Third Law of Motion and Momentum and Impulse Questions in English

(B) The impulse $I$ is defined as the change in momentum,$\Delta p = I$. Given $I = 20 t^2 - 40 t$.
To find the time at which the change in momentum is minimum,we differentiate $I$ with respect to $t$ and set it to zero:
$\frac{dI}{dt} = \frac{d}{dt}(20 t^2 - 40 t) = 40 t - 40$.
Setting $\frac{dI}{dt} = 0$ gives $40 t - 40 = 0$,which implies $t = 1 \text{ s}$.
To confirm this is a minimum,we check the second derivative: $\frac{d^2I}{dt^2} = 40$. Since $40 > 0$,the function has a minimum at $t = 1 \text{ s}$.

(C) According to Newton's Second Law of Motion,the net external force acting on a system is equal to the rate of change of linear momentum,given by $F_{net} = \frac{dp}{dt}$.
If the final momentum $(p_f)$ is equal to the initial momentum $(p_i)$,then the change in momentum $\Delta p = p_f - p_i = 0$.
This implies that the average net force over the time interval $\Delta t$ is $\frac{\Delta p}{\Delta t} = 0$.
However,this does not necessarily mean the instantaneous net force is zero at every moment during the process. The net force could be non-zero at various instants such that the total impulse (integral of force over time) is zero.
Therefore,it is possible that there is a net force on the system at some point,even if the total change in momentum is zero.
Thus,the correct option is $(c)$.

(A) Given:
Mass $m = 5 \,kg$
Initial velocity $\vec{v}_i = (2 \hat{i} + 6 \hat{j}) \,m/s$
Final velocity $\vec{v}_f = (10 \hat{i} + 6 \hat{j}) \,m/s$
The change in momentum $\Delta \vec{p}$ is given by the formula:
$\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i)$
Substitute the given values:
$\Delta \vec{p} = 5 \times [(10 \hat{i} + 6 \hat{j}) - (2 \hat{i} + 6 \hat{j})]$
$\Delta \vec{p} = 5 \times [(10 - 2) \hat{i} + (6 - 6) \hat{j}]$
$\Delta \vec{p} = 5 \times [8 \hat{i} + 0 \hat{j}]$
$\Delta \vec{p} = 40 \hat{i} \,kg \cdot m/s$
Therefore,the change in momentum is $40 \hat{i} \,kg \cdot m/s$.

(A) In an elastic collision with a smooth wall,the component of velocity parallel to the wall remains unchanged.
Let the velocity of the particle be $\vec{v}$. The angle with the wall is $60^{\circ}$,so the angle with the normal to the wall is $90^{\circ} - 60^{\circ} = 30^{\circ}$.
The component of velocity parallel to the wall is $v_{\parallel} = v \cos(60^{\circ})$.
Since the wall is smooth,there is no impulsive force acting along the wall,meaning the parallel component of velocity remains constant throughout the collision.
Initial momentum along the wall: $p_{i, \parallel} = m v \cos(60^{\circ})$.
Final momentum along the wall: $p_{f, \parallel} = m v \cos(60^{\circ})$.
Change in momentum along the wall: $\Delta p_{\parallel} = p_{f, \parallel} - p_{i, \parallel} = m v \cos(60^{\circ}) - m v \cos(60^{\circ}) = 0$.
Therefore,the magnitude of the change in momentum along the wall is $0$.

(A) According to Newton's second law of motion,the average force $F$ is equal to the rate of change of momentum.
$F = \frac{\Delta p}{\Delta t}$
Here,the initial momentum is $p$ and the final momentum is $0$ (since the hammer comes to rest).
Change in momentum $\Delta p = p - 0 = p$.
The time interval $\Delta t = 0.5 \; s$.
Therefore,the average force $F = \frac{p}{0.5} = 2p \; N$.
Thus,the average force required is $2p \; N$.

