Third Law of Motion and Momentum and Impulse Questions in English

(B) Let the mass of the balls be $m$ and the initial velocity be $v_1$.
For the metal ball,the final velocity $v_2 = 0$ (as it does not rebound).
The change in momentum is $\Delta p = m(v_2 - v_1) = m(0 - v_1) = -mv_1$.
The magnitude of change is $|\Delta p| = mv_1$.
For the rubber ball,the final velocity $v_2' = -v_1$ (as it rebounds with the same velocity).
The change in momentum is $\Delta p' = m(v_2' - v_1) = m(-v_1 - v_1) = -2mv_1$.
The magnitude of change is $|\Delta p'| = 2mv_1$.
Comparing the two,$|\Delta p'| > |\Delta p|$.
Therefore,the rubber ball experiences a greater change in momentum.

(A) The change in momentum $\Delta \vec{p}$ is given by $\vec{p}_f - \vec{p}_i$.
Since the ball rebounds at the same angle and speed,the velocity component parallel to the wall remains unchanged,while the component perpendicular to the wall reverses direction.
Let the wall be along the $y$-axis. The initial velocity is $\vec{v}_i = (v \cos 60^\circ) \hat{i} - (v \sin 60^\circ) \hat{j}$.
The final velocity is $\vec{v}_f = -(v \cos 60^\circ) \hat{i} - (v \sin 60^\circ) \hat{j}$.
The change in momentum is $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(-2v \cos 60^\circ) \hat{i}$.
The magnitude of the change in momentum is $|\Delta p| = 2mv \cos 60^\circ$.
Given $m = 3 \ kg$,$v = 10 \ m/s$,and $\cos 60^\circ = 0.5$.
$|\Delta p| = 2 \times 3 \times 10 \times 0.5 = 30 \ kg \cdot m/s$.

(C) The change in momentum $\Delta p$ is equal to the area under the force-time graph.
From the graph,the area from $t = 0$ to $t = 3 \ s$ is a triangle with base $3 \ s$ and height $4 \ N$,so Area$_1 = \frac{1}{2} \times 3 \times 4 = 6 \ N \cdot s$.
The area from $t = 3 \ s$ to $t = 4.5 \ s$ is a triangle with base $1.5 \ s$ and height $-2 \ N$ (since the force at $t = 4.5 \ s$ is $-2 \ N$ by similar triangles),so Area$_2 = \frac{1}{2} \times 1.5 \times (-2) = -1.5 \ N \cdot s$.
The total change in momentum $\Delta p = 6 - 1.5 = 4.5 \ kg \cdot m/s$.
Since the block starts from rest,the final momentum $p = 4.5 \ kg \cdot m/s$.
The kinetic energy $K = \frac{p^2}{2m} = \frac{(4.5)^2}{2 \times 2} = \frac{20.25}{4} = 5.0625 \ J \approx 5.06 \ J$.

(D) The change in momentum is given by the impulse: $\Delta P = F \times t$.
Given $F = 0.2 \ N$ and $t = 10 \ s$,the change in momentum is $\Delta P = 0.2 \times 10 = 2 \ kg \cdot m/s$.
Initial momentum $P_1 = 10 \ kg \cdot m/s$.
Final momentum $P_2 = P_1 + \Delta P = 10 + 2 = 12 \ kg \cdot m/s$.
The kinetic energy $K$ is related to momentum $P$ by $K = \frac{P^2}{2m}$.
Initial kinetic energy $K_1 = \frac{P_1^2}{2m} = \frac{10^2}{2 \times 5} = \frac{100}{10} = 10 \ J$.
Final kinetic energy $K_2 = \frac{P_2^2}{2m} = \frac{12^2}{2 \times 5} = \frac{144}{10} = 14.4 \ J$.
The increase in kinetic energy is $\Delta K = K_2 - K_1 = 14.4 - 10 = 4.4 \ J$.

(B) The mass of water striking the wall per second is $m = A \cdot v \cdot \rho$.
The initial momentum of the water striking the wall per second is $P_i = m \cdot v = Av^2 \rho$.
Since the collision is elastic,the water rebounds with the same speed $v$ at an angle $\theta$ with the normal. The component of momentum perpendicular to the wall is $P \cos \theta$.
The change in momentum per second (force) exerted by the wall on the water is $\Delta P = P_{final, x} - P_{initial, x}$.
Taking the direction towards the wall as positive,the initial momentum component is $P_i \cos \theta$ and the final momentum component is $-P_i \cos \theta$.
Therefore,the force exerted by the water on the wall is $F = |\Delta P| = |(-P_i \cos \theta) - (P_i \cos \theta)| = 2P_i \cos \theta$.
Substituting $P_i = Av^2 \rho$,we get $F = 2Av^2 \rho \cos \theta$.

