CBSE Class 11 Physics Newton laws of Motion Solved Examples

SOLVED EXAMPLES - NEWTON'S LAWS OF MOTION

Class 12 Physics solved examples on Newton's laws of motion, covering free body diagrams, equilibrium of a sphere against a wall, friction on blocks, pulley and string constraints, connected masses on inclines, wedge problems, impulse from a force-time graph, circular motion of a block on springs and a block in a groove of a rotating disc.

Ex.1: A spherical shell is resting against the vertical wall with the help of a string which makes an angle 30° with the vertical as shown in the figure. Determine the normal reaction at the wall and tension in the string.

[Figure: Spherical shell of mass m = 5 kg hanging against a vertical wall, held by a string attached to the wall that makes an angle 30° with the vertical. The free body diagram shows tension T along the string, normal reaction N from the wall horizontal, and weight mg vertically downwards.]

Answer:
1. Since the sphere is in equilibrium, the forces balance in both directions.
2. Horizontal balance: N - T sin 30° = 0, so N = T sin 30°. ........(i)
3. Vertical balance: T cos 30° = mg. ........(ii)
4. From (ii), T = mg / cos 30° = (5 × 10 × 2) / √3 = 100 / √3 N.
5. From (i), N = T sin 30° = (100 / √3) × (1/2) = 50 / √3 N.
6. Hence, tension in the string = 100 / √3 N (about 57.7 N) and normal reaction at the wall = 50 / √3 N (about 28.9 N).

Teacher's Note:
a) Draw the free body diagram first; the sphere has only three forces acting on it.
b) Resolve the tension into horizontal and vertical components before writing equations.
c) Check the answer: the wall reaction must be smaller than the tension, as N = T sin 30° is less than T.

Ex.2: A block is placed on the ground under the effect of the force F. Draw the free body diagram.

[Figure: A rectangular block of mass m rests on the horizontal ground. A force F acts vertically downwards on its top face.]

Answer:
1. The block is acted upon by three forces only.
2. Force F acts vertically downwards on the block.
3. Weight mg acts vertically downwards at the centre of the block.
4. Normal reaction N from the ground acts vertically upwards on the block.
5. Hence, the free body diagram shows F and mg downwards and N upwards.

Teacher's Note:
a) The applied force F is not the weight; it is an external push and must be shown separately.
b) Friction is absent here, because the question gives no horizontal force.

Ex.3: When a body kept on rough inclined plane is F.B.D is shown in the figure.

[Figure: Block resting on a rough inclined plane at angle θ with the horizontal. The free body diagram shows the weight mg acting vertically downwards, the normal reaction N perpendicular to the plane, and the friction force f acting up along the plane.]

Answer: The contact reaction is the resultant of two forces due to contact, that is, the normal reaction N and the friction force f.

Teacher's Note:
a) Friction on a rough incline acts opposite to the direction of impending or actual motion.
b) The normal reaction and friction together form the total contact reaction.

Ex.4: Figure shows blocks A and B weighing 4N and 8N respectively and coefficient of sliding friction between any two surfaces is 0.25. Find the force neccessary to drag block B to the left with a constant velocity in all the three cases when (a) A is kept over B (b) A is held firmly over B

[Figure: Block A of weight 4 N placed on block B of weight 8 N, with B resting on a horizontal rough surface. A horizontal force F pulls B to the left. In case (b) A is tied to a fixed vertical wall on the right by a horizontal string, so A does not move.]

Answer:
1. Weight of A, W_A = 4 N and weight of B, W_B = 8 N.
2. Case (a): Normal reaction on B from the ground N_B = W_A + W_B = 4 + 8 = 12 N.
3. Friction from the ground on B = μ N_B = 0.25 × 12 = 3 N.
4. Since B moves with constant velocity, F = 3 N.
5. Case (b): A is held firmly, so B slides under A as well.
6. Friction at the top of B from A = μ N_A = μ W_A = 0.25 × 4 = 1 N, and friction from the ground = μ(W_A + W_B) = 0.25 × 12 = 3 N.
7. Total force F = 3 N + 1 N = 4 N, which can be written as F = μ(2W_A + W_B) = 0.25 (8 + 8) = 4 N.
8. Hence, the force needed is 3 N in case (a) and 4 N in case (b).

