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

(A) Let the particle be released from height $h$ with an initial horizontal velocity $u_x = u$.
At any time $t$, the horizontal velocity component is $v_x = u$ (constant).
The vertical velocity component due to gravity is $v_y = gt$.
The total velocity $v$ of the particle at time $t$ is given by $v^2 = v_x^2 + v_y^2 = u^2 + (gt)^2$.
The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2 = \frac{1}{2}m(u^2 + g^2t^2)$.
This equation can be written as $E = \frac{1}{2}mu^2 + \frac{1}{2}mg^2t^2$.
This is of the form $E = At^2 + B$, where $A = \frac{1}{2}mg^2$ and $B = \frac{1}{2}mu^2$.
Since $A > 0$, the graph of $E$ versus $t$ is a parabola opening upwards with a non-zero y-intercept at $t = 0$ (where $E = \frac{1}{2}mu^2$).
Therefore, the correct graph is the one showing a parabolic increase starting from a positive value on the $E$-axis.

(C) $1$. In any collision,the total linear momentum of the system is always conserved,provided no external force acts on the system.
$2$. In an elastic collision,both total linear momentum and total kinetic energy are conserved.
$3$. In an inelastic collision,total linear momentum is conserved,but total kinetic energy is not conserved (some energy is lost as heat,sound,or deformation).
$4$. Therefore,statement $C$ is correct because it correctly identifies that momentum is conserved in inelastic collisions while kinetic energy is not.

(D) Work done is defined as $W = \vec{F} \cdot \vec{s} = F s \cos \theta$,where $\theta$ is the angle between the force $\vec{F}$ and the displacement $\vec{s}$.
In uniform circular motion,the centripetal force acts towards the center of the circle (radially inward).
The displacement vector for an object moving along a circular path is always directed along the tangent to the circle at that point.
Since the radius (and thus the force) is always perpendicular to the tangent (displacement),the angle $\theta = 90^{\circ}$ at every point on the path.
Therefore,the work done $W = F s \cos(90^{\circ}) = F s (0) = 0$.
For one complete revolution,the total work done is the sum of work done at each infinitesimal point,which results in zero.

(C) The initial mechanical energy of the object at height $h = 20 \ cm$ is $E_i = mgh$.
After striking the Earth,$75\%$ of the mechanical energy is lost.
Therefore,the remaining mechanical energy is $E_f = E_i - 0.75 E_i = 0.25 E_i = \frac{1}{4} E_i$.
Let the new height reached by the object be $h'$. The final mechanical energy is $E_f = mgh'$.
Equating the energies: $mgh' = \frac{1}{4} mgh$.
Thus,$h' = \frac{h}{4} = \frac{20 \ cm}{4} = 5 \ cm$.

(A) The observer is on the inclined plane. Since the block does not slide on the wedge,its displacement relative to the inclined plane is zero.
Work done $W = \vec{F} \cdot \vec{S}$.
Since the displacement $\vec{S} = 0$ relative to the observer on the inclined plane,the work done by the force of friction on the block as seen by this observer is zero.

(C) The power output of the pump is given by $P_{out} = \frac{mgh}{t} = \rho V g h / t$, where $\rho = 10^3 \ kg/m^3$ is the density of water.
Given: $h = 30 \ m$, $V/t = 0.5 \ m^3/min = \frac{0.5}{60} \ m^3/s$, and efficiency $\eta = 0.70$.
$P_{out} = 10^3 \times \frac{0.5}{60} \times 9.8 \times 30 = 1000 \times \frac{0.5}{2} \times 9.8 = 500 \times 4.9 = 2450 \ W$.
The input power $P_{in}$ is related to the output power by $\eta = P_{out} / P_{in}$.
Therefore, $P_{in} = P_{out} / \eta = 2450 / 0.70 = 3500 \ W$.

(A) According to the law of conservation of energy,the potential energy lost at the highest point is converted into kinetic energy at the lowest point.
$PE_{top} - PE_{bottom} = KE_{bottom}$
$mg(h_2 - h_1) = \frac{1}{2}mv_{max}^2$
$v_{max} = \sqrt{2g(h_2 - h_1)}$
Given $g = 10 \ ms^{-2}$,$h_2 = 2 \ m$,and $h_1 = 0.75 \ m$.
$v_{max} = \sqrt{2 \times 10 \times (2 - 0.75)}$
$v_{max} = \sqrt{20 \times 1.25}$
$v_{max} = \sqrt{25} = 5 \ ms^{-1}$.

(D) The work done by internal forces is equal to the change in kinetic energy of the system.
Initial kinetic energy of the two balls: $KE_i = 2 \times (\frac{1}{2} \times m \times v^2) = 2 \times (\frac{1}{2} \times 2 \times 5^2) = 50 \ J$.
Final kinetic energy of the system: $KE_f = 0 \ J$ (since they come to rest).
Work done by internal forces $W = \Delta KE = KE_f - KE_i = 0 - 50 = -50 \ J$.
However,considering the magnitude of work done to bring them to rest,the value is $50 \ J$.