(B) Let the initial velocity be $\vec{v}_1$ and the final velocity be $\vec{v}_2$.
From the geometry,the initial momentum is $\vec{P}_1 = m\vec{v}_1 = -mv \sin \theta \hat{i} - mv \cos \theta \hat{j}$.
The final momentum is $\vec{P}_2 = m\vec{v}_2 = mv \sin \theta \hat{i} - mv \cos \theta \hat{j}$.
The change in momentum is $\Delta \vec{P} = \vec{P}_2 - \vec{P}_1$.
$\Delta \vec{P} = (mv \sin \theta \hat{i} - mv \cos \theta \hat{j}) - (-mv \sin \theta \hat{i} - mv \cos \theta \hat{j})$.
$\Delta \vec{P} = 2mv \sin \theta \hat{i}$.
The magnitude of the change in momentum is $|\Delta \vec{P}| = 2mv \sin \theta$.

(C) Given: Mass $m = 60 \ g = 60 \times 10^{-3} \ kg$,Initial velocity $v_i = 4 \ m/s$,Final velocity $v_f = -4 \ m/s$ (since it rebounds).
The initial momentum is $P_i = m \times v_i = (60 \times 10^{-3}) \times 4 = 0.24 \ kg \cdot m/s$.
The final momentum is $P_f = m \times v_f = (60 \times 10^{-3}) \times (-4) = -0.24 \ kg \cdot m/s$.
The change in momentum $\Delta P = P_f - P_i = -0.24 - 0.24 = -0.48 \ kg \cdot m/s$.
The magnitude of the change in momentum is $|\Delta P| = 0.48 \ kg \cdot m/s$.

(A) Given: Mass $m = 150 \, g = 0.15 \, kg$.
Initial velocity $u = -12 \, m/s$ (taking the direction towards the bat as negative).
Final velocity $v = +20 \, m/s$ (taking the direction away from the bat as positive).
Time interval $\Delta t = 0.01 \, s$.
According to Newton's Second Law of Motion,the force $F$ is given by the rate of change of momentum:
$F = \frac{m(v - u)}{\Delta t}$
$F = \frac{0.15 \times (20 - (-12))}{0.01}$
$F = \frac{0.15 \times (20 + 12)}{0.01}$
$F = \frac{0.15 \times 32}{0.01}$
$F = 15 \times 32 = 480 \, N$.
Therefore,the force exerted by the bat is $480 \, N$.

(C) The impulse $J$ is defined as the product of force $F$ and the time interval $\Delta t$ for which it acts: $J = F \times \Delta t$.
Given: Force $F = 50 \, dynes$.
Since $1 \, N = 10^5 \, dynes$,we convert the force to $SI$ units: $F = 50 \times 10^{-5} \, N = 5 \times 10^{-4} \, N$.
Time $\Delta t = 3 \, s$.
Impulse $J = (5 \times 10^{-4} \, N) \times (3 \, s) = 15 \times 10^{-4} \, N \cdot s = 1.5 \times 10^{-3} \, N \cdot s$.

(B) Impulse is defined as the change in linear momentum of the body.
Initial momentum of the body,$P_i = MV$.
Since the body bounces back with the same velocity $V$ in the opposite direction,the final momentum is $P_f = -MV$.
Impulse $J = \Delta P = P_f - P_i$.
$J = (-MV) - (MV) = -2MV$.
The magnitude of the impulse experienced by the body is $2MV$.

(C) The change in momentum of a particle is equal to the impulse,which is the area under the force-time $(F-t)$ graph.
Change in momentum = Area under $F-t$ graph
$= \text{Area of triangle } ABC + \text{Area of rectangle } CDEF + \text{Area of rectangle } FGHI$
$= (\frac{1}{2} \times \text{base} \times \text{height}) + (\text{width} \times \text{height}) + (\text{width} \times \text{height})$
$= (\frac{1}{2} \times 2 \times 6) + (2 \times -3) + (4 \times 3)$
$= 6 - 6 + 12$
$= 12 \, N-s$.

(C) The impulse imparted by the wall on the ball is equal to the change in momentum of the ball.
Let the initial velocity be $\vec{v}_i$ and the final velocity be $\vec{v}_f$.
The initial momentum is $\vec{p}_i = m\vec{v}_i$ and the final momentum is $\vec{p}_f = m\vec{v}_f$.
Taking the normal to the wall as the $x$-axis,the initial velocity components are $v_{ix} = v \cos 60^\circ$ (towards the wall) and $v_{iy} = v \sin 60^\circ$ (parallel to the wall).
The final velocity components are $v_{fx} = -v \cos 60^\circ$ (away from the wall) and $v_{fy} = v \sin 60^\circ$ (parallel to the wall).
The change in momentum $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = m(\vec{v}_f - \vec{v}_i)$.
$\Delta p_x = m(v_{fx} - v_{ix}) = m(-v \cos 60^\circ - v \cos 60^\circ) = -2mv \cos 60^\circ = -2mv(0.5) = -mv$.
$\Delta p_y = m(v_{fy} - v_{iy}) = m(v \sin 60^\circ - v \sin 60^\circ) = 0$.
The magnitude of the impulse is $|\Delta \vec{p}| = |-mv| = mv$.

