Perfectly Inelastic Collision Questions in English

(D) Step $1$: Apply the law of conservation of linear momentum during the collision.
Let $m_1 = 10 \text{ g} = 0.01 \text{ kg}$ be the mass of the bullet and $m_2 = 10 \text{ g} = 0.01 \text{ kg}$ be the mass of the block.
Initial velocity of the bullet $u_1 = 1 \text{ m/s}$ and initial velocity of the block $u_2 = 0 \text{ m/s}$.
Let $v$ be the common velocity of the system (bullet + block) immediately after the collision.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
$0.01 \times 1 + 0.01 \times 0 = (0.01 + 0.01) v$
$0.01 = 0.02 v$
$v = 0.5 \text{ m/s}$.
Step $2$: Apply the law of conservation of mechanical energy to find the height $h$.
The kinetic energy of the combined system is converted into potential energy at the maximum height $h$.
$\frac{1}{2} (m_1 + m_2) v^2 = (m_1 + m_2) g h$
$h = \frac{v^2}{2g}$
$h = \frac{(0.5)^2}{2 \times 10} = \frac{0.25}{20} = 0.0125 \text{ m}$.
Step $3$: Convert the height into centimeters.
$h = 0.0125 \text{ m} = 0.0125 \times 100 \text{ cm} = 1.25 \text{ cm}$.

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

(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 $\vec{v}'$ be the final velocity of the combined mass.
The initial momentum of mass $m$ is $\vec{p}_1 = m v \hat{i}$.
The initial momentum of mass $3m$ is $\vec{p}_2 = (3m)(2v) \hat{j} = 6mv \hat{j}$.
The total initial momentum is $\vec{p}_{initial} = m v \hat{i} + 6mv \hat{j}$.
The total mass after the collision is $M = m + 3m = 4m$.
The final momentum is $\vec{p}_{final} = (4m) \vec{v}'$.
Equating the two,we get:
$4m \vec{v}' = m v \hat{i} + 6mv \hat{j}$
$\vec{v}' = \frac{mv \hat{i} + 6mv \hat{j}}{4m}$
$\vec{v}' = \frac{1}{4}v \hat{i} + \frac{6}{4}v \hat{j} = \frac{1}{4}v \hat{i} + \frac{3}{2}v \hat{j}$.

(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 of the system = $m_1 \times 40 + m_2 \times 0 = 40m_1$.
Final momentum of the system = $(m_1 + m_2) \times 30 = 30m_1 + 30m_2$.
Equating the initial and final momentum:
$40m_1 = 30m_1 + 30m_2$
$40m_1 - 30m_1 = 30m_2$
$10m_1 = 30m_2$
$\frac{m_1}{m_2} = \frac{30}{10} = 3$.
Therefore,the ratio of their masses is $3$.

(D) Initial kinetic energy of the bullet is $K_i = \frac{1}{2}mv^2$.
After the perfectly inelastic collision,the bullet gets embedded in the sandbag,and the system moves with a common velocity $V$.
By the principle of conservation of linear momentum:
$mv + M(0) = (m + M)V$
$V = \frac{mv}{m + M}$
The final kinetic energy of the system is $K_f = \frac{1}{2}(m + M)V^2$.
Substituting the value of $V$:
$K_f = \frac{1}{2}(m + M) \left( \frac{mv}{m + M} \right)^2 = \frac{1}{2} \frac{m^2v^2}{m + M}$.
The loss in kinetic energy is $\Delta K = K_i - K_f$.
$\Delta K = \frac{1}{2}mv^2 - \frac{1}{2} \frac{m^2v^2}{m + M}$
$\Delta K = \frac{1}{2}mv^2 \left( 1 - \frac{m}{m + M} \right)$
$\Delta K = \frac{1}{2}mv^2 \left( \frac{m + M - m}{m + M} \right)$
$\Delta K = \frac{1}{2}mv^2 \left( \frac{M}{m + M} \right)$.

(B) Let the initial velocity of the particle with mass $2m$ be $u$. The particle with mass $m$ is at rest.
By the law of conservation of linear momentum, the final velocity $v$ of the combined mass $(2m + m = 3m)$ is:
$(2m)u + (m)(0) = (3m)v$
$v = \frac{2mu}{3m} = \frac{2u}{3}$
Initial kinetic energy $K_i = \frac{1}{2}(2m)u^2 = mu^2$.
Final kinetic energy $K_f = \frac{1}{2}(3m)v^2 = \frac{1}{2}(3m)\left(\frac{2u}{3}\right)^2 = \frac{3m}{2} \cdot \frac{4u^2}{9} = \frac{2}{3}mu^2$.
The loss in kinetic energy is $\Delta K = K_i - K_f = mu^2 - \frac{2}{3}mu^2 = \frac{1}{3}mu^2$.
The fraction of initial kinetic energy lost is $\frac{\Delta K}{K_i} = \frac{\frac{1}{3}mu^2}{mu^2} = \frac{1}{3}$.

