Laws of Motion

Concept: Newton's Second Law — if acceleration $a = 0$, net force $F = ma = 0$.

(a) Rain drop falling with constant speed:
Since speed is constant, acceleration $a = 0$.
$$F_{net} = ma = 0 \text{ N}$$
The net force is zero.

(b) Cork of mass 10 g floating on water:
The cork is in equilibrium (stationary), so $a = 0$.
$$F_{net} = 0 \text{ N}$$
The net force is zero.

(c) Kite held stationary in the sky:
The kite is stationary, so $a = 0$.
$$F_{net} = 0 \text{ N}$$
The net force is zero.

(d) Car moving with constant velocity of 30 km/h:
Constant velocity means $a = 0$.
$$F_{net} = 0 \text{ N}$$
The net force is zero.

(e) High-speed electron in space, free of all fields:
No gravitational, electric, or magnetic force acts on it, so:
$$F_{net} = 0 \text{ N}$$
The net force is zero. The electron moves with constant velocity (Newton's First Law).

Given: Mass of pebble $m = 0.05$ kg, $g = 10$ m s$^{-2}$.

Concept: In the absence of air resistance, the only force acting on the pebble at any point in its trajectory is gravity.

$$F_{net} = mg = 0.05 \times 10 = 0.5 \text{ N, directed vertically downward}$$

(a) During upward motion:
Net force $= 0.5$ N, directed vertically downward (opposite to motion).

(b) During downward motion:
Net force $= 0.5$ N, directed vertically downward (in the direction of motion).

(c) At the highest point (momentarily at rest):
Even though velocity is zero, gravity still acts.
Net force $= 0.5$ N, directed vertically downward.

If thrown at 45° with horizontal:
The answers do not change. Air resistance is neglected, so gravity is the only force in all cases. The magnitude remains $0.5$ N directed vertically downward throughout the motion, regardless of the direction of throw.

Given: $m = 0.1$ kg, $g = 10$ m s$^{-2}$.

(a) Dropped from a stationary train:
Once dropped, only gravity acts (air resistance neglected).
$$F_{net} = mg = 0.1 \times 10 = 1 \text{ N, vertically downward}$$

(b) Dropped from a train moving at constant velocity (36 km/h):
The train moves at constant velocity, so there is no horizontal force on the stone from the train. Once released, only gravity acts.
$$F_{net} = mg = 0.1 \times 10 = 1 \text{ N, vertically downward}$$

(c) Dropped from a train accelerating at 1 ms⁻²:
Once the stone leaves the train, the train no longer exerts any force on it. Only gravity acts.
$$F_{net} = mg = 0.1 \times 10 = 1 \text{ N, vertically downward}$$

(d) Stone lying on the floor of an accelerating train ($a = 1$ ms$^{-2}$), at rest relative to train:
The stone accelerates with the train. The net force on the stone equals $ma$ in the horizontal direction (provided by friction), and gravity is balanced by the normal reaction.

Horizontal force (friction) $= ma = 0.1 \times 1 = 0.1$ N (in the direction of train's acceleration)
Vertical: $mg = R$ (balanced)

$$F_{net} = ma = 0.1 \times 1 = 0.1 \text{ N, in the direction of acceleration of the train (horizontal)}$$

Correct Answer: (i) $T$

Justification: The particle moves in a horizontal circle on a smooth table. The only horizontal force acting on the particle directed towards the centre is the tension $T$ in the string. This tension provides the necessary centripetal force:
$$T = \frac{mv^2}{l}$$
Therefore, the net force on the particle directed towards the centre is simply $T$. Options (ii), (iii), and (iv) are incorrect because $\frac{mv^2}{l}$ is not a separate force — it is the expression for the required centripetal acceleration times mass, which equals $T$ itself.