(C) $1$. Calculate the velocity of the ball just before hitting the bat $(v_1)$: Using $v^2 = u^2 + 2gh$,where $u = 0$,$h = 20 \,m$,and $g = 10 \,m/s^2$,we get $v_1 = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \,m/s$ (downwards).
$2$. Calculate the velocity of the ball just after being hit by the bat $(v_2)$: Using $v^2 = u^2 + 2gh$ for the upward motion,where $v = 0$ at the maximum height $h = 45 \,m$,we get $0 = u^2 - 2 \times 10 \times 45$,so $u = v_2 = \sqrt{900} = 30 \,m/s$ (upwards).
$3$. Apply the impulse-momentum theorem: $\text{Impulse} = F \Delta t = m(v_{\text{final}} - v_{\text{initial}})$.
$4$. Taking upward direction as positive: $v_{\text{final}} = +30 \,m/s$ and $v_{\text{initial}} = -20 \,m/s$.
$5$. The force exerted by the bat is upward,so $F = +200 \,N$. The weight of the ball $(mg = 0.05 \times 10 = 0.5 \,N)$ acts downwards. The net average force is $F_{\text{net}} = F_{\text{bat}} - mg = 200 - 0.5 \approx 200 \,N$.
$6$. $\Delta t = \frac{m(v_2 - v_1)}{F_{\text{net}}} = \frac{0.05 \times (30 - (-20))}{200} = \frac{0.05 \times 50}{200} = \frac{2.5}{200} = \frac{1}{80} \,s$.

(D) The correct answer is $(D)$.
$(1)$ Action and reaction forces act on different bodies,so they cannot cancel each other out.
$(2)$ Physical contact is not required for action-reaction pairs; for example,gravitational force and electrostatic (Coulomb) force act at a distance.
$(3)$ Newton's $3^{rd}$ law is universal and applies regardless of whether the bodies are at rest or in motion.
Therefore,all the given statements are correct.

(C) According to Newton's third law of motion,for every action,there is an equal and opposite reaction. The force exerted by the rifle on the bullet is equal in magnitude to the force exerted by the bullet on the rifle.
First,convert the given values into $SI$ units:
Mass of the bullet,$m = 10 \,g = 0.01 \,kg$.
Acceleration of the bullet,$a = 3 \times 10^6 \,cm/s^2 = 3 \times 10^6 \times 10^{-2} \,m/s^2 = 3 \times 10^4 \,m/s^2$.
The force acting on the bullet is $F = m \times a$.
$F = 0.01 \,kg \times 3 \times 10^4 \,m/s^2 = 300 \,N$.
Since the force acting on the rifle is equal in magnitude to the force acting on the bullet,the magnitude of the force acting on the rifle is $300 \,N$.

(B) The force exerted by the machine gun on the man is equal to the rate of change of momentum of the bullets fired.
Let $n$ be the number of bullets fired per second.
The force $F$ is given by the formula $F = n \cdot m \cdot v$,where $m$ is the mass of one bullet and $v$ is its velocity.
Given: $m = 65 \, g = 0.065 \, kg$,$v = 1300 \, m/s$,and $F = 169 \, N$.
Substituting the values into the equation:
$169 = n \times 0.065 \times 1300$
$169 = n \times 84.5$
$n = \frac{169}{84.5} = 2$.
Therefore,the man can fire $2$ bullets per second.

(D) The impulse imparted to the particle by the wall is equal to the change in its linear momentum.
Let the wall be along the $y$-axis. The velocity of the particle before collision is $\vec{v}_i = v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$.
Since the collision is elastic,the component of velocity parallel to the wall remains unchanged,and the component perpendicular to the wall reverses its direction.
Thus,the velocity after collision is $\vec{v}_f = -v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$.
The change in momentum $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(-2v \cos 30^{\circ} \hat{i}) = -2mv \cos 30^{\circ} \hat{i}$.
The magnitude of impulse is $|\Delta \vec{p}| = 2mv \cos 30^{\circ} = 2mv \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} mv$.

(B) The change in momentum $\Delta \vec{p}$ is equal to the impulse,which is the area under the $F-t$ graph.
For the body to regain its initial momentum $\vec{p}$,the total change in momentum must be zero,meaning the net area under the $F-t$ graph must be zero.
Let the graph cross the $t$-axis at $t = 4 \ s$. The area of the positive triangle (from $t=0$ to $t=4$) is $A_1 = \frac{1}{2} \times 4 \times 1 = 2 \ N \cdot s$.
For the total area to be zero,the area of the negative triangle (from $t=4$ to $t_0$) must be equal in magnitude to the positive area.
Let the force at time $t_0$ be $-F_0$. From the slope of the line,$\tan \theta = \frac{1}{2} = \frac{F_0}{t_0-4}$,so $F_0 = \frac{t_0-4}{2}$.
The area of the negative triangle is $A_2 = \frac{1}{2} \times (t_0-4) \times F_0 = \frac{1}{2} \times (t_0-4) \times \frac{t_0-4}{2} = \frac{(t_0-4)^2}{4}$.
Setting $A_1 = A_2$,we get $2 = \frac{(t_0-4)^2}{4}$,which implies $(t_0-4)^2 = 8$.
Thus,$t_0-4 = \sqrt{8} = 2\sqrt{2}$,so $t_0 = 4 + 2\sqrt{2} \ s$.