Teacher's Note:
a) Separate A and B into two free body diagrams before writing friction forces.
b) The normal reaction on B includes the weight of A when A rests on B.
c) Holding A fixed adds the friction between A and B, so the force increases.

Ex.5: A frictionless block 3 carries two other frictionless blocks 1 and 2 connected by a light string passing over a weightless and frictionless pulley as shown in the figure. What horizontal force must continuously be applied to block 3 so that m₁ and m₂ do not move relative to 3?

[Figure: Block 3 is a large block on a horizontal floor. Block 1 rests on top of block 3. Block 2 hangs against the vertical side of block 3. A light string runs from block 1 over a pulley at the top right corner of block 3 and down to block 2. A horizontal force F acts on block 3 to the right.]

Answer:
1. Let a_1x, a_2x and a_3x be the horizontal accelerations of blocks 1, 2 and 3 to the right, and a_1y, a_2y, a_3y their downward accelerations.
2. Block 1 slides on top of block 3 and does not move vertically, so a_1y = 0 and a_3y = 0.
3. Blocks 1 and 3 do not move relative to each other horizontally, so a_1x = a_3x.
4. Block 2 does not move relative to 3 vertically, so a_2y = a_3y = 0, and it does not move relative to 3 horizontally, so a_2x = a_3x.
5. Hence, a_1x = a_2x = a_3x = a, and all vertical accelerations are zero.
6. Equations for block 1: T = m₁a ........(i) and N - m₁g = 0 ........(ii).
7. Equations for block 2: R = m₂a ........(iii) and T - m₂g = 0 ........(iv).
8. Equation for block 3: F - R - T = m₃a ........(v).
9. Adding (i), (iii) and (v): F = (m₁ + m₂ + m₃)a.
10. From (i) and (iv), T = m₂g and T = m₁a, so a = (m₂ / m₁) g.
11. Hence, F = (m₁ + m₂ + m₃) × (m₂ / m₁) × g.

Teacher's Note:
a) Vertical constraints give zero vertical acceleration for blocks 1 and 3.
b) Equal horizontal accelerations come from the requirement of no relative motion.
c) Block 2 needs the string tension to be exactly m₂g, so the string does not slip.

Ex.6: In the system of two pulleys connected as shown in the figure, M₁ = 4M₂ and mass M₁ is 20 cm above the ground whereas mass M₂ is lying on the ground. Find the distance covered by M₂ when the system is released. (g = 10 m s^-2)

[Figure: Two pulleys A and B. Pulley A is fixed to the ceiling, and pulley B is a movable pulley hanging below it. Mass M₁ hangs from one end of the string over pulley A and is 20 cm above the ground. Mass M₂ hangs from pulley B and rests on the ground. The other string end is attached to the ceiling pulley arrangement in the figure.]

Answer:
1. M₁ is heavier, so M₁ moves down with acceleration a and M₂ moves up with acceleration 2a, because the string around pulley B moves through half the distance moved by the string around pulley A.
2. For M₁: M₁g - 2T = M₁a.
3. For M₂: T - M₂g = M₂ × 2a.
4. Substituting M₁ = 4M₂ and solving gives a = g / 4 = 10 / 4 = 2.5 m/s².
5. Time for M₁ to fall 20 cm = 0.2 m: t = √(2s / a) = √(2 × 0.2 / 2.5) = √0.16 = 0.4 s.
6. Distance covered by M₂ in this time: x₁ = ½ × (2a) × t² = ½ × 5 × (0.4)² = 0.4 m.
7. Velocity of M₂ after 0.4 s: v = u + 2at = 0 + 5 × 0.4 = 2 m/s, upwards.
8. After the string goes slack, M₂ still moves up with 2 m/s and comes to rest under gravity after x₂ = v² / (2g) = 2² / (2 × 10) = 0.2 m.
9. Total distance covered by M₂ = x₁ + x₂ = 0.4 + 0.2 = 0.6 m.