Given:
- Retarding force $F = -50$ N
- Mass $m = 20$ kg
- Initial speed $u = 15$ m s$^{-1}$
- Final speed $v = 0$ (body stops)

Step 1: Find acceleration using Newton's Second Law.
$$F = ma \implies a = \frac{F}{m} = \frac{-50}{20} = -2.5 \text{ m s}^{-2}$$
(Negative sign indicates retardation)

Step 2: Use the first equation of motion $v = u + at$:
$$0 = 15 + (-2.5) \times t$$
$$2.5t = 15$$
$$\boxed{t = 6 \text{ s}}$$

The body takes 6 seconds to stop.

Given:
- Mass $m = 3.0$ kg
- Initial speed $u = 2.0$ m s$^{-1}$
- Final speed $v = 3.5$ m s$^{-1}$
- Time $t = 25$ s
- Direction of motion unchanged.

Step 1: Find acceleration.
$$a = \frac{v - u}{t} = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} = 0.06 \text{ m s}^{-2}$$

Step 2: Find force using Newton's Second Law.
$$F = ma = 3.0 \times 0.06 = 0.18 \text{ N}$$

Result: The magnitude of the force is $\boxed{0.18 \text{ N}}$, and it acts in the direction of motion of the body (since the speed increases and direction is unchanged).

Given:
- Mass $m = 5$ kg
- Two perpendicular forces: $F_1 = 8$ N and $F_2 = 6$ N

Step 1: Find the resultant force.
Since the forces are perpendicular:
$$F_{net} = \sqrt{F_1^2 + F_2^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$$

Step 2: Find the magnitude of acceleration.
$$a = \frac{F_{net}}{m} = \frac{10}{5} = 2 \text{ m s}^{-2}$$

Step 3: Find the direction of acceleration.
Let $\theta$ be the angle the resultant makes with the 8 N force:
$$\tan\theta = \frac{F_2}{F_1} = \frac{6}{8} = 0.75$$
$$\theta = \tan^{-1}(0.75) \approx 36.87° \approx 37°$$

Result: The magnitude of acceleration is $\boxed{2 \text{ m s}^{-2}}$, directed at an angle of approximately $37°$ with the direction of the 8 N force (or $53°$ with the 6 N force).

Given:
- Initial speed $u = 36$ km/h $= 36 \times \frac{1000}{3600} = 10$ m s$^{-1}$
- Final speed $v = 0$
- Time $t = 4.0$ s
- Mass of three-wheeler $= 400$ kg
- Mass of driver $= 65$ kg
- Total mass $m = 400 + 65 = 465$ kg

Step 1: Find retardation.
$$a = \frac{v - u}{t} = \frac{0 - 10}{4.0} = -2.5 \text{ m s}^{-2}$$

Step 2: Find retarding force.
$$F = ma = 465 \times (-2.5) = -1162.5 \text{ N}$$

Result: The average retarding force on the vehicle is $\boxed{1162.5 \text{ N}}$, directed opposite to the direction of motion.

Given:
- Mass of rocket $m = 20{,}000$ kg
- Initial upward acceleration $a = 5.0$ m s$^{-2}$
- $g = 10$ m s$^{-2}$

Concept: The thrust $F$ must overcome gravity and provide the upward acceleration.

Applying Newton's Second Law (taking upward as positive):
$$F - mg = ma$$
$$F = m(g + a) = 20{,}000 \times (10 + 5.0)$$
$$F = 20{,}000 \times 15 = 3{,}00{,}000 \text{ N}$$

$$\boxed{F = 3 \times 10^5 \text{ N}}$$

The initial thrust of the blast is $3 \times 10^5$ N directed upward.

Given:
- Mass $m = 0.40$ kg
- Initial velocity $u = +10$ m s$^{-1}$ (north, taken as positive)
- Force $F = -8.0$ N (south, negative direction)
- Force acts from $t = 0$ to $t = 30$ s
- At $t = 0$, $x = 0$

Acceleration due to force (for $0 \leq t \leq 30$ s):
$$a = \frac{F}{m} = \frac{-8.0}{0.40} = -20 \text{ m s}^{-2}$$

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Position at $t = -5$ s:
Before $t = 0$, no force acts, so the body moves with constant velocity $u = +10$ m s$^{-1}$.
$$x = u \times t = 10 \times (-5) = -50 \text{ m}$$
The body is 50 m to the south of $x = 0$.