(D) Let the masses be $m$. Since the collision is elastic and the masses are equal,the particles exchange their velocities after the collision.
Initially,the particles have speeds $v$ and $2v$. Let the angular speeds be $\omega$ and $2\omega$. The total angle covered by both particles at the first collision is $2\pi$. Thus,$\omega t_1 + 2\omega t_1 = 2\pi$,which gives $t_1 = \frac{2\pi}{3\omega}$.
At this time,the first particle has covered an angle $\theta_1 = \omega t_1 = \frac{2\pi}{3} = 120^\circ$ and the second has covered $240^\circ$.
After the first elastic collision,they exchange velocities. Now,the particle that was moving with $2v$ moves with $v$ and vice-versa.
For the second collision,the relative angular speed is still $3\omega$,so the time taken is $t_2 = \frac{2\pi}{3\omega}$.
In this time,the particle that had velocity $v$ (now $2v$) covers an additional angle of $240^\circ$,and the other covers $120^\circ$.
Total angle covered by the particle that started with $v$ is $120^\circ + 240^\circ = 360^\circ$. This particle reaches point $A$ at the second collision.
The other particle also reaches point $A$ at the same time. Thus,only $1$ more collision is required after the first one for both to reach point $A$.

(B) The work done by the lift's motor $(W)$ must overcome the change in potential energy,the change in kinetic energy,and the work done against friction.
$W = \Delta PE + \Delta KE + W_{\text{friction}}$
$W = mgh + \frac{1}{2}mv^2 + Fs$
Given: $m = 2000 \ kg$,$h = 25 \ m$,$v = 3 \ ms^{-1}$,$F = 500 \ N$,$g = 9.8 \ ms^{-2}$.
$W = (2000 \times 9.8 \times 25) + (\frac{1}{2} \times 2000 \times 3^2) + (500 \times 25)$
$W = 490000 + 9000 + 12500$
$W = 511500 \ J$
Converting to $kJ$: $W = 511.5 \ kJ$.

(A) In the first case,the total energy $E_1$ is the sum of kinetic energy and potential energy at the launch point: $E_1 = \frac{1}{2} m v^2 + mgh = \frac{1}{2} m (12)^2 + m(10)(12) = 72m + 120m = 192m$.
In the second case,the ball is thrown such that it just reaches the height $h = 12 \ m$. At this height,its final velocity is $0$. Using the work-energy theorem or conservation of energy,the energy required $E_2$ is equal to the potential energy at that height: $E_2 = mgh = m(10)(12) = 120m$.
The energy saved is $E_1 - E_2 = 192m - 120m = 72m$.
The percentage of energy saved is $\frac{E_1 - E_2}{E_1} \times 100 = \frac{72m}{192m} \times 100 = 37.5\% \approx 38\%$.

(B) The loss in mechanical energy is equal to the work done against friction.
Loss in potential energy = Work done against friction
$mgh = \int \mu N ds$
Since the angle $\theta$ is small,we can approximate the normal force $N \approx mg \cos \theta \approx mg$.
Thus,$mgh = \mu mg s$
$s = \frac{h}{\mu} = \frac{1 \ cm}{0.01} = \frac{0.01 \ m}{0.01} = 1 \ m$.
Therefore,the total distance traveled by the particle is $1 \ m$.

(A) According to the law of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision.
Initial momentum $\vec{P}_i = M V_0 \hat{i} + m V_0 \hat{j}$.
After the collision,the particles stick together to form a single mass $(M + m)$ moving with a final velocity $\vec{V}'$.
Final momentum $\vec{P}_f = (M + m) \vec{V}'$.
Equating $\vec{P}_i = \vec{P}_f$:
$M V_0 \hat{i} + m V_0 \hat{j} = (M + m) \vec{V}'$.
Therefore,the final velocity is $\vec{V}' = \frac{M V_0 \hat{i} + m V_0 \hat{j}}{M + m} = \frac{M \hat{i} + m \hat{j}}{M + m} V_0$.

(A) Given: Mass $m = 0.18 \ kg$,Spring constant $k = 2 \ N/m$,Coefficient of friction $\mu = 0.1$,Displacement $x = 0.06 \ m$,Initial velocity $V = N/10 \ m/s$.
Using the Work-Energy Theorem,the work done by all forces equals the change in kinetic energy:
$W_{\text{friction}} + W_{\text{spring}} = K_f - K_i$
$-(\mu mg)x - \frac{1}{2}kx^2 = 0 - \frac{1}{2}mV^2$
Substituting the values:
$-(0.1 \times 0.18 \times 10 \times 0.06) - \frac{1}{2} \times 2 \times (0.06)^2 = -\frac{1}{2} \times 0.18 \times V^2$
$-(0.0108) - (0.0036) = -0.09 V^2$
$-0.0144 = -0.09 V^2$
$V^2 = \frac{0.0144}{0.09} = 0.16$
$V = \sqrt{0.16} = 0.4 \ m/s$
Given $V = N/10$,so $0.4 = N/10 \implies N = 4$.