Teacher's Note:
a) The movable pulley B halves the string length, so M₂ has twice the acceleration of M₁, in the opposite sense.
b) Convert 20 cm to 0.2 m before using s = ½at².
c) The printed working writes the velocity as "μ + 2at"; the initial velocity is zero, so the correct form is u + 2at.
d) Add the rising distance and the distance travelled while decelerating under gravity to get the total distance.

Ex.7: The figure shows a block of mass m₁ on a smooth incline, connected by strings over pulleys to a hanging mass m₂.

[Figure: Block m₁ lies on a smooth inclined plane. A string from m₁ runs up the incline over a pulley at the top, then down to a pulley at the bottom, which carries mass m₂ hanging below it. Another string from the bottom pulley is fixed to a ceiling support at height y₃. Lengths x₁, y₁, y₂ and y₃ are marked. Accelerations a₁ (along the incline) and a₂ (downwards for m₂) are marked.]

Answer:
1. Length of each string is constant, so x₁ + (y₂ - y₃) + y₂ = c₁. ........(i)
2. Since the second string has constant length, y₁ - y₂ = c₂, so 2y₁ - 2y₂ = 2c₂. ........(ii)
3. Adding (i) and (ii): x₁ + 2y₁ - y₃ = c₁ + 2c₂, so x₁ + 2y₁ = c, a constant, since y₃ is constant. ........(iii)
4. Differentiating (iii) twice with respect to time: d²x₁/dt² + 2 d²y₁/dt² = 0.
5. Hence, -a₁ + 2a₂ = 0, so a₁ = 2a₂. ........(iv)

Teacher's Note:
a) Write one length equation for each string, then combine them to remove the fixed lengths.
b) The sign convention for each acceleration must match the direction of the coordinate.
c) The acceleration of the block on the incline is always twice that of the hanging mass.

Ex.8: Three blocks of masses M₁, M₂ and M₃ are connected by inextensible strings passing over three massless pulleys as shown in the figure. The coefficient of friction between the masses and horizontal surfaces is μ. Assume that M₁ and M₂ are sliding. Find (a) Relation between accelerations a₁, a₂ and a₃ (b) Tension T in the strings

[Figure: Blocks M₁ and M₂ rest on horizontal surfaces on the left and right, connected to pulleys at points A and B. Block M₃ hangs vertically below a pulley at the centre. Friction forces f act on M₁ and M₂ opposing their motion. Tension T acts in every string. The string lengths are marked x₁, x₂ and x₃, and the accelerations a₁, a₂ and a₃ are marked.]

Answer:
1. Let x₁, x₂ and x₃ be the lengths of the strings, and L the constant total length, so x₁ + x₂ + 2x₃ = L.
2. Differentiating twice: a₁ + a₂ + 2a₃ = 0.
3. Hence, (a) the relation between the accelerations is a₁ + a₂ = 2a₃, taking a₃ downwards as positive.
4. Equation of motion for M₃: M₃g - 2T = M₃a₃. ........(i)
5. Equation of motion for M₁: T - μM₁g = M₁a₁. ........(ii)
6. Equation of motion for M₂: T - μM₂g = M₂a₂. ........(iii)
7. Dividing (i) by M₃: g - 2T / M₃ = a₃.
8. Dividing (ii) by M₁: T / M₁ - μg = a₁, and dividing (iii) by M₂: T / M₂ - μg = a₂.
9. Substituting in a₁ + a₂ = 2a₃: T / M₁ + T / M₂ - 2μg = 2g - 4T / M₃.
10. Hence, T (1 / M₁ + 1 / M₂ + 4 / M₃) = 2g (μ + 1).
11. So (b) T = 2g(μ + 1) / (1 / M₁ + 1 / M₂ + 4 / M₃) = \( \frac{(\mu+1)g}{\frac{1}{2M_1}+\frac{1}{2M_2}+\frac{2}{M_3}} \).