(A) The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum.
$J = \int F dt = \Delta p = m(v - v_0)$
The impulse is equal to the area under the $F-t$ curve.
The given curve is a semi-ellipse with semi-major axis $a = T_0/2$ and semi-minor axis $b = F_0$. The area of a semi-ellipse is given by $\frac{1}{2} \pi a b$.
Area $= \frac{1}{2} \pi \left(\frac{T_0}{2}\right) F_0 = \frac{\pi F_0 T_0}{4}$
Equating impulse to the change in momentum:
$m(v - v_0) = \frac{\pi F_0 T_0}{4}$
$v - v_0 = \frac{\pi F_0 T_0}{4m}$
$v = v_0 + \frac{\pi F_0 T_0}{4m}$
Thus,the correct option is $A$.

(A) According to Newton's second law of motion,the force $F$ is given by the rate of change of momentum:
$F = \frac{dp}{dt}$
This means the force $F$ is equal to the slope of the $p-t$ graph.
$1$. From $t = 0$ to $t = 2 \ s$,the momentum increases linearly from $0$ to $10 \ kg \ m/s$. The slope is constant and positive:
$F = \frac{10 - 0}{2 - 0} = 5 \ N$
$2$. From $t = 2$ to $t = 6 \ s$,the momentum is constant at $10 \ kg \ m/s$. The slope is zero:
$F = \frac{10 - 10}{6 - 2} = 0 \ N$
$3$. From $t = 6$ to $t = 8 \ s$,the momentum decreases linearly from $10$ to $0 \ kg \ m/s$. The slope is constant and negative:
$F = \frac{0 - 10}{8 - 6} = -5 \ N$
Comparing these values with the given options,the graph in Option $A$ correctly represents these force values over time.

(A) The velocity of the ball just before hitting the floor is $v_i = \sqrt{2gh_i} = \sqrt{2 \times 10 \times 9.8} = \sqrt{196} = 14\,m/s$ (downwards).
Taking the downward direction as negative,$v_i = -14\,m/s$.
The velocity of the ball just after rebounding is $v_f = \sqrt{2gh_f} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\,m/s$ (upwards).
Taking the upward direction as positive,$v_f = +10\,m/s$.
The change in velocity is $\Delta v = v_f - v_i = 10 - (-14) = 24\,m/s$.
The average acceleration is $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120\,m/s^2$.

(B) The initial momentum of $100$ balls is $P_i = 100mv$ (taking the direction towards the wall as positive).
After reflection,the final momentum of the balls is $P_f = -100mv$ (as they move in the opposite direction with the same speed).
The change in momentum of the balls is $\Delta P = P_f - P_i = -100mv - 100mv = -200mv$.
The force exerted by the wall on the balls is $F_{wall} = \frac{\Delta P}{t} = -\frac{200mv}{t}$.
According to Newton's third law,the force exerted by the balls on the wall is equal in magnitude and opposite in direction to the force exerted by the wall on the balls.
Therefore,the magnitude of the force exerted by the balls on the wall is $|F| = \frac{200mv}{t}$.

(B) The impulse is defined as the area under the force-time $(F-t)$ curve.
$(a)$ Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1.0 \times 0.5 = 0.25 \, N \cdot s$
$(b)$ Area $= \text{length} \times \text{breadth} = 2.0 \times 0.5 = 1.0 \, N \cdot s$
$(c)$ Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1.0 \times 0.75 = 0.375 \, N \cdot s$
$(d)$ Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0 \times 0.5 = 0.5 \, N \cdot s$
Comparing the values, the impulse is highest in figure $(b)$.

(B) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.
$F = \frac{dp}{dt} = n \cdot m \cdot v$
where $n$ is the number of bullets fired per second,$m$ is the mass of each bullet,and $v$ is the velocity of the bullets.
Given:
Force $F = 125\,N$
Mass $m = 10\,g = 10 \times 10^{-3}\,kg = 0.01\,kg$
Velocity $v = 250\,m/s$
Substituting the values into the formula:
$125 = n \times 0.01 \times 250$
$125 = n \times 2.5$
$n = \frac{125}{2.5} = 50$
Therefore,the number of bullets fired per second is $50$.