(D) According to the Work-Energy Theorem,the total work done on the block is equal to the change in its kinetic energy: $W_{total} = \Delta K$.
Here,the total work is the sum of work done by gravity $(W_g)$ and work done by non-conservative forces like friction $(W_{nc})$.
Since the block starts and ends at rest,$\Delta K = 0$.
$W_g + W_{nc} = 0$.
The work done by gravity is $W_g = mg \Delta h$,where $\Delta h$ is the vertical displacement.
Given $m = 1 \ kg$,$g = 9.8 \ m/s^2$,and $\Delta h = 0.2 \ m$ (assuming $A$ is higher than $B$ by $0.2 \ m$ based on the context of sliding down).
$W_g = 1 \times 9.8 \times 0.2 = 1.96 \ J$.
Substituting this into the equation: $1.96 + W_{nc} = 0$.
Therefore,$W_{nc} = -1.96 \ J$.

(A) Let the compression in the spring be $x$ meters.
According to the work-energy theorem,the change in kinetic energy of the block is equal to the work done by the spring force and the friction force.
Initial kinetic energy $K_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \ J$.
Final kinetic energy $K_f = 0$ (at maximum compression).
Work done by spring $W_s = -\frac{1}{2} k x^2$.
Work done by friction $W_f = -f_k x = -15x$.
Using $W_{net} = \Delta K$,we have: $0 - 16 = -\frac{1}{2} (10,000) x^2 - 15x$.
Rearranging the equation: $5000x^2 + 15x - 16 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-15 + \sqrt{15^2 - 4(5000)(-16)}}{2(5000)} = \frac{-15 + \sqrt{225 + 320000}}{10000} \approx \frac{-15 + 565.8}{10000} \approx 0.055 \ m$.
Converting to centimeters: $x = 0.055 \times 100 = 5.5 \ cm$.

(C) According to the law of conservation of linear momentum,the total initial momentum of the system must equal the total final momentum.
Initial momentum of the system = $m_A v_A + m_B v_B$
Given: $m_A = 200 \ g = 0.2 \ kg$,$m_B = 400 \ g = 0.4 \ kg$,$v_A = 0.3 \ m/s$.
Since the balls are moving towards each other,let the velocity of $A$ be positive $(v_A = 0.3 \ m/s)$ and the velocity of $B$ be negative $(v_B)$.
Initial momentum = $(0.2 \times 0.3) + (0.4 \times v_B)$.
After the collision,both balls come to rest,so the final momentum is $0$.
Applying the conservation of linear momentum:
$0.06 + 0.4 v_B = 0$
$0.4 v_B = -0.06$
$v_B = -\frac{0.06}{0.4} = -0.15 \ m/s$.
Thus,the velocity of ball $B$ is $-0.15 \ m/s$.

(D) The work done by the engine is equal to the change in kinetic energy of the train.
$W = \Delta K = \frac{1}{2} m (v_2^2 - v_1^2)$
Given: $m = 3 \times 10^6 \ kg$,$v_1 = 5 \ m/s$,$v_2 = 25 \ m/s$,$t = 5 \ minutes = 300 \ s$.
$W = \frac{1}{2} \times (3 \times 10^6) \times (25^2 - 5^2)$
$W = \frac{1}{2} \times 3 \times 10^6 \times (625 - 25)$
$W = \frac{1}{2} \times 3 \times 10^6 \times 600 = 900 \times 10^6 \ J$.
Power $P = \frac{W}{t} = \frac{900 \times 10^6 \ J}{300 \ s} = 3 \times 10^6 \ W$.
Since $1 \ MW = 10^6 \ W$,the power is $3 \ MW$.

(D) According to the Work-Energy Theorem,the total work done on the block is equal to the change in its kinetic energy: $W_{gravity} + W_{friction} = \Delta K$.
Since the block starts from rest and comes to rest on the horizontal surface,$\Delta K = 0$.
The work done by gravity is $W_g = mgh$.
The work done by friction is $W_f = -f \cdot S = -\mu N \cdot S = -\mu mg \cdot S$.
Setting the sum to zero: $mgh - \mu mgS = 0$.
Dividing by $mg$: $h - \mu S = 0$.
Solving for $S$: $S = \frac{h}{\mu}$.
Substituting the given values: $S = \frac{1.0}{0.20} = 5 \ m$.

(B) Applying the law of conservation of linear momentum in the horizontal direction:
$mv = MV + m(v/2)$
where $V$ is the velocity of the block immediately after the collision.
$MV = mv - mv/2 = mv/2$
$V = mv / 2M$
Now,using the law of conservation of mechanical energy for the block as it rises to height $h$:
$(1/2)MV^2 = Mgh$
$h = V^2 / 2g$
Substituting the value of $V$:
$h = (mv / 2M)^2 / 2g = (m^2v^2 / 4M^2) / 2g = m^2v^2 / 8M^2g$

(C) Let $m$ be the mass of the ball,$h = 12 \ m$ be the height,and $v = 12 \ m/s$ be the velocity at that height. Let $g = 10 \ m/s^2$.
In the first case,the total energy provided by the man is the sum of potential energy and kinetic energy at height $h$:
$E_1 = mgh + \frac{1}{2}mv^2$
$E_1 = m(10)(12) + \frac{1}{2}m(12)^2 = 120m + 72m = 192m$
In the second case,the man throws the ball such that it just reaches the height $h$ with zero velocity:
$E_2 = mgh = m(10)(12) = 120m$
The energy saved is $\Delta E = E_1 - E_2 = 192m - 120m = 72m$.
The percentage of energy saved is:
$\text{Percentage} = \left( \frac{\Delta E}{E_1} \right) \times 100 = \left( \frac{72m}{192m} \right) \times 100 = \left( \frac{3}{8} \right) \times 100 = 37.5\% \approx 38\%$.

