Laws Of MotionClass 11 Physics NCERT Solutions

According to Newton's first law of motion, if a body is at rest or moving with a constant velocity, its acceleration is zero, and therefore the net external force acting on it is zero.

(a) The rain drop is falling with a constant speed, meaning its velocity is constant. Therefore, its acceleration is zero. The net force on the drop is zero. The downward gravitational force is balanced by the upward forces of buoyancy and viscous drag.

(b) The cork is floating on water, which means it is at rest. Its acceleration is zero, and the net force on it is zero. The downward gravitational force (weight) is balanced by the upward buoyant force from the water.

(c) The kite is stationary, so its acceleration is zero. The net force on the kite is zero. The forces acting on it, such as the tension in the string, the lift from the wind, and its weight, all sum up to zero.

(d) The car is moving with a constant velocity, so its acceleration is zero. The net force on the car is zero. The forward force from the engine is balanced by the opposing forces of friction and air resistance.

(e) The electron is in space far from any material objects and fields, which means there are no external forces acting on it. Therefore, the net force on the electron is zero. It will continue to move with its constant high-speed velocity.

The problem states to ignore air resistance. The only external force acting on the pebble in all cases is the force of gravity, which is its weight ($W = mg$) acting vertically downwards.

Given: Mass of the pebble, $m = 0.05 \text{ kg}$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Calculation of force: Net force $F = mg = 0.05 \text{ kg} \times 10 \text{ m/s}^2 = 0.5 \text{ N}$. The direction is always vertically downwards.

(a) During its upward motion: The net force is $0.5 \text{ N}$ directed vertically downwards.

(b) During its downward motion: The net force is $0.5 \text{ N}$ directed vertically downwards.

(c) At the highest point: The pebble is momentarily at rest ($v=0$), but acceleration is still acting on it ($a=g$). The net force is $0.5 \text{ N}$ directed vertically downwards.

If the pebble was thrown at an angle of 45°: The answers do not change. As long as the pebble is in the air and we ignore air resistance, the only force acting on it is the force of gravity ($0.5 \text{ N}$), which always acts vertically downwards, regardless of the pebble's velocity or direction of motion.

In all cases where the stone is dropped, we neglect air resistance. The only force acting on it is gravity.

Given: Mass of the stone, $m = 0.1 \text{ kg}$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$

(a) Dropped from a stationary train: The only force acting on the stone is gravity. Force $F = mg = 0.1 \text{ kg} \times 10 \text{ m/s}^2 = 1.0 \text{ N}$. The direction is vertically downwards.

(b) Dropped from a train with constant velocity: Just after being dropped, the stone has a horizontal velocity, but the only force acting on it is gravity. The horizontal motion does not affect the net force. Force $F = mg = 1.0 \text{ N}$. The direction is vertically downwards.

(c) Dropped from an accelerating train: The moment the stone is dropped, it is no longer in contact with the train. The force that was accelerating it horizontally is gone. The only force acting on it is gravity. Force $F = mg = 1.0 \text{ N}$. The direction is vertically downwards.

(d) Lying on the floor of an accelerating train: The stone is at rest relative to the train, which means it is accelerating with the train at $a = 1 \text{ m/s}^2$. This acceleration is caused by a net force. The force is the static friction between the stone and the floor of the train. Force $F = ma = 0.1 \text{ kg} \times 1 \text{ m/s}^2 = 0.1 \text{ N}$. The direction is horizontal, in the direction of the train's acceleration.

The particle is moving in a circle on a smooth horizontal table. The forces acting on the particle are:

Since the table is horizontal, the weight and normal reaction are perpendicular to the motion and cancel each other out ($N=mg$). They do not contribute to the horizontal motion.

The net force on the particle is the force that provides the centripetal acceleration required for circular motion. In this case, the only horizontal force is the tension $T$ in the string, which acts towards the center.

Therefore, the net force on the particle is equal to the tension $T$.

The expression for centripetal force is $\frac{mv^2}{l}$. So, we have $T = \frac{mv^2}{l}$. The question asks for the net force on the particle, which is simply $T$.