Teacher's Note:
a) Use one length constraint for the whole system, then differentiate it twice.
b) Note the factor 2 in front of x₃; it gives the factor 2 in the acceleration relation.
c) Keep the friction term with the correct sign for each block that is sliding.

Ex.9: Masses M and 2M are connected through pulleys A and B with strings as shown in figure. Both the pulleys and the strings are light and all the surfaces are frictionless. (a) Find the acceleration of the mass M, (b) Find the tension in the string, (c) Calculate the force exerted by the clamp on the pulley A.

[Figure: Mass 2M rests on a frictionless horizontal surface on the left. A light string runs horizontally from 2M over a small pulley B and then to pulley A, which is fixed to a clamp on the wall at the right. The string over pulley A then runs vertically down to mass M, which hangs below pulley A. The horizontal string lengths are x₂ and the vertical length is x₁. Accelerations a₁ (of M, downwards) and a₂ (of 2M, to the right) are marked.]

Answer:
1. Let L be the length of the string, x₁ the vertical length and x₂ the horizontal length of each string, so x₁ + 2x₂ = L.
2. Differentiating twice: a₁ + 2a₂ = 0, so a₂ = a₁ / 2 in magnitude. Let a₁ = a be the acceleration of M, and a/2 the acceleration of 2M.
3. For M: Mg - T = Ma. ........(i)
4. For 2M: 2T = 2M × (a / 2) = Ma, so T = Ma / 2. ........(ii)
5. Substituting (ii) in (i): Mg - Ma / 2 = Ma, so 3Ma / 2 = Mg.
6. Hence, (a) a = 2g / 3.
7. (b) T = Ma / 2 = M × (2g / 3) / 2 = Mg / 3.
8. (c) The string pulls pulley A with 2T horizontally and T vertically, so the force on the clamp = √((2T)² + T²) = √5 T = √5 Mg / 3.

Teacher's Note:
a) The constraint x₁ + 2x₂ = L gives the acceleration relation.
b) The clamp force is the vector sum of the two string pulls, not their simple total.
c) Use Pythagoras only when the two pulls are at right angles.

Ex.10: Two blocks each having mass 20 kg rest on frictionless surfaces are shown in the figure. Assume the pulleys to be light and frictionless. Find: (a) the time required for the block A to move 1m down the plane, starting from rest; (b) the tension in the cord connecting the blocks.

[Figure: Block A of mass 20 kg rests on a smooth inclined plane at angle θ. A cord runs from A up the incline over a pulley at the top edge, then horizontally to block B of mass 20 kg, which rests on a smooth horizontal surface. The free body diagram of A shows weight m_Ag, its components m_Ag sin θ down the incline and m_Ag cos θ perpendicular to the incline, the normal reaction N and the tension T. The free body diagram of B shows T, m_Bg and the normal reaction N'.]

Answer:
1. For block A along the incline: m_Ag sin θ - T = m_Aa. ........(i)
2. Perpendicular to the incline: N = m_Ag cos θ.
3. For block B: T = m_Ba. ........(ii)
4. Adding (i) and (ii): m_Ag sin θ = (m_A + m_B)a.
5. a = (m_A / (m_A + m_B)) g sin θ = (20 / 40) × 10 × (3 / 5) = 3 m/s².
6. (a) For 1 m down the plane from rest: s = ½at², so t = √(2s / a) = √(2 / 3) = 0.82 s.
7. (b) T = m_Ba = 20 × 3 = 60 N.

Teacher's Note:
a) Take sin θ = 3/5 from the figure, which gives the incline component of weight.
b) Both blocks have the same acceleration, because they are joined by the cord.
c) The tension found from block B's equation must match the value from block A's equation.
d) Use t = √(2s / a) with s in metres and a in m/s², and round the time only at the end.