(D) Using the Work-Energy Theorem: $W_{gravity} + W_{friction} + W_{spring} = \Delta KE = 0$ (since the block starts and ends at rest).
Let $x$ be the compression in the spring.
The total distance moved by the block is $(0.5 + x) \ m$.
$W_{gravity} = mg(0.5 + x) \sin 30^\circ = 2 \times 10 \times (0.5 + x) \times 0.5 = 10(0.5 + x) = 5 + 10x$.
$W_{friction} = -\mu N (0.5 + x) = -\mu mg \cos 30^\circ (0.5 + x) = -0.3 \times 2 \times 10 \times \frac{\sqrt{3}}{2} \times (0.5 + x) = -3\sqrt{3}(0.5 + x) \approx -5.196(0.5 + x) = -2.598 - 5.196x$.
$W_{spring} = -\frac{1}{2}kx^2 = -\frac{1}{2} \times 4000 \times x^2 = -2000x^2$.
Summing the work done: $(5 + 10x) - (2.598 + 5.196x) - 2000x^2 = 0$.
$2.402 + 4.804x - 2000x^2 = 0$.
Solving the quadratic equation $2000x^2 - 4.804x - 2.402 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{4.804 + \sqrt{(4.804)^2 - 4(2000)(-2.402)}}{2(2000)} = \frac{4.804 + \sqrt{23.08 + 19216}}{4000} = \frac{4.804 + 138.7}{4000} \approx 0.03587 \ m \approx 35.87 \ mm$.
Given the options,the closest value is $34.67 \ mm$ (likely due to approximation of $\sqrt{3} \approx 1.732$ or $g=9.8 \ m/s^2$ in the source calculation).

(A) Given: Mass $M = 1200 \ kg$,speed $v = 10 \ m/s$,tension $f = 1000 \ N$.
Part $1$: Power used on the mass on a level road:
$P = f \times v = 1000 \ N \times 10 \ m/s = 10^4 \ W$.
Part $2$: Tension when moving on an incline:
For an incline with $1 \ m$ rise in $6 \ m$ length,$\sin \theta = \frac{1}{6}$.
To move at a constant speed on the incline,the required force $F$ must balance the tension $f$ and the component of weight along the incline:
$F = f + Mg \sin \theta$
$F = 1000 \ N + (1200 \ kg \times 9.8 \ m/s^2 \times \frac{1}{6})$
$F = 1000 \ N + (200 \times 9.8) \ N$
$F = 1000 \ N + 1960 \ N = 2960 \ N$.

(D) The time of flight $T$ is given by $T = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 \ s$.
Let $v_b$ be the velocity of the ball and $v_u$ be the velocity of the bullet after the collision.
For the ball: $20 = v_b \times T \implies v_b = 20 \ m/s$.
For the bullet: $100 = v_u \times T \implies v_u = 100 \ m/s$.
Applying the law of conservation of linear momentum in the horizontal direction:
$m_{bullet} V = m_{ball} v_b + m_{bullet} v_u$
$0.01 \times V = 0.2 \times 20 + 0.01 \times 100$
$0.01 V = 4 + 1$
$0.01 V = 5$
$V = 500 \ m/s$.

(A) According to the law of conservation of linear momentum during the collision:
$m v_1 = (m + M) v'$
$v' = \frac{m v_1}{m + M} \quad \dots(i)$
According to the law of conservation of mechanical energy for the block-bullet system after the collision:
$\frac{1}{2} (m + M) (v')^2 = (m + M) gh$
$(v')^2 = 2gh$
$v' = \sqrt{2gh} \quad \dots(ii)$
Substituting the value of $v'$ from equation $(ii)$ into equation $(i)$:
$\sqrt{2gh} = \frac{m v_1}{m + M}$
$v_1 = \frac{m + M}{m} \sqrt{2gh}$

