Third Law of Motion and Momentum and Impulse Questions in English

(A) Let the normal to the wall be along the $y$-axis and the wall be along the $x$-axis.
The initial velocity is $\vec{v}_1 = v \sin \theta \hat{i} - v \cos \theta \hat{j}$.
The final velocity is $\vec{v}_2 = v \sin \theta \hat{i} + v \cos \theta \hat{j}$.
The initial momentum is $\vec{P}_1 = m\vec{v}_1 = m v \sin \theta \hat{i} - m v \cos \theta \hat{j}$.
The final momentum is $\vec{P}_2 = m\vec{v}_2 = m v \sin \theta \hat{i} + m v \cos \theta \hat{j}$.
The change in momentum is $\Delta \vec{P} = \vec{P}_2 - \vec{P}_1$.
$\Delta \vec{P} = (m v \sin \theta \hat{i} + m v \cos \theta \hat{j}) - (m v \sin \theta \hat{i} - m v \cos \theta \hat{j}) = 2 m v \cos \theta \hat{j}$.
The magnitude of the change in momentum is $|\Delta \vec{P}| = 2 m v \cos \theta$.

(C) Given the linear momentum vector $\overrightarrow{P} = P_x \hat{i} + P_y \hat{j} = (2\cos t) \hat{i} + (2\sin t) \hat{j}$.
The force $\overrightarrow{F}$ is the rate of change of linear momentum: $\overrightarrow{F} = \frac{d\overrightarrow{P}}{dt}$.
Calculating the derivative: $\overrightarrow{F} = \frac{d}{dt}(2\cos t) \hat{i} + \frac{d}{dt}(2\sin t) \hat{j} = (-2\sin t) \hat{i} + (2\cos t) \hat{j}$.
To find the angle $\theta$ between $\overrightarrow{F}$ and $\overrightarrow{P}$,we use the dot product formula: $\overrightarrow{F} \cdot \overrightarrow{P} = |\overrightarrow{F}| |\overrightarrow{P}| \cos \theta$.
Calculating the dot product: $\overrightarrow{F} \cdot \overrightarrow{P} = (-2\sin t)(2\cos t) + (2\cos t)(2\sin t) = -4\sin t \cos t + 4\sin t \cos t = 0$.
Since the dot product is $0$,the vectors are perpendicular,which means $\cos \theta = 0$,so $\theta = 90^o$.

(C) Impulse is defined as the product of force and time interval.
$\text{Impulse} = \text{Force} \times \text{time}$
Since the unit of force is $kg \cdot m/s^2$ (Newton) and the unit of time is $s$:
$\text{Unit of Impulse} = (kg \cdot m/s^2) \times s = kg \cdot m/s$
Therefore,the correct unit is $kg \cdot m/s$.

(A) The dimension of momentum is given by $p = mv = [M][L][T^{-1}] = [MLT^{-1}]$.
Impulse is defined as the product of force and time,$I = F \times \Delta t$.
The dimensions of force are $[MLT^{-2}]$ and time are $[T]$.
Therefore,the dimensions of impulse are $[MLT^{-2}] \times [T] = [MLT^{-1}]$.
Since the dimensions of impulse and momentum are both $[MLT^{-1}]$,they are equal.

(B) When a sudden jerk is given to string $C$,an impulsive force is applied to it. Due to the inertia of the $1 \ kg$ mass,it does not immediately move.
This impulsive force creates a very high tension in string $C$ that exceeds its breaking strength.
Since the impulse is applied directly to $C$ and the mass acts as a buffer,the tension does not propagate to string $A$ instantaneously.
Therefore,string $C$ breaks before the impulse can reach string $A$.

(B) The impulse $J$ acting on an object is equal to the change in its linear momentum,given by $J = \Delta p = m(v_f - v_i)$.
Given: mass $m = 10 \, kg$,initial velocity $v_i = 10 \, m/s$,and final velocity $v_f = -2 \, m/s$ (since it is in the opposite direction).
Substituting the values: $J = 10 \times (-2 - 10) = 10 \times (-12) = -120 \, N \cdot s$.
Thus,the impulse acting on the object is $-120 \, N \cdot s$.

