Perfectly Inelastic Collision Questions in English

(A) Initial momentum of the particle = $m V_0$.
Final momentum of the system (particle + pendulum) = $(m + m)v = 2mv$.
By the law of conservation of momentum:
$m V_0 = 2mv$
$v = \frac{V_0}{2}$
Initial kinetic energy of the system = $\frac{1}{2}(2m)v^2 = \frac{1}{2}(2m)\left(\frac{V_0}{2}\right)^2 = \frac{1}{2}(2m)\frac{V_0^2}{4} = \frac{m V_0^2}{4}$.
If the system rises to a maximum height $h$,the potential energy gained is $P.E. = (2m)gh$.
By the law of conservation of energy,the initial kinetic energy is converted into potential energy at the maximum height:
$\frac{m V_0^2}{4} = 2mgh$
$h = \frac{m V_0^2}{4 \cdot 2mg} = \frac{V_0^2}{8g}$.

(C) Initial momentum of the system is given by:
$\vec{P}_i = mv\hat{i} + mv\hat{j}$
Magnitude of initial momentum:
$|\vec{P}_i| = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2}mv$
Final momentum of the system after coalescence is:
$\vec{P}_f = (2m)\vec{V}$
By the law of conservation of linear momentum,$\vec{P}_i = \vec{P}_f$:
$\sqrt{2}mv = 2mV$
Solving for $V$:
$V = \frac{\sqrt{2}mv}{2m} = \frac{v}{\sqrt{2}}$

(B) The coefficient of restitution $e$ is defined as the ratio of the relative velocity of separation to the relative velocity of approach of the colliding bodies.
In a perfectly inelastic collision,the two bodies stick together after the collision and move with a common velocity.
Since the bodies move together,the relative velocity of separation is $0$.
Therefore,the coefficient of restitution $e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{0}{v_{approach}} = 0$.

(C) In a perfectly inelastic collision,the two colliding bodies stick together and move with a common velocity after the impact.
For a perfectly inelastic collision,the coefficient of restitution $e = 0$.
In such a collision,the total linear momentum of the system remains conserved just before and just after the collision,but there is a maximum loss of kinetic energy during the process.

(B) Initially,the bullet moves with velocity $b$ and the block is at rest. After the collision,the bullet gets embedded in the block,and both move together with a common velocity $V$.
By the principle of conservation of linear momentum:
Initial momentum = Final momentum
$a \cdot b + c \cdot 0 = (a + c) \cdot V$
$a \cdot b = (a + c) \cdot V$
$V = \frac{a \cdot b}{a + c}$

(B) completely inelastic collision is defined as a collision in which the coefficient of restitution $(e)$ is equal to $0$.
The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
Since $e = 0$,the relative velocity of separation must be $0$.
This implies that the two particles move with the same velocity after the collision,meaning they remain together after the collision.

(C) When a bullet hits and gets embedded in a block,the collision is perfectly inelastic.
In any collision where no external horizontal force acts on the system,the total linear momentum of the system is conserved.
However,in a perfectly inelastic collision,kinetic energy is not conserved because some of the energy is dissipated as heat,sound,and deformation energy during the collision process.
Therefore,only momentum is conserved.

(C) According to the law of conservation of linear momentum,the total momentum of the system remains constant if no external force acts on it.
Initial momentum of the system = $m(v) + m(-v) = mv - mv = 0$.
Let the final velocity of the combined mass be $V$.
Since the bodies stick together,the final mass of the system is $m + m = 2m$.
Final momentum of the system = $(2m)V$.
By the law of conservation of momentum: Initial momentum = Final momentum.
$0 = 2mV$.
Therefore,$V = 0$.

(A) According to the law of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision.
Let $m$ be the mass of each body.
Initial momentum of the first body = $mv$.
Initial momentum of the second body = $m \times 0 = 0$.
Total initial momentum = $mv + 0 = mv$.
After the collision,the two bodies stick together to form a compound body of mass $M = m + m = 2m$.
Let $V$ be the velocity of the compound body after the collision.
Total final momentum = $(2m)V$.
By the law of conservation of momentum: $mv = 2mV$.
Solving for $V$,we get $V = \frac{mv}{2m} = \frac{v}{2}$.