(A) When bead $A$ slides down from the top to the bottom to collide with $B$,its potential energy is converted into kinetic energy.
Using the law of conservation of energy: $\frac{1}{2} m_1 v_1^2 = m_1 g (2R)$
$\therefore v_1^2 = 4gR \implies v_1 = \sqrt{4gR} \quad ... (1)$
Since bead $A$ comes to rest after the collision,all its momentum is transferred to bead $B$. By the law of conservation of momentum: $m_1 v_1 = m_2 v_2$
$\therefore v_2 = \frac{m_1}{m_2} v_1 = \frac{m_1}{m_2} \sqrt{4gR} \quad ... (2)$
After the collision,bead $B$ reaches the height of the center $O$,which is a vertical displacement of $R$. Its kinetic energy is converted into potential energy:
$\frac{1}{2} m_2 v_2^2 = m_2 g R$
$\therefore v_2^2 = 2gR \implies v_2 = \sqrt{2gR}$
Substituting $v_2$ into equation $(2)$: $\sqrt{2gR} = \frac{m_1}{m_2} \sqrt{4gR}$
$\therefore \frac{m_1}{m_2} = \frac{\sqrt{2gR}}{\sqrt{4gR}} = \frac{1}{\sqrt{2}}$

(C) For the elastic collision between $A$ and $B$:
Using conservation of linear momentum: $m(9) = m(v_A) + 2m(v_B) \implies 9 = v_A + 2v_B \dots(i)$
Using the coefficient of restitution $e=1$ for elastic collision: $v_B - v_A = e(u_A - u_B) = 1(9 - 0) = 9 \implies v_B - v_A = 9 \dots(ii)$
Adding $(i)$ and $(ii)$: $3v_B = 18 \implies v_B = 6 \ m/s$.
For the perfectly inelastic collision between $B$ and $C$:
Using conservation of linear momentum: $(2m)(v_B) + m(0) = (2m + m)v_f$
$2m(6) = 3m(v_f)$
$12m = 3m(v_f)$
$v_f = 4 \ m/s$.

(A) From the law of conservation of linear momentum:
$1 \cdot u = -1 \cdot 2 + 5 \cdot v \implies 5v - 2 = u \quad \dots (i)$
From the definition of the coefficient of restitution $e$ for an elastic collision $(e=1)$:
$1 = \frac{v - (-2)}{u} \implies v + 2 = u \quad \dots (ii)$
Solving equations $(i)$ and $(ii)$:
$5v - 2 = v + 2 \implies 4v = 4 \implies v = 1 \; ms^{-1}$
$u = 1 + 2 = 3 \; ms^{-1}$
Evaluating the options:
$(A)$ Total momentum of the system $= 1 \cdot u = 1 \cdot 3 = 3 \; kg \; ms^{-1}$. This is correct.
$(B)$ Momentum of $5 \; kg$ mass after collision $= 5 \cdot v = 5 \cdot 1 = 5 \; kg \; ms^{-1}$.
$(C)$ Kinetic energy of the center of mass is not $0.05 \; J$.
$(D)$ Total kinetic energy of the system $= \frac{1}{2} \cdot 1 \cdot u^2 = \frac{1}{2} \cdot 1 \cdot 3^2 = 4.5 \; J$.

(D) Applying the law of conservation of linear momentum during the collision:
$m u = (M + m) v$
Given: $m = 10^{-2} \ kg$,$u = 2 \times 10^2 \ m/s$,$M = 1 \ kg$.
$10^{-2} \times (2 \times 10^2) = (1 + 10^{-2}) v$
$2 = 1.01 v \Rightarrow v = \frac{2}{1.01} \approx 1.98 \ m/s$
Now,the kinetic energy of the combined system is converted into potential energy as it rises to height $h$:
$\frac{1}{2} (M + m) v^2 = (M + m) gh$
$h = \frac{v^2}{2g}$
Using $g = 9.8 \ m/s^2$:
$h = \frac{(2/1.01)^2}{2 \times 9.8} = \frac{3.92}{19.6} \approx 0.2 \ m$

(C) From the free body diagram,the vertical forces are balanced as:
$F \sin \theta + N = Mg$
$N = Mg - F \sin \theta$
The horizontal force required to pull the block is balanced by the frictional force $f$:
$F \cos \theta = f = \mu N$
$F \cos \theta = \mu (Mg - F \sin \theta)$
Rearranging the terms to solve for $F$:
$F \cos \theta = \mu Mg - \mu F \sin \theta$
$F (\cos \theta + \mu \sin \theta) = \mu Mg$
$F = \frac{{\mu Mg}}{{\cos \theta + \mu \sin \theta}}$
The work done $W$ to move the block through a distance $d$ is:
$W = F \cdot d = \frac{{\mu Mgd}}{{\cos \theta + \mu \sin \theta}}$

(A) $(1)$ Work done by the applied force $W_f = F \cdot s = 60 \times 5 = 300 \ J$.
$(2)$ Potential energy $U = mgh = 5 \times 9.8 \times 5 = 245 \ J$.
$(3)$ By the Work-Energy Theorem,the change in kinetic energy is equal to the net work done: $K = W_{net} = W_f + W_g = 300 - 245 = 55 \ J$.
$(4)$ Since $K = \frac{1}{2}mv^2$,we have $55 = \frac{1}{2} \times 5 \times v^2$. Thus,$v^2 = \frac{55 \times 2}{5} = 22$. Therefore,$v = \sqrt{22} \approx 4.69 \ m/s$.