(D) Given: Mass $m = 250\, g = 0.25\, kg$,initial velocity $u = 10\, m/s$,final velocity $v = -10\, m/s$ (since it returns in the opposite direction),and time interval $\Delta t = 0.01\, s$.
According to Newton's second law,the force $F$ is given by the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
Substituting the values:
$F = \frac{0.25 \times (-10 - 10)}{0.01}$
$F = \frac{0.25 \times (-20)}{0.01}$
$F = \frac{-5}{0.01} = -500\, N$
The magnitude of the force exerted on the ball is $500\, N$.

(B) According to Newton's second law of motion,the impulse applied to an object is equal to the change in its momentum.
Impulse $(J)$ is defined as the product of force $(F)$ and the time interval $(\Delta t)$ over which it acts.
Given:
Force $(F)$ = $10 \, N$
Time interval $(\Delta t)$ = $10 \, s$
Change in momentum $(\Delta p)$ = $F \times \Delta t$
$\Delta p = 10 \, N \times 10 \, s = 100 \, kg \cdot m/s$.
Therefore,the change in momentum is $100 \, kg \cdot m/s$.

(B) Given:
Mass of the ball,$m = 150 \, g = 0.15 \, kg$.
Initial velocity,$u = 20 \, m/s$.
Final velocity,$v = 0 \, m/s$ (since the ball is caught).
Time taken,$\Delta t = 0.1 \, s$.
According to Newton's second law of motion,the force exerted is given by the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
$F = \frac{0.15 \times (0 - 20)}{0.1}$
$F = \frac{0.15 \times (-20)}{0.1} = \frac{-3}{0.1} = -30 \, N$.
The magnitude of the force exerted by the ball on the hands is $30 \, N$.

(D) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets fired.
$F = \frac{dp}{dt} = v \cdot \frac{dm}{dt}$
Here,$v = 1200\,m/s$ is the velocity of the bullet.
Let $n$ be the number of bullets fired per second.
The mass of one bullet is $m = 40\,g = 40 \times 10^{-3}\,kg = 0.04\,kg$.
The total mass fired per second is $\frac{dm}{dt} = n \times m$.
Given the maximum force $F = 144\,N$,we have:
$144 = 1200 \times (n \times 0.04)$
$144 = 1200 \times 0.04 \times n$
$144 = 48 \times n$
$n = \frac{144}{48} = 3$.
Therefore,the man can fire at most $3$ bullets per second.

(B) The momentum $p$ of a body is defined as the product of its mass $m$ and its velocity $v$,given by the formula $p = mv$.
Since the mass $m$ of the body remains constant,the momentum $p$ is directly proportional to the speed $v$ $(p \propto v)$.
When the speed is doubled $(v' = 2v)$,the new momentum $p'$ becomes $p' = m(2v) = 2(mv) = 2p$.
Therefore,the momentum of the body is doubled.
Kinetic energy $K = \frac{1}{2}mv^2$ would increase by a factor of $4$,not $2$.

(A) The initial momentum of the body is $p_i = mv$ (taking the direction towards the wall as positive).
After the collision,the body rebounds with the same speed $v$ in the opposite direction,so the final momentum is $p_f = -mv$.
The change in momentum $\Delta p$ is defined as the final momentum minus the initial momentum:
$\Delta p = p_f - p_i$
$\Delta p = (-mv) - (mv) = -2\,mv$.
Since the magnitude of the change in momentum is requested,we consider the absolute value: $|\Delta p| = 2\,mv$.

(A) The force exerted by the wall on the bullets is equal to the rate of change of momentum of the bullets.
Each bullet has a mass $m$ and velocity $v$,so the momentum of one bullet is $p = mv$.
Since $n$ bullets strike the wall per second,the total momentum change per second is $\Delta p = n \times (mv - 0) = nmv$.
According to Newton's second law,the force $F$ is equal to the rate of change of momentum: $F = \frac{dp}{dt} = nmv$.
Thus,the reaction force offered by the wall is $nmv$.