(D) Initial momentum of the system = $mv$.
Final momentum of the system = $(m + M)V$,where $V$ is the common velocity of the system after collision.
By the law of conservation of linear momentum,$mv = (m + M)V$.
Therefore,the common velocity $V = \frac{mv}{M + m}$.
The kinetic energy of the combined system is $K = \frac{1}{2}(m + M)V^2$.
Substituting the value of $V$,we get $K = \frac{1}{2}(m + M) \left( \frac{mv}{M + m} \right)^2$.
$K = \frac{1}{2}(m + M) \frac{m^2v^2}{(M + m)^2} = \frac{m^2v^2}{2(M + m)}$.
Thus,option $D$ is correct.

(B) Initial kinetic energy of the system $(K_i)$ = Kinetic energy of the bullet = $\frac{1}{2} m_B v_B^2 = \frac{1}{2} \times 0.05 \, kg \times (10 \, m/s)^2 = 2.5 \, J$.
By the law of conservation of linear momentum,$m_B v_B = (m_B + M) v_{sys}$.
$0.05 \times 10 = (0.05 + 0.95) \times v_{sys} \implies 0.5 = 1 \times v_{sys} \implies v_{sys} = 0.5 \, m/s$.
Final kinetic energy of the system $(K_f)$ = $\frac{1}{2} (m_B + M) v_{sys}^2 = \frac{1}{2} \times 1 \, kg \times (0.5 \, m/s)^2 = 0.125 \, J$.
Loss in kinetic energy = $K_i - K_f = 2.5 - 0.125 = 2.375 \, J$.
Percentage loss = $\frac{K_i - K_f}{K_i} \times 100 = \frac{2.375}{2.5} \times 100 = 95\%$.

(B) Let the two balls have mass $m$ and move along the $x$ and $y$ axes respectively.
Initial velocity of ball $1$: $\vec{v}_1 = 45\sqrt{2} \hat{i} \, m/s$.
Initial velocity of ball $2$: $\vec{v}_2 = 45\sqrt{2} \hat{j} \, m/s$.
Initial total momentum: $\vec{P}_i = m\vec{v}_1 + m\vec{v}_2 = m(45\sqrt{2} \hat{i} + 45\sqrt{2} \hat{j})$.
The magnitude of initial momentum is: $|\vec{P}_i| = m \sqrt{(45\sqrt{2})^2 + (45\sqrt{2})^2} = m \sqrt{4050 + 4050} = m \sqrt{8100} = 90m \, kg \cdot m/s$.
After collision,the balls stick together to form a mass $2m$ moving with velocity $\vec{V}$.
By the law of conservation of linear momentum: $\vec{P}_i = \vec{P}_f$.
$90m = (2m)V$.
$V = \frac{90m}{2m} = 45 \, m/s$.

(C) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Initial momentum = $m \times v + 2m \times 0 = mv$.
Final momentum = $(m + 2m) \times V = 3mV$,where $V$ is the final velocity of the combined system.
Equating initial and final momentum: $mv = 3mV$.
Therefore,the speed of the system is $V = v/3$.

(C) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Initial momentum = $(m \times 3) + (2m \times 0) = 3m \, kg \cdot km/h$.
Final mass of the combined body = $m + 2m = 3m$.
Let the final velocity of the combined mass be $V$.
Final momentum = $(3m) \times V$.
Equating initial and final momentum:
$3m = 3m \times V$
$V = \frac{3m}{3m} = 1 \, km/h$.
Therefore,the combined velocity is $1 \, km/h$.

