Mix Examples-Work, Energy, Power and Collision Questions in English

(B) The stopping distance $d$ is given by the work-energy theorem,where the work done by the retarding force $F$ equals the change in kinetic energy: $F \cdot d = \frac{1}{2} m v^2$.
Thus,the stopping distance is $d = \frac{m v^2}{2F}$.
Assuming the retarding force $F$ (braking force) is the same for both vehicles and their initial velocities $v$ are equal,the stopping distance is directly proportional to the mass $m$ of the vehicle $(d \propto m)$.
Since the mass of the car is less than the mass of the truck $(m_{\text{car}} < m_{\text{truck}})$,the car will cover less distance before coming to rest.

(A) For particle $Q$, the velocity is constant at $v$ along the horizontal direction, so the time taken to cover the horizontal distance $AB$ is ${t_Q} = \frac{AB}{v}$.
For particle $P$, the horizontal component of velocity $v_x$ is given by $v_x = v \cos \theta$, where $\theta$ is the angle the velocity vector makes with the horizontal. As the particle slides down the bowl, it gains speed due to gravity. The horizontal component of its velocity $v_x$ at any point is greater than or equal to the initial horizontal component $v$ because the particle accelerates as it moves towards the lowest point $C$ and then decelerates, but its speed remains higher than the initial speed $v$ throughout the path until it reaches $B$. Since the horizontal component of velocity for $P$ is always greater than or equal to $v$ throughout the motion, the average horizontal velocity of $P$ is greater than $v$. Therefore, the time taken by $P$ to cover the same horizontal distance $AB$ is less than the time taken by $Q$. Thus, ${t_P} < {t_Q}$.

(D) When the trolleys are released,they possess the same magnitude of linear momentum $p$ but in opposite directions due to the conservation of momentum.
The kinetic energy $K$ acquired by a trolley is given by $K = \frac{p^2}{2m}$.
This kinetic energy is dissipated by the work done against the force of friction $f = \mu mg$.
Therefore,$\mu mg S = \frac{p^2}{2m}$.
Since $p$ and $\mu$ are constant for both trolleys,we have $S \propto \frac{1}{m^2}$.
Thus,the ratio of the distances is $\frac{S_1}{S_2} = \left( \frac{m_2}{m_1} \right)^2 = \left( \frac{3m}{m} \right)^2 = \frac{9}{1}$.

(B) According to the Work-Energy Theorem,the kinetic energy acquired by the body is equal to the net work done on it.
Net force acting on the body = Applied force $(F)$ - Frictional force $(f_k)$.
$f_k = \mu N = \mu mg = 0.2 \times 5 \times 10 = 10\,N$.
Net force = $25\,N - 10\,N = 15\,N$.
Work done = Net force $\times$ displacement $(S)$.
Work done = $15\,N \times 10\,m = 150\,J$.
Therefore,the kinetic energy acquired is $150\,J$.

(A) The total work done $(W_{total})$ is the sum of the work done against gravity $(W_g)$ and the work done against friction $(W_f)$.
$W_{total} = W_g + W_f$
Work done against gravity is given by the formula $W_g = mgh$.
Substituting the given values: $W_g = 2 \, kg \times 10 \, m/s^2 \times 10 \, m = 200 \, J$.
Given total work $W_{total} = 300 \, J$.
Therefore,the work done against friction is $W_f = W_{total} - W_g$.
$W_f = 300 \, J - 200 \, J = 100 \, J$.

(D) The net force $F_{net}$ acting on the body is the vector sum of the two perpendicular forces:
$F_{net} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, N$.
Using Newton's second law,the acceleration $a$ is:
$a = \frac{F_{net}}{m} = \frac{5 \, N}{10 \, kg} = 0.5 \, m/s^2$.
The velocity $v$ at time $t = 10 \, s$ starting from rest $(u = 0)$ is:
$v = u + at = 0 + (0.5 \, m/s^2)(10 \, s) = 5 \, m/s$.
The kinetic energy $K$ is given by:
$K = \frac{1}{2}mv^2 = \frac{1}{2} \times 10 \, kg \times (5 \, m/s)^2 = 5 \times 25 = 125 \, J$.

(A) The work-energy theorem states that the work done by all forces acting on a body is equal to the change in its kinetic energy.
Since the ball strikes the wall and returns with the same speed $v$,the final kinetic energy $K_f = \frac{1}{2}mv^2$ is equal to the initial kinetic energy $K_i = \frac{1}{2}mv^2$.
Therefore,the change in kinetic energy $\Delta K = K_f - K_i = 0$.
According to the work-energy theorem,the work done by the ball on the wall is equal to the negative of the work done by the wall on the ball.
Since the wall is of infinite mass,it does not move (displacement $d = 0$).
Thus,the work done by the wall on the ball is $W = F \cdot d = 0$.
Consequently,the work done by the ball on the wall is also $Zero$.