(A) The force exerted by the ball on the wall is equal to the rate of change of momentum of the ball.
Initial momentum of the ball,$p_i = mv$.
Final momentum of the ball after reflection,$p_f = -mv$ (since it moves in the opposite direction).
Change in momentum,$\Delta p = p_f - p_i = -mv - mv = -2mv$.
The magnitude of the change in momentum is $|\Delta p| = 2mv$.
The force exerted on the wall is given by $F = \frac{|\Delta p|}{t} = \frac{2mv}{t}$.

(A) Given: Mass $m = 0.5\, kg$,initial velocity $u = 2\, m/s$,final velocity $v = -2\, m/s$ (since it bounces back),and time of contact $\Delta t = 1\, ms = 10^{-3}\, s$.
According to Newton's second law,the average force $F_{av}$ is the rate of change of momentum:
$F_{av} = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
$F_{av} = \frac{0.5 \times (-2 - 2)}{10^{-3}}$
$F_{av} = \frac{0.5 \times (-4)}{10^{-3}} = \frac{-2}{10^{-3}} = -2000\, N$.
The magnitude of the average force exerted by the wall on the ball is $2000\, N$.

(A) Given the linear momentum vector is $\vec{p} = p_x \hat{i} + p_y \hat{j} = (2\cos t) \hat{i} + (2\sin t) \hat{j}$.
The force $\vec{F}$ is the rate of change of linear momentum,so $\vec{F} = \frac{d\vec{p}}{dt}$.
Calculating the derivative: $\vec{F} = \frac{d}{dt}(2\cos t) \hat{i} + \frac{d}{dt}(2\sin t) \hat{j} = (-2\sin t) \hat{i} + (2\cos t) \hat{j}$.
To find the angle between $\vec{F}$ and $\vec{p}$,we calculate their dot product: $\vec{F} \cdot \vec{p} = [(-2\sin t)(2\cos t)] + [(2\cos t)(2\sin t)]$.
$\vec{F} \cdot \vec{p} = -4\sin t \cos t + 4\sin t \cos t = 0$.
Since the dot product $\vec{F} \cdot \vec{p} = |\vec{F}| |\vec{p}| \cos \theta = 0$,and neither vector is a null vector,we must have $\cos \theta = 0$,which implies $\theta = 90^o$.

(A) $1$. Velocity of the ball just before hitting the bat:
$v_1 = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 5} = 10\, m/s$ (downwards).
$2$. Velocity of the ball just after leaving the bat:
$v_2 = \sqrt{2gh_2} = \sqrt{2 \times 10 \times 20} = 20\, m/s$ (upwards).
$3$. Using the impulse-momentum theorem:
$F_{avg} \times \Delta t = m(v_{final} - v_{initial})$
Taking the upward direction as positive (+),we have $v_{final} = +20\, m/s$ and $v_{initial} = -10\, m/s$.
Given $m = 400\, g = 0.4\, kg$ and $F_{avg} = 100\, N$.
$100 \times \Delta t = 0.4 \times (20 - (-10))$
$100 \times \Delta t = 0.4 \times 30$
$100 \times \Delta t = 12$
$\Delta t = 0.12\, s$.

(A) According to Newton's second law of motion,the force applied is equal to the rate of change of momentum.
$\vec{F} = \frac{\Delta \vec{p}}{\Delta t}$
Given:
Force,$F = 2 \, N$
Change in momentum,$\Delta p = 0.4 \, kg \cdot m/s$
Rearranging the formula to solve for time,$\Delta t$:
$\Delta t = \frac{\Delta p}{F}$
$\Delta t = \frac{0.4}{2} = 0.2 \, s$
Therefore,the time required is $0.2 \, s$.

(C) Swimming is possible due to Newton's $Third$ law of motion.
When a swimmer pushes the water backward with their hands and feet, the water exerts an equal and opposite force on the swimmer in the forward direction.
This action-reaction pair allows the swimmer to move forward through the water.