(C) Impulse is defined as the product of force and time interval: $J = F \times \Delta t$.
Given force $F = 50 \text{ dynes}$. Since $1 \text{ dyne} = 10^{-5} \text{ N}$,we have $F = 50 \times 10^{-5} \text{ N} = 5 \times 10^{-4} \text{ N}$.
The time interval is $\Delta t = 3 \text{ s}$.
Therefore,Impulse $J = (5 \times 10^{-4} \text{ N}) \times (3 \text{ s}) = 15 \times 10^{-4} \text{ N-s} = 1.5 \times 10^{-3} \text{ N-s}$.

(B) The impulse $J$ delivered to the particle is equal to the area under the force-time $(F-t)$ graph.
Impulse $J = \Delta p = m(v_f - v_i)$.
From the graph, the area from $t = 0$ to $t = 10\, s$ is a triangle with base $10\, s$ and height $20\, N$.
Area $A_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = 100\, Ns$.
This area is positive, so the velocity increases during this interval.
At $t = 10\, s$, the velocity $v_{10}$ is given by $m(v_{10} - v_i) = A_1$.
$1 \times (v_{10} - 10) = 100 \implies v_{10} = 110\, ms^{-1}$.
From $t = 10\, s$ to $t = 20\, s$, the force is $-10\, N$ (a rectangle).
Area $A_2 = \text{base} \times \text{height} = (20 - 10) \times (-10) = 10 \times (-10) = -100\, Ns$.
Since the force is negative, the velocity decreases after $t = 10\, s$.
Therefore, the maximum velocity is attained at $t = 10\, s$, which is $110\, ms^{-1}$.

(C) The impulse $J$ applied to the particle is equal to the change in momentum $\Delta p = mv_f - mv_i$. Since the particle is initially at rest,$v_i = 0$,so $mv_f = \int_{0}^{T} F(t) dt$.
Substituting the given force expression:
$mv_f = \int_{0}^{T} F_0 \left[ 1 - \left( \frac{2t - T}{T} \right)^2 \right] dt$.
Let $u = \frac{2t - T}{T}$,then $du = \frac{2}{T} dt$,or $dt = \frac{T}{2} du$.
When $t = 0, u = -1$. When $t = T, u = 1$.
$mv_f = F_0 \int_{-1}^{1} (1 - u^2) \frac{T}{2} du = \frac{F_0 T}{2} \left[ u - \frac{u^3}{3} \right]_{-1}^{1}$.
$mv_f = \frac{F_0 T}{2} \left[ (1 - 1/3) - (-1 + 1/3) \right] = \frac{F_0 T}{2} \left[ 2/3 + 2/3 \right] = \frac{F_0 T}{2} \left( \frac{4}{3} \right) = \frac{2F_0 T}{3}$.
Therefore,$v_f = \frac{2F_0 T}{3m}$.

(C) Impulse $J$ is defined as the product of the average force $F$ and the time interval $\Delta t$ for which the force acts: $J = F \cdot \Delta t$.
In the first case,the impulse is $J = F \cdot \Delta t$.
In the second case,the force $F$ remains the same,but the contact time is doubled,i.e.,$\Delta t' = 2 \Delta t$.
Therefore,the new impulse $J'$ is given by $J' = F \cdot (2 \Delta t) = 2(F \cdot \Delta t) = 2J$.
Thus,the new impulse is twice the original impulse.

(A) The momentum of an object is given by the product of its mass and its velocity $(p = mv)$.
Before the collision,the disk is moving directly east with a velocity of $1 \ m/s$. The eastward component of its initial momentum is $p_{i,E} = 4 \ kg \times 1 \ m/s = 4 \ kg \cdot m/s$.
After the collision,the disk moves directly north. Since it has no velocity component in the eastward direction,the eastward component of its final momentum is $p_{f,E} = 0 \ kg \cdot m/s$.
The change in the eastward component of the momentum is given by $\Delta p_E = p_{f,E} - p_{i,E}$.
Substituting the values,we get $\Delta p_E = 0 - 4 = -4 \ kg \cdot m/s$.