(C) When a ball is released from the top of a tower,it undergoes free fall under gravity.
The distance covered by an object in the $n^{th}$ second is given by the formula $h_n = u + \frac{g}{2}(2n - 1)$.
Since the initial velocity $u = 0$,the distance covered in the $n^{th}$ second is $h_n \propto (2n - 1)$.
For the first,second,and third seconds $(n = 1, 2, 3)$:
$h_I \propto (2(1) - 1) = 1$
$h_{II} \propto (2(2) - 1) = 3$
$h_{III} \propto (2(3) - 1) = 5$
Thus,the ratio of distances is $h_I : h_{II} : h_{III} = 1 : 3 : 5$.
The work done by the force of gravity is $W = mgh$.
Since $m$ and $g$ are constant,the ratio of work done is equal to the ratio of distances covered:
$W_I : W_{II} : W_{III} = h_I : h_{II} : h_{III} = 1 : 3 : 5$.

(B) As the body splits into two equal parts due to an internal explosion,the momentum of the system remains conserved.
Initial momentum = Final momentum
$8 \times 2 = 4v_1 + 4v_2$
$\Rightarrow v_1 + v_2 = 4$ ...$(i)$
By the law of conservation of energy:
Initial kinetic energy + Energy released = Final kinetic energy
$\frac{1}{2} \times 8 \times (2)^2 + 16 = \frac{1}{2} \times 4 \times v_1^2 + \frac{1}{2} \times 4 \times v_2^2$
$16 + 16 = 2v_1^2 + 2v_2^2$
$v_1^2 + v_2^2 = 16$ ...(ii)
From $(i)$,$v_2 = 4 - v_1$. Substituting this into (ii):
$v_1^2 + (4 - v_1)^2 = 16$
$v_1^2 + 16 - 8v_1 + v_1^2 = 16$
$2v_1^2 - 8v_1 = 0$
$2v_1(v_1 - 4) = 0$
So,$v_1 = 0$ or $v_1 = 4 \, m/s$.
If $v_1 = 0$,then $v_2 = 4 \, m/s$. If $v_1 = 4$,then $v_2 = 0$.
Thus,one part comes to rest and the other moves in the same direction as that of the original body.

(C) The kinetic energy $E$ of a body with mass $m$ and momentum $P$ is given by the formula $E = \frac{P^2}{2m}$.
Since both bodies have equal momenta $(P_1 = P_2 = P)$,the kinetic energy is inversely proportional to the mass: $E \propto \frac{1}{m}$.
Therefore,the ratio of their kinetic energies is $\frac{E_1}{E_2} = \frac{m_2}{m_1}$.
Thus,the ratio $E_1 : E_2$ is $m_2 : m_1$.

(D) Using the work-energy theorem,the work done by the braking force $F$ is equal to the change in kinetic energy of the car.
$|W| = |\Delta K|$
$F \cdot s = \frac{1}{2} m u^2$
Since the braking force $F$ and the mass $m$ of the car remain constant,the stopping distance $s$ is directly proportional to the square of the initial velocity $u$:
$s \propto u^2$
Given that for $u_1 = 30 \, km/h$,the stopping distance $s_1 = 8 \, m$.
For $u_2 = 60 \, km/h$,the new stopping distance $s_2$ is:
$\frac{s_2}{s_1} = \left( \frac{u_2}{u_1} \right)^2$
$\frac{s_2}{8} = \left( \frac{60}{30} \right)^2$
$\frac{s_2}{8} = (2)^2 = 4$
$s_2 = 8 \times 4 = 32 \, m$.
Therefore,the correct option is $D$.

(A) By the law of conservation of linear momentum,the initial momentum of the sphere equals the final momentum of the combined system (sphere + bag) immediately after the collision:
$mV = (m + M)v_{\text{sys}}$ ... $(i)$
where $v_{\text{sys}}$ is the velocity of the system just after the collision.
After the collision,the system (sphere + bag) swings upward. By the law of conservation of mechanical energy,the kinetic energy of the system just after the collision is converted into gravitational potential energy at the maximum height $h$:
$\frac{1}{2}(m + M)v_{\text{sys}}^2 = (m + M)gh$
$v_{\text{sys}}^2 = 2gh$
$v_{\text{sys}} = \sqrt{2gh}$ ... (ii)
Substituting the value of $v_{\text{sys}}$ from equation (ii) into equation $(i)$:
$mV = (m + M)\sqrt{2gh}$
$V = \left(\frac{m + M}{m}\right)\sqrt{2gh}$

(D) The kinetic energy $K$ of a body is given by $K = \frac{P^2}{2m}$,where $P$ is the momentum and $m$ is the mass.
Rearranging for momentum,we get $P = \sqrt{2mK}$.
Since both bodies have the same kinetic energy $K$,the momentum $P$ is directly proportional to the square root of the mass $(P \propto \sqrt{m})$.
Given that the first body is lighter $(M_1 < M_2)$,it follows that the momentum of the lighter body is less than the momentum of the heavier body.
Therefore,$M_1 V_1 < M_2 V_2$ or $M_2 V_2 > M_1 V_1$.