Ex.11: Masses 4kg and 8kg are attached to free end of an inextensible light string passing over two fixed pulleys as shown in the figure. The movable pulley carries a mass of 16 kg. Assuming frictionless motion, calculate the acceleration of three masses.

[Figure: Two fixed pulleys P and Q are mounted on a horizontal support at the top. The 8 kg mass A hangs from a string over pulley P, and the 4 kg mass B hangs from a string over pulley Q. The two strings go down to a movable pulley which carries the 16 kg mass C below it. The distances x₁, x₂ and x₃ are marked, and the accelerations a (of A, downwards), b (of B, downwards) and c (of C, upwards) are marked.]

Answer:
1. Let a, b be the downward accelerations of the 8 kg and 4 kg masses, and c the upward acceleration of the 16 kg mass.
2. The constraint relation for the string is x₁ + x₂ + 2x₃ = L (constant).
3. Differentiating twice gives a + b = 2c. ........(iii)
4. Since the tension T is the same in all parts of the string over smooth pulleys: for 8 kg, 8g - T = 8a ........(i); for 4 kg, 4g - T = 4b ........(ii).
5. For the movable pulley carrying 16 kg, with the 2T upward pull: 2T - 16g = 16c ........(iv).
6. From (iii) and (iv): 2T - 16g = 8(a + b) = 8g - T + 8g - 2T = 16g - 3T, so 5T = 32g and T = 32g / 5.
7. From (i): 8a = 8g - 32g / 5 = 8g / 5, so a = g / 5 = 2 m/s² downwards.
8. From (ii): 4b = 4g - 32g / 5 = -12g / 5, so b = -3g / 5 = -6 m/s², that is, the 4 kg mass moves upwards with 6 m/s².
9. From (iii): c = (a + b) / 2 = (g/5 - 3g/5) / 2 = -g / 5 = -2 m/s², that is, the 16 kg mass moves downwards with 2 m/s².
10. Hence, the 8 kg and 16 kg masses move downwards, and the 4 kg mass moves upwards.

Teacher's Note:
a) Write the constraint with the correct number of segments: the movable pulley counts twice.
b) Assign a sign to each acceleration first, and let the result show the true direction.
c) The tension is 32g / 5 = 64 N with g = 10 m/s², the same in every string.
d) A negative value for an acceleration only means the motion is opposite to the assumed direction.

Ex.12: The figure shows a rod of mass 2m, restricted horizontally by two vertical supports which rests on a wedge of mass m. When the rod is released from rest, find the acceleration of the rod and that of wedge.

[Figure: A rod of mass 2m is held between two vertical supports and rests on the sloping face of a wedge of mass m. The wedge OXY rests on a smooth horizontal floor, with its sloping face inclined at angle θ to the horizontal. The rod moves vertically downwards and the wedge moves horizontally to the right. Free body diagrams show the rod with weight 2mg, normal reaction N and accelerations 2ma_y (downwards); and the wedge with N sin θ and N cos θ components and acceleration ma_x (to the right).]

Answer:
1. Let the rod move down by y when the wedge moves right by x. Then y / x = tan θ, so x = y cot θ.
2. Let a_y be the downward acceleration of the rod and a_x the acceleration of the wedge to the right. Then a_x = a_y cot θ.
3. For the rod (vertical): 2ma_y = 2mg - N cos θ. ........(i)
4. For the wedge (horizontal): ma_x = N sin θ, so ma_y cot θ = N sin θ. ........(ii)
5. From (ii), N = ma_y cot θ / sin θ. Substituting in (i): 2ma_y = 2mg - ma_y cot²θ.
6. Hence, a_y (2 + cot²θ) = 2g, so a_y = 2g / (2 + cot²θ).
7. a_x = a_y cot θ = 2g cot θ / (2 + cot²θ) = \( \frac{2g}{2\tan\theta + \cot\theta} \).