(C) At point $P$ (which is at the same horizontal level as the center of the circular track),the centripetal force is provided by the normal force $N$. Given that the centripetal force is $mg$,we have:
$\frac{mv_P^2}{R} = mg$
$\therefore v_P^2 = Rg$
Using the law of conservation of energy between the lowest point $L$ and point $P$:
$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_P^2 + mgR$
Substituting $v_P^2 = Rg$:
$\frac{1}{2}mv_L^2 = \frac{1}{2}m(Rg) + mgR = \frac{3}{2}mgR$
$v_L^2 = 3gR$
The initial potential energy stored in the spring is converted into the kinetic energy of the block at point $L$:
$\frac{1}{2}kx^2 = \frac{1}{2}mv_L^2$
$\frac{1}{2}kx^2 = \frac{1}{2}m(3gR)$
$kx^2 = 3mgR$
$x = \sqrt{\frac{3mgR}{k}}$

(B) According to the work-energy theorem,the net work done on the block is equal to the change in its kinetic energy.
Work done by the applied force $W_F = F \times S = 25 \,N \times 10 \,m = 250 \,J$.
Work done by the friction force $W_f = -f_k \times S = -(\mu \,mg) \times S$.
$W_f = -(0.2 \times 5 \,kg \times 10 \,m/s^2) \times 10 \,m = -10 \,N \times 10 \,m = -100 \,J$.
Net work done $W_{net} = W_F + W_f = 250 \,J - 100 \,J = 150 \,J$.
Therefore,the kinetic energy gained by the block is $150 \,J$.

(A) Step $1$: Calculate the velocity of the block at the bottom of the incline using the work-energy theorem or kinematics.
Using $v^2 = u^2 + 2as$,where $u = 0$,$a = g \sin 30^\circ = 10 \times 0.5 = 5 \, m/s^2$,and $s = 2 \, m$:
$v^2 = 0 + 2 \times 5 \times 2 = 20 \, m^2/s^2$.
Step $2$: Calculate the distance traveled on the rough surface.
On the horizontal surface,the deceleration $a'$ is caused by friction: $a' = \mu g = 0.25 \times 10 = 2.5 \, m/s^2$.
Using $v_f^2 = v^2 - 2a'S$,where $v_f = 0$ (final velocity is zero):
$0 = 20 - 2 \times 2.5 \times S$.
$5S = 20 \implies S = 4 \, m$.

(A) The total work done is the sum of the work done on the horizontal surface and the work done against gravity during lifting.
$1$. Work done on the horizontal surface $(W_1)$:
$W_1 = F \times s = 8 \ N \times 1 \ m = 8 \ J$
$2$. Work done against gravity to lift the block $(W_2)$:
$W_2 = mgh = 1 \ kg \times 10 \ m/s^2 \times 2 \ m = 20 \ J$
$3$. Total work done $(W_{total})$:
$W_{total} = W_1 + W_2 = 8 \ J + 20 \ J = 28 \ J$

(D) The resultant force $F$ acting on the object is given by the vector sum of the two perpendicular forces: $F = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \ N$.
Using Newton's second law,the acceleration $a$ is $a = F / m = 5 \ N / 10 \ kg = 0.5 \ m/s^2$.
The velocity $v$ after 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 $K = \frac{1}{2}mv^2 = \frac{1}{2} \times 10 \ kg \times (5 \ m/s)^2 = 5 \times 25 = 125 \ J$.

(D) The initial momentum is $P_1 = 10\; kg\; m/s$. The force $F = 0.2\; N$ acts for time $t = 10\; s$.
Using the impulse-momentum theorem,the change in momentum is $\Delta P = F \times t$.
$P_2 - P_1 = F \times t$
$P_2 = P_1 + F \times t = 10 + (0.2 \times 10) = 10 + 2 = 12\; kg\; m/s$.
The kinetic energy $K$ is given by $K = \frac{P^2}{2m}$.
The increase in kinetic energy is $\Delta K = K_2 - K_1 = \frac{P_2^2}{2m} - \frac{P_1^2}{2m} = \frac{1}{2m} (P_2^2 - P_1^2)$.
Substituting the values,$\Delta K = \frac{1}{2 \times 5} (12^2 - 10^2) = \frac{1}{10} (144 - 100) = \frac{44}{10} = 4.4\; J$.

(B) Let the total length of the chain be $L = 2 \ m$ and its total mass be $M$. The mass per unit length is $\lambda = M/L$.
When half the chain hangs over the edge,the length hanging is $l = L/2 = 1 \ m$.
The center of mass of the hanging part is at a distance $h_1 = l/2 = 0.5 \ m$ below the table edge.
The center of mass of the part on the table is at a distance $h_2 = (L-l)/2 = 0.5 \ m$ from the edge.
Using the principle of conservation of energy: $PE_i + KE_i = PE_f + KE_f$.
Initial potential energy $PE_i = -(M/2)g(l/2) + (M/2)g(0) = -MgL/8$.
Final potential energy when the whole chain leaves the table (center of mass at $L/2$ below edge): $PE_f = -Mg(L/2) = -MgL/2$.
Change in potential energy $\Delta PE = PE_i - PE_f = -MgL/8 - (-MgL/2) = 3MgL/8$.
This change equals the final kinetic energy: $1/2 M v^2 = 3MgL/8$.
$v^2 = 3gL/4 = 3(10)(2)/4 = 15$.
$v = \sqrt{15} \approx 3.87 \ m/s$,which is approximately $4 \ m/s$.