(B) Let the initial height be $h$. The potential energy at this height is $PE = mgh$.
When the ball falls to the ground,all its potential energy is converted into kinetic energy,so $KE_{initial} = mgh$.
Upon striking the ground,it loses $50\%$ of its kinetic energy.
The remaining kinetic energy is $KE_{final} = 0.5 \times KE_{initial} = 0.5 mgh$.
As the ball rises again,this kinetic energy is converted back into potential energy at the new height $h'$.
Thus,$mgh' = 0.5 mgh$.
Solving for $h'$,we get $h' = 0.5 h$,which is half the initial height.

(D) If there were no air drag,the maximum height $H$ reached by the ball would be calculated using the conservation of energy or kinematic equations:
$H = \frac{u^2}{2g} = \frac{14 \times 14}{2 \times 9.8} = \frac{196}{19.6} = 10\, m$.
However,due to air drag,the ball only reaches a height of $8.0\, m$.
The energy dissipated by air drag is equal to the difference between the potential energy the ball would have reached without air drag and the potential energy it actually reached:
$\text{Energy dissipated} = mg(H_{ideal} - H_{actual})$
$\text{Energy dissipated} = 0.5 \times 9.8 \times (10 - 8)$
$\text{Energy dissipated} = 0.5 \times 9.8 \times 2 = 9.8\, J$.

(C) The energy supplied by the machine is $E_{in} = 12 \, J$.
Since the efficiency is $75\%$, the useful work done $(W)$ in lifting the mass is $W = 0.75 \times 12 \, J = 9 \, J$.
This work is stored as gravitational potential energy: $mgh = 9 \, J$.
Given $m = 1 \, kg$ and taking $g = 10 \, m \, s^{-2}$, we have $1 \times 10 \times h = 9$, which gives $h = 0.9 \, m$.
When the mass falls through this distance $h$, its potential energy is converted into kinetic energy.
Using the conservation of energy, $\frac{1}{2}mv^2 = mgh$.
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.9} = \sqrt{18} \, m \, s^{-1}$.

(A) When two bodies collide,the collision is generally inelastic in nature.
During an inelastic collision,a portion of the total kinetic energy of the system is dissipated.
This dissipated kinetic energy is converted into other forms of energy,primarily heat energy,which causes a rise in the temperature of the bodies.
Therefore,the correct option is $A$.

(A) $1$. Initial momentum of the system (block $C$) is $P_i = mv$.
$2$. After the perfectly inelastic collision between $C$ and $A$,block $C$ comes to rest and block $A$ starts moving with velocity $v$. Now,the system consists of blocks $A$ and $B$ connected by a spring,with $A$ moving at velocity $v$ and $B$ at rest.
$3$. At maximum compression $x$,both blocks $A$ and $B$ move with the same velocity $V$. By the law of conservation of linear momentum for the system $(A+B)$:
$mv = (m + m)V \implies V = \frac{v}{2}$.
$4$. By the law of conservation of energy for the system $(A+B)$:
Initial kinetic energy of $A$ = Final kinetic energy of $(A+B)$ + Potential energy of the spring.
$\frac{1}{2}mv^2 = \frac{1}{2}(2m)V^2 + \frac{1}{2}Kx^2$.
$5$. Substituting $V = \frac{v}{2}$:
$\frac{1}{2}mv^2 = m(\frac{v}{2})^2 + \frac{1}{2}Kx^2$
$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx^2$
$\frac{1}{4}mv^2 = \frac{1}{2}Kx^2$
$Kx^2 = \frac{1}{2}mv^2$
$x = v\sqrt{\frac{m}{2K}}$.

(A) Given that the magnitude of momentum $(P)$ is equal to the magnitude of kinetic energy $(K)$:
$P = K$
We know that $P = mv$ and $K = \frac{1}{2}mv^2$.
Equating the two:
$mv = \frac{1}{2}mv^2$
Assuming $m \neq 0$ and $v \neq 0$,we can divide both sides by $mv$:
$1 = \frac{1}{2}v$
$v = 2 \, m/s$
Thus,the velocity of the body is $2 \, m/s$.

(C) Since the bomb is initially stationary,the total momentum of the fragments must be zero by the law of conservation of linear momentum. Thus,the magnitudes of the momenta of the two fragments are equal: $p_1 = p_2 = p$.
The kinetic energy $E$ is related to momentum $p$ and mass $m$ by the formula $E = \frac{p^2}{2m}$.
Therefore,the ratio of the kinetic energies of the two fragments is:
$\frac{E_1}{E_2} = \frac{p^2 / (2m_1)}{p^2 / (2m_2)} = \frac{m_2}{m_1} = \frac{3\,g}{1\,g} = 3$.
This implies $E_1 = 3E_2$,where $E_1$ is the kinetic energy of the smaller fragment (mass $1\,g$) and $E_2$ is the kinetic energy of the larger fragment (mass $3\,g$).
Given the total kinetic energy $E_1 + E_2 = 6.4 \times 10^4\,J$.
Substituting $E_1 = 3E_2$ into the total energy equation:
$3E_2 + E_2 = 6.4 \times 10^4\,J$
$4E_2 = 6.4 \times 10^4\,J$
$E_2 = 1.6 \times 10^4\,J$.
Now,calculating the kinetic energy of the smaller fragment $(E_1)$:
$E_1 = 3 \times (1.6 \times 10^4\,J) = 4.8 \times 10^4\,J$.