Teacher's Note:
a) The geometric link x = y cot θ turns the two accelerations into one unknown.
b) Mixing the wedge and rod equations needs both to use the same N.
c) Multiply top and bottom by tan θ to change the form of the answer.

Ex.13: Two blocks connected by a massless string slide down an inclined plane, having an angle of inclination of 37º. The masses of the two blocks are M₁ = 4kg and M₂ = 2kg respectively and the coefficients of friction of M₁ and M₂ with the inclined plane are 0.75 and 0.25 respectively. Assuming the string to be taut, find (a) the common acceleration of the two masses and (b) the tension in the string. (sin 37º = 0.6, cos37º = 0.8)

[Figure: Two blocks M₁ (4 kg) and M₂ (2 kg) are joined by a string and slide down an inclined plane making an angle 37° with the horizontal. Block M₁ is higher on the incline and M₂ is below it. Free body diagrams show the weight components M g sin 37° down the incline, friction μ M g cos 37° up the incline and the tension T in the string.]

Answer:
1. Coefficients of friction: μ₁ = 3/4 for M₁ and μ₂ = 1/4 for M₂. Take g = 9.8 m/s².
2. For M₁: M₁a = M₁g sin 37° + T - μ₁M₁g cos 37°. ........(i)
3. For M₂: M₂a = M₂g sin 37° - T - μ₂M₂g cos 37°. ........(ii)
4. Adding (i) and (ii): (M₁ + M₂)a = (M₁ + M₂)g sin 37° - (μ₁M₁ + μ₂M₂)g cos 37°.
5. Substituting values: (4 + 2)a = (4 + 2) × 9.8 × 0.6 - (\( \frac{3}{4} \) × 4 + \( \frac{1}{4} \) × 2) × 9.8 × 0.8.
6. 6a = 9.8 × [3.6 - 2.8] = 9.8 × 0.8 = 7.84.
7. (a) a = 7.84 / 6 = 1.31 m/s².
8. (b) From (ii): T = M₂g sin 37° - μ₂M₂g cos 37° - M₂a = 11.76 - 3.92 - 2.61 = 5.23 N.
9. Check with (i): 4 × 1.31 = 23.52 + T - 23.52, so T = 5.23 N.

Teacher's Note:
a) Friction on each block acts up the incline, because the blocks slide down.
b) Use g = 9.8 m/s² throughout, so that the values agree with the printed data.
c) The printed working gives a = 1.27 m/s² and T = 5.46 N; the arithmetic 9.8 × 0.8 = 7.84 gives a = 1.31 m/s² and T = 5.23 N.
d) Check the tension from both equations; the two values must match.

Ex.14: Masses M₁, M₂ and M₃ are connected by strings of negligible mass which pass over massless and frictionless pulleys P₁ and P₂ as shown in figure. The masses move such that the portion of the string between P₁ and P₂ is parallel to the incline and the portion of the string between P₂ and M₃ is horizontal. The masses M₂ and M₃ are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37º with the horizontal. If the mass M₁ moves with uniform velocity downwards, find: (a) the tension in the horizontal portion of the string (b) the mass M₁ (g = 9.8 ms^-2, sin 37º = 3/5).

[Figure: Mass M₁ hangs vertically on the left, connected over pulley P₁ at the top of an incline of angle 37°. Mass M₂ rests on the incline and is connected to M₁ by the string parallel to the incline. The string then passes over pulley P₂ at the foot of the incline and runs horizontally to mass M₃ on a horizontal rough surface. Tension T₁ acts in the string between M₁ and M₂, and T₂ in the horizontal string between M₂ and M₃. Friction μ = 0.25 acts on M₂ and M₃.]