(D) Given: $m_1 = 0.50 \ kg$,$u_1 = 2.00 \ m/s$,$m_2 = 1.00 \ kg$,$u_2 = 0 \ m/s$.
Since the bodies move together after the collision,it is a perfectly inelastic collision.
The formula for energy loss in a perfectly inelastic collision is given by:
$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$
Substituting the values:
$\Delta K = \frac{0.50 \times 1.00}{2(0.50 + 1.00)} (2.00 - 0)^2$
$\Delta K = \frac{0.50}{2(1.50)} (4)$
$\Delta K = \frac{0.50}{3.00} \times 4 = \frac{2}{3} \approx 0.67 \ J$.

(B) Initial momentum of the system in $x$-direction: $p_x = m(2v) = 2mv$.
Initial momentum of the system in $y$-direction: $p_y = (2m)v = 2mv$.
Total initial kinetic energy: $K_i = \frac{1}{2}m(2v)^2 + \frac{1}{2}(2m)v^2 = 2mv^2 + mv^2 = 3mv^2$.
Since the collision is perfectly inelastic,the particles stick together. Let the final velocity be $V_f$ at an angle $\theta$.
By conservation of linear momentum: $p_f = \sqrt{p_x^2 + p_y^2} = \sqrt{(2mv)^2 + (2mv)^2} = 2\sqrt{2}mv$.
Also,$p_f = (m + 2m)V_f = 3mV_f$.
Equating the two: $3mV_f = 2\sqrt{2}mv \Rightarrow V_f = \frac{2\sqrt{2}}{3}v$.
Final kinetic energy: $K_f = \frac{1}{2}(3m)V_f^2 = \frac{3m}{2} \left( \frac{8v^2}{9} \right) = \frac{4}{3}mv^2$.
Loss in energy: $\Delta K = K_i - K_f = 3mv^2 - \frac{4}{3}mv^2 = \frac{5}{3}mv^2$.
Percentage loss: $\frac{\Delta K}{K_i} \times 100 = \frac{(5/3)mv^2}{3mv^2} \times 100 = \frac{5}{9} \times 100 \approx 55.55\% \approx 56\%$.

(A) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum.
$mv + M(0) = (M + m)V_{\text{common}}$
$\therefore V_{\text{common}} = \frac{mv}{M + m}$
Now,the final kinetic energy $(KE_{\text{final}})$ of the combined system is given by:
$KE_{\text{final}} = \frac{1}{2}(M + m)V_{\text{common}}^2$
Substituting the value of $V_{\text{common}}$:
$KE_{\text{final}} = \frac{1}{2}(M + m) \left( \frac{mv}{M + m} \right)^2$
$KE_{\text{final}} = \frac{1}{2}(M + m) \frac{m^2v^2}{(M + m)^2}$
$KE_{\text{final}} = \frac{1}{2}mv^2 \times \left( \frac{m}{M + m} \right)$

(A) Step $1$: Apply the principle of conservation of linear momentum during the collision.
Since the bullet gets embedded in the block,the collision is perfectly inelastic.
Let $V'$ be the velocity of the combined system $(M+m)$ immediately after the collision.
$m V_0 = (M + m) V'$
$V' = \frac{m V_0}{M + m}$
Step $2$: Apply the law of conservation of energy for the system after the collision.
The kinetic energy of the system is converted into the potential energy of the spring at maximum compression $X_{\max}$.
$\frac{1}{2} (M + m) (V')^2 = \frac{1}{2} k X_{\max}^2$
$(M + m) \left( \frac{m V_0}{M + m} \right)^2 = k X_{\max}^2$
$\frac{m^2 V_0^2}{M + m} = k X_{\max}^2$
$X_{\max}^2 = \frac{m^2 V_0^2}{(M + m)k}$
$X_{\max} = \sqrt{\frac{m^2 V_0^2}{(M + m)k}} = \left( \frac{m^2 V_0^2}{(M + m)k} \right)^{1/2}$