Correct alternative: (i) T

Given: Retarding force, $F = -50 \text{ N}$ (negative sign indicates retardation) Mass of the body, $m = 20 \text{ kg}$ Initial speed, $u = 15 \text{ m/s}$ Final speed, $v = 0 \text{ m/s}$ (since the body stops)

To Find: Time taken to stop, $t$

Formula: First, we find the acceleration using Newton's second law: $F = ma$ Then, we use the first equation of motion: $v = u + at$

Calculation: Step 1: Calculate acceleration ($a$). $a = \frac{F}{m} = \frac{-50 \text{ N}}{20 \text{ kg}} = -2.5 \text{ m/s}^2$

Step 2: Calculate time ($t$). $v = u + at$ $0 = 15 + (-2.5)t$ $2.5t = 15$ $t = \frac{15}{2.5} = 6 \text{ s}$

Final Answer: The body takes $6$ seconds to stop.

Given: Mass of the body, $m = 3.0 \text{ kg}$ Initial speed, $u = 2.0 \text{ m/s}$ Final speed, $v = 3.5 \text{ m/s}$ Time interval, $t = 25 \text{ s}$

To Find: Magnitude and direction of the force, $F$

Formula: First, we find the acceleration using the first equation of motion: $v = u + at$ Then, we use Newton's second law: $F = ma$

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

Step 2: Calculate force ($F$). $F = ma = 3.0 \text{ kg} \times 0.06 \text{ m/s}^2 = 0.18 \text{ N}$

Since the speed of the body increases, the acceleration is positive and in the direction of motion. Therefore, the force is also in the direction of motion.

Final Answer: The magnitude of the force is $0.18 \text{ N}$, and its direction is along the direction of motion of the body.

Given: Mass of the body, $m = 5 \text{ kg}$ Force 1, $F_1 = 8 \text{ N}$ Force 2, $F_2 = 6 \text{ N}$ The forces are perpendicular to each other.

To Find: Magnitude and direction of the acceleration, $a$

Formula: First, find the net force $F_{net}$ using the Pythagorean theorem for perpendicular vectors: $F_{net} = \sqrt{F_1^2 + F_2^2}$ Then, find the acceleration using Newton's second law: $F_{net} = ma$ The direction of acceleration will be the same as the direction of the net force.

Calculation: Step 1: Calculate the magnitude of the net force ($F_{net}$). $F_{net} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$

Step 2: Calculate the magnitude of the acceleration ($a$). $a = \frac{F_{net}}{m} = \frac{10 \text{ N}}{5 \text{ kg}} = 2 \text{ m/s}^2$

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

Final Answer: The magnitude of the acceleration is $2 \text{ m/s}^2$. Its direction is at an angle of approximately $36.87^\circ$ with the direction of the $8 \text{ N}$ force.

Given: Initial speed, $u = 36 \text{ km/h}$ Final speed, $v = 0 \text{ m/s}$ Time taken, $t = 4.0 \text{ s}$ Mass of three-wheeler, $m_1 = 400 \text{ kg}$ Mass of driver, $m_2 = 65 \text{ kg}$

To Find: Average retarding force, $F$

Formula: First, convert the initial speed to m/s. Then, calculate the acceleration using the first equation of motion: $v = u + at$ Then, calculate the force using Newton's second law on the total mass: $F = (m_1 + m_2)a$

Calculation: Step 1: Convert initial speed to m/s. $u = 36 \text{ km/h} = 36 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 10 \text{ m/s}$

Step 2: Calculate acceleration ($a$). $a = \frac{v - u}{t} = \frac{0 - 10}{4.0} = -2.5 \text{ m/s}^2$ The negative sign indicates retardation.

Step 3: Calculate total mass ($m_{total}$). $m_{total} = m_1 + m_2 = 400 \text{ kg} + 65 \text{ kg} = 465 \text{ kg}$

Step 4: Calculate the retarding force ($F$). $F = m_{total} \times a = 465 \text{ kg} \times (-2.5 \text{ m/s}^2) = -1162.5 \text{ N}$ The magnitude of the retarding force is $1162.5 \text{ N}$.