(B) The impulse supplied to the gun is equal to the change in momentum of the bullet.
Given:
Mass of the bullet,$m = 10\,g = 0.01\,kg$
Velocity of the bullet,$v = 600\,m/s$
Mass of the gun,$M = 3\,kg$
Impulse $J = \Delta p = m \times v$
$J = 0.01\,kg \times 600\,m/s = 6\,Ns$
Since the system is initially at rest,the impulse imparted to the gun is equal in magnitude to the momentum gained by the bullet.

(D) The impulse $\vec{I}$ is equal to the change in momentum $\Delta \vec{P} = \vec{P}_f - \vec{P}_i$.
Mass $m = 100 \,g = 0.1 \,kg$.
Velocity just before hitting the ground: $v_i = -\sqrt{2gh_1} = -\sqrt{2 \times 9.8 \times 10} = -\sqrt{196} = -14 \,m/s$.
Velocity just after rebounding: $v_f = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 7\sqrt{2} \approx 9.9 \,m/s$.
Impulse $I = m(v_f - v_i) = 0.1 \times (9.9 - (-14)) = 0.1 \times (23.9) = 2.39 \,kg \,m/s$.

(A) Given linear momentum $\vec{P}(t) = \cos(kt) \hat{i} - \sin(kt) \hat{j}$.
Force $\vec{F}$ is the rate of change of momentum: $\vec{F} = \frac{d\vec{P}}{dt}$.
$\vec{F} = \frac{d}{dt} [\cos(kt) \hat{i} - \sin(kt) \hat{j}] = -k \sin(kt) \hat{i} - k \cos(kt) \hat{j}$.
To find the angle $\theta$ between $\vec{F}$ and $\vec{P}$,we use the dot product: $\vec{F} \cdot \vec{P} = |\vec{F}| |\vec{P}| \cos \theta$.
$\vec{F} \cdot \vec{P} = (-k \sin(kt))(\cos(kt)) + (-k \cos(kt))(-\sin(kt))$.
$\vec{F} \cdot \vec{P} = -k \sin(kt) \cos(kt) + k \sin(kt) \cos(kt) = 0$.
Since the dot product is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(C) Given: Mass $m = 150 \ g = 0.15 \ kg$,Initial velocity $u = 20 \ m/s$,Final velocity $v = 0 \ m/s$,Time interval $\Delta t = 0.1 \ s$.
According to Newton's second law of motion,the force $F$ exerted is equal to 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 is $|F| = 30 \ N$.

(C) The impulse imparted to the block is equal to the area under the $F-t$ graph.
Impulse $J = \int F \ dt = \text{Area of triangle from } t=0 \text{ to } 3 \ s - \text{Area of triangle from } t=3 \text{ to } 4.5 \ s$.
At $t=0$,$F=4 \ N$. The slope of the line is $m = \frac{0-4}{3-0} = -\frac{4}{3} \ N/s$.
So,$F(t) = 4 - \frac{4}{3}t$.
At $t=4.5 \ s$,$F(4.5) = 4 - \frac{4}{3}(4.5) = 4 - 6 = -2 \ N$.
Impulse $J = \left(\frac{1}{2} \times 3 \times 4\right) - \left(\frac{1}{2} \times (4.5-3) \times 2\right) = 6 - 1.5 = 4.5 \ kg \cdot m/s$.
Since the block starts from rest,the final momentum $p = J = 4.5 \ kg \cdot m/s$.
The kinetic energy $K.E. = \frac{p^2}{2m} = \frac{(4.5)^2}{2 \times 2} = \frac{20.25}{4} = 5.0625 \ J \approx 5.06 \ J$.

(B) Given,$m = 0.4 \ kg$,Impulse $J = 1.0 \ N \ s$,$\tau = 4 \ s$,and $v(t) = v_0 e^{-t/\tau}$.
At $t = 0$,the velocity is $v(0) = v_0 e^0 = v_0$.
From the impulse-momentum theorem,$J = \Delta p = m \Delta v = m(v(0) - 0) = m v_0$.
Therefore,$v_0 = J/m = 1.0 / 0.4 = 2.5 \ m/s$.
The displacement $S$ at $t = \tau$ is given by the integral of velocity:
$S = \int_0^{\tau} v(t) \ dt = \int_0^{\tau} v_0 e^{-t/\tau} \ dt$.
$S = v_0 \left[ -\tau e^{-t/\tau} \right]_0^{\tau} = v_0 \tau (1 - e^{-1})$.
Substituting the values: $S = (2.5) \times (4) \times (1 - 0.37)$.
$S = 10 \times 0.63 = 6.3 \ m$.