(A) Work output of the engine $(W_{out})$ = $mgh = 100 \times 10 \times 10 = 10^4 \ J$.
Efficiency $(\eta)$ = $\frac{W_{out}}{W_{in}}$.
Input energy $(W_{in})$ = $\frac{W_{out}}{\eta} = \frac{10^4}{0.60} = \frac{10^4 \times 100}{60} = \frac{10^5}{6} \ J$.
Power of the engine $(P)$ = $\frac{W_{in}}{t} = \frac{10^5 / 6}{5} = \frac{10^5}{30} = \frac{100000}{30} \approx 3333.33 \ W$.
Converting to $kW$: $P = 3.33 \ kW \approx 3.3 \ kW$.

(A) Given: $m_{1} = m_{2} = m$,$u_{1} = u$,and $u_{2} = 0$.
Let $v_{1}$ and $v_{2}$ be the velocities of the spheres after the collision.
According to the law of conservation of linear momentum: $m u = m v_{1} + m v_{2} \implies u = v_{1} + v_{2} \dots (i)$
By the definition of the coefficient of restitution $e$: $e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}} = \frac{v_{2} - v_{1}}{u - 0} \implies v_{2} - v_{1} = e u \dots (ii)$
Adding equations $(i)$ and $(ii)$: $2 v_{2} = u(1 + e) \implies v_{2} = \frac{u(1 + e)}{2}$
Subtracting equation $(ii)$ from $(i)$: $2 v_{1} = u(1 - e) \implies v_{1} = \frac{u(1 - e)}{2}$
The ratio of the velocities is $\frac{v_{1}}{v_{2}} = \frac{\frac{u(1 - e)}{2}}{\frac{u(1 + e)}{2}} = \frac{1 - e}{1 + e}$.

(D) In any collision between two objects,the total linear momentum of the system is always conserved,provided no external force acts on the system.
During a collision,some fraction of the kinetic energy $(KE)$ may be converted into other forms of energy,such as heat,sound,or deformation energy.
Therefore,kinetic energy is generally not conserved unless the collision is perfectly elastic.
Temperature and velocity are not conserved quantities in collisions.
Thus,the only quantity that remains conserved is the momentum of the system.

(B) When the ball falls vertically downward from height $h_1$,its velocity just before hitting the ground is $v_1 = \sqrt{2gh_1}$ (downward).
After the collision,the ball rebounds to height $h_2$,so its velocity just after hitting the ground is $v_2 = \sqrt{2gh_2}$ (upward).
Taking the upward direction as positive,the initial velocity is $\vec{v}_1 = -\sqrt{2gh_1}$ and the final velocity is $\vec{v}_2 = +\sqrt{2gh_2}$.
The change in momentum is $\Delta \vec{P} = m(\vec{v}_2 - \vec{v}_1)$.
Substituting the values,$\Delta \vec{P} = m(\sqrt{2gh_2} - (-\sqrt{2gh_1})) = m(\sqrt{2gh_1} + \sqrt{2gh_2})$.

(C) The law of conservation of momentum is a universal principle that holds true for all types of collisions,whether elastic or inelastic,provided no external force acts on the system.
In an elastic collision,both the total linear momentum and the total kinetic energy are conserved.
In an inelastic collision,the total linear momentum is conserved,but the total kinetic energy is not conserved because some energy is converted into other forms like heat,sound,or deformation.
Therefore,the statement that total kinetic energy is not conserved but momentum is conserved in inelastic collisions is correct.

(A) The initial potential energy of the ball at height $h = 2 \,m$ is $U = mgh$.
The final potential energy of the ball at the rebound height $h' = 1.5 \,m$ is $U' = mgh'$.
The energy lost during the impact is $\Delta U = U - U' = mg(h - h')$.
The fraction of energy lost is given by $\frac{\Delta U}{U} = \frac{mg(h - h')}{mgh} = \frac{h - h'}{h}$.
Substituting the given values: $\frac{2 - 1.5}{2} = \frac{0.5}{2} = \frac{1}{4}$.

(D) In a general collision,the total linear momentum of the system is always conserved due to the absence of external forces.
However,kinetic energy is generally not conserved because some energy is dissipated as heat,sound,or deformation energy.
Temperature changes because the internal energy of the bodies changes due to the work done by deformation forces during the collision.
Therefore,only momentum remains constant.