Answer:
1. Let T₁ be the tension between M₁ and M₂, and T₂ the tension between M₂ and M₃.
2. (a) M₃ moves with uniform velocity, so T₂ = μM₃g = 0.25 × 4.0 × 9.8 = 9.8 N.
3. (b) M₁ moves with uniform velocity, so T₁ = M₁g.
4. For M₂ moving up the incline with uniform velocity: T₁ - T₂ - μM₂g cos 37° - M₂g sin 37° = 0.
5. Substituting: M₁g - 9.8 - 0.25 × 4 × 9.8 × (4/5) - 4 × 9.8 × (3/5) = 0.
6. M₁ × 9.8 = 9.8 + 7.84 + 23.52 = 41.16 N.
7. Hence, M₁ = 41.16 / 9.8 = 4.2 kg.

Teacher's Note:
a) Uniform velocity means zero net force on each mass.
b) Friction on M₂ acts down the incline, because M₂ moves up it.
c) Add the three terms carefully, because the printed sum mixes 4/5 and 12/5 factors.

Ex.15: A block of mass m is pulled up by means of a thread up an inclined plane forming an angle α with the horizontal. The coefficient of friction is equal to μ. Find the angle β which the thread must form with the inclined plane for the tension of the thread to be minimum. Also find the value of minimum tension.

[Figure: A block of mass m on an inclined plane at angle α with the horizontal is pulled up the incline by a thread. The thread makes an angle β with the inclined plane.]

Answer:
1. When the block is about to move up, friction f = μN acts down the incline.
2. Along the incline: T cos β = mg sin α + μN. ........(i)
3. Perpendicular to the incline: N + T sin β = mg cos α, so N = mg cos α - T sin β.
4. Substituting N in (i): T cos β + μT sin β = mg (sin α + μ cos α).
5. Hence, T = mg (sin α + μ cos α) / (cos β + μ sin β).
6. For T to be minimum, (cos β + μ sin β) must be maximum, so its derivative is zero: -sin β + μ cos β = 0.
7. Hence, tan β = μ, so β = tan^-1 μ. The second derivative is negative, so this is a maximum.
8. cos β = 1 / √(1 + μ²) and sin β = μ / √(1 + μ²).
9. Substituting: cos β + μ sin β = (1 + μ²) / √(1 + μ²) = √(1 + μ²).
10. Hence, T_min = mg (sin α + μ cos α) / √(1 + μ²).

Teacher's Note:
a) Maximising the denominator is the key step to a minimum tension.
b) A right triangle with sides 1 and μ gives both cos β and sin β quickly.
c) Check that μ = 0 gives β = 0 and T = mg sin α, which is the frictionless case.

Ex.16: A body of mass = 2 kg starts from rest whose force time graph is given by following graph. (a) What is momentum of body at t = 4 seconds. (b) What is velocity of body at t = 3 seconds.

[Figure: Force (N) against time t (s) graph. The force is constant at 2 N from t = 0 to t = 2 s, then falls in a straight line from 2 N at t = 2 s to -3 N at t = 4 s. The dashed lines mark t = 2 s and t = 4 s, and the level -3 N. The line crosses the time axis at about t = 3 s.]

Answer:
1. Impulse equals the area under the force-time graph.
2. From t = 0 to 2 s: area = 2 × 2 = 4 N s.
3. From t = 2 s to 3 s: A₂ = ½ × 1 × 2 = 1 N s.
4. From t = 3 s to 4 s: A₃ = -½ × 1 × 3 = -1.5 N s.
5. (a) Net impulse = 4 + 1 - 1.5 = 3.5 N s. Since impulse = P_f - P_i, P_f = 3.5 + 0 = 3.5 N s or kg m/s.
6. (b) Impulse from 0 to 3 s = 4 + ∫₂³ (2 - 2.5(t - 2)) dt = 4 + (2 - 1.25) = 4.75 N s.
7. Momentum at t = 3 s = 5 N s (P = 0 at t = 0), so mv = 5 and v = 5 / 2 = 2.5 m/s.