(A) When the ball hits the sand and embeds itself,it undergoes a perfectly inelastic collision with the sand-earth system.
In any collision,the total linear momentum of the system (ball + earth) remains conserved,provided no external impulsive force acts on the system.
However,kinetic energy is not conserved in an inelastic collision because a significant portion of the energy is dissipated as heat,sound,and work done against the resistive force of the sand.
Therefore,only momentum remains conserved.

(B) Initial velocity of the first ball,$u_1 = 36 \,km/h = 36 \times \frac{5}{18} = 10 \,m/s$.
Initial velocity of the second ball,$u_2 = 0 \,m/s$.
Masses are $m_1 = 2 \,kg$ and $m_2 = 3 \,kg$.
By the law of conservation of linear momentum,$m_1 u_1 + m_2 u_2 = (m_1 + m_2)V$,where $V$ is the common velocity after the collision.
$2 \times 10 + 3 \times 0 = (2 + 3)V \implies 20 = 5V \implies V = 4 \,m/s$.
Initial kinetic energy,$K_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} \times 2 \times (10)^2 + 0 = 100 \,J$.
Final kinetic energy,$K_f = \frac{1}{2} (m_1 + m_2) V^2 = \frac{1}{2} \times 5 \times (4)^2 = \frac{1}{2} \times 5 \times 16 = 40 \,J$.
Loss in kinetic energy,$\Delta K = K_i - K_f = 100 - 40 = 60 \,J$.

(D) Let the mass of each body be $m = 2 \, kg$ and the velocity be $v = 10 \, m/s$.
Initial momentum of the first body: $\vec{P}_1 = mv \hat{i} = 2 \times 10 \hat{i} = 20 \hat{i} \, kg \cdot m/s$.
Initial momentum of the second body: $\vec{P}_2 = mv \hat{j} = 2 \times 10 \hat{j} = 20 \hat{j} \, kg \cdot m/s$.
Total initial momentum: $\vec{P}_{total} = \vec{P}_1 + \vec{P}_2 = 20 \hat{i} + 20 \hat{j}$.
Magnitude of total initial momentum: $|\vec{P}_{total}| = \sqrt{20^2 + 20^2} = 20\sqrt{2} \, kg \cdot m/s$.
After collision,the bodies coalesce into a single mass $M = m + m = 4 \, kg$.
Let the final velocity be $V$. By the law of conservation of momentum: $\vec{P}_{final} = \vec{P}_{total}$.
$MV = |\vec{P}_{total}| \implies 4V = 20\sqrt{2}$.
$V = \frac{20\sqrt{2}}{4} = 5\sqrt{2} \, m/s$.

(B) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let $m_1 = 20\, kg$,$v_1 = 10\, m/s$,$m_2 = 5\, kg$,and $v_2 = 0\, m/s$.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{sys}$
$20 \times 10 + 5 \times 0 = (20 + 5) v_{sys}$
$200 = 25 v_{sys}$
$v_{sys} = \frac{200}{25} = 8\, m/s$
The kinetic energy of the composite mass is given by $K.E. = \frac{1}{2} (m_1 + m_2) v_{sys}^2$.
$K.E. = \frac{1}{2} \times (20 + 5) \times (8)^2$
$K.E. = \frac{1}{2} \times 25 \times 64$
$K.E. = 25 \times 32 = 800\, J$.

(C) According to the law of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.
Let $m_1 = 1.67 \times 10^{-27} \ kg$ be the mass of the neutron and $v_1 = 10^8 \ m/s$ be its velocity.
Let $m_2 = 3.34 \times 10^{-27} \ kg$ be the mass of the deuteron,which is initially at rest $(v_2 = 0)$.
Since the particles stick together,they move with a common velocity $v$.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v$
$(1.67 \times 10^{-27}) \times 10^8 + 0 = (1.67 \times 10^{-27} + 3.34 \times 10^{-27}) \times v$
$1.67 \times 10^{-19} = (5.01 \times 10^{-27}) \times v$
$v = \frac{1.67 \times 10^{-19}}{5.01 \times 10^{-27}}$
$v = \frac{1}{3} \times 10^8 \ m/s = 3.33 \times 10^7 \ m/s$.