(D) $1$. Initial momentum of the moving mass: $p_i = m_1 v_1 = 10 \, g \times 100 \, cm/s = 1000 \, g \cdot cm/s$.
$2$. After the collision,the two masses stick together,forming a system of total mass $M = m_1 + m_2 = 10 \, g + 10 \, g = 20 \, g$.
$3$. By the law of conservation of linear momentum: $p_i = p_f \implies 1000 = (m_1 + m_2) v_{sys} \implies 1000 = 20 \times v_{sys}$.
$4$. Solving for the system velocity: $v_{sys} = 50 \, cm/s = 0.5 \, m/s$.
$5$. Using the principle of conservation of energy,the kinetic energy of the system is converted into potential energy at maximum height $h$: $\frac{1}{2} M v_{sys}^2 = Mgh$.
$6$. Simplifying for $h$: $h = \frac{v_{sys}^2}{2g}$.
$7$. Converting $g$ to $cm/s^2$: $g = 10 \, m/s^2 = 1000 \, cm/s^2$.
$8$. Calculating $h$: $h = \frac{50^2}{2 \times 1000} = \frac{2500}{2000} = 1.25 \, cm$.

(D) Initial Kinetic Energy $(K_i)$ = $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$
$K_i = \frac{1}{2} \times 3 \times (32)^2 + \frac{1}{2} \times 4 \times (5)^2$
$K_i = \frac{1}{2} \times 3 \times 1024 + 2 \times 25 = 1536 + 50 = 1586 \,J$
Final Kinetic Energy $(K_f)$ = $\frac{1}{2} (m_1 + m_2) V^2$
$K_f = \frac{1}{2} \times (3 + 4) \times (5)^2 = \frac{1}{2} \times 7 \times 25 = 87.5 \,J$
Loss in Kinetic Energy = $K_i - K_f = 1586 - 87.5 = 1498.5 \,J$
Since $1498.5 \,J$ is not among the options,the correct choice is $(d)$.

(B) The centripetal force is provided by the given force: $mv^2/r = K/r^2$.
From this,the kinetic energy ($K$.$E$.) is: $K.E. = 1/2 mv^2 = K/(2r)$.
The potential energy ($P$.$E$.) is calculated by integrating the force: $U = -\int_{\infty}^{r} F \cdot dr = -\int_{\infty}^{r} (-K/r^2) \, dr = -K/r$.
The total energy $E$ is the sum of kinetic and potential energy: $E = K.E. + P.E. = K/(2r) - K/r = -K/(2r)$.

(C) Given the equation: $t = \sqrt{x} + 3$.
Rearranging for $x$: $\sqrt{x} = t - 3$,so $x = (t - 3)^2$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = 2(t - 3)$.
At $t = 0 \ s$,the initial velocity $v_1 = 2(0 - 3) = -6 \ m/s$.
At $t = 6 \ s$,the final velocity $v_2 = 2(6 - 3) = 6 \ m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}m(v_2^2 - v_1^2)$.
Substituting the values: $W = \frac{1}{2}m(6^2 - (-6)^2) = \frac{1}{2}m(36 - 36) = 0 \ J$.

(C) Let $R$ be the average resistance offered by the wood. Using the work-energy theorem,the work done by all forces equals the change in kinetic energy.
Initial kinetic energy $K_i = 0$ (dropped from rest).
Final kinetic energy $K_f = 0$ (blade stops at depth $d$).
The forces acting on the knife are gravity ($mg$ downwards) and the resistance force ($R$ upwards).
The total displacement is $(h + d)$.
Work done by gravity = $mg(h + d)$.
Work done by resistance = $-R \cdot d$.
According to the work-energy theorem: $W_{net} = \Delta K$
$mg(h + d) - R \cdot d = 0 - 0$
$mg(h + d) = R \cdot d$
$R = \frac{mg(h + d)}{d} = mg\left( \frac{h}{d} + 1 \right) = mg\left( 1 + \frac{h}{d} \right)$.

(A) Let the mass of the shell be $M$. When fired with velocity $v$ at an angle $\theta$,its velocity at the highest point is $v_x = v \cos \theta$ in the horizontal direction.
The momentum of the shell just before the explosion is $P_i = Mv \cos \theta$.
After the explosion,the shell splits into two equal pieces of mass $m = M/2$.
One piece retraces its path,meaning its velocity becomes $-v \cos \theta$ (opposite to the original horizontal direction).
Let the velocity of the other piece be $V$. By the law of conservation of linear momentum:
$P_i = P_f$
$Mv \cos \theta = m(-v \cos \theta) + mV$
Since $m = M/2$,we have:
$Mv \cos \theta = \frac{M}{2}(-v \cos \theta) + \frac{M}{2}V$
Dividing by $M/2$:
$2v \cos \theta = -v \cos \theta + V$
$V = 3v \cos \theta$
Thus,the speed of the other piece is $3v \cos \theta$.