Teacher's Note:
a) Impulse is the area under the F-t graph, and it is positive or negative with the force.
b) The areas are taken as triangles: A₂ = ½ × 1 × 2 and A₃ = -½ × 1 × 3.
c) Area below the time axis subtracts from the impulse.

Ex.17: A block of mass 0.4 kg is attached to a vertical rotating spindle of length 1.6 m by two springs each of length 1 m of equal lengths as shown in the figure. The time period of rotation is 1.2 second. Find the tension in the springs.

[Figure: A vertical spindle of length 1.6 m rotates about its axis. Two springs, each 1 m long, join the top point C and the bottom point A of the spindle to a block B of mass 0.4 kg. The block moves in a horizontal circle of radius R = AB = 0.6 m. Each spring makes angle θ with the horizontal, with sin θ = 0.8 and cos θ = 0.6. Tension T₁ is in the lower spring and T₂ in the upper spring.]

Answer:
1. Radius of the circular path R = √(1 - 0.8²) = 0.6 m, with sin θ = 0.8 and cos θ = 0.6.
2. Angular velocity ω = 2π / T = 2π / 1.2 = π / 0.6 rad/s, so ω² = π² / 0.36.
3. Vertical balance: T₂ sin θ = T₁ sin θ + mg. ........(i)
4. Horizontal balance (towards the centre): (T₁ + T₂) cos θ = mω²R. ........(ii)
5. Dividing (i) by sin θ and (ii) by cos θ gives T₂ - T₁ = mg / sin θ and T₁ + T₂ = mω²R / cos θ.
6. mω²R = 0.4 × (π² / 0.36) × 0.6 = (2/3)π² ≈ 6.58 N, so mω²R / cos θ = 6.58 / 0.6 ≈ 10.97 N.
7. mg / sin θ = 0.4 × 10 / 0.8 = 5 N.
8. 2T₂ = 10.97 + 5 = 15.97, so T₂ = 7.99 N, about 8 N.
9. 2T₁ = 10.97 - 5 = 5.97, so T₁ = 2.99 N, about 3 N.

Teacher's Note:
a) The upper spring has the larger tension because it must also support the weight.
b) Use the horizontal radius R = 0.6 m, not the spindle length, in the centripetal term.
c) Keep π² exact until the last step to avoid rounding errors.

Ex.18: A groove is cut in disc placed on horizontal plane and a mass of 1 kg is put in the groove as shown in the figure. What is the acceleration of block w.r.t. disc if disc is provided acceleration of 25 m/sec² (co-efficient of friction between block and disc is 2/5)

[Figure: A disc on a horizontal plane with a straight groove cut across it at angle θ to the direction of the disc's acceleration. A block of mass 1 kg sits in the groove. The disc acceleration a = 25 m/s² is marked to the left. The angle θ has sin θ = 3/5 and cos θ = 4/5. Free body diagram of the block with respect to the disc shows the pseudo force ma, the normal reaction N from the groove wall, the friction force, and the weight mg. The question is marked as an IIT-JEE 2006 question.]

Answer:
1. In the frame of the disc, a pseudo force ma acts on the block opposite to the disc's acceleration.
2. Component of ma perpendicular to the groove: N = ma sin θ = 25 × 3/5 = 15 N.
3. Maximum friction = μN + μmg = (2/5) × 15 + (2/5) × 10 = 6 + 4 = 10 N.
4. Component of ma along the groove: ma cos θ = 25 × 4/5 = 20 N.
5. Since ma cos θ = 20 N is greater than f_max = 10 N, the block slides along the groove.
6. Net force along the groove (relative to the disc) = 20 - 10 = 10 N.
7. Hence, acceleration of the block with respect to the disc, a' = 10 / 1 = 10 m/s², along the groove.

Teacher's Note:
a) Work in the disc frame and add the pseudo force, as the disc is accelerating.
b) Friction acts opposite to the relative motion, so it is subtracted from the driving force.
c) The block slides only when the driving component exceeds the maximum friction.
d) Use the component of ma along the groove, not the full value of ma.