(A) The velocity of the ball just before hitting the ground is $v_i = -\sqrt{2gh_1} = -\sqrt{2 \times 9.8 \times 40} = -\sqrt{784} = -28 \ m/s$.
The velocity of the ball just after rebounding is $v_f = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \ m/s$.
The impulse imparted to the ball is equal to the change in momentum: $\vec{I} = \Delta \vec{P} = m(\vec{v}_f - \vec{v}_i)$.
Substituting the values: $\vec{I} = 0.5 \times [14 - (-28)] = 0.5 \times [14 + 28] = 0.5 \times 42 = 21 \ Ns$.

(C) The force acting on the football can be calculated using Newton's second law in terms of impulse: $F = \frac{\Delta p}{\Delta t}$.
Here,the mass of the football $m = 0.5 \ kg$.
The initial velocity $u = 0 \ m/s$ and the final velocity $v = 10 \ m/s$.
The time of contact $\Delta t = \frac{1}{50} \ s$.
The change in momentum $\Delta p = m(v - u) = 0.5 \times (10 - 0) = 5 \ kg \cdot m/s$.
Therefore,the force $F = \frac{5}{1/50} = 5 \times 50 = 250 \ N$.

(D) The impulse applied to the particle is equal to the area under the $F-t$ graph.
Impulse $J = \int F \, dt = \Delta p = m(v_f - v_i)$.
Given $m = 2 \ kg$ and $v_i = 0 \ ms^{-1}$.
Area under the graph = Area of triangle $(0-2 \ s)$ + Area of rectangle $(2-4 \ s)$ + Area of trapezium $(4-6 \ s)$ + Area of triangle $(6-10 \ s)$.
Area $= (\frac{1}{2} \times 2 \times 10) + (2 \times 10) + (\frac{1}{2} \times (10 + 20) \times 2) + (\frac{1}{2} \times 4 \times 20)$.
Area $= 10 + 20 + 30 + 40 = 100 \ N \cdot s$.
Since $J = m \cdot v_f$,we have $100 = 2 \times v_f$.
Therefore,$v_f = 50 \ ms^{-1}$.

(C) The force exerted on the wall is given by the rate of change of momentum.
Initial momentum of $N$ balls per second = $Nmu \hat{i}$.
Since the collision is elastic,the balls return with the same speed in the opposite direction.
Final momentum of $N$ balls per second = $-Nmu \hat{i}$.
Change in momentum per second = $\text{Final momentum} - \text{Initial momentum} = -Nmu \hat{i} - (Nmu \hat{i}) = -2Nmu \hat{i}$.
The magnitude of the change in momentum per second is $2Nmu$.
Since force is the rate of change of momentum,$F = \frac{\Delta p}{\Delta t} = \frac{2Nmu}{1} = 2Nmu \ N$.
Therefore,the correct option is $C$.

(D) The impulse applied to the body is equal to the area under the force-time $(F-t)$ graph.
Impulse $(J)$ = Change in momentum $(\Delta p)$ = $m(v - u)$.
Given mass $m = 3 \ kg$, initial velocity $u = 0 \ m/s$.
Area under the graph = Area of the first rectangle + Area of the second rectangle.
Area = $(8 \ N \times 0.5 \ s) + (4 \ N \times (1.0 - 0.5) \ s)$.
Area = $(4 \ Ns) + (4 \ N \times 0.5 \ s) = 4 \ Ns + 2 \ Ns = 6 \ Ns$.
Since Impulse = $\Delta p = m(v - u)$,
$6 = 3 \times (v - 0)$,
$6 = 3v$,
$v = 2 \ m/s$.
Therefore, the speed of the body after $1 \ s$ is $2 \ m/s$.

(C) The impulse applied to the particle is equal to the area under the force-time graph.
Impulse $J = \text{Area of triangle} = \frac{1}{2} \times 4 \, s \times 6 \, N = 12 \, N \cdot s$.
According to the impulse-momentum theorem, Impulse $J = \Delta p = m(v_f - v_i)$.
Given mass $m = 2 \, kg$, initial velocity $v_i = 6 \, m/s$, and impulse $J = 12 \, N \cdot s$.
$12 = 2(v_f - 6)$.
$6 = v_f - 6$.
$v_f = 12 \, m/s$.
Thus, the final velocity of the particle is $12 \, m/s$.