(B) According to Newton's $Third$ $Law$ $of$ $Motion$,for every action,there is an equal and opposite reaction.
When a person jumps forward out of a boat,the person exerts a force on the boat in the forward direction (action).
In response,the boat exerts an equal and opposite force on the person,and the person exerts an equal and opposite force on the boat in the backward direction (reaction).
Therefore,the boat moves backward.

(C) According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
On a perfectly smooth (frictionless) surface,a man cannot walk normally because there is no friction to provide the necessary reaction force.
However,if the man throws an object in the direction opposite to the shore,the object exerts an equal and opposite force on the man.
This reaction force pushes the man toward the shore,allowing him to move despite the lack of friction.

(C) When a cannon fires a cannonball,it exerts a large forward force on the ball,causing it to travel a large distance. This force is the action.
At the same time,the cannonball exerts an equal and opposite force on the cannon,which causes the cannon to recoil. This force is the reaction.
Since this is an action-reaction pair,it is governed by Newton's third law of motion.
Therefore,option $C$ is correct.

(B) According to Newton's third law of motion,for every action,there is an equal and opposite reaction. These two forces always act on different bodies simultaneously. Therefore,they do not cancel each other out.

(D) The force required to hold the gun is equal to the rate of change of momentum of the bullets.
Given:
Number of bullets per second $(n)$ = $20 \, s^{-1}$
Mass of each bullet $(m)$ = $150 \, g = 0.15 \, kg$
Velocity of each bullet $(v)$ = $800 \, m/s$
The force $(F)$ is given by the formula:
$F = n \times m \times v$
$F = 20 \times 0.15 \times 800$
$F = 20 \times 120$
$F = 2400 \, N.$

(D) According to Newton's $Third$ $Law$ $of$ $Motion$,for every action,there is an equal and opposite reaction.
The action force exerted by the book on the table acts downwards.
The reaction force exerted by the table on the book acts upwards.
Since these two forces act along the same line but in exactly opposite directions,the angle between them is $180^o$.

(D) According to Newton's third law,for every action,there is an equal and opposite reaction. When the student pulls his hair,he exerts a force on his hair,and his hair exerts an equal and opposite force on his hand. Since both the student and his hair are part of the same system,these forces are internal forces. Internal forces cannot change the state of motion of the center of mass of a system. Therefore,the student cannot lift himself up.

(A) The impulse applied to a body is equal to the change in its momentum.
According to the impulse-momentum theorem:
$Impulse = F \times \Delta t = \Delta p$
Given:
Force $(F)$ = $250\, N$
Change in momentum $(\Delta p)$ = $125\, kg \cdot m/s$
We need to find the time interval $(\Delta t)$:
$\Delta t = \frac{\Delta p}{F}$
$\Delta t = \frac{125}{250} = 0.5\, s$
Therefore,the period for which the force acts on the body is $0.5\, s$.

(C) Given: Mass $m = 150\,g = 0.15\,kg$,Acceleration $a = 20\,m/s^2$,Time $t = 0.1\,s$.
Impulse is defined as the change in momentum,which is equal to the product of force and time.
Impulse $J = F \times t = (m \times a) \times t$.
Substituting the values: $J = 0.15\,kg \times 20\,m/s^2 \times 0.1\,s$.
$J = 3\,N \times 0.1\,s = 0.3\,N-s$.
Therefore,the impulsive force is $0.3\,N-s$.

(B) The impulse is defined as the change in momentum of an object. Mathematically,it is given by $J = \Delta p = F \times \Delta t$.
Since impulse is directly equal to the change in momentum,it is the most closely related quantity among the given options.
Force is the rate of change of momentum $(F = \frac{dp}{dt})$,which involves time as a divisor.
Kinetic energy $(K.E. = \frac{p^2}{2m})$ depends on both momentum and mass.
Power is the rate of doing work and is not directly related to momentum in the same fundamental way as impulse.
Therefore,impulse is the correct answer.