(C) For a perfectly inelastic collision, the loss in kinetic energy $(\Delta K)$ is given by the formula:
$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_1 - u_2)^2$
Given:
$m_1 = 40\,kg$, $u_1 = 4\,m/s$
$m_2 = 60\,kg$, $u_2 = 2\,m/s$
Substituting the values into the formula:
$\Delta K = \frac{1}{2} \left( \frac{40 \times 60}{40 + 60} \right) (4 - 2)^2$
$\Delta K = \frac{1}{2} \left( \frac{2400}{100} \right) (2)^2$
$\Delta K = \frac{1}{2} \times 24 \times 4$
$\Delta K = 12 \times 4 = 48\,J$
Therefore, the loss in kinetic energy is $48\,J$.

(B) According to the law of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision.
Let $v$ be the final velocity of the combined system.
Initial momentum = $m_1 V + m_2 \times 0 = m_1 V$.
Final momentum = $(m_1 + m_2)v$.
Equating the two: $m_1 V = (m_1 + m_2)v$.
Therefore,$v = \frac{m_1}{m_1 + m_2} V$.
Since $\frac{m_1}{m_1 + m_2} < 1$,the final velocity $v$ is less than the initial velocity $V$ but greater than zero (assuming $m_1, m_2 > 0$ and $V > 0$).

(A) By the law of conservation of linear momentum,the initial momentum of the system equals the final momentum of the system.
$mv + M(0) = (m + M)V$
Where $V$ is the velocity of the composite block after the collision.
$V = \frac{mv}{m + M}$
The kinetic energy $(KE)$ of the composite block is given by $KE = \frac{1}{2}(m + M)V^2$.
Substituting the value of $V$:
$KE = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2$
$KE = \frac{1}{2}(m + M) \frac{m^2 v^2}{(m + M)^2}$
$KE = \frac{1}{2} m v^2 \left( \frac{m}{m + M} \right)$

(C) Initial kinetic energy of the system $(K_i)$ = $\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} \times 4 \times (12)^2 + 0 = 288\, J$.
Since the bodies stick together,it is a perfectly inelastic collision.
By the law of conservation of linear momentum,$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$.
$4 \times 12 + 6 \times 0 = (4 + 6) v \implies 48 = 10v \implies v = 4.8\, m/s$.
Final kinetic energy of the system $(K_f)$ = $\frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 10 \times (4.8)^2 = 5 \times 23.04 = 115.2\, J$.
Loss in kinetic energy = $K_i - K_f = 288 - 115.2 = 172.8\, J$.

(D) In a perfectly inelastic collision,the colliding bodies stick together and move with a common velocity after the impact.
$A$. $A$ bullet embedded in a block means they move together. This is a perfectly inelastic collision.
$B$. The capture of an electron by an atom results in the electron becoming part of the atomic system. This is a perfectly inelastic collision.
$C$. $A$ man jumping onto a moving boat results in the man moving with the boat. This is a perfectly inelastic collision.
$D$. $A$ ball bearing striking another ball bearing usually results in them bouncing off each other,which is an elastic or partially inelastic collision,not a perfectly inelastic one.
Therefore,the correct option is $D$.

(C) $1$. In this scenario,the bullet and the block form an isolated system because there are no external horizontal forces acting on them (the table is frictionless).
$2$. According to the law of conservation of linear momentum,the total momentum of an isolated system remains constant.
$3$. Since the bullet gets embedded in the block,this is a perfectly inelastic collision.
$4$. In a perfectly inelastic collision,kinetic energy is not conserved; it is partially converted into heat,sound,and deformation energy.
$5$. Therefore,only the linear momentum is conserved.