(B) The force exerted on the wall is equal to the rate of change of momentum of the balls.
For an elastic collision,the ball hits the wall with velocity $u$ and rebounds with velocity $-u$.
The change in momentum for a single ball is $\Delta p = m(u - (-u)) = 2mu$.
Since $n$ balls hit the wall per second,the total change in momentum per second is $\frac{dp}{dt} = n \times \Delta p = n(2mu) = 2mnu$.
According to Newton's second law,the force $F$ is equal to the rate of change of momentum,so $F = 2mnu$.

(A) Given: Mass of the bullet $m = 20 \,g = 0.02 \,kg$. Initial velocity $u = 200 \,m/s$. Final velocity $v = 0 \,m/s$. Time taken $\Delta t = \frac{1}{50} \,s = 0.02 \,s$.
Impulse $(J)$ is equal to the change in momentum:
$J = \Delta p = m(v - u)$
$J = 0.02 \,kg \times (0 - 200) \,m/s = -4 \,kg \cdot m/s$.
The magnitude of impulse is $4 \,Ns$.
Average force $(F_{avg})$ is given by the rate of change of momentum:
$F_{avg} = \frac{J}{\Delta t} = \frac{4 \,Ns}{0.02 \,s} = 200 \,N$.
Thus,the impulse is $4 \,Ns$ and the average force is $200 \,N$.

(A) The impulse imparted to an object is equal to the change in its linear momentum.
Let the initial velocity of the ball be $u = 6 \ m/s$ (towards the batsman).
After being hit,the ball moves towards the bowler with the same speed,so the final velocity is $v = -6 \ m/s$.
The mass of the ball is $m = 0.2 \ kg$.
Impulse $J = \Delta p = m(v - u)$.
$J = 0.2 \times (-6 - 6) = 0.2 \times (-12) = -2.4 \ Ns$.
The magnitude of the impulse imparted to the ball is $2.4 \ Ns$.

(B) Impulse is defined as the change in momentum of the object.
Let the initial velocity be $u = 30 \,ms^{-1}$ and the final velocity be $v = -20 \,ms^{-1}$ (since it rebounds in the opposite direction).
The mass of the ball is $m = 0.1 \,kg$.
Impulse $J = \Delta p = m(v - u)$.
Taking the direction of the initial velocity as positive:
$J = m(v_{final} - v_{initial}) = 0.1 \times (-20 - 30) = 0.1 \times (-50) = -5 \,N-s$.
The magnitude of the impulse exerted by the wall on the ball is $|J| = 5 \,N-s$.

(D) The external force acting on the system of two particles is the gravitational force: $F_{ext} = (m_1 + m_2)g$ (acting downwards).
According to Newton's second law,the rate of change of momentum is equal to the external force: $F_{ext} = \frac{\Delta P}{\Delta t}$.
Here,$\Delta P = (m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)$ and $\Delta t = 2t - 0 = 2t$.
Substituting these into the equation: $(m_1 + m_2)g = \frac{(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)}{2t}$.
Therefore,the change in momentum is: $[(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)] = 2(m_1 + m_2)gt$.

(B) Given that:
Number of balls $N = 1000 = 10^3$
Mass of each ball $m = 1 \text{ g} = 10^{-3} \text{ kg}$
Area $A = 1 \text{ cm}^2 = 10^{-4} \text{ m}^2$
Velocity $v = 50 \text{ m/s}$
The change in momentum for each collision is $\Delta p = m[v - (-v)] = 2mv$.
Since there are $N$ collisions per second,the total force $F$ exerted on the surface is:
$F = N \times \Delta p = N \times 2mv$
$F = 10^3 \times 2 \times 10^{-3} \text{ kg} \times 50 \text{ m/s} = 100 \text{ N}$.
Pressure $P$ is defined as force per unit area:
$P = \frac{F}{A} = \frac{100 \text{ N}}{10^{-4} \text{ m}^2} = 10^6 \text{ N/m}^2$.

(C) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets fired.
Let $n$ be the number of bullets fired per second.
The mass of each bullet is $m = 30 \text{ g} = 0.03 \text{ kg}$.
The velocity of each bullet is $v = 1000 \text{ m/s}$.
The force $F$ exerted by the gun is given by the formula:
$F = n \times m \times v$
Given $F = 300 \text{ N}$,$m = 0.03 \text{ kg}$,and $v = 1000 \text{ m/s}$.
Substituting the values:
$300 = n \times 0.03 \times 1000$
$300 = n \times 30$
$n = \frac{300}{30} = 10$
Therefore,the man can fire at most $10$ bullets per second.