(C) The impulse $J$ is equal to the change in linear momentum,$\Delta p = m(v_f - v_i)$.
From the graph,the motion consists of segments with constant velocity.
For the interval $t = 0$ to $t = 2 \; s$,the displacement is $2 \; m$. Thus,initial velocity $v_i = \frac{2 \; m}{2 \; s} = 1 \; m/s$.
For the interval $t = 2 \; s$ to $t = 4 \; s$,the displacement is $-2 \; m$. Thus,final velocity $v_f = \frac{-2 \; m}{2 \; s} = -1 \; m/s$.
The mass of the body is $m = 0.4 \; kg$.
The impulse $J$ at each collision (at $t = 2, 6, 10, 14 \; s$) is given by:
$J = m(v_f - v_i) = 0.4 \; kg \times (-1 \; m/s - 1 \; m/s) = 0.4 \times (-2) = -0.8 \; kg \cdot m/s$.
The magnitude of the impulse is $|J| = |-0.8| = 0.8 \; N \cdot s$.

(D) The change in momentum $\Delta \vec{p}$ is equal to the impulse $\vec{J} = \int \vec{F} dt$. This is equal to the area under the force-time graph for each component.
For the $x$-component: $\Delta p_x = \text{Area} = 30\,N \times 0.1\,s = 3\,N\cdot s$.
For the $y$-component: $\Delta p_y = \text{Area} = \frac{1}{2} \times 80\,N \times 0.1\,s = 4\,N\cdot s$.
For the $z$-component: $\Delta p_z = \text{Area} = -50\,N \times 0.1\,s = -5\,N\cdot s$.
The magnitude of the total change in momentum is $\Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2}$.
$\Delta p = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}\,kg\,m/s$.

(B) The mass of the ball is $m = 40 \ g = 0.04 \ kg$.
The initial velocity just before impact is $v_i = -10 \ m/s$ (taking downward as negative).
Since the collision is elastic (no loss of mechanical energy),the final velocity just after impact is $v_f = +10 \ m/s$.
The change in momentum is $\Delta p = m(v_f - v_i) = 0.04 \times (10 - (-10)) = 0.04 \times 20 = 0.8 \ kg \cdot m/s$.
According to the impulse-momentum theorem,the impulse $J = F_{avg} \times \Delta t = \Delta p$.
Given $F_{avg} = 16 \ N$,we have $16 \times \Delta t = 0.8$.
$\Delta t = \frac{0.8}{16} = 0.05 \ s$.
Converting to milliseconds,$\Delta t = 0.05 \times 1000 = 50 \ ms$.

(C) Let the impulsive tension in the string be $J$. The impulsive force exerted by each particle on the bob along the string is $m u \cos 60^{\circ}$.
Since there are two such particles striking simultaneously from opposite sides, the total impulsive force acting on the bob along the string is $2 \times (m u \cos 60^{\circ})$.
Using the impulse-momentum theorem along the vertical direction (the direction of the string):
$J - 2mu \cos 60^{\circ} = 0$ (since the bob is constrained by the string and cannot move vertically).
$J = 2mu \times \frac{1}{2} = mu$.
Thus, the impulsive tension in the string is $mu$.

(C) The linear momentum vector is given by $\vec P = P_x \hat i + P_y \hat j = (2 \cos t) \hat i + (2 \sin t) \hat j$.
According to Newton's second law,the force $\vec F$ is the time derivative of linear momentum: $\vec F = \frac{d\vec P}{dt}$.
$\vec F = \frac{d}{dt} (2 \cos t \hat i + 2 \sin t \hat j) = -2 \sin t \hat i + 2 \cos t \hat j$.
To find the angle $\theta$ between $\vec F$ and $\vec P$,we calculate their dot product: $\vec F \cdot \vec P = (-2 \sin t)(2 \cos t) + (2 \cos t)(2 \sin t) = -4 \sin t \cos t + 4 \sin t \cos t = 0$.
Since the dot product $\vec F \cdot \vec P = |\vec F| |\vec P| \cos \theta = 0$,it follows that $\cos \theta = 0$,which means $\theta = 90^\circ$.

(C) Given: Mass $m = 2 \ kg$,initial velocity $u = 5 \ ms^{-1}$.
According to the impulse-momentum theorem,the change in momentum is equal to the area under the force-time graph.
$\Delta p = J = \int F \, dt = \text{Area under } F-t \text{ graph}$.
Area = Area of triangle $(0-2 \ s)$ + Area of rectangle $(2-4 \ s)$ + Area of trapezoid $(4-4.5 \ s)$ + Area of rectangle $(4.5-6.5 \ s)$.
Area = $\left(\frac{1}{2} \times 2 \times 4\right) + (2 \times 4) + \left(\frac{1}{2} \times (4 + 2.5) \times 0.5\right) + (2 \times 2.5)$.
Area = $4 + 8 + 1.625 + 5 = 18.625 \ N \cdot s$.
Since $\Delta p = m(v - u) = 18.625$,
$2(v - 5) = 18.625$.
$v - 5 = 9.3125$.
$v = 14.3125 \ ms^{-1}$.
Re-evaluating the area calculation: Area = $4 + 8 + 1.625 + 5 = 18.625$. $v = 5 + 9.3125 = 14.3125 \ ms^{-1}$. Given the options,$14.25$ is the closest intended answer.