(A) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Initial momentum $P_i = m_1 \times 5 + m_2 \times 0 = 5m_1$.
Final momentum $P_f = (m_1 + m_2) \times 4 = 4m_1 + 4m_2$.
Equating $P_i$ and $P_f$:
$5m_1 = 4m_1 + 4m_2$.
$5m_1 - 4m_1 = 4m_2$.
$m_1 = 4m_2$.
Therefore,the ratio $\frac{m_1}{m_2} = \frac{4}{1}$.

(B) perfectly inelastic collision is defined as a collision in which the two colliding bodies stick together after the impact.
In such a collision,the kinetic energy is not conserved,but the total linear momentum of the system remains conserved.
Since the bodies move with a common velocity after the collision,the final kinetic energy is less than the initial kinetic energy.
Therefore,the correct description is that the particles stick together and move as a single unit.

(A) In a perfectly inelastic collision,the two particles stick together after the collision and move with a common velocity $v_f$.
By the law of conservation of linear momentum: $m_1 u_1 + m_2(0) = (m_1 + m_2)v_f$.
Thus,$v_f = \frac{m_1 u_1}{m_1 + m_2}$.
The initial kinetic energy is $K_i = \frac{1}{2} m_1 u_1^2$.
The final kinetic energy is $K_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} (m_1 + m_2) \left( \frac{m_1 u_1}{m_1 + m_2} \right)^2 = \frac{1}{2} \frac{m_1^2 u_1^2}{m_1 + m_2}$.
The loss in kinetic energy (which is converted to heat) is $\Delta K = K_i - K_f = \frac{1}{2} m_1 u_1^2 - \frac{1}{2} \frac{m_1^2 u_1^2}{m_1 + m_2}$.
$\Delta K = \frac{1}{2} m_1 u_1^2 \left( 1 - \frac{m_1}{m_1 + m_2} \right) = \frac{1}{2} m_1 u_1^2 \left( \frac{m_2}{m_1 + m_2} \right)$.
The fraction of initial kinetic energy converted to heat is $\frac{\Delta K}{K_i} = \frac{m_2}{m_1 + m_2}$.

(A) This is a case of a perfectly inelastic collision where the bullet gets embedded in the sandbag on the truck.
According to the law of conservation of linear momentum, the total momentum before the collision must equal the total momentum after the collision.
Let $V$ be the final velocity of the truck (with the bullet and sandbag) after the collision.
Initial momentum of the system = $m \cdot u + M \cdot 0 = \mu$.
Final momentum of the system = $(M + m) \cdot V$.
Equating the two: $\mu = (M + m)V$.
Solving for $V$: $V = \frac{\mu}{M + m}$.

(B) Statement $1$ is true: In a perfectly inelastic collision,the particles stick together after the collision. While kinetic energy is lost (converted into heat,sound,or deformation energy),the total energy of the system remains conserved. Since the particles continue to move with a common velocity,the system retains some kinetic energy,so the total energy is not completely lost.
Statement $2$ is true: The law of conservation of linear momentum is a fundamental principle derived from Newton's laws and is applicable to all types of collisions (elastic,inelastic,or perfectly inelastic) provided no external force acts on the system.
Conclusion: Both statements are true,but Statement $2$ does not explain why the energy is not completely lost in Statement $1$.

(D) In a perfectly inelastic collision,the particles stick together and move with a common velocity $V_f = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$.
Since the system retains a common velocity after the collision,the final kinetic energy $K_f = \frac{1}{2}(m_1 + m_2)V_f^2$ is not zero.
Thus,Statement-$1$ is true because the system does not lose all its energy.
Statement-$2$ is also true because the law of conservation of linear momentum is a fundamental principle that applies to all collisions in the absence of external forces.
However,the conservation of momentum (Statement-$2$) is not the reason why some kinetic energy is retained (Statement-$1$); the retention of energy is due to the requirement of momentum conservation in a system that must move together.
Therefore,Statement-$1$ and Statement-$2$ are both true,but Statement-$2$ is not the correct explanation of Statement-$1$.