(C) The force exerted by the machine gun is equal to the rate of change of momentum of the bullets.
Let $n$ be the number of bullets fired per second.
The mass of each bullet is $m = 35 \ g = 0.035 \ kg$.
The velocity of each bullet is $v = 600 \ m/s$.
The force $F$ exerted by the gun is given by the formula $F = n \times (m \times v)$.
Given $F = 147 \ N$,$m = 0.035 \ kg$,and $v = 600 \ m/s$.
Substituting the values: $147 = n \times (0.035 \times 600)$.
$147 = n \times 21$.
$n = 147 / 21 = 7$.
Therefore,the number of bullets that can be fired per second is $7$.

(A) The area under the $F-t$ graph represents the impulse, which is equal to the change in momentum of the body.
Since the body is initially at rest, the impulse equals the final momentum of the body after $1 \,s$.
$\text{Impulse} = \text{Area under } F-t \text{ graph}$
$\text{Impulse} = (10 \,N \times 0.5 \,s) + (20 \,N \times 0.5 \,s)$
$\text{Impulse} = 5 \,N-s + 10 \,N-s = 15 \,N-s$
Since $\text{Impulse} = \Delta p = m \times v - m \times u$ and $u = 0$, we have:
$15 = 2 \,kg \times v$
$v = \frac{15}{2} = 7.5 \,m/s$

(C) When a sudden jerk is given to string $C$,an impulsive force is applied to it.
Because the mass has inertia,it resists the sudden change in motion.
This impulsive tension develops in string $C$ first,exceeding its breaking strength.
Since the impulse takes finite time to propagate through the mass to string $A$,string $C$ breaks before the tension in string $A$ can increase significantly.

(B) The change in momentum $\Delta p$ is given by the final momentum minus the initial momentum.
Initial momentum $p_i = m \times v = 2 \,kg \times 100 \,ms^{-1} = 200 \,kg \cdot ms^{-1}$.
Since the body rebounds with the same speed in the opposite direction,the final momentum $p_f = m \times (-v) = 2 \,kg \times (-100 \,ms^{-1}) = -200 \,kg \cdot ms^{-1}$.
The change in momentum $\Delta p = p_f - p_i = -200 - 200 = -400 \,kg \cdot ms^{-1}$.
The magnitude of the change in momentum is $|\Delta p| = 400 \,kg \cdot ms^{-1}$.
The force exerted on the wall is given by Newton's Second Law: $F = \frac{|\Delta p|}{\Delta t}$.
Given $\Delta t = 1/50 \,s$,we have $F = \frac{400}{1/50} = 400 \times 50 = 20,000 \,N = 2 \times 10^{4} \,N$.

(A) When objects are released from height $h$ under gravity,they undergo free fall.
According to the equations of motion,the final velocity $v$ just before striking the ground is given by $v = \sqrt{2gh}$.
Since $g$ and $h$ are the same for all objects,the velocity $v$ is the same for all objects regardless of their mass.
However,the momentum $p$ of an object is given by $p = mv$.
Since the masses $m$ of the five objects are different,their momenta $p$ will be different at the instant they strike the ground.
Therefore,the physical quantity that depends on the mass and changes for different objects is momentum.

(C) The mass of each ball is $m = 250 \ g = 0.25 \ kg$.
The initial velocity of the first ball is $v_i = 16 \ m \ s^{-1}$.
After the collision,the ball rebounds with the same speed in the opposite direction,so the final velocity is $v_f = -16 \ m \ s^{-1}$.
Impulse is defined as the change in momentum: $J = \Delta p = m(v_f - v_i)$.
Substituting the values: $J = 0.25 \ kg \times (-16 \ m \ s^{-1} - 16 \ m \ s^{-1})$.
$J = 0.25 \ kg \times (-32 \ m \ s^{-1}) = -8 \ kg \ m \ s^{-1}$.
The magnitude of the impulse imparted is $|J| = 8 \ kg \ m \ s^{-1}$.