Final Answer: The average retarding force on the vehicle is $1162.5 \text{ N}$.

Given: Mass of the rocket, $m = 20,000 \text{ kg}$ Initial upward acceleration, $a = 5.0 \text{ m/s}^2$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$

To Find: Initial thrust of the blast, $F_{thrust}$

Formula: The net force on the rocket is the upward thrust minus the downward force of gravity (weight). According to Newton's second law: $F_{net} = F_{thrust} - mg = ma$

Calculation: Rearranging the formula to solve for $F_{thrust}$: $F_{thrust} = ma + mg = m(a + g)$ $F_{thrust} = 20000 \text{ kg} \times (5.0 \text{ m/s}^2 + 10 \text{ m/s}^2)$ $F_{thrust} = 20000 \times 15.0 = 300000 \text{ N}$ $F_{thrust} = 3.0 \times 10^5 \text{ N}$

Final Answer: The initial thrust of the blast is $3.0 \times 10^5 \text{ N}$.

Given: Mass, $m = 0.40 \text{ kg}$ Initial velocity (north), $u = +10 \text{ m/s}$ (Let's take north as the positive direction) Force (south), $F = -8.0 \text{ N}$ Duration of force, from $t=0$ to $t=30 \text{ s}$ Position at $t=0$ is $x=0$.

Step 1: Analyze motion for t < 0 For $t < 0$, the force has not been applied yet. The body moves with a constant velocity of $10 \text{ m/s}$. Position at $t = -5 \text{ s}$: $x = u \times t = 10 \text{ m/s} \times (-5 \text{ s}) = -50 \text{ m}$. So, the position is $50 \text{ m}$ south of the origin.

Step 2: Analyze motion for 0 ≤ t ≤ 30 s During this interval, the constant force is acting. Acceleration, $a = \frac{F}{m} = \frac{-8.0 \text{ N}}{0.40 \text{ kg}} = -20 \text{ m/s}^2$. Position is given by $x(t) = ut + \frac{1}{2}at^2$. Position at $t = 25 \text{ s}$: $x(25) = (10)(25) + \frac{1}{2}(-20)(25)^2$ $x(25) = 250 - 10 \times 625 = 250 - 6250 = -6000 \text{ m}$. So, the position is $6000 \text{ m}$ south of the origin.

Step 3: Analyze motion for t > 30 s First, find the velocity and position at $t=30 \text{ s}$. Velocity at $t=30 \text{ s}$: $v_{30} = u + at = 10 + (-20)(30) = 10 - 600 = -590 \text{ m/s}$. Position at $t=30 \text{ s}$: $x_{30} = ut + \frac{1}{2}at^2 = (10)(30) + \frac{1}{2}(-20)(30)^2 = 300 - 10 \times 900 = 300 - 9000 = -8700 \text{ m}$. For $t > 30 \text{ s}$, the force is removed, so the body moves with a constant velocity of $v_{30} = -590 \text{ m/s}$. The position for $t > 30 \text{ s}$ is given by: $x(t) = x_{30} + v_{30}(t - 30)$. Position at $t = 100 \text{ s}$: $x(100) = -8700 + (-590)(100 - 30)$ $x(100) = -8700 - 590 \times 70 = -8700 - 41300 = -50000 \text{ m}$. So, the position is $50000 \text{ m}$ south of the origin.

Final Answers: Position at $t = -5 \text{ s}$ is $-50 \text{ m}$. Position at $t = 25 \text{ s}$ is $-6000 \text{ m}$. Position at $t = 100 \text{ s}$ is $-50000 \text{ m}$.

Given: Initial velocity of truck, $u_{truck} = 0$ Acceleration of truck, $a_{truck} = 2.0 \text{ m/s}^2$ Height of truck, $h = 6 \text{ m}$ Time of drop, $t_{drop} = 10 \text{ s}$ Time of observation, $t_{obs} = 11 \text{ s}$

Step 1: Find the velocity of the truck at t = 10 s When the stone is dropped, it will have the same horizontal velocity as the truck. $v_x = u_{truck} + a_{truck} \times t_{drop} = 0 + 2.0 \times 10 = 20 \text{ m/s}$. This is the initial horizontal velocity of the stone, $u_{stone, x}$. The initial vertical velocity of the stone is zero, $u_{stone, y} = 0$.