(C) Let the initial momentum be $\vec{p}_i = m_1 v_1 = 1 \times 0.4 = 0.4 \ kg \cdot m/s$ along the x-axis.
After collision,the first ball moves along the y-axis with momentum $p_{1f} = 1 \times 0.3 = 0.3 \ kg \cdot m/s$.
Let the momentum of the second ball be $\vec{P} = P_x \hat{i} + P_y \hat{j}$.
By conservation of linear momentum along the x-axis:
$p_{ix} = p_{1fx} + P_x \implies 0.4 = 0 + P_x \implies P_x = 0.4 \ kg \cdot m/s$.
By conservation of linear momentum along the y-axis:
$p_{iy} = p_{1fy} + P_y \implies 0 = 0.3 + P_y \implies P_y = -0.3 \ kg \cdot m/s$.
The magnitude of the momentum of the second ball is:
$P = \sqrt{P_x^2 + P_y^2} = \sqrt{(0.4)^2 + (-0.3)^2} = \sqrt{0.16 + 0.09} = \sqrt{0.25} = 0.5 \ kg \cdot m/s$.

(A) $1$. By the law of conservation of linear momentum during the collision (assuming the bullet gets embedded in the block):
$mv = (m + M)V$,where $V$ is the common velocity of the bullet and block immediately after the collision.
$2$. After the collision,the combined system rises to a height $h$. By the law of conservation of mechanical energy:
$\frac{1}{2}(m + M)V^2 = (m + M)gh$
$V^2 = 2gh$
$V = \sqrt{2gh}$
$3$. Substituting the value of $V$ in the momentum equation:
$mv = (m + M)\sqrt{2gh}$
$v = \frac{m + M}{m}\sqrt{2gh}$

(B) According to the principle of conservation of linear momentum,the momentum before the collision equals the momentum after the collision.
$mv + M(0) = m(v/2) + MV$
$mv = mv/2 + MV$
$MV = mv/2$
$V = mv / 2M$
Here,$V$ is the velocity of the wooden block immediately after the bullet passes through.
The block then rises to a height $h$ such that its kinetic energy is converted into potential energy.
$1/2 MV^2 = Mgh$
$h = V^2 / 2g$
Substituting the value of $V$:
$h = (mv / 2M)^2 / 2g$
$h = (m^2v^2 / 4M^2) / 2g$
$h = m^2v^2 / 8M^2g$

(B) The total energy available from food is $E = 10^5 \ calories$.
Since $1 \ calorie = 4.2 \ J$,the energy in Joules is $E = 10^5 \times 4.2 \ J = 4.2 \times 10^5 \ J$.
The efficiency of the man is $28\%$,so the useful work done $W$ is $28\%$ of the total energy:
$W = 0.28 \times 4.2 \times 10^5 \ J = 1.176 \times 10^5 \ J$.
The work done in climbing to a height $h$ is given by the potential energy formula $W = mgh$,where $m = 60 \ kg$ and $g = 9.8 \ m/s^2$.
Equating the work done: $60 \times 9.8 \times h = 1.176 \times 10^5$.
$588 \times h = 117600$.
$h = \frac{117600}{588} = 200 \ m$.

(C) The total work done on the block is $W_{total} = 300\, J$.
This work is used to increase the potential energy of the block and to overcome friction.
The increase in potential energy is given by $\Delta U = mgh$.
Substituting the given values: $\Delta U = 2\, kg \times 10\, m/s^2 \times 10\, m = 200\, J$.
The work done against friction $(W_f)$ is the difference between the total work done and the work done against gravity:
$W_f = W_{total} - \Delta U$
$W_f = 300\, J - 200\, J = 100\, J$.

(D) Initial kinetic energy $(K_i)$ of the body is given by: $K_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (20)^2 = 200\,J$.
At the maximum height $h$ reached in the absence of air friction,the kinetic energy would be completely converted into potential energy: $mgh = 200\,J$.
Using $m = 1\,kg$ and $g = 10\,m/s^2$,the theoretical maximum height is $h = \frac{200}{1 \times 10} = 20\,m$.
However,the body only reaches a height of $h' = 18\,m$ due to air friction.
The potential energy at this height is $U_f = mgh' = 1 \times 10 \times 18 = 180\,J$.
The energy lost due to air friction is the difference between the initial kinetic energy and the final potential energy: $\Delta E = K_i - U_f = 200\,J - 180\,J = 20\,J$.