(C) Given: Mass $m = 250 \text{ g} = 0.25 \text{ kg} = \frac{1}{4} \text{ kg}$.
Initial velocity $\vec{u} = -22 \hat{i} \text{ ms}^{-1}$.
Final velocity $\vec{v} = 30 \cos 53^{\circ} \hat{i} + 30 \sin 53^{\circ} \hat{j} = 30(\frac{3}{5}) \hat{i} + 30(\frac{4}{5}) \hat{j} = 18 \hat{i} + 24 \hat{j} \text{ ms}^{-1}$.
Change in momentum $\Delta \vec{P} = m(\vec{v} - \vec{u}) = \frac{1}{4} [(18 \hat{i} + 24 \hat{j}) - (-22 \hat{i})] = \frac{1}{4} [40 \hat{i} + 24 \hat{j}] = 10 \hat{i} + 6 \hat{j} \text{ Ns}$.
Magnitude of change in momentum $|\Delta \vec{P}| = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \text{ Ns}$.
Average force $\vec{F}_{avg} = \frac{\Delta \vec{P}}{\Delta t} = \frac{\sqrt{136}}{0.01} = 100 \sqrt{136} \approx 100 \times 11.66 = 1166 \text{ N}$.

(A) The angle given is with the wall,so the angle with the normal is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Change in momentum $\Delta p$ occurs only in the direction perpendicular to the wall.
The component of velocity perpendicular to the wall is $v_{\perp} = v \sin(60^{\circ}) = v \cos(30^{\circ})$.
Initial momentum component perpendicular to the wall: $p_i = m v \cos(30^{\circ})$.
Final momentum component perpendicular to the wall: $p_f = -m v \cos(30^{\circ})$.
Change in momentum: $\Delta p = |p_f - p_i| = 2 m v \cos(30^{\circ})$.
Substituting the values: $\Delta p = 2 \times 3 \times 100 \times \frac{\sqrt{3}}{2} = 300\sqrt{3} \ kg \cdot m/s$.
The force exerted is $F = \frac{\Delta p}{\Delta t} = \frac{300\sqrt{3}}{0.2} = 1500\sqrt{3} \ N$.

(B) Given, mass of the body, $m = 2 \, kg$.
Impulse is defined as the change in momentum: $\text{Impulse} = \Delta p = p_f - p_i = m(v_f - v_i)$.
From the position-time graph, the velocity is the slope of the $x-t$ graph $(v = \frac{dx}{dt})$.
For $t < 4 \, s$, the velocity $v_i$ is the slope of the line from $(0,0)$ to $(4,3)$:
$v_i = \frac{3 - 0}{4 - 0} = 0.75 \, m/s$.
For $t > 4 \, s$, the velocity $v_f$ is the slope of the horizontal line:
$v_f = 0 \, m/s$.
Therefore, the impulse at $t = 4 \, s$ is:
$\text{Impulse} = m(v_f - v_i) = 2 \, kg \times (0 - 0.75 \, m/s) = -1.5 \, kg \cdot m/s$.
Thus, the correct option is $B$.

(A) Linear momentum is a vector quantity.
Let the direction towards the wall be positive.
Initial momentum $\vec{p}_i = m \times 10 \hat{i} = 0.5 \times 10 \hat{i} = 5 \hat{i} \ kg \ m \ s^{-1}$.
After rebounding,the ball moves in the opposite direction.
Final momentum $\vec{p}_f = -m \times v \hat{i} = -0.5 \times v \hat{i} \ kg \ m \ s^{-1}$.
The change in linear momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i$.
$\Delta \vec{p} = (-0.5v \hat{i}) - (5 \hat{i}) = -(0.5v + 5) \hat{i}$.
The magnitude of the change in linear momentum is given as $8.0 \ kg \ m \ s^{-1}$.
$|\Delta \vec{p}| = 0.5v + 5 = 8.0$.
$0.5v = 8.0 - 5 = 3.0$.
$v = \frac{3.0}{0.5} = 6.0 \ m \ s^{-1}$.

(B) The change in linear momentum $(\Delta p)$ is equal to the area under the force-time $(F-t)$ graph.
$\Delta p = \int_0^8 F \,dt = \text{Area under the } F-t \text{ curve}$.
From the graph,the area from $t=0$ to $t=2$ is a quarter-circle below the $t$-axis (negative area).
The area from $t=2$ to $t=6$ is a semi-circle above the $t$-axis (positive area).
The area from $t=6$ to $t=8$ is a quarter-circle below the $t$-axis (negative area).
Given the radius of the circular segments is $r=2$ units (from $F=0$ to $F=2$ or $F=-2$):
Area of quarter-circle $= \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (2)^2 = \pi$.
Area of semi-circle $= \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (2)^2 = 2\pi$.
Total area $= -(\text{Area}_{0-2}) + (\text{Area}_{2-6}) - (\text{Area}_{6-8})$
$\Delta p = -\pi + 2\pi - \pi = 0$.
Thus,the linear momentum gained between $0$ and $8 \,s$ is $0$.