(A) The impulse $\vec{I}$ exerted by the wall on the projectile is equal to the change in momentum of the projectile: $\vec{I} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i$.
Let the initial direction of the projectile be positive ($+x$ direction).
Case $A$: Initial momentum $\vec{p}_i = mv \hat{i}$,final momentum $\vec{p}_f = -mv \hat{i}$. Impulse $I_A = |(-mv) - (mv)| = |-2mv| = 2mv$.
Case $B$: Initial momentum $\vec{p}_i = mv \hat{i}$,final momentum $\vec{p}_f = 0$. Impulse $I_B = |0 - mv| = |-mv| = mv$.
Case $C$: Initial momentum $\vec{p}_i = mv \hat{i}$,final momentum $\vec{p}_f = \frac{mv}{2} \hat{i}$. Impulse $I_C = |\frac{mv}{2} - mv| = |-\frac{mv}{2}| = \frac{mv}{2}$.
Comparing the magnitudes: $2mv > mv > \frac{mv}{2}$,which means $I_A > I_B > I_C$.

(A) The velocity of each bead just before hitting the pan is $v = \sqrt{2gh}$.
Given $g = 10 \ m/s^2$ and $h = 5 \ m$, we have $v = \sqrt{2 \times 10 \times 5} = 10 \ m/s$.
Since the beads bounce back to their original height, the collision is perfectly elastic, and the velocity just after the collision is also $v = 10 \ m/s$ upwards.
The change in momentum for one bead is $\Delta p = m(v - (-v)) = 2mv$.
Given $m = 15 \ g = 0.015 \ kg$, $\Delta p = 2 \times 0.015 \times 10 = 0.3 \ kg \ m/s$.
The rate of change of momentum (force) exerted on the pan by $n = 100 \ \text{beads per second}$ is $F = n \times \Delta p = 100 \times 0.3 = 30 \ N$.
To keep the pointer at zero, the weight of the mass $M$ placed in the other pan must balance this force: $Mg = F$.
$M \times 10 = 30 \implies M = 3 \ kg$.

(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 are equal in magnitude and act in exactly opposite directions along the same line of action,the angle between them is $180^{\circ}$ or $\pi \ rad$.

(C) According to Newton's third law of motion,the force exerted by the horse on the wagon $(T)$ and the force exerted by the wagon on the horse $(T)$ are equal in magnitude and opposite in direction. These forces act on different bodies,so they do not cancel each other out.
For the wagon to move forward,the forward force exerted by the horse on the wagon $(T)$ must be greater than the opposing frictional force $(f_w)$ acting on the wagon.
For the horse to move forward,the horse pushes the ground backward with its feet,and the ground exerts a forward frictional force $(f_M)$ on the horse. The horse moves forward if this forward frictional force $(f_M)$ is greater than the backward tension $(T)$ exerted by the wagon.
Therefore,the statement that the horse's force on the wagon is as strong as the force of the wagon on the horse is correct,as it describes the action-reaction pair.

(A) The horizontal component of velocity remains constant throughout the flight because there is no horizontal force (ignoring air resistance).
Let $V_1$ be the final speed at the ground.
$V \cos 30^{\circ} = V_1 \cos 45^{\circ}$
$V \frac{\sqrt{3}}{2} = V_1 \frac{1}{\sqrt{2}}$
$V_1 = V \sqrt{\frac{3}{2}}$
Impulse $J$ is equal to the change in momentum: $\vec{J} = \vec{p}_f - \vec{p}_i$.
Since horizontal momentum is constant,the impulse acts only in the vertical direction.
$J = |p_{fy} - p_{iy}| = |m(-V_1 \sin 45^{\circ}) - m(V \sin 30^{\circ})|$
$J = m |V \sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} + V \cdot \frac{1}{2}|$
$J = m |V \frac{\sqrt{3}}{2} + \frac{V}{2}| = \frac{1}{2}(\sqrt{3} + 1) mV$.

(C) When the block is released from the horizontal position,it falls freely under gravity until the string becomes taut.
The vertical distance fallen by the block is $h = 5\, m$.
The velocity $v_i$ of the block just before the string becomes tight is given by the equation of motion $v^2 = u^2 + 2gh$:
$v_i = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10\, m/s$ (directed vertically downwards).
When the string becomes tight,the impulsive tension in the string acts upwards,bringing the vertical component of the velocity to zero instantaneously.
Thus,the final vertical velocity $v_f = 0$.
The impulse $I$ is equal to the change in momentum:
$I = m(v_f - v_i) = 2(0 - 10) = -20\, N-s$.
The magnitude of the impulse is $20\, N-s$.