(C) Impulse $(I)$ is defined as the product of force $(F)$ and the time interval $(\Delta t)$ for which it acts.
Given:
Force $(F)$ = $50\, dynes = 50 \times 10^{-5}\, N$ (since $1\, dyne = 10^{-5}\, N$)
Time interval $(\Delta t)$ = $3\, s$
Impulse $(I)$ = $F \times \Delta t$
$I = (50 \times 10^{-5}\, N) \times (3\, s)$
$I = 150 \times 10^{-5}\, Ns$
$I = 1.5 \times 10^{-3}\, Ns$

(C) The force on the bullet is given by $F = 600 - 2 \times 10^5 t$.
When the bullet leaves the barrel,the force becomes zero,so $F = 0$.
$600 - 2 \times 10^5 t = 0$
$2 \times 10^5 t = 600$
$t = \frac{600}{2 \times 10^5} = 3 \times 10^{-3} \ s$.
Impulse $I$ is defined as the integral of force with respect to time: $I = \int_{0}^{t} F \ dt$.
$I = \int_{0}^{3 \times 10^{-3}} (600 - 2 \times 10^5 t) \ dt$
$I = [600t - 10^5 t^2]_{0}^{3 \times 10^{-3}}$
$I = 600(3 \times 10^{-3}) - 10^5(3 \times 10^{-3})^2$
$I = 1.8 - 10^5(9 \times 10^{-6})$
$I = 1.8 - 0.9 = 0.9 \ N-s$.

(C) The correct relation is $\overrightarrow{F} \Delta t = m \Delta \overrightarrow{v}$,which represents the impulse-momentum theorem.
From this,we can write $F = \frac{m \Delta \overrightarrow{v}}{\Delta t}$.
When you bend your knees upon landing,the time interval $\Delta t$ during which your velocity changes to zero increases.
Since the change in momentum $m \Delta \overrightarrow{v}$ is constant,increasing the time $\Delta t$ results in a decrease in the impulsive force $F$ exerted on your knees,thereby reducing the risk of injury.

(B) The impulse acting on a particle is equal to the change in its momentum.
Impulse $J = \Delta p = m(v_f - v_i)$.
From the position-time graph,the velocity $v$ is the slope of the line.
For $t = 0$ to $t = 2 \,s$,the velocity $v_i = \frac{\Delta x}{\Delta t} = \frac{4 - 0}{2 - 0} = 2 \,m/s$.
For $t > 2 \,s$,the position is constant $(x = 4 \,m)$,so the velocity $v_f = 0 \,m/s$.
The mass of the particle is $m = 0.1 \,kg$.
Therefore,the impulse at $t = 2 \,s$ is $J = 0.1 \,kg \times (0 \,m/s - 2 \,m/s) = -0.2 \,kg \,m \,s^{-1}$.

(D) The change in momentum $(\Delta p)$ of a particle is equal to the impulse,which is given by the area under the force-time $(F-t)$ graph.
$\Delta p = \int F \, dt = \text{Area under } F-t \text{ curve}$.
In the given graph,the area from $t = 2 \ s$ to $t = 6 \ s$ is above the time axis,representing a positive impulse.
The area from $t = 0 \ s$ to $t = 2 \ s$ and $t = 6 \ s$ to $t = 8 \ s$ is below the time axis,representing a negative impulse.
Due to the symmetry of the curve,the positive area (between $2 \ s$ and $6 \ s$) is equal in magnitude to the total negative area (from $0$ to $2 \ s$ and $6$ to $8 \ s$).
Therefore,the net area is $0$,which implies the net change in momentum acquired by the particle in the interval from $0$ to $8 \ s$ is $0 \ N-s$.