Step 2: Analyze the motion of the stone after being dropped (for t > 10 s) Once dropped, the stone is a projectile. Air resistance is neglected. Horizontal acceleration, $a_x = 0$. Vertical acceleration (due to gravity), $a_y = -g = -10 \text{ m/s}^2$ (taking upward as positive).

The time elapsed since the drop is $\Delta t = t_{obs} - t_{drop} = 11 - 10 = 1 \text{ s}$.

(a) Velocity of the stone at t = 11 s Horizontal velocity at $t=11\text{s}$: $v_x = u_{stone, x} + a_x \Delta t = 20 + 0 \times 1 = 20 \text{ m/s}$. Vertical velocity at $t=11\text{s}$: $v_y = u_{stone, y} + a_y \Delta t = 0 + (-10) \times 1 = -10 \text{ m/s}$.

The magnitude of the resultant velocity is: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + (-10)^2} = \sqrt{400 + 100} = \sqrt{500} = 10\sqrt{5} \approx 22.36 \text{ m/s}$. The direction $\theta$ below the horizontal is: $\tan{\theta} = \frac{|v_y|}{|v_x|} = \frac{10}{20} = 0.5 \implies \theta = \tan^{-1}(0.5) \approx 26.57^\circ$.

(b) Acceleration of the stone at t = 11 s After the stone is dropped, the only force acting on it is gravity (neglecting air resistance). Therefore, its acceleration is the acceleration due to gravity, $g$. $a = g = 10 \text{ m/s}^2$, directed vertically downwards.

Final Answer: (a) The velocity of the stone at $t=11\text{s}$ is approximately $22.36 \text{ m/s}$ at an angle of about $26.57^\circ$ below the horizontal. (b) The acceleration of the stone at $t=11\text{s}$ is $10 \text{ m/s}^2$ vertically downwards.

(a) At one of its extreme positions: At an extreme position of its oscillation, the bob is momentarily at rest. Its velocity is zero. If the string is cut at this instant, the bob has zero initial velocity. The only force acting on it will be gravity. Therefore, the bob will fall vertically downwards from that height. Trajectory: A straight line, vertically downwards.

(b) At its mean position: The mean position is the lowest point of the oscillation. At this point, the bob has its maximum speed, which is given as $1 \text{ m/s}$. The direction of this velocity is horizontal (tangential to the circular path). If the string is cut at this instant, the bob becomes a projectile with an initial horizontal velocity of $1 \text{ m/s}$ and an initial vertical velocity of zero. Under the influence of gravity, it will follow a parabolic path. Trajectory: A parabolic path.

A weighing scale measures the normal reaction force ($R$) exerted on the person. The reading is this force divided by $g$. Let's call the reading $M_{apparent} = R/g$.

Given: Mass of the man, $m = 70 \text{ kg}$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$ Weight of the man, $W = mg = 70 \times 10 = 700 \text{ N}$.

The equation of motion is $F_{net} = R - mg = ma$, where upward direction is positive.

(a) Moving upwards with uniform speed: Uniform speed means acceleration $a = 0$. $R - mg = m(0) \implies R = mg = 700 \text{ N}$. Reading on the scale = $\frac{700 \text{ N}}{10 \text{ m/s}^2} = 70 \text{ kg}$.

(b) Moving downwards with uniform acceleration of 5 m s⁻²: Acceleration $a = -5 \text{ m/s}^2$. $R - mg = ma$ $R - 700 = 70 \times (-5) = -350$ $R = 700 - 350 = 350 \text{ N}$. Reading on the scale = $\frac{350 \text{ N}}{10 \text{ m/s}^2} = 35 \text{ kg}$.