(D) According to the law of conservation of linear momentum,the initial momentum of the system equals the final momentum.
$mv + M(0) = (m + M)v'$
$v' = \frac{mv}{m + M}$
where $v'$ is the common velocity of the system after the collision.
The kinetic energy of the (bag + bullet) system after the collision is given by:
$K = \frac{1}{2}(m + M)(v')^2$
$K = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2$
$K = \frac{1}{2}(m + M) \frac{m^2v^2}{(m + M)^2}$
$K = \frac{m^2v^2}{2(m + M)}$

(C) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ m/s$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ m/s$.
By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$.
$(0.50)(2.00) + (1.00)(0) = (0.50 + 1.00)v$.
$1.00 = 1.50v \implies v = \frac{1.00}{1.50} = \frac{2}{3} \ m/s$.
Initial kinetic energy $K_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (0.50) (2.00)^2 = 1.00 \ J$.
Final kinetic energy $K_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (1.50) (\frac{2}{3})^2 = 0.75 \times \frac{4}{9} = 0.333... \ J$.
Energy loss $\Delta K = K_i - K_f = 1.00 - 0.333 = 0.667 \ J \approx 0.67 \ J$.

(C) The loss in kinetic energy during a perfectly inelastic collision is given by the formula: $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_1 - u_2)^2$.
Given: $m_1 = 40 \ kg$,$u_1 = 4 \ m/s$,$m_2 = 60 \ kg$,$u_2 = 2 \ m/s$.
Substituting the values: $\Delta K = \frac{1}{2} \left( \frac{40 \times 60}{40 + 60} \right) (4 - 2)^2$.
$\Delta K = \frac{1}{2} \left( \frac{2400}{100} \right) (2)^2$.
$\Delta K = \frac{1}{2} \times 24 \times 4$.
$\Delta K = 12 \times 4 = 48 \ J$.

(B) According to the law of conservation of linear momentum,$m_1v_1 + m_2v_2 = (m_1 + m_2)v$.
Given $m_1 = 20 \ kg$,$v_1 = 10 \ m/s$,$m_2 = 5 \ kg$,$v_2 = 0 \ m/s$.
Substituting the values: $20 \times 10 + 5 \times 0 = (20 + 5)v$.
$200 = 25v$,which gives $v = 8 \ m/s$.
The combined kinetic energy $K = \frac{1}{2}(m_1 + m_2)v^2$.
$K = \frac{1}{2}(20 + 5) \times (8)^2 = \frac{1}{2} \times 25 \times 64 = 25 \times 32 = 800 \ J$.

(D) Let the mass of each ball be $m$. When the first ball is released from a vertical height $h$,its velocity $v$ just before the collision is given by the conservation of energy: $v = \sqrt{2gh}$.
Since the collision is perfectly inelastic,the two balls stick together and move with a common velocity $V$ after the collision. Applying the law of conservation of linear momentum:
$m \cdot v + m \cdot 0 = (m + m) \cdot V$
$m \sqrt{2gh} = 2m \cdot V$
$V = \frac{\sqrt{2gh}}{2} = \sqrt{\frac{2gh}{4}}$
Let the combined system rise to a new vertical height $h'$. Using the conservation of energy for the combined system:
$\frac{1}{2} (2m) V^2 = (2m) g h'$
$V^2 = 2g h'$
Substituting $V^2 = \frac{2gh}{4}$:
$\frac{2gh}{4} = 2g h'$
$h' = \frac{h}{4}$

(A) Using the law of conservation of linear momentum,the initial momentum is $\vec{p}_i = m v \hat{i} + m v \hat{j}$.
After the collision,the particles stick together to form a mass of $2m$ moving with velocity $\vec{v}'$.
By conservation of momentum: $m v \hat{i} + m v \hat{j} = (2m) \vec{v}'$.
Dividing by $2m$,we get $\vec{v}' = \frac{v}{2} \hat{i} + \frac{v}{2} \hat{j}$.
The magnitude of the velocity is $v' = \sqrt{(\frac{v}{2})^2 + (\frac{v}{2})^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}$.
The direction is along the resultant of $\hat{i}$ and $\hat{j}$,which is the North-East direction.