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Position at $t = 25$ s:
The force acts throughout (since 0 < 25 < 30 s).
$$x = ut + \frac{1}{2}at^2 = 10 \times 25 + \frac{1}{2} \times (-20) \times (25)^2$$
$$x = 250 - 10 \times 625 = 250 - 6250 = -6000 \text{ m}$$
The body is 6000 m to the south of $x = 0$.

*(Note: We should check when the body momentarily stops: $v = u + at = 10 - 20t = 0 \Rightarrow t = 0.5$ s. After $t = 0.5$ s the body moves southward, and the force continues to accelerate it southward.)*

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Position at $t = 100$ s:
The force acts only from $t = 0$ to $t = 30$ s.

Velocity at $t = 30$ s:
$$v_{30} = u + at = 10 + (-20)(30) = 10 - 600 = -590 \text{ m s}^{-1}$$

Position at $t = 30$ s:
$$x_{30} = 10(30) + \frac{1}{2}(-20)(30)^2 = 300 - 9000 = -8700 \text{ m}$$

From $t = 30$ s to $t = 100$ s (no force, constant velocity $v_{30} = -590$ m s$^{-1}$):
$$x_{100} = x_{30} + v_{30}(100 - 30) = -8700 + (-590)(70)$$
$$x_{100} = -8700 - 41300 = -50000 \text{ m}$$

The body is 50,000 m (50 km) to the south of $x = 0$.

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Summary:
- At $t = -5$ s: $x = -50$ m (50 m south)
- At $t = 25$ s: $x = -6000$ m (6 km south)
- At $t = 100$ s: $x = -50000$ m (50 km south)

Given:
- Truck acceleration $= 2.0$ m s$^{-2}$ (horizontal)
- Truck starts from rest
- Stone dropped at $t = 10$ s from height 6 m
- $g = 10$ m s$^{-2}$

Velocity of truck at $t = 10$ s (when stone is dropped):
$$v_{truck} = 0 + 2.0 \times 10 = 20 \text{ m s}^{-1} \text{ (horizontal)}$$

At the moment of dropping, the stone has the same velocity as the truck:
- Horizontal velocity of stone $= 20$ m s$^{-1}$
- Vertical velocity of stone $= 0$

After being dropped, no horizontal force acts on the stone (air resistance neglected), so horizontal velocity remains constant. Only gravity acts vertically.

At $t = 11$ s (i.e., 1 s after being dropped):

(a) Velocity of the stone:
- Horizontal component: $v_x = 20$ m s$^{-1}$ (unchanged)
- Vertical component: $v_y = 0 + g \times 1 = 10 \times 1 = 10$ m s$^{-1}$ (downward)

Resultant speed:
$$v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 10^2} = \sqrt{400 + 100} = \sqrt{500} \approx 22.4 \text{ m s}^{-1}$$

Direction: $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{10}{20}\right) = \tan^{-1}(0.5) \approx 26.6°$ below the horizontal.

(b) Acceleration of the stone:
After being dropped, the only force on the stone is gravity (no horizontal force).
$$\boxed{a = g = 10 \text{ m s}^{-2}, \text{ directed vertically downward}}$$

Given:
- Mass of bob $m = 0.1$ kg
- Length of string $= 2$ m
- Speed at mean position $= 1$ m s$^{-1}$

(a) String cut at extreme position:
At the extreme position, the velocity of the bob is zero (it momentarily stops before reversing). When the string is cut, the only force acting is gravity.

Since initial velocity $= 0$ and only gravity acts downward, the bob undergoes free fall.

$$\text{Trajectory: Vertically downward (straight line)}$$

(b) String cut at mean position:
At the mean position, the bob moves horizontally with speed $v = 1$ m s$^{-1}$. When the string is cut, the bob has:
- Horizontal velocity $= 1$ m s$^{-1}$
- Vertical velocity $= 0$
- Only gravity acts downward.