(C) The impulse applied to the body is equal to the change in its momentum: $J = \Delta p = m(v_f - v_i)$ ... $(i)$
The impulse is also equal to the area under the force-time graph:
Area $= \text{Area of triangle} (0-2s) + \text{Area of rectangle} (2-4s) + \text{Area of trapezoid} (4-4.5s) + \text{Area of rectangle} (4.5-6.5s)$
Area $= (\frac{1}{2} \times 2 \times 4) + (2 \times 4) + (\frac{1}{2} \times (4 + 2.5) \times 0.5) + (2 \times 2.5)$
Area $= 4 + 8 + 1.625 + 5 = 18.625\, Ns$
Equating $(i)$ and the calculated impulse:
$m(v_f - v_i) = 18.625$
$2(v_f - 5) = 18.625$
$v_f - 5 = 9.3125$
$v_f = 14.3125\, ms^{-1}$
*Note: Based on the provided options,the intended calculation likely used a slightly different area approximation or rounding. Re-evaluating the area: $4 + 8 + 1.625 + 5 = 18.625$. $v_f = 5 + (18.625/2) = 5 + 9.3125 = 14.3125$. Given the options,$14.25$ is the closest match.*

(C) The impulse applied to the particle is equal to the change in its momentum.
Since the particle is initially at rest, the final momentum $p = mu$ is equal to the impulse, which is the area under the $F-t$ graph.
The graph is a semicircle with radius $r_1 = F_0$ (height) and $r_2 = T/2$ (base width).
The area of a semicircle is given by $\frac{1}{2} \pi r_1 r_2$.
Therefore, $mu = \frac{1}{2} \pi (F_0) (T/2)$.
$mu = \frac{\pi F_0 T}{4}$.
Solving for velocity $u$, we get $u = \frac{\pi F_0 T}{4m}$.

(D) The impulse applied to a body is equal to the change in its momentum. According to the impulse-momentum theorem,the impulse is equal to the area under the force-time $(F-t)$ graph.
Impulse = Area under the $F-t$ graph
The graph consists of a triangle from $t = 0$ to $t = 2\,s$ and a rectangle from $t = 2\,s$ to $t = 6\,s$.
Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\,s \times 10\,N = 10\,N-s$.
Area of rectangle = $\text{length} \times \text{width} = (6 - 2)\,s \times 10\,N = 4\,s \times 10\,N = 40\,N-s$.
Total momentum acquired = Total area = $10\,N-s + 40\,N-s = 50\,N-s$.
Therefore,the correct option is $D$.

(C) Impulse is defined as the area under the force-time $(F-t)$ graph.
For figure $(I)$ (Rectangle): $\text{Area} = \text{base} \times \text{height} = 1.0 \times 0.25 = 0.25 \text{ units}$.
For figure $(II)$ (Triangle): $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0 \times 0.3 = 0.3 \text{ units}$.
For figure $(III)$ (Triangle): $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1.0 \times 1.0 = 0.5 \text{ units}$.
For figure $(IV)$ (Triangle): $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1.0 \times 1.0 = 0.5 \text{ units}$.
Comparing the areas,the impulse is highest for figures $(III)$ and $(IV)$.

(D) Initial momentum $P_1 = 10 \, kg \cdot m/s$.
Force $F = 0.2 \, N$ acts for time $t = 10 \, s$.
Change in momentum $\Delta P = F \times t = 0.2 \times 10 = 2 \, kg \cdot m/s$.
Final momentum $P_2 = P_1 + \Delta P = 10 + 2 = 12 \, kg \cdot m/s$.
Kinetic energy $K = \frac{P^2}{2m}$.
Increase in kinetic energy $\Delta K = \frac{P_2^2 - P_1^2}{2m}$.
$\Delta K = \frac{12^2 - 10^2}{2 \times 5} = \frac{144 - 100}{10} = \frac{44}{10} = 4.4 \, J$.