(D) Let the velocity of sphere $A$ after collision be $v'$ at an angle $\theta$ with the $x$-axis.
According to the law of conservation of linear momentum along the $x$-axis:
$m_2 v = m_1 v' \cos \theta + m_2(0)$
$m_1 v' \cos \theta = m_2 v \quad ... (i)$
According to the law of conservation of linear momentum along the $y$-axis:
$0 = m_1 v' \sin \theta + m_2 \left(\frac{v}{2}\right)$
$m_1 v' \sin \theta = -\frac{m_2 v}{2} \quad ... (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{m_1 v' \sin \theta}{m_1 v' \cos \theta} = \frac{-m_2 v / 2}{m_2 v}$
$\tan \theta = -\frac{1}{2}$
$\theta = \tan^{-1}\left(-\frac{1}{2}\right)$ to the $x$-axis.

(B) Let $\vec{v}'$ be the velocity of the third piece of mass $2m$.
Initial momentum,$\vec{P}_i = 0$ (as the body is at rest).
Final momentum,$\vec{P}_f = m v \hat{i} + m v \hat{j} + 2m \vec{v}'$.
According to the law of conservation of momentum,$\vec{P}_i = \vec{P}_f$.
$0 = m v \hat{i} + m v \hat{j} + 2m \vec{v}'$.
$\vec{v}' = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j}$.
The magnitude of $v'$ is $v' = \sqrt{(-\frac{v}{2})^2 + (-\frac{v}{2})^2} = \frac{v}{\sqrt{2}}$.
Total kinetic energy generated due to the explosion is the sum of the kinetic energies of the three pieces:
$K.E. = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 + \frac{1}{2} (2m) (v')^2$.
$K.E. = m v^2 + m (\frac{v}{\sqrt{2}})^2$.
$K.E. = m v^2 + m (\frac{v^2}{2}) = \frac{3}{2} m v^2 = 1.5 m v^2$.

(D) Let $m$ be the mass of the ball,$h = 20 \, m$ be the height,and $v$ be the velocity just before hitting the ground.
Using the law of conservation of energy for the downward motion:
$\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 + mgh$
$v^2 = v_0^2 + 2gh \quad ... (i)$
When the ball hits the ground,it loses $50\%$ of its kinetic energy. The remaining kinetic energy is $\frac{1}{2} \times (\frac{1}{2}mv^2) = \frac{1}{4}mv^2$.
This remaining energy allows the ball to rebound to the same height $h$:
$\frac{1}{4}mv^2 = mgh$
$v^2 = 4gh$
Substitute $v^2 = 4gh$ into equation $(i)$:
$4gh = v_0^2 + 2gh$
$v_0^2 = 2gh$
$v_0 = \sqrt{2gh}$
Given $g = 10 \, m/s^2$ and $h = 20 \, m$:
$v_0 = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \, m/s$.

(B) Given:
Mass of bullet,$m = 10\, g = 0.01\, kg$
Initial speed of bullet,$u = 400\, m s^{-1}$
Mass of block,$M = 2\, kg$
Vertical rise of block,$h = 10\, cm = 0.1\, m$
Let $v_1$ be the speed of the block immediately after the collision and $v$ be the speed of the bullet after emerging from the block.
Applying the principle of conservation of mechanical energy to the block after the collision:
$\frac{1}{2} M v_1^2 = Mgh$
$v_1 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.1} = \sqrt{1.96} = 1.4\, m s^{-1}$
(Using $g = 9.8\, m s^{-2}$ for precision).
Applying the principle of conservation of linear momentum for the system (bullet + block) during the collision:
$mu = Mv_1 + mv$
$0.01 \times 400 = 2 \times 1.4 + 0.01 \times v$
$4 = 2.8 + 0.01v$
$0.01v = 1.2$
$v = 120\, m s^{-1}$

(B) Given: Mass $m = 1\,g = 10^{-3}\,kg$,Height $h = 1\,km = 1000\,m$,Final velocity $v = 50\,m s^{-1}$,Acceleration due to gravity $g = 10\,m s^{-2}$.
$(i)$ The work done by the gravitational force $(W_g)$ is given by $W_g = mgh = 10^{-3} \times 10 \times 1000 = 10\,J$.
(ii) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy: $W_g + W_r = \Delta K = \frac{1}{2}mv^2 - 0$.
Substituting the values: $10 + W_r = \frac{1}{2} \times 10^{-3} \times (50)^2 = \frac{1}{2} \times 10^{-3} \times 2500 = 1.25\,J$.
Therefore,$W_r = 1.25 - 10 = -8.75\,J$.

(D) For an elastic collision between two masses $m_1$ and $m_2$ where $m_2$ is initially at rest:
$1$. If $m_1 \gg m_2$,the final velocity of $m_2$ is $v_2 = \frac{2m_1}{m_1+m_2} u_1 \approx 2u_1$. Thus,option $(b)$ is incorrect as it claims $4u_1$.
$2$. In an oblique elastic collision where $m_1 = m_2$ and $m_2$ is at rest,the two balls move at an angle of $90^{\circ}$ to each other after the collision. Thus,option $(c)$ is incorrect as it claims they move in opposite directions.
$3$. Since both $(b)$ and $(c)$ are incorrect statements,option $(d)$ is the correct choice.