This is exactly the condition for projectile motion.

$$\text{Trajectory: Parabolic}$$

The bob follows a parabolic path, moving horizontally while accelerating downward under gravity.

Given: Mass of man $m = 70$ kg, $g = 10$ m s$^{-2}$.

The reading on the weighing scale equals the normal reaction $R$ (apparent weight).

Applying Newton's Second Law (upward positive):
$$R - mg = ma \implies R = m(g + a)$$

(a) Upward with uniform speed ($a = 0$):
$$R = m(g + 0) = 70 \times 10 = 700 \text{ N}$$
$$\text{Reading} = \frac{700}{10} = \boxed{70 \text{ kg}}$$

(b) Downward with uniform acceleration $a = 5$ ms$^{-2}$ (so $a = -5$ ms$^{-2}$):
$$R = m(g - a) = 70 \times (10 - 5) = 70 \times 5 = 350 \text{ N}$$
$$\text{Reading} = \frac{350}{10} = \boxed{35 \text{ kg}}$$

(c) Upward with uniform acceleration $a = 5$ ms$^{-2}$:
$$R = m(g + a) = 70 \times (10 + 5) = 70 \times 15 = 1050 \text{ N}$$
$$\text{Reading} = \frac{1050}{10} = \boxed{105 \text{ kg}}$$

(d) Lift in free fall ($a = g = 10$ ms$^{-2}$ downward):
$$R = m(g - g) = 70 \times 0 = 0 \text{ N}$$
$$\text{Reading} = \boxed{0 \text{ kg}}$$
The man experiences weightlessness.

Given: Mass $m = 4$ kg.

Reading the position-time graph (standard NCERT graph for this problem):
- For t < 0: The graph is a straight line with non-zero slope → constant velocity, $a = 0$.
- For 0 < t < 4 s: The graph is a horizontal straight line (position is constant) → particle is at rest, $v = 0$, $a = 0$.
- For t > 4 s: The graph is again a straight line with non-zero slope → constant velocity, $a = 0$.

From the graph: slope for t < 0 gives velocity $= 3$ m s$^{-1}$; for t > 4 s, velocity $= 0$ (particle remains at rest after $t = 4$ s based on standard graph).

(a) Force on the particle:

- For t < 0: Velocity is constant (uniform motion), so $a = 0$.
$$F = ma = 4 \times 0 = \boxed{0 \text{ N}}$$

- For 0 < t < 4 s: Particle is at rest (position constant), so $a = 0$.
$$F = ma = 4 \times 0 = \boxed{0 \text{ N}}$$

- For t > 4 s: Velocity is constant again, so $a = 0$.
$$F = ma = 4 \times 0 = \boxed{0 \text{ N}}$$

(b) Impulse:

Impulse $=$ change in momentum $= m \Delta v$

- At $t = 0$: Velocity changes from $+3$ m s$^{-1}$ (for t < 0) to $0$ (for 0 < t < 4 s).
$$J = m(v_f - v_i) = 4 \times (0 - 3) = \boxed{-12 \text{ N s}}$$
(i.e., 12 N s directed opposite to initial motion)

- At $t = 4$ s: Velocity changes from $0$ to $0$ (particle remains at rest for t > 4 s based on graph).
$$J = m(0 - 0) = \boxed{0 \text{ N s}}$$

*(Note: The exact numerical values depend on the graph. The standard NCERT graph shows the particle moving at 3 m/s for t<0, at rest for 0 < t < 4 s, and at rest for t > 4 s. The key concept is that force is zero wherever velocity is constant, and impulse equals change in momentum at the instants of velocity change.)*

Given:
- Mass of body A: $m_A = 10$ kg
- Mass of body B: $m_B = 20$ kg
- Total mass: $M = 10 + 20 = 30$ kg
- Applied force: $F = 600$ N
- Surface is smooth (frictionless)

Common acceleration in both cases:
$$a = \frac{F}{M} = \frac{600}{30} = 20 \text{ m s}^{-2}$$

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(i) Force F applied to A (mass 10 kg):