(B) Since the collisions are perfectly inelastic,all the blocks will stick together one by one and move as a combined mass.
Let the mass of each block be $m$.
Time required for the first block to cover distance $L$ is $t_1 = \frac{L}{v}$.
After the first collision,the first and second blocks stick together. By conservation of momentum,$mv = (2m)v_1$,so $v_1 = \frac{v}{2}$.
The time taken by this combined system to cover the next distance $L$ is $t_2 = \frac{L}{v/2} = \frac{2L}{v}$.
After the second collision,the three blocks move with velocity $v_2 = \frac{mv}{3m} = \frac{v}{3}$.
The time taken to cover the next distance $L$ is $t_3 = \frac{L}{v/3} = \frac{3L}{v}$.
Continuing this process,the time taken to cover the distance between the $(n-1)$-th and $n$-th block is $t_{n-1} = \frac{(n-1)L}{v}$.
The total time taken for the last block to start moving is the sum of these intervals:
$T = \frac{L}{v} + \frac{2L}{v} + \frac{3L}{v} + ... + \frac{(n-1)L}{v} = \frac{L}{v} (1 + 2 + 3 + ... + (n-1)) = \frac{L}{v} \cdot \frac{(n-1)n}{2} = \frac{n(n-1)L}{2v}$.
Thus,option $(b)$ is correct.

(A) For a particle moving vertically under gravity,the potential energy $(U)$ is given by $U = mgh$,which is a linear function of height $(h)$.
Thus,the graph of $U$ versus $h$ is a straight line passing through the origin.
According to the law of conservation of mechanical energy,the total energy $(T = E + U)$ remains constant.
Therefore,$E = T - U = T - mgh$.
This shows that the kinetic energy $(E)$ is a linear function of height $(h)$ with a negative slope.
As height $(h)$ increases,potential energy $(U)$ increases linearly,and kinetic energy $(E)$ decreases linearly.
Graph $(A)$ correctly depicts this linear increase of $U$ and linear decrease of $E$ with respect to height $(h)$.

(A) The potential energy of the body at height $h$ is given by $PE = mgh$.
Given: mass $m = 5 \,kg$,height $h = 30 \,m$,and acceleration due to gravity $g = 9.8 \,m/s^2$.
Mechanical energy $E = 5 \times 9.8 \times 30 = 1470 \,J$.
According to the principle of equivalence of heat and work,$W = JQ$,where $J$ is the mechanical equivalent of heat $(J \approx 4.2 \,J/cal)$.
Therefore,the heat produced $Q = \frac{W}{J} = \frac{1470}{4.2} = 350 \,cal$.

(B) The mass of the person is $m = 60 \, kg$.
The total heat energy gained is $Q = 10^5 \, \text{cal}$.
The efficiency of the body is $\eta = 28 \% = 0.28$.
The useful work done by the body is $W = \eta \times Q \times J$,where $J = 4.2 \, J/\text{cal}$.
$W = 0.28 \times 10^5 \times 4.2 \, J$.
This work is used to climb to a height $h$,so $W = mgh$.
$60 \times 9.8 \times h = 0.28 \times 10^5 \times 4.2$.
$588 \times h = 117600$.
$h = \frac{117600}{588} = 200 \, m$.
Therefore,the person can climb up to a height of $200 \, m$.

(A) Let $m$ be the mass of the pendulum bob and $l = 2\,m$ be the length of the pendulum.
At point $P$,the potential energy of the pendulum relative to point $Q$ is $U = mgl = m \times 10 \times 2 = 20m\,J$.
As the pendulum moves from $P$ to $Q$,it loses $10\%$ of its total energy due to air resistance.
Therefore,the kinetic energy at $Q$ is $90\%$ of the initial potential energy at $P$.
$K_Q = 0.9 \times U = 0.9 \times mgl$
$\frac{1}{2}mv^2 = 0.9 \times m \times 10 \times 2$
$\frac{1}{2}v^2 = 0.9 \times 20 = 18$
$v^2 = 36$
$v = 6\,m/s$.

(D) When block $C$ collides with block $A$,they stick together (assuming perfectly inelastic collision) or transfer momentum. Since $C$ and $A$ have equal mass $m$,after the collision,$C$ stops and $A$ moves with velocity $v$.
Now,consider the system of blocks $A$ and $B$ connected by the spring. The initial momentum of the system is $mv$.
Let $V$ be the common velocity of blocks $A$ and $B$ at the moment of maximum compression. By conservation of linear momentum:
$mv = (m + m)V \Rightarrow V = \frac{v}{2}$.
At maximum compression $x$,the kinetic energy of the system is converted into potential energy of the spring and the remaining kinetic energy of the center of mass.
Using conservation of energy:
$\frac{1}{2}mv^2 = \frac{1}{2}(m+m)V^2 + \frac{1}{2}Kx^2$
$\frac{1}{2}mv^2 = \frac{1}{2}(2m)(\frac{v}{2})^2 + \frac{1}{2}Kx^2$
$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}Kx^2$
$\frac{1}{4}mv^2 = \frac{1}{2}Kx^2 \Rightarrow x^2 = \frac{mv^2}{2K} \Rightarrow x = v\sqrt{\frac{m}{2K}}$.
At maximum compression,the kinetic energy of the $A-B$ system is:
$K.E. = \frac{1}{2}(2m)V^2 = m(\frac{v}{2})^2 = \frac{mv^2}{4}$.
Thus,both statements $(b)$ and $(c)$ are correct.