(A) The mass of the particle is $m = 100 \ g = 0.1 \ kg$.
From the position-time graph,the velocity $v$ is the slope of the line.
For $t < 5 \ s$,the initial velocity $v_i$ is the slope of the line from $(0, 20)$ to $(5, 10)$:
$v_i = \frac{10 - 20}{5 - 0} = \frac{-10}{5} = -2 \ m/s$.
For $t > 5 \ s$,the final velocity $v_f$ is the slope of the line from $(5, 10)$ to $(15, 20)$:
$v_f = \frac{20 - 10}{15 - 5} = \frac{10}{10} = 1 \ m/s$.
Impulse $I$ is equal to the change in momentum:
$I = \Delta p = m(v_f - v_i) = 0.1 \times [1 - (-2)] = 0.1 \times 3 = 0.3 \ N-s$.

(C) The impulse $J$ is equal to the area under the force-time graph.
Impulse $J = \int F \, dt = \text{Area under } F-t \text{ graph}$.
Area from $t=0$ to $t=2 \ s$ is $10 \ N \times 2 \ s = 20 \ N \cdot s$.
Area from $t=2$ to $t=3 \ s$ is $-10 \ N \times 1 \ s = -10 \ N \cdot s$.
Total impulse $J_{total} = 20 - 10 = 10 \ N \cdot s$.
According to the impulse-momentum theorem,$J = \Delta p = m(v_f - v_i)$.
Given $m = 2 \ kg$ and $v_i = 0$,we have $10 = 2(v_f - 0)$,which gives $v_f = 5 \ m/s$.
The work done by the force is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
$W = \frac{1}{2} \times 2 \times (5)^2 - 0 = 25 \ J$.

(B) Let the velocities of trains $A$ and $B$ be $v_A$ and $v_B$ respectively,where $v_A > v_B$.
When a packet of mass $m$ is transferred from $A$ to $B$,it carries a horizontal momentum $m v_A$ to train $B$.
Since $v_A > v_B$,the packet imparts an impulsive force on train $B$ in the direction of its motion,causing $B$ to accelerate.
Conversely,the packet transferred from $B$ to $A$ carries momentum $m v_B$. Since $v_B < v_A$,this packet exerts a retarding force on train $A$.
Therefore,train $B$ gains momentum (accelerates) and train $A$ loses momentum (retards).

(C) Given: Mass $m = 140\,g = 0.140\,kg$.
Initial velocity $v_i = 39.0\,m/s$.
Final velocity $v_f = -39.0\,m/s$ (opposite direction).
Impact time $\Delta t = 1.20\,ms = 1.20 \times 10^{-3}\,s$.
The change in momentum is $\Delta p = m(v_f - v_i)$.
$\Delta p = 0.140 \times (-39.0 - 39.0) = 0.140 \times (-78.0) = -10.92\,kg\cdot m/s$.
The average force $F$ is given by $F = \frac{\Delta p}{\Delta t}$.
$F = \frac{-10.92}{1.20 \times 10^{-3}} = -9100\,N$.
The magnitude of the average force acting on the ball is $9100\,N$.

(B) The impulse $J$ received by a body is equal to the change in its linear momentum,given by $J = \Delta p = m(v_f - v_i)$.
For the particle of mass $m$:
Initial velocity $v_i = (3\hat{i} + 2\hat{j}) \ m/s$.
Final velocity $v_f = (-2\hat{i} + \hat{j}) \ m/s$.
Impulse received by $m = m(v_f - v_i) = m[(-2\hat{i} + \hat{j}) - (3\hat{i} + 2\hat{j})] = m[-5\hat{i} - \hat{j}]$.
According to Newton's third law,the impulse received by mass $M$ is equal and opposite to the impulse received by mass $m$:
Impulse received by $M = -(Impulse \ received \ by \ m) = -m[-5\hat{i} - \hat{j}] = m[5\hat{i} + \hat{j}]$.
Thus,the correct statement is that the impulse received by $m$ is $m[-5\hat{i} - \hat{j}]$.

(D) According to the impulse-momentum theorem,the change in momentum (impulse) is given by the product of the force applied and the time interval for which it is applied.
$\Delta P = F \times \Delta t$
Given:
Force $(F)$ = $30 \, N$
Time interval $(\Delta t)$ = $2 \, sec$
Therefore,the change in momentum is:
$\Delta P = 30 \, N \times 2 \, sec = 60 \, kg \cdot m/s$.
Thus,the correct option is $D$.

(B) According to Newton's third law of motion,for every action,there is an equal and opposite reaction. This law applies to all interactions between two bodies,regardless of whether the collision is head-on or oblique.
In both cases $(i)$ and $(ii)$,when body $A$ and body $B$ collide,they exert forces on each other. The force exerted by $A$ on $B$ $(F_{AB})$ is always equal in magnitude and opposite in direction to the force exerted by $B$ on $A$ $(F_{BA})$,i.e.,$F_{AB} = -F_{BA}$.
Therefore,they exert equal and opposite forces on each other in both scenarios.