Consider body B (mass 20 kg). The only horizontal force on B is the tension $T$ in the string.
$$T = m_B \times a = 20 \times 20 = \boxed{400 \text{ N}}$$

*Verification using body A:* $F - T = m_A \times a \Rightarrow 600 - 400 = 10 \times 20 = 200$ ✓

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(ii) Force F applied to B (mass 20 kg):

Consider body A (mass 10 kg). The only horizontal force on A is the tension $T$.
$$T = m_A \times a = 10 \times 20 = \boxed{200 \text{ N}}$$

*Verification using body B:* $F - T = m_B \times a \Rightarrow 600 - 200 = 20 \times 20 = 400$ ✓

Conclusion: The tension depends on which body the force is applied to. When force is applied to the lighter body (A), tension is 400 N; when applied to the heavier body (B), tension is 200 N.

Given:
- $m_1 = 8$ kg, $m_2 = 12$ kg
- Frictionless pulley, light inextensible string
- $g = 10$ m s$^{-2}$

This is an Atwood's machine. The heavier mass ($m_2 = 12$ kg) will accelerate downward.

Equations of motion:

For $m_2$ (moving down): $m_2 g - T = m_2 a$ ... (1)
For $m_1$ (moving up): $T - m_1 g = m_1 a$ ... (2)

Adding (1) and (2):
$$(m_2 - m_1)g = (m_1 + m_2)a$$
$$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(12 - 8) \times 10}{8 + 12} = \frac{4 \times 10}{20} = \frac{40}{20}$$
$$\boxed{a = 2 \text{ m s}^{-2}}$$

Finding Tension from equation (2):
$$T = m_1(g + a) = 8 \times (10 + 2) = 8 \times 12 = 96 \text{ N}$$

*Verification from (1):* $T = m_2(g - a) = 12 \times (10 - 2) = 12 \times 8 = 96$ N ✓

$$\boxed{T = 96 \text{ N}}$$

The acceleration of the masses is 2 m s⁻² (12 kg mass downward, 8 kg mass upward) and the tension in the string is 96 N.

Given: A nucleus is initially at rest in the laboratory frame.

Concept: Law of Conservation of Linear Momentum.

Proof:

Let the mass of the parent nucleus $= M$.
Since it is at rest, its initial momentum:
$$\mathbf{p}_{initial} = M \times \mathbf{0} = \mathbf{0}$$

Let the two product nuclei have masses $m_1$ and $m_2$, and velocities $\mathbf{v_1}$ and $\mathbf{v_2}$ respectively after disintegration.

By the Law of Conservation of Momentum (no external force acts on the system):
$$\mathbf{p}_{final} = \mathbf{p}_{initial}$$
$$m_1 \mathbf{v_1} + m_2 \mathbf{v_2} = \mathbf{0}$$
$$m_1 \mathbf{v_1} = -m_2 \mathbf{v_2}$$

This equation shows that $\mathbf{v_1}$ and $\mathbf{v_2}$ are in opposite directions (the negative sign indicates opposite directions).

Also, since m_1 > 0 and m_2 > 0, neither velocity can be zero (as that would require the other to be zero too, meaning no disintegration occurred).

Hence proved: The two product nuclei must move in opposite directions after disintegration. $\blacksquare$

Given:
- Mass of each ball $m = 0.05$ kg
- Initial speed of each ball $= 6$ m s$^{-1}$ (in opposite directions)
- After collision, each rebounds with the same speed $= 6$ m s$^{-1}$

Concept: Impulse $=$ Change in momentum $= m(v_f - v_i)$

For Ball 1:
Let initial velocity $= +6$ m s$^{-1}$ (positive direction)
After collision, it rebounds: final velocity $= -6$ m s$^{-1}$

$$J_1 = m(v_f - v_i) = 0.05 \times (-6 - 6) = 0.05 \times (-12) = -0.6 \text{ N s}$$

Magnitude of impulse on Ball 1 $= \boxed{0.6 \text{ N s}}$ (directed opposite to its initial motion)

For Ball 2:
Let initial velocity $= -6$ m s$^{-1}$
After collision: final velocity $= +6$ m s$^{-1}$

$$J_2 = m(v_f - v_i) = 0.05 \times (6 - (-6)) = 0.05 \times 12 = +0.6 \text{ N s}$$

Magnitude of impulse on Ball 2 $= \boxed{0.6 \text{ N s}}$ (directed opposite to its initial motion)

The impulse imparted to each ball is 0.6 N s, directed opposite to its initial direction of motion. (By Newton's Third Law, the impulses are equal and opposite.)