(c) Moving upwards with uniform acceleration of 5 m s⁻²: Acceleration $a = +5 \text{ m/s}^2$. $R - mg = ma$ $R - 700 = 70 \times (5) = 350$ $R = 700 + 350 = 1050 \text{ N}$. Reading on the scale = $\frac{1050 \text{ N}}{10 \text{ m/s}^2} = 105 \text{ kg}$.

(d) Lift hurtles down freely under gravity: This is a state of free fall, so the acceleration is $a = -g = -10 \text{ m/s}^2$. $R - mg = m(-g)$ $R - mg = -mg \implies R = 0 \text{ N}$. Reading on the scale = $\frac{0 \text{ N}}{10 \text{ m/s}^2} = 0 \text{ kg}$. This is the condition of weightlessness.

Final Answers: (a) $70 \text{ kg}$ (b) $35 \text{ kg}$ (c) $105 \text{ kg}$ (d) $0 \text{ kg}$

Given: Mass of the particle, $m = 4 \text{ kg}$ The graph is a position-time graph.

(a) Force on the particle Force is given by $F = ma$. Acceleration is the second derivative of position with respect to time, and velocity is the first derivative (slope of the x-t graph).

(b) Impulse on the particle Impulse is the change in momentum ($J = \Delta p = m\Delta v$).

Final Answer: (a) The force is $0$ in all three intervals ($t < 0$, $t > 4$ s, and $0 < t < 4$ s). (b) The impulse at $t=0$ is $3 \text{ N s}$, and the impulse at $t=4$ s is $-3 \text{ N s}$.

This is an Atwood machine setup.

Given: Mass 1, $m_1 = 8 \text{ kg}$ Mass 2, $m_2 = 12 \text{ kg}$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Since $m_2 > m_1$, the 12 kg mass will move downwards and the 8 kg mass will move upwards with the same acceleration, $a$.

To Find: Acceleration of the masses, $a$ Tension in the string, $T$

Formula: For the heavier mass ($m_2$), moving down: $m_2g - T = m_2a \quad (1)$ For the lighter mass ($m_1$), moving up: $T - m_1g = m_1a \quad (2)$

Calculation: Step 1: Find acceleration ($a$). Add equations (1) and (2): $(m_2g - T) + (T - m_1g) = m_2a + m_1a$ $g(m_2 - m_1) = a(m_1 + m_2)$ $a = g \frac{m_2 - m_1}{m_1 + m_2}$ $a = 10 \times \frac{12 - 8}{12 + 8} = 10 \times \frac{4}{20} = 10 \times 0.2 = 2 \text{ m/s}^2$

Step 2: Find tension ($T$). Substitute the value of $a$ into equation (2): $T = m_1a + m_1g = m_1(a + g)$ $T = 8 \text{ kg} \times (2 \text{ m/s}^2 + 10 \text{ m/s}^2)$ $T = 8 \times 12 = 96 \text{ N}$

Final Answer: The acceleration of the masses is $2 \text{ m/s}^2$. The tension in the string is $96 \text{ N}$.

Let the initial nucleus be at rest. Its initial momentum is zero. $p_{initial} = 0$ Let the nucleus disintegrate into two smaller nuclei with masses $m_1$ and $m_2$. Let their velocities after disintegration be $v_1$ and $v_2$ respectively.

The final momentum of the system is the vector sum of the momenta of the two products: $p_{final} = m_1v_1 + m_2v_2$

The disintegration is an internal process, and there are no external forces acting on the system. Therefore, according to the law of conservation of linear momentum, the total momentum of the system must be conserved. $p_{initial} = p_{final}$ $0 = m_1v_1 + m_2v_2$ $m_1v_1 = -m_2v_2$ $v_1 = -\frac{m_2}{m_1}v_2$

Since masses $m_1$ and $m_2$ are positive scalars, the equation shows that the velocity vector $v_1$ is in the opposite direction to the velocity vector $v_2$. The negative sign indicates that the two velocities are directed oppositely.

Thus, the two product nuclei must move in opposite directions to conserve momentum.