(C) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.
Let $m_1 = 1.67 \times 10^{-27} \ kg$ be the mass of the neutron and $v_1 = 10^8 \ m/s$ be its velocity.
Let $m_2 = 3.34 \times 10^{-27} \ kg$ be the mass of the deuteron,which is initially at rest $(v_2 = 0)$.
Let $v$ be the final velocity of the combined system of mass $(m_1 + m_2)$.
Applying the conservation of momentum:
$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v$
$(1.67 \times 10^{-27}) \times 10^8 + (3.34 \times 10^{-27}) \times 0 = (1.67 \times 10^{-27} + 3.34 \times 10^{-27})v$
$1.67 \times 10^{-19} = (5.01 \times 10^{-27})v$
$v = \frac{1.67 \times 10^{-19}}{5.01 \times 10^{-27}}$
$v = 0.333 \times 10^8 \ m/s = 3.33 \times 10^7 \ m/s$.

(B) Given: $m_1 = 2 \ kg$,$u_1 = 36 \ km/hr = 36 \times \frac{5}{18} = 10 \ m/s$,$m_2 = 3 \ kg$,$u_2 = 0 \ m/s$.
According to the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$.
$(2)(10) + (3)(0) = (2 + 3)v \implies 20 = 5v \implies v = 4 \ m/s$.
Initial kinetic energy $K_i = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2 \times (10)^2 = 100 \ J$.
Final kinetic energy $K_f = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times (2 + 3) \times (4)^2 = \frac{1}{2} \times 5 \times 16 = 40 \ J$.
Loss in kinetic energy $\Delta K = K_i - K_f = 100 \ J - 40 \ J = 60 \ J$.

(C) For a perfectly inelastic collision,the loss in kinetic energy $\Delta K$ is given by the formula: $\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2$.
Given: $m_1 = 40 \ kg$,$u_1 = 4 \ m/s$,$m_2 = 60 \ kg$,$u_2 = 2 \ m/s$.
Substituting the values: $\Delta K = \frac{1}{2} \times \frac{40 \times 60}{40 + 60} \times (4 - 2)^2$.
$\Delta K = \frac{1}{2} \times \frac{2400}{100} \times (2)^2$.
$\Delta K = \frac{1}{2} \times 24 \times 4$.
$\Delta K = 12 \times 4 = 48 \ J$.

(A) Let the two cars have mass $m$ and velocity $v$. They collide at an angle $\theta$ between their paths.
By the principle of conservation of linear momentum,the component of momentum along the direction of the combined motion must be conserved.
The initial momentum of each car along the direction of the resultant motion is $mv \cos(\theta/2)$.
Since there are two cars,the total initial momentum is $P_i = mv \cos(\theta/2) + mv \cos(\theta/2) = 2mv \cos(\theta/2)$.
After the collision,the two cars stick together and move with a common velocity $V$. The total mass is $2m$.
The final momentum is $P_f = (2m)V$.
Equating initial and final momentum: $2mV = 2mv \cos(\theta/2)$.
Therefore,the final velocity is $V = v \cos(\theta/2)$.

(B) Given: $m_1 = 2 \ kg$,$u_1 = 36 \ km/h = 36 \times \frac{5}{18} = 10 \ m/s$,$m_2 = 3 \ kg$,$u_2 = 0 \ m/s$.
For a perfectly inelastic collision,the loss in kinetic energy is given by the formula:
$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_1 - u_2)^2$
Substituting the values:
$\Delta K = \frac{1}{2} \left( \frac{2 \times 3}{2 + 3} \right) (10 - 0)^2$
$\Delta K = \frac{1}{2} \times \frac{6}{5} \times 100$
$\Delta K = 0.6 \times 100 = 60 \ J$.