(C) The force applied on the mass is given by the rate of change of momentum: $\vec{F} = \frac{\Delta \vec{P}}{\Delta t}$.
Let the wall be along the $y$-axis. The initial velocity vector is $\vec{v}_i = v \sin \theta \hat{i} + v \cos \theta \hat{j}$.
The final velocity vector is $\vec{v}_f = -v \sin \theta \hat{i} + v \cos \theta \hat{j}$.
The change in momentum is $\Delta \vec{P} = m(\vec{v}_f - \vec{v}_i) = m(-2v \sin \theta \hat{i}) = -2mv \sin \theta \hat{i}$.
Given $m = 0.1\,kg$,$v = 5\,m/s$,$\theta = 60^\circ$,and $\Delta t = 2 \times 10^{-3}\,s$.
Substituting the values: $\Delta P_x = -2 \times 0.1 \times 5 \times \sin(60^\circ) = -1 \times \frac{\sqrt{3}}{2} = -0.5\sqrt{3}\,kg\cdot m/s$.
The force $\vec{F} = \frac{-0.5\sqrt{3}}{2 \times 10^{-3}} = -250\sqrt{3}\,N$.
The negative sign indicates that the force exerted by the wall on the mass is directed to the left.

(C) Change in momentum is defined as $\Delta \vec{P} = \vec{P}_{final} - \vec{P}_{initial}$.
For the lead ball,it strikes the wall and falls down,so its final velocity is $0$. Thus,$\Delta \vec{P}_{lead} = 0 - m\vec{v} = -m\vec{v}$. The magnitude of change is $m\vec{v}$.
For the tennis ball,it bounces back with the same velocity $\vec{v}$ in the opposite direction,so its final velocity is $-\vec{v}$. Thus,$\Delta \vec{P}_{tennis} = m(-\vec{v}) - m\vec{v} = -2m\vec{v}$. The magnitude of change is $2m\vec{v}$.
Comparing the magnitudes,$|\Delta \vec{P}_{tennis}| > |\Delta \vec{P}_{lead}|$. Therefore,the tennis ball suffers a greater change in momentum.

(D) The initial momentum of the body is $p_i = Mv$ (taking the initial direction as positive).
Since the body retraces its path with the same speed,the final velocity is $-v$.
Therefore,the final momentum is $p_f = M(-v) = -Mv$.
The change in momentum $\Delta p$ is given by $\Delta p = p_f - p_i$.
Substituting the values: $\Delta p = -Mv - (Mv) = -2Mv$.

(B) The force exerted on the wall is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$.
Initial momentum $p_i = m \times v = 2 \, kg \times 100 \, m/s = 200 \, kg \cdot m/s$.
Final momentum $p_f = m \times (-v) = 2 \, kg \times (-100 \, m/s) = -200 \, kg \cdot m/s$ (assuming the wall is in the positive direction).
Change in momentum $\Delta p = p_f - p_i = -200 - 200 = -400 \, kg \cdot m/s$.
The magnitude of the change in momentum is $|\Delta p| = 400 \, kg \cdot m/s$.
Given time of contact $\Delta t = 1/50 \, s$.
Force $F = \frac{|\Delta p|}{\Delta t} = \frac{400}{1/50} = 400 \times 50 = 20,000 \, N = 2 \times 10^4 \, N$.

(B) Impulse is defined as the change in momentum of an object.
Formula: $J = \Delta p = m(v_f - v_i)$
Given:
Mass $m = 0.1 \, kg$
Final velocity $v_f = 40 \, m/s$ (taken as positive)
Initial velocity $v_i = -30 \, m/s$ (since it is in the opposite direction to the final velocity)
Calculation:
$J = 0.1 \times (40) - 0.1 \times (-30)$
$J = 4 + 3 = 7 \, N \cdot s$
Therefore,the correct expression is $0.1 \times (40) - 0.1 \times (-30)$.

(A) Impulse is defined as the change in momentum of an object.
Let the mass of each ball be $m = 0.06 \,kg$.
The initial velocity of the first ball is $v_i = 4 \,m/s$ and the second ball is $-4 \,m/s$.
After the collision,the balls rebound with the same speed,so the final velocity of the first ball is $v_f = -4 \,m/s$ and the second ball is $4 \,m/s$.
The change in momentum for one ball is $\Delta p = m(v_f - v_i)$.
$\Delta p = 0.06 \times (-4 - 4) = 0.06 \times (-8) = -0.48 \,kg \cdot m/s$.
The magnitude of the impulse imparted to each ball is $|\Delta p| = 0.48 \,kg \cdot m/s$.