Given: Mass of each ball, $m = 0.05 \text{ kg}$ Initial speed of each ball, $u = 6 \text{ m/s}$ Final speed of each ball, $v = 6 \text{ m/s}$

Let's consider ball 1. Let its initial direction of motion be positive. Initial velocity of ball 1, $u_1 = +6 \text{ m/s}$. After collision, it rebounds, so its final velocity is in the opposite direction. Final velocity of ball 1, $v_1 = -6 \text{ m/s}$.

To Find: Impulse imparted to each ball.

Formula: Impulse ($J$) is the change in momentum ($\Delta p$). $J = \Delta p = m(v - u)$

Calculation for ball 1: Impulse on ball 1, $J_1 = m(v_1 - u_1)$ $J_1 = 0.05 \text{ kg} \times (-6 \text{ m/s} - 6 \text{ m/s})$ $J_1 = 0.05 \times (-12) = -0.6 \text{ kg m/s}$ or $-0.6 \text{ N s}$. The impulse on ball 1 is $0.6 \text{ N s}$ in the direction opposite to its initial motion.

Calculation for ball 2: Ball 2 was initially moving in the opposite direction. Initial velocity of ball 2, $u_2 = -6 \text{ m/s}$. After collision, it rebounds. Final velocity of ball 2, $v_2 = +6 \text{ m/s}$. Impulse on ball 2, $J_2 = m(v_2 - u_2)$ $J_2 = 0.05 \text{ kg} \times (6 \text{ m/s} - (-6 \text{ m/s}))$ $J_2 = 0.05 \times (12) = +0.6 \text{ kg m/s}$ or $+0.6 \text{ N s}$. The impulse on ball 2 is $0.6 \text{ N s}$ in the positive direction (opposite to its initial motion).

According to Newton's third law, the impulse imparted by ball 2 on ball 1 is equal and opposite to the impulse imparted by ball 1 on ball 2. Our results ($J_1 = -J_2$) confirm this.

Final Answer: The impulse imparted to each ball is $0.6 \text{ N s}$ in magnitude, directed opposite to its initial velocity.

Given: Mass of the shell, $m_s = 0.020 \text{ kg}$ Mass of the gun, $m_g = 100 \text{ kg}$ Muzzle speed of the shell, $v_s = 80 \text{ m/s}$ Initially, the gun and shell are at rest, so the total initial momentum is zero.

To Find: Recoil speed of the gun, $v_g$

Formula: According to the law of conservation of linear momentum: Total initial momentum = Total final momentum $0 = m_s v_s + m_g v_g$

Calculation: $m_g v_g = -m_s v_s$ $v_g = -\frac{m_s v_s}{m_g}$ $v_g = -\frac{0.020 \text{ kg} \times 80 \text{ m/s}}{100 \text{ kg}}$ $v_g = -\frac{1.6}{100} = -0.016 \text{ m/s}$ The negative sign indicates that the gun recoils in the direction opposite to the shell's motion. The recoil speed is the magnitude of this velocity.

Final Answer: The recoil speed of the gun is $0.016 \text{ m/s}$.

Given: Mass of the ball, $m = 0.15 \text{ kg}$ Initial speed, $u = 54 \text{ km/h}$ Final speed, $v = 54 \text{ km/h}$ Angle of deflection, $\theta = 45^\circ$

First, convert speed to m/s: $u = 54 \times \frac{1000}{3600} = 15 \text{ m/s}$. So, $|\vec{u}| = |\vec{v}| = 15 \text{ m/s}$.

Let's set up a coordinate system. Let the initial momentum $\vec{p_i}$ be along the x-axis. The final momentum $\vec{p_f}$ makes an angle of $45^\circ$ with the negative x-axis. Initial momentum vector: $\vec{p_i} = mu \hat{i} = (0.15)(15) \hat{i} = 2.25 \hat{i} \text{ N s}$. Final momentum vector: $\vec{p_f} = mv(\cos{45^\circ}(-\hat{i}) + \sin{45^\circ}(\hat{j}))$ (Assuming the deflection is in the xy-plane, perpendicular to the bowler's direction) This interpretation is tricky. A more common interpretation is that the angle between the initial and final velocity vectors is $180^\circ - 45^\circ = 135^\circ$. Let's assume the question implies the ball is hit straight back, but deflected by $45^\circ$ from the straight path. Let the initial velocity be $\vec{u}$. Let the final velocity be $\vec{v}$. The angle between $\vec{u}$ and $-\vec{v}$ is $45^\circ$. So the angle between $\vec{u}$ and $\vec{v}$ is $180^\circ - 45^\circ = 135^\circ$.