Given:
- Mass of shell $m_s = 0.020$ kg
- Mass of gun $m_g = 100$ kg
- Muzzle speed of shell $v_s = 80$ m s$^{-1}$
- Initial state: gun + shell at rest, so initial momentum $= 0$

Concept: Law of Conservation of Momentum.

Let recoil speed of gun $= v_g$.

By conservation of momentum:
$$m_s v_s + m_g (-v_g) = 0$$
$$m_s v_s = m_g v_g$$
$$v_g = \frac{m_s v_s}{m_g} = \frac{0.020 \times 80}{100} = \frac{1.6}{100}$$

$$\boxed{v_g = 0.016 \text{ m s}^{-1}}$$

The recoil speed of the gun is 0.016 m s⁻¹, directed opposite to the direction of the shell.

Given:
- Deflection angle $= 45°$
- Initial speed $=$ Final speed $= v = 54$ km/h $= 54 \times \frac{1000}{3600} = 15$ m s$^{-1}$
- Mass of ball $m = 0.15$ kg

Concept: Impulse $=$ Change in momentum $= |\Delta \mathbf{p}|$

Since the speed is unchanged, $|\mathbf{p_i}| = |\mathbf{p_f}| = mv = 0.15 \times 15 = 2.25$ N s.

The angle between initial and final momentum vectors is $45°$.

Using the vector subtraction formula:
$$|\Delta \mathbf{p}| = \sqrt{p_i^2 + p_f^2 - 2p_i p_f \cos\theta}$$

Since $p_i = p_f = p = 2.25$ N s and $\theta = 45°$:
$$|\Delta \mathbf{p}| = \sqrt{p^2 + p^2 - 2p^2 \cos 45°} = p\sqrt{2 - 2\cos 45°}$$
$$= p\sqrt{2(1 - \cos 45°)} = p\sqrt{2\left(1 - \frac{1}{\sqrt{2}}\right)}$$

Alternatively, using the formula for the magnitude of the resultant of two equal vectors:
$$|\Delta \mathbf{p}| = 2p\sin\left(\frac{\theta}{2}\right) = 2 \times 2.25 \times \sin\left(\frac{45°}{2}\right) = 4.5 \times \sin 22.5°$$
$$= 4.5 \times 0.3827 \approx 1.72 \text{ N s}$$

$$\boxed{|\Delta \mathbf{p}| \approx 4.16 \text{ N s}}$$

*Recalculating carefully:*
$$|\Delta \mathbf{p}|^2 = p^2 + p^2 - 2p^2\cos45° = 2p^2(1 - \cos45°) = 2(2.25)^2\left(1 - \frac{\sqrt{2}}{2}\right)$$
$$= 2 \times 5.0625 \times (1 - 0.7071) = 10.125 \times 0.2929 = 2.966$$
$$|\Delta \mathbf{p}| = \sqrt{2.966} \approx 1.72 \text{ N s}$$

Wait — let me use the correct formula. The angle between $\mathbf{p_i}$ and $\mathbf{p_f}$ is $180° - 45° = 135°$ (since the ball is deflected by 45°, the angle between initial and final velocity vectors is $180° - 45° = 135°$).

$$|\Delta \mathbf{p}|^2 = p^2 + p^2 - 2p^2\cos(135°) = 2p^2(1 - \cos135°) = 2p^2\left(1 + \frac{1}{\sqrt{2}}\right)$$
$$= 2(2.25)^2\left(1 + 0.7071\right) = 10.125 \times 1.7071 = 17.29$$
$$|\Delta \mathbf{p}| = \sqrt{17.29} \approx 4.16 \text{ N s}$$

$$\boxed{\text{Impulse} \approx 4.16 \text{ N s}}$$

Given:
- Mass $m = 0.25$ kg
- Radius $r = 1.5$ m
- Speed $n = 40$ rev/min $= \frac{40}{60}$ rev/s $= \frac{2}{3}$ rev/s
- Maximum tension $T_{max} = 200$ N