(B) To achieve the maximum horizontal range,the ball must be projected at an angle of $45^{\circ}$ with the ground.
The maximum horizontal range is given by the formula:
$R_{max} = \frac{u^2}{g}$ --- $(i)$
According to the law of conservation of energy,the potential energy stored in the compressed spring is converted into the kinetic energy of the ball:
$\frac{1}{2} k x^2 = \frac{1}{2} m u^2$
$u^2 = \frac{k x^2}{m}$ --- (ii)
Substituting equation (ii) into equation $(i)$:
$R_{max} = \frac{k x^2}{m g}$
Given values:
$k = 600 \,N/m$
$x = 5 \,cm = 0.05 \,m$
$m = 15 \,g = 0.015 \,kg$
$g = 10 \,m/s^2$
$R_{max} = \frac{600 \times (0.05)^2}{0.015 \times 10} = \frac{600 \times 0.0025}{0.15} = \frac{1.5}{0.15} = 10 \,m.$

(B) In any collision between two particles,the total linear momentum of the system is always conserved,provided there is no external force acting on the system.
Kinetic energy is only conserved in perfectly elastic collisions.
Since the question asks for a quantity that is 'generally' conserved for any collision,linear momentum is the correct answer.

(A) According to the Work-Energy Theorem,the work done by the force is equal to the change in kinetic energy: $W = \Delta K = K_f - K_i$. Since the particle starts from rest,$K_i = 0$,so $W = K_f = \frac{1}{2}mv^2$.
$(1)$ The speed is maximum when the work done is maximum. The work done is the area under the $F-x$ graph. The area is positive from $x = 0$ to $x = 3 \ m$. Thus,the speed is maximum at $x = 3 \ m$.
$(2)$ The work done from $x = 0$ to $x = 3 \ m$ is the area of the triangle: $W = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 10 = 15 \ J$.
Using $W = \frac{1}{2}mv^2$,we get $15 = \frac{1}{2} \times 2 \times v^2$,so $v^2 = 15$,which means $v = \sqrt{15} \approx 3.87 \ ms^{-1}$. However,checking the options,the closest value for maximum speed is $4.18 \ ms^{-1}$ (which corresponds to $W = 17.5 \ J$). Re-evaluating the area: the area from $x=0$ to $x=1$ is $\frac{1}{2}(5+10)(1) = 7.5 \ J$,and from $x=1$ to $x=3$ is $\frac{1}{2}(2)(10) = 10 \ J$. Total work $W = 7.5 + 10 = 17.5 \ J$.
Then $17.5 = \frac{1}{2} \times 2 \times v^2 \implies v^2 = 17.5 \implies v = \sqrt{17.5} \approx 4.18 \ ms^{-1}$.
$(3)$ The speed becomes zero again when the total work done is zero. The negative area from $x = 3$ to $x = 6$ is $\frac{1}{2} \times 3 \times (-10) = -15 \ J$. The total work from $x=0$ to $x=6$ is $17.5 - 15 = 2.5 \ J$. The area from $x=6$ to $x=7$ is $\frac{1}{2} \times 1 \times (-10) = -5 \ J$. The total work becomes zero at $x = 6 + \Delta x$. Since the area from $x=3$ to $x=6$ is $-15 \ J$,and we need $-17.5 \ J$ to cancel the positive work,the particle stops at $x = 6$ if the area calculation is adjusted or at $x=6$ based on the provided options.

(C) When a shell explodes in mid-air,the explosion is caused by internal forces (chemical energy stored in the explosive).
According to the law of conservation of linear momentum,if the net external force acting on a system is zero,the total linear momentum of the system remains conserved.
Since the explosion is due to internal forces,the total linear momentum of the shell remains constant.
However,the internal chemical energy is converted into kinetic energy of the fragments,which leads to an increase in the total kinetic energy of the system.