Impulse $\vec{J} = \vec{p_f} - \vec{p_i} = m(\vec{v} - \vec{u})$. The magnitude of the impulse is given by: $|\vec{J}| = \sqrt{p_f^2 + p_i^2 - 2p_f p_i \cos(135^\circ)}$ Since $|p_i| = |p_f| = p = mu = 2.25 \text{ N s}$, $|\vec{J}| = \sqrt{p^2 + p^2 - 2p^2 \cos(135^\circ)} = \sqrt{2p^2(1 - \cos(135^\circ))}$ $\cos(135^\circ) = -\cos(45^\circ) = -\frac{1}{\sqrt{2}}$ $|\vec{J}| = p \sqrt{2(1 - (-\frac{1}{\sqrt{2}}))} = p \sqrt{2 + \sqrt{2}}$ $|\vec{J}| = 2.25 \times \sqrt{2 + 1.414} = 2.25 \times \sqrt{3.414} \approx 2.25 \times 1.848 \approx 4.16 \text{ N s}$

Alternative interpretation: Let's assume the initial velocity is along the x-axis, and the final velocity vector makes an angle of $45^\circ$ with the x-axis. $\vec{p_i} = 2.25 \hat{i}$ $\vec{p_f} = (2.25 \cos 45^\circ) \hat{i} + (2.25 \sin 45^\circ) \hat{j} = (2.25 \frac{1}{\sqrt{2}}) \hat{i} + (2.25 \frac{1}{\sqrt{2}}) \hat{j}$ $\Delta \vec{p} = \vec{p_f} - \vec{p_i} = 2.25 ((\frac{1}{\sqrt{2}} - 1) \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$ $|\Delta \vec{p}| = 2.25 \sqrt{(\frac{1}{\sqrt{2}}-1)^2 + (\frac{1}{\sqrt{2}})^2} = 2.25 \sqrt{\frac{1}{2} - \sqrt{2} + 1 + \frac{1}{2}} = 2.25 \sqrt{2 - \sqrt{2}} \approx 2.25 \times 0.765 \approx 1.72 \text{ N s}$

The most standard interpretation for this problem is the change in momentum along the bisector of the angle between initial and final directions. The magnitude of change in momentum is $2p \sin(\theta/2)$. Here, the angle of deflection is $45^\circ$, so the angle between the two velocity vectors is $45^\circ$. $|\vec{J}| = |\Delta \vec{p}| = 2mv \sin(\frac{\theta}{2})$ $|\vec{J}| = 2 \times 0.15 \text{ kg} \times 15 \text{ m/s} \times \sin(\frac{45^\circ}{2})$ $|\vec{J}| = 4.5 \times \sin(22.5^\circ) \approx 4.5 \times 0.3827 \approx 1.72 \text{ N s}$. Let's assume the question meant deflected from the straight path towards the bowler. Initial velocity $u$, final velocity $-v$, angle between them $45^\circ$. Then the angle between $u$ and $v$ is $180-45 = 135^\circ$. Then $|\vec{J}| = 2mv \sin(\frac{135^\circ}{2}) = 4.5 \sin(67.5^\circ) \approx 4.5 \times 0.9239 \approx 4.16 \text{ N s}$. Given the context of Example 4.4, the deflection is likely from the initial path. Let's stick with the first reasonable calculation.

Final Answer: The impulse imparted to the ball is approximately $4.16 \text{ N s}$.