(C) The velocity of the ball changes from $v_i = 10 \ m/s$ to $v_f = -10 \ m/s$ (taking the direction towards the wall as positive).
Since velocity is a vector quantity,a change in direction implies a change in velocity.
Change in momentum $\Delta p = m(v_f - v_i) = m(-10 - 10) = -20m$.
Since there is a change in momentum over a finite time interval during the collision,a force $F = \frac{\Delta p}{\Delta t}$ acts on the ball.
According to Newton's second law,$F = ma$,therefore,the ball experiences an acceleration during the impact.

(B) The force experienced by the ball is equal to the rate of change of momentum.
Given: mass $m = 100 \, g = 0.1 \, kg$,velocity $v = 10 \, m/s$,angle with the wall $\theta = 30^\circ$,and time $t = 0.1 \, s$.
The change in momentum $\Delta p$ is given by the component perpendicular to the wall: $\Delta p = 2mv \sin \theta$.
Substituting the values: $\Delta p = 2 \times 0.1 \times 10 \times \sin 30^\circ = 2 \times 0.1 \times 10 \times 0.5 = 1 \, kg \cdot m/s$.
The force $F = \frac{\Delta p}{t} = \frac{1}{0.1} = 10 \, N$.

(C) The initial momentum of the two-particle system at $t = 0$ is $\vec{P}_i = m_1\vec{v}_1 + m_2\vec{v}_2$.
Collision between the two particles is an internal interaction and does not affect the total momentum of the system.
The only external force acting on the system is gravity,which is $\vec{F}_{ext} = (m_1 + m_2)g$ acting downwards.
According to the impulse-momentum theorem,the change in momentum $\Delta \vec{P}$ is equal to the impulse of the external force: $\Delta \vec{P} = \int_{0}^{2t_0} \vec{F}_{ext} dt$.
Since the force is constant,$\Delta \vec{P} = (m_1 + m_2)g \times (2t_0 - 0) = 2(m_1 + m_2)gt_0$.
Therefore,the magnitude of the change in momentum is $|(m_1\vec{v}_1' + m_2\vec{v}_2') - (m_1\vec{v}_1 + m_2\vec{v}_2)| = 2(m_1 + m_2)gt_0$.

(B) The change in linear momentum $\Delta p$ is equal to the impulse,which is the area under the force-time $(F-t)$ graph.
$\Delta p = \int F \, dt = \text{Area under } F-t \text{ graph}$.
The graph consists of circular segments. From $t = 0$ to $t = 2$,the area is a quarter-circle below the time axis with radius $r = 2$. Area $= -\frac{1}{4} \pi r^2 = -\frac{1}{4} \pi (2)^2 = -\pi$.
From $t = 2$ to $t = 6$,the area is a semi-circle above the time axis with radius $r = 2$. Area $= +\frac{1}{2} \pi r^2 = +\frac{1}{2} \pi (2)^2 = +2\pi$.
From $t = 6$ to $t = 8$,the area is a quarter-circle below the time axis with radius $r = 2$. Area $= -\frac{1}{4} \pi r^2 = -\frac{1}{4} \pi (2)^2 = -\pi$.
The total area (net impulse) $= -\pi + 2\pi - \pi = 0$.
Therefore,the linear momentum gained between $0$ and $8$ seconds is $0 \, N \cdot s$.

(C) The momentum of an object is given by the formula $p = mv$,where $m$ is the mass and $v$ is the velocity (or speed).
If the speed is doubled,the new velocity becomes $v' = 2v$.
Substituting this into the momentum formula: $p' = m(2v) = 2(mv) = 2p$.
Therefore,the momentum becomes double.
Kinetic energy is given by $K.E. = \frac{1}{2}mv^2$. If $v$ is doubled,$K.E.' = \frac{1}{2}m(2v)^2 = 4(\frac{1}{2}mv^2) = 4K.E.$,so kinetic energy becomes four times.