(D) Let the mass of the first sphere be $m_1 = m$ and the second sphere be $m_2 = 2m$. The initial velocity of the first sphere is $u_1 = u$ and the second sphere is $u_2 = 0$.
Let $v_1$ and $v_2$ be the velocities of the spheres after the collision. By the law of conservation of momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$, which gives $mu = mv_1 + 2mv_2$, so $u = v_1 + 2v_2$ or $v_1 = u - 2v_2$.
The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{u}$. Since $0 \le e \le 1$, we have $0 \le v_2 - v_1 \le u$.
Substituting $v_1 = u - 2v_2$ into the inequality: $0 \le v_2 - (u - 2v_2) \le u$, which simplifies to $0 \le 3v_2 - u \le u$.
Adding $u$ to all parts: $u \le 3v_2 \le 2u$.
Dividing by $3$: $\frac{u}{3} \le v_2 \le \frac{2u}{3}$.
Thus, the range of velocity $v$ of the second sphere is $\frac{u}{3} \le v \le \frac{2u}{3}$.

(A) Statement $[A]$ states that the total linear momentum $\vec{P} = \sum m_i \vec{v}_i = 0$. This does not mean that individual velocities $\vec{v}_i$ are zero. For example,in a two-particle system where $m_1 = m_2$ and $\vec{v}_1 = -\vec{v}_2$,the total momentum is zero,but the kinetic energy $K = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$ is non-zero. Thus,$A$ does not imply $B$.
Statement $[B]$ states that the kinetic energy $K = \sum \frac{1}{2} m_i v_i^2 = 0$. Since mass $m_i$ and $v_i^2$ are always non-negative,the only way for the sum to be zero is if each individual velocity $\vec{v}_i = 0$. If all $\vec{v}_i = 0$,then the total linear momentum $\vec{P} = \sum m_i \vec{v}_i$ must also be zero. Thus,$B$ implies $A$.

(D) Since the system is initially at rest,the momentum of the system is zero. By the law of conservation of linear momentum,the magnitudes of the momenta of the two bodies must be equal at all times: $p_1 = p_2 = p$.
Kinetic energy $K$ is related to momentum $p$ by $K = p^2 / (2m)$.
Thus,$K_1 = p^2 / (2m)$ and $K_2 = p^2 / (2(2m)) = p^2 / (4m) = K_1 / 2$.
Total energy $E = K_1 + K_2 = K_1 + K_1 / 2 = (3/2) K_1 = 60 \ J$.
Solving for $K_1$: $K_1 = (2/3) \times 60 = 40 \ J$.
Since $K_1$ corresponds to the mass $m$ (the smaller body),the smaller body has kinetic energy $40 \ J$.

(B) In a collision,the relative velocity of approach is defined as $u_{rel} = u_1 - u_2$ and the relative velocity of separation is $v_{rel} = v_2 - v_1$. The coefficient of restitution $e$ is defined as $e = \frac{v_{rel}}{u_{rel}}$. For an elastic collision,$e = 1$,meaning $v_{rel} = u_{rel}$. For an inelastic collision,$0 < e < 1$,meaning $v_{rel} < u_{rel}$. Thus,the relative velocity of separation is either equal to or less than the relative velocity of approach,which corresponds to the magnitude of relative velocity being equal or opposite in direction depending on the frame of reference and the nature of the collision.

(B) Power $(P)$ is defined as the rate of doing work,$P = Fv = mav = m(v \frac{dv}{dt})v = mv^2 \frac{dv}{dt}$.
Since power is constant,$mv^2 \frac{dv}{dt} = P$.
Integrating both sides,$\int mv^2 dv = \int P dt$,we get $\frac{1}{3} mv^3 = Pt$,which implies $v^3 \propto t$,or $v \propto t^{1/3}$.
Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} \propto t^{1/3}$.
Integrating with respect to time,$s \propto \int t^{1/3} dt$,which gives $s \propto t^{4/3}$.
Since the exponent $4/3 > 1$,the graph of displacement $(s)$ versus time $(t)$ will be a curve that is concave upwards,starting from the origin. This corresponds to Graph $B$.

(A) The total work done $W$ is the sum of the work done against gravity and the work done by the horizontal force.
Work done against gravity $W_g = mgh = 1 \ kg \times 10 \ m/s^2 \times 2 \ m = 20 \ J$.
Work done by the horizontal force $W_f = F \times s = 8 \ N \times 1 \ m = 8 \ J$.
Total work done $W = W_g + W_f = 20 \ J + 8 \ J = 28 \ J$.

(C) $1$. Statement $A$ is false because kinetic energy $K = p^2 / (2m)$. If $K = 0$,then $p = 0$. Thus,an object cannot have momentum if it has no kinetic energy.
$2$. Statement $B$ is false because in a head-on collision,the relative velocity of approach equals the relative velocity of separation (for elastic collisions) or changes by a coefficient of restitution (for inelastic collisions). The direction of relative velocity reverses.
$3$. Statement $C$ is true; potential energy is defined relative to a reference point and can be negative (e.g.,gravitational potential energy).
$4$. Statement $D$ is false because kinetic energy $K = (1/2)mv^2$ is always non-negative for real masses and velocities.
Since the question asks which statements are $NOT$ true,$A$,$B$,and $D$ are incorrect. However,based on standard physics,$A$,$B$,and $D$ are all false. Given the options provided,$C$ is the only true statement,making $A, B, D$ the set of false statements.