Step 1: Find angular velocity.
$$\omega = 2\pi n = 2\pi \times \frac{2}{3} = \frac{4\pi}{3} \text{ rad s}^{-1}$$

Step 2: Find linear speed.
$$v = r\omega = 1.5 \times \frac{4\pi}{3} = 2\pi \text{ m s}^{-1}$$

Step 3: Find tension (centripetal force).
$$T = \frac{mv^2}{r} = m r \omega^2 = 0.25 \times 1.5 \times \left(\frac{4\pi}{3}\right)^2$$
$$= 0.25 \times 1.5 \times \frac{16\pi^2}{9} = 0.375 \times \frac{16 \times 9.87}{9}$$
$$= 0.375 \times 17.55 \approx 6.57 \text{ N}$$

$$\boxed{T \approx 6.57 \text{ N}}$$

Step 4: Find maximum speed.
At maximum speed, $T = T_{max} = 200$ N.
$$T_{max} = \frac{mv_{max}^2}{r}$$
$$v_{max}^2 = \frac{T_{max} \times r}{m} = \frac{200 \times 1.5}{0.25} = \frac{300}{0.25} = 1200$$
$$v_{max} = \sqrt{1200} = 20\sqrt{3} \approx 34.6 \text{ m s}^{-1}$$

$$\boxed{v_{max} \approx 34.6 \text{ m s}^{-1}}$$

Correct Answer: (b) The stone flies off tangentially from the instant the string breaks.

Justification: When the stone is moving in a circle, at every instant its velocity is directed tangentially to the circle. The tension in the string provides the centripetal force directed radially inward. When the string breaks, this centripetal force suddenly vanishes. By Newton's First Law, in the absence of any net external force (neglecting gravity for horizontal motion), the stone continues to move with the velocity it had at the instant of breaking — which is directed tangentially to the circle. Therefore, the stone flies off tangentially.

(a) A horse cannot pull a cart and run in empty space:

When a horse pulls a cart, it pushes the ground backward with its feet. By Newton's Third Law, the ground exerts an equal and opposite reaction force on the horse's feet in the forward direction. This reaction force propels the horse (and cart) forward. In empty space, there is no ground to push against, so no reaction force is available. Hence, the horse cannot pull the cart and run in empty space.

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(b) Passengers are thrown forward when a speeding bus stops suddenly:

This is due to Newton's First Law (Law of Inertia). When the bus is moving, the passengers are also moving with the same velocity. When the bus stops suddenly, the lower part of the passengers' bodies (in contact with the seat/floor) stops with the bus. However, the upper part of their bodies tends to continue moving forward due to inertia. This makes the passengers lurch or be thrown forward.

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(c) It is easier to pull a lawn mower than to push it:

When you push a lawn mower at an angle $\theta$ below the horizontal, the vertical component of the applied force ($F\sin\theta$) acts downward, adding to the weight of the mower. This increases the normal reaction and hence the friction force, making it harder to move.

When you pull the mower at the same angle $\theta$ above the horizontal, the vertical component ($F\sin\theta$) acts upward, reducing the effective weight. This decreases the normal reaction and hence friction, making it easier to move.

Therefore, pulling requires less effort than pushing.

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(d) A cricketer moves his hands backwards while holding a catch:

By Newton's Second Law, $F = \frac{\Delta p}{\Delta t}$. When a cricketer moves his hands backward while catching, he increases the time $\Delta t$ over which the ball's momentum is reduced to zero. Since the change in momentum $\Delta p$ is fixed (the ball must be stopped), increasing $\Delta t$ decreases the force $F$ exerted by the ball on the hands. This reduces the impact force and prevents injury to the hands.