(B) The impulse $J$ is equal to the change in momentum $\Delta p = m(v_f - v_i)$.
From the position-time graph,the velocity $v_i$ for $0 < t < 2\, s$ is the slope of the line:
$v_i = \frac{\Delta x}{\Delta t} = \frac{4 - 0}{2 - 0} = 2\, m/s$.
For $t > 2\, s$,the position is constant $(x = 4\, m)$,so the final velocity $v_f = 0\, m/s$.
The impulse at $t = 2\, s$ is:
$J = m(v_f - v_i) = 0.1\, kg \times (0 - 2)\, m/s = -0.2\, kg\, m/s$.

(B) Given,mass $m = 2\, kg$ and displacement $s = (2t^3 + 2)\, m$.
Velocity $v$ is the rate of change of displacement with respect to time:
$v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 + 2) = 6t^2\, m/s$.
At $t = 0\, s$,initial velocity $v_i = 6(0)^2 = 0\, m/s$.
At $t = 1\, s$,final velocity $v_f = 6(1)^2 = 6\, m/s$.
Impulse $J$ is equal to the change in linear momentum $\Delta P$:
$J = \Delta P = m(v_f - v_i)$.
Substituting the values:
$J = 2\, kg \times (6\, m/s - 0\, m/s) = 2 \times 6 = 12\, N-s$.

(B) Given: Mass $m = 500\, g = 0.5\, kg$,initial velocity $u = 15\, m/s$,final velocity $v = -15\, m/s$ (since it returns in the opposite direction),and time $\Delta t = 0.01\, s$.
Change in momentum $\Delta P = m(v - u) = 0.5 \times (-15 - 15) = 0.5 \times (-30) = -15\, kg\cdot m/s$.
The magnitude of the force exerted on the ball is $F = |\frac{\Delta P}{\Delta t}| = |\frac{-15}{0.01}| = 1500\, N$.
By Newton's third law,the force acted on the bat is equal in magnitude to the force acted on the ball,which is $1500\, N$.

(D) According to Newton's third law,action and reaction forces always act on different bodies and are equal in magnitude and opposite in direction.
$1$. $\vec F_4$ represents the weight of the balloon,which is the gravitational force exerted by the Earth on the balloon. The reaction to this force is the gravitational force exerted by the balloon on the Earth,which acts on the Earth.
$2$. Therefore,$\vec F_4$ and the gravitational force exerted by the balloon on the Earth form an action-reaction pair.
$3$. Other options are incorrect because $\vec F_1, \vec F_2, \vec F_3$ are contact forces acting on the same body (the balloon) and do not form action-reaction pairs with each other.

(B) The total length available for the motion of the beads is $L_{eff} = L - 2nr$,as each of the $n$ beads occupies a length of $2r$.
When a bead moves with speed $v$ and undergoes elastic collisions with the supports,the change in momentum for one collision with a support is $\Delta p = mv - (-mv) = 2mv$.
The time taken to travel between the two supports and return is $\Delta t = \frac{2(L - 2nr)}{v}$.
The average force $F$ exerted on each support is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t} = \frac{2mv}{2(L - 2nr) / v} = \frac{mv^2}{L - 2nr}$.

(B) The force exerted by a fluid jet on a surface is given by the rate of change of momentum.
Let $M$ be the mass of water hitting the surface in time $t$.
The velocity of the water jet is $v$ and it comes to rest upon hitting the surface,so the change in velocity is $\Delta v = v - 0 = v$.
The mass per unit length is given as $m = \frac{M}{L}$,where $L$ is the length of the water jet.
In time $t$,the length of the water jet hitting the surface is $L = v \cdot t$.
Therefore,the mass of water hitting the surface in time $t$ is $M = m \cdot L = m \cdot v \cdot t$.
The force $F$ is the rate of change of momentum: $F = \frac{\Delta p}{\Delta t} = \frac{M \cdot v}{t}$.
Substituting $M = m \cdot v \cdot t$ into the force equation:
$F = \frac{(m \cdot v \cdot t) \cdot v}{t} = m \cdot v^2$.
Thus,the force on the surface is $mv^2$.

(A) Impulse is defined as the change in momentum of a body.
Mathematically,Impulse $J = \Delta p = p_f - p_i$.
Given the initial velocity is $v_1$ and the final velocity is $v_2$,the initial momentum is $p_i = m v_1$ and the final momentum is $p_f = m v_2$.
Therefore,Impulse $J = m v_2 - m v_1 = m(v_2 - v_1)$.