(B) According to the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2)V$
Given: $m_1 = 20 \ kg$,$u_1 = 10 \ m/s$,$m_2 = 5 \ kg$,$u_2 = 0 \ m/s$.
Substituting the values:
$20 \times 10 + 5 \times 0 = (20 + 5)V$
$200 = 25V$
$V = \frac{200}{25} = 8 \ m/s$
The combined kinetic energy $K$ is given by:
$K = \frac{1}{2}(m_1 + m_2)V^2$
$K = \frac{1}{2}(20 + 5) \times (8)^2$
$K = \frac{1}{2} \times 25 \times 64$
$K = 25 \times 32 = 800 \ J$.

(C) According to the law of conservation of linear momentum,the initial momentum of the system must equal the final momentum.
Initial momentum in the East direction: $P_x = mv$.
Initial momentum in the North direction: $P_y = mv$.
The magnitude of the total initial momentum is $P_i = \sqrt{P_x^2 + P_y^2} = \sqrt{(mv)^2 + (mv)^2} = \sqrt{2}mv$.
The final momentum of the combined body of mass $2m$ moving with velocity $V$ is $P_f = 2mV$.
Equating initial and final momentum: $\sqrt{2}mv = 2mV$.
Solving for $V$: $V = \frac{\sqrt{2}mv}{2m} = \frac{v}{\sqrt{2}}$.

(C) According to the law of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let $V$ be the combined velocity of the system after the collision.
Initial momentum = $m \times v + 2m \times 0 = mv$.
Final momentum = $(m + 2m) \times V = 3mV$.
Equating the two: $mv = 3mV$.
Therefore,$V = \frac{v}{3}$.

(B) According to the principle of conservation of linear momentum,the total momentum before the collision is equal to the total momentum after the collision.
Let $V$ be the final velocity of the block-bullet system.
Initial momentum of the system = (mass of bullet $\times$ velocity of bullet) + (mass of block $\times$ velocity of block) = $a \times b + c \times 0 = ab$.
Final momentum of the system = (mass of bullet + mass of block) $\times$ final velocity = $(a + c)V$.
Equating the initial and final momentum: $ab = (a + c)V$.
Therefore,the velocity of the block is $V = \frac{ab}{a + c}$.

(D) According to the law of conservation of linear momentum for a perfectly inelastic collision:
$m_1\vec{u}_1 + m_2\vec{u}_2 = (m_1 + m_2)\vec{V}$
Given:
$m_1 = 1 \ g, \vec{u}_1 = 3\hat{i} - 2\hat{j} \ m/s$
$m_2 = 2 \ g, \vec{u}_2 = 4\hat{j} - 6\hat{k} \ m/s$
Substituting the values:
$\vec{V} = \frac{m_1\vec{u}_1 + m_2\vec{u}_2}{m_1 + m_2} = \frac{1(3\hat{i} - 2\hat{j}) + 2(4\hat{j} - 6\hat{k})}{1 + 2}$
$\vec{V} = \frac{3\hat{i} - 2\hat{j} + 8\hat{j} - 12\hat{k}}{3} = \frac{3\hat{i} + 6\hat{j} - 12\hat{k}}{3}$
$\vec{V} = 1\hat{i} + 2\hat{j} - 4\hat{k} \ m/s$
The magnitude of the combined velocity is:
$|\vec{V}| = \sqrt{(1)^2 + (2)^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \approx 4.58 \ m/s \approx 4.6 \ m/s$.

(A) According to the law of conservation of linear momentum,the initial momentum of the bullet equals the final momentum of the combined system:
$mv = (m + M)V$
where $V$ is the final velocity of the combined system.
$V = \frac{mv}{m + M}$
The kinetic energy $K$ of the combined system is given by:
$K = \frac{1}{2}(m + M)V^2$
Substituting the value of $V$:
$K = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2$
$K = \frac{1}{2}(m + M) \frac{m^2v^2}{(m + M)^2}$
$K = \frac{1}{2} \frac{m^2v^2}{m + M}$
$K = \left( \frac{1}{2}mv^2 \right) \left( \frac{m}{m + M} \right)$