mu_k\) = 0.20, the particle is released at point A which is at a height h = 1.5 m above the flat part of the track. Where does the particle finally come to rest? [4 Marks]

[Figure: A track with curved elevated ends A and D, and a flat horizontal central part of length l between B and C. Point A is at height h.]

Answer:
1. Given mgh = initial potential energy at A. Let total distance traveled on the flat rough part be d.
2. By work - energy theorem, the loss in potential energy is equal to the work done against friction on the flat track.
3. mgh = \(\mu_k\) mg d.
4. Cancelling mass m on both sides: gh = \(\mu_k\) g d, which simplifies to d = \(\frac{h}{\mu_k}\).
5. Substituting the given values: h = 1.5 m and \(\mu_k\) = 0.20, we get d = \(\frac{1.5}{0.20}\) = 7.5 m.
6. The length of the flat track is l = 3.0 m. The particle crosses the flat track multiple times before coming to rest.
7. Number of full crossings n = \(\frac{d}{l}\) = \(\frac{7.5}{3.0}\) = 2.5.
8. This means the particle traverses the flat part 2 full times and goes an extra 1.5 m into the third crossing, coming to rest at a distance of 1.5 m from point B (or C) on the flat track.

Teacher's Note:
a) Friction only performs work on the flat horizontal portion of the track.
b) Total distance calculation relies entirely on initial potential energy equating to total work done by friction.
c) Interpreting the decimal remainder helps locate the exact final stopping position on the flat segment.
d) Verify that curved tracks are frictionless, meaning no energy loss occurs during the descent and ascent on the sides.

Basic Level Assignment

1. Answer key item: mgh\(\left(\frac{M}{M+m}\right)\) [1 Mark]

Answer:
1. This corresponds to the final expression for the result of problem 1 in the assignment.
2. It represents energy conservation or work done in a system of combined masses M and m.

Teacher's Note:
a) Represents derived mechanical energy expression.
b) Useful for problems involving inelastic collisions and combined masses.

2. Answer key item: 4 R [1 Mark]

Answer:
1. This corresponds to the numerical or algebraic result of problem 2.
2. It is expressed in terms of radius R.

Teacher's Note:
a) Often related to critical velocity or vertical circular motion radii.
b) Check dimensional consistency.

3. Answer key item: 0.25 m [1 Mark]

Answer:
1. This corresponds to the length or displacement value for problem 3.
2. The value is 0.25 meters.

Teacher's Note:
a) Standard SI unit meters used for length.
b) Verify conversions if centimetres were given originally.

5. Answer key item: 3.96 m [1 Mark]

Answer:
1. This corresponds to the solution value for problem 5.
2. The distance or height calculated is 3.96 m.

Teacher's Note:
a) Obtained by applying conservation of energy equations.
b) Keep appropriate significant figures.

6. Answer key item: (a) \(\sqrt{2.5}\) m/s (b) 1 m [1 Mark]

Answer:
1. Part (a) gives the velocity as \(\sqrt{2.5}\) m/s.
2. Part (b) gives the distance or height as 1 m.

Teacher's Note:
a) Multi-part answer combining kinematics and energy.
b) Use \(v = \sqrt{2gh}\) or similar relations.

8. Answer key item: m k² r² t [1 Mark]

Answer:
1. This is the algebraic expression for problem 8.
2. It combines mass m, radius of gyration k, radius r, and time t terms.

Teacher's Note:
a) Dimensional analysis can verify correctness.
b) Associated with rotational dynamics or power.

10. Answer key item: \(\frac{h}{4} \sqrt{\frac{k}{m}}\) [1 Mark]

Answer:
1. This represents the expression from problem 10.
2. It involves height h, spring constant k, and mass m.

Teacher's Note:
a) Combines potential energy and simple harmonic motion concepts.
b) Ensure square root terms cover the correct fraction.

11. Answer key item: \(\sqrt{\frac{3mg R}{k}}\) [1 Mark]

Answer:
1. This is the resulting formula for problem 11.
2. It involves mass m, acceleration due to gravity g, radius R, and spring/restoring constant k.

Teacher's Note:
a) Derived using work-energy theorem with elastic forces.
b) Pay close attention to the factor of 3 inside the numerator.

12. Answer key item: 8.9 × 10⁴ dynes, 37 cm/s [1 Mark]

Answer:
1. Force value is 8.9 × 10⁴ dynes.
2. Velocity value is 37 cm/s.

Teacher's Note:
a) Uses CGS system units consistently.
b) Convert to SI units only if explicitly required by instructions.

13. Answer key item: (a) (\(\sqrt{65}\)) mg (b) 3 R [1 Mark]

Answer:
1. Part (a) gives the force or tension as (\(\sqrt{65}\)) mg.
2. Part (b) gives the distance or radius parameter as 3 R.

Teacher's Note:
a) Typical circular motion problem involving tension at an arbitrary angle.
b) Retain radical form for exact values.

14. Answer key item: At the centre of flat part [1 Mark]

Answer:
1. This is the positional answer for problem 14.
2. The equilibrium or specific point lies at the centre of the flat part.

Teacher's Note:
a) Conceptual question on potential energy curves and stable equilibrium.
b) Centre of flat region denotes zero force/derivative point.