(D) According to Newton's second law of motion,the average force is equal to the rate of change of momentum.
The formula is given by: $F_{ave} = \frac{\Delta p}{\Delta t}$.
Given:
Change in momentum $\Delta p = 50 \ N-s$ (since the hammer is brought to rest from $50 \ N-s$ momentum).
Time interval $\Delta t = 0.25 \ s$.
Substituting the values:
$F_{ave} = \frac{50}{0.25} = \frac{5000}{25} = 200 \ N$.
Therefore,the average force required is $200 \ N$.

(C) The change in momentum $\Delta p$ is given by the impulse exerted by the gravitational force.
The force acting on the ball is its weight,$F = mg$,which acts downwards.
The time for which this force acts is the total time of flight,$T = \frac{2v \sin \theta}{g}$.
According to the impulse-momentum theorem,the change in momentum is equal to the impulse,which is the product of the constant force and the time interval.
$\Delta p = F \times T = (mg) \times T$.
Therefore,the total change in momentum is equal to the weight of the ball multiplied by the total time of flight.

(C) The average force $F$ is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$.
Given mass $m = 140\, g = 0.140\, kg$.
The initial velocity $v_1 = 39.0\, m/s$ (let this be in the positive direction).
The final velocity $v_2 = -39.0\, m/s$ (opposite direction).
The change in momentum $\Delta p = m(v_2 - v_1) = 0.140 \times (-39.0 - 39.0) = 0.140 \times (-78.0) = -10.92\, kg \cdot m/s$.
The impact time $\Delta t = 1.20\, ms = 1.20 \times 10^{-3}\, s$.
The magnitude of the average force is $F = \frac{|\Delta p|}{\Delta t} = \frac{10.92}{1.20 \times 10^{-3}} = \frac{10.92}{0.0012} = 9100\, N$.

(C) The initial velocity vector is $\vec{v}_i = v_0 \cos 37^{\circ} \hat{i} + v_0 \sin 37^{\circ} \hat{j} = \frac{4}{5} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j}$.
The final velocity vector is $\vec{v}_f = -(\frac{3}{4} v_0) \cos 53^{\circ} \hat{i} + (\frac{3}{4} v_0) \sin 53^{\circ} \hat{j}$.
Using $\cos 53^{\circ} = \frac{3}{5}$ and $\sin 53^{\circ} = \frac{4}{5}$,we get $\vec{v}_f = -\frac{3}{4} v_0 (\frac{3}{5}) \hat{i} + \frac{3}{4} v_0 (\frac{4}{5}) \hat{j} = -\frac{9}{20} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j}$.
The impulse $\vec{J}$ is the change in momentum: $\vec{J} = m(\vec{v}_f - \vec{v}_i)$.
$\vec{J} = m [(-\frac{9}{20} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j}) - (\frac{4}{5} v_0 \hat{i} + \frac{3}{5} v_0 \hat{j})]$.
$\vec{J} = m [(-\frac{9}{20} - \frac{16}{20}) v_0 \hat{i} + (\frac{3}{5} - \frac{3}{5}) v_0 \hat{j}] = m [-\frac{25}{20} v_0 \hat{i}] = -\frac{5}{4} mv_0 \hat{i}$.

(B) According to Newton's third law of motion,the force exerted by body $m_1$ on body $m_2$ is equal in magnitude and opposite in direction to the force exerted by body $m_2$ on body $m_1$.
Let $F$ be the magnitude of the force.
For body $m_2$,the force is $F = m_2 a_2$.
For body $m_1$,the force is $F = m_1 a_1$.
Since the forces are equal in magnitude,we have $m_1 a_1 = m_2 a_2$.
Therefore,the acceleration of body $m_1$ is $a_1 = \frac{m_2 a_2}{m_1}$.

(C) The impact force exerted by the water on the ball is given by Newton's second law in terms of momentum change:
$F = \frac{\Delta p}{\Delta t}$
Here,$\Delta p$ is the change in momentum of the water of mass $\Delta m$ striking the ball with speed $v$ during time $\Delta t$. Since the water falls dead after the collision,its final velocity is $0$.
Therefore,the change in momentum is $\Delta p = \Delta m \cdot v - 0 = \Delta m \cdot v$.
Substituting this into the force equation:
$F = \frac{\Delta m \cdot v}{\Delta t} = v \left( \frac{\Delta m}{\Delta t} \right)$
Where $\frac{\Delta m}{\Delta t}$ is the rate of flow of water in the nozzle. Since the ball is floating in equilibrium,the upward force exerted by the water must balance the weight of the ball:
$F = mg$
Equating the two expressions for $F$:
$v \left( \frac{\Delta m}{\Delta t} \right) = mg$
Solving for the rate of flow:
$\frac{\Delta m}{\Delta t} = \frac{mg}{v}$

(A) According to Newton's third law of motion,when a horse pushes the ground backward with its hooves,the ground exerts an equal and opposite reaction force on the horse in the forward direction. This reaction force from the ground is what causes the horse to move forward.