Given: Mass of stone, $m = 0.25 \text{ kg}$ Radius of circle, $r = 1.5 \text{ m}$ Angular speed, $\omega = 40 \text{ rev/min}$ Maximum tension, $T_{max} = 200 \text{ N}$

Part 1: Tension in the string First, convert angular speed to SI units (rad/s). $\omega = 40 \frac{\text{rev}}{\text{min}} = 40 \times \frac{2\pi \text{ rad}}{60 \text{ s}} = \frac{4\pi}{3} \text{ rad/s}$ Linear speed, $v = r\omega = 1.5 \times \frac{4\pi}{3} = 2\pi \text{ m/s}$.

The tension in the string provides the necessary centripetal force. $T = F_c = \frac{mv^2}{r}$ $T = \frac{0.25 \text{ kg} \times (2\pi \text{ m/s})^2}{1.5 \text{ m}} = \frac{0.25 \times 4\pi^2}{1.5} = \frac{\pi^2}{1.5} \approx \frac{(3.14)^2}{1.5} \approx \frac{9.86}{1.5} \approx 6.57 \text{ N}$

Part 2: Maximum speed The maximum tension the string can withstand corresponds to the maximum speed. $T_{max} = \frac{mv_{max}^2}{r}$ $v_{max}^2 = \frac{T_{max} \times r}{m}$ $v_{max}^2 = \frac{200 \text{ N} \times 1.5 \text{ m}}{0.25 \text{ kg}} = \frac{300}{0.25} = 1200$ $v_{max} = \sqrt{1200} = \sqrt{400 \times 3} = 20\sqrt{3} \approx 34.64 \text{ m/s}$

Final Answer: The tension in the string is approximately $6.57 \text{ N}$. The maximum speed with which the stone can be whirled is approximately $34.64 \text{ m/s}$.

According to Newton's first law of motion (the law of inertia), a body in motion continues to move in a straight line with constant velocity unless acted upon by an external force.

When the stone is being whirled, the tension in the string provides the centripetal force that constantly changes the direction of the stone's velocity, keeping it in a circular path. The instantaneous velocity of the stone at any point is always directed along the tangent to the circle at that point.

If the string breaks, the centripetal force (tension) suddenly becomes zero. At that instant, there is no longer a force to change the direction of the stone's motion. Due to its inertia, the stone will continue to move in the direction of its velocity at the moment the string broke. This direction is tangential to the circular path.

Therefore, the stone will fly off tangentially from the point where the string breaks.

Correct alternative: (b) the stone flies off tangentially from the instant the string breaks

(a) A horse cannot pull a cart and run in empty space: To move forward, the horse must push against something. On the ground, the horse pushes the ground backward with its hooves. According to Newton's third law, the ground exerts an equal and opposite force (the reaction force) on the horse in the forward direction. This forward reaction force is what allows the horse and the cart to accelerate. In empty space, there is no ground or surface to push against, so the horse cannot generate a forward reaction force and cannot move.

(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly: This is due to inertia of motion. When the bus is moving, the passengers are also moving with the same velocity as the bus. When the driver applies the brakes, the bus slows down and comes to a stop. However, due to inertia, the upper bodies of the passengers tend to continue moving forward with the original velocity. As a result, they are thrown forward relative to the bus.

(c) It is easier to pull a lawn mower than to push it: When pulling a lawn mower, the force is applied at an angle upwards. This force has a horizontal component that moves the mower forward and a vertical component that lifts the mower slightly. This upward vertical component reduces the normal force between the mower and the ground. Since the force of friction is proportional to the normal force ($f = \mu N$), reducing the normal force reduces friction, making it easier to pull. When pushing a lawn mower, the force is applied at an angle downwards. This force has a horizontal component and a downward vertical component. The downward component increases the normal force, which in turn increases the frictional force, making it harder to push.

(d) A cricketer moves his hands backwards while holding a catch: This action is to increase the time of impact. According to Newton's second law, force is the rate of change of momentum ($F = \frac{\Delta p}{\Delta t}$). The change in momentum ($\Delta p$) of the ball is fixed (from its initial high momentum to zero). By moving his hands backward, the cricketer increases the time interval ($\Delta t$) over which the ball's momentum is brought to zero. Since $\Delta t$ is larger, the magnitude of the force ($F$) exerted by the ball on his hands is smaller, reducing the sting and preventing injury.