(C) For a perfectly inelastic collision,the coefficient of restitution $e = 0$.
Using the law of conservation of linear momentum:
$M\mu_1 + M\mu_2 = (M + M)V$
$M(\mu_1 + \mu_2) = 2MV$
$V = \frac{\mu_1 + \mu_2}{2}$
Initial kinetic energy $(K_i)$:
$K_i = \frac{1}{2}M\mu_1^2 + \frac{1}{2}M\mu_2^2 = \frac{M}{2}(\mu_1^2 + \mu_2^2)$
Final kinetic energy $(K_f)$:
$K_f = \frac{1}{2}(2M)V^2 = M \left( \frac{\mu_1 + \mu_2}{2} \right)^2 = \frac{M}{4}(\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2)$
Loss in energy $(\Delta K = K_i - K_f)$:
$\Delta K = \frac{M}{2}(\mu_1^2 + \mu_2^2) - \frac{M}{4}(\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2)$
$\Delta K = \frac{M}{4} [2\mu_1^2 + 2\mu_2^2 - \mu_1^2 - \mu_2^2 - 2\mu_1\mu_2]$
$\Delta K = \frac{M}{4}(\mu_1^2 + \mu_2^2 - 2\mu_1\mu_2)$
$\Delta K = \frac{M}{4}(\mu_1 - \mu_2)^2$

(A) Let the mass of each particle be $M$ and their speed be $v$. After the collision,they form a single particle $C$ of mass $2M$ moving with velocity $v'$ at an angle $\theta$ with the $X$-axis.
Applying the law of conservation of linear momentum along the $X$ and $Y$ axes:
Along the $X$-axis:
$P_{ix} = P_{fx}$
$Mv \cos(60^{\circ}) - Mv \cos(45^{\circ}) = (2M)v' \cos \theta$
$v(\frac{1}{2} - \frac{1}{\sqrt{2}}) = 2v' \cos \theta \quad ... (i)$
Along the $Y$-axis:
$P_{iy} = P_{fy}$
$Mv \sin(60^{\circ}) + Mv \sin(45^{\circ}) = (2M)v' \sin \theta$
$v(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}) = 2v' \sin \theta \quad ... (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\tan \theta = \frac{v(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}})}{v(\frac{1}{2} - \frac{1}{\sqrt{2}})} = \frac{\sqrt{3} + \sqrt{2}}{1 - \sqrt{2}}$

(D) $1$. Velocity of the $1\,kg$ body just before hitting the platform: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 100} = \sqrt{2000} = 20\sqrt{5}\,m/s.$
$2$. Using the Conservation of Linear Momentum $(COLM)$ during the perfectly inelastic collision: $m_1 v = (m_1 + m_2) v',$
where $v'$ is the velocity of the combined mass $(1+3 = 4\,kg)$ immediately after the collision.
$1 \times 20\sqrt{5} = 4 \times v' \implies v' = 5\sqrt{5}\,m/s.$
$3$. Using the Conservation of Mechanical Energy for the system from the point of collision to the point of maximum compression $x$:
The initial kinetic energy of the combined mass is converted into the potential energy of the spring and the gravitational potential energy change.
Let the equilibrium position of the spring be the reference level. The platform is already compressed by $x_0 = \frac{m_2 g}{k} = \frac{3 \times 10}{1.25 \times 10^6} = 2.4 \times 10^{-5}\,m,$ which is negligible.
Applying energy conservation: $\frac{1}{2} M (v')^2 + M g x = \frac{1}{2} k x^2,$
where $M = 4\,kg.$
$\frac{1}{2} \times 4 \times (5\sqrt{5})^2 + 4 \times 10 \times x = \frac{1}{2} \times 1.25 \times 10^6 \times x^2.$
$2 \times 125 + 40x = 6.25 \times 10^5 x^2.$
$6.25 \times 10^5 x^2 - 40x - 250 = 0.$
Since $x$ is very small,$40x$ is negligible compared to $250$ and $6.25 \times 10^5 x^2.$
$6.25 \times 10^5 x^2 \approx 250 \implies x^2 \approx \frac{250}{6.25 \times 10^5} = 40 \times 10^{-5} = 4 \times 10^{-4}.$
$x = 0.02\,m = 2\,cm.$

(C) Initial kinetic energy $K_i = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (40)(4)^2 + \frac{1}{2} (60)(2)^2 = 320 + 120 = 440\,J$.
By the Law of Conservation of Linear Momentum,$m_1 v_1 + m_2 v_2 = (m_1 + m_2) V_c$.
$40(4) + 60(2) = (40 + 60) V_c \implies 160 + 120 = 100 V_c \implies 280 = 100 V_c \implies V_c = 2.8\,m/s$.
Final kinetic energy $K_f = \frac{1}{2} (m_1 + m_2) V_c^2 = \frac{1}{2} (100) (2.8)^2 = 50 \times 7.84 = 392\,J$.
Loss in kinetic energy $\Delta K = K_i - K_f = 440 - 392 = 48\,J$.
Alternatively,for a perfectly inelastic collision: $(\Delta K)_{\text{loss}} = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (v_1 - v_2)^2 = \frac{1}{2} \times \frac{40 \times 60}{100} \times (4 - 2)^2 = \frac{1}{2} \times 24 \times 4 = 48\,J$.

(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 the final speed of the combined system be $v_f$.
The initial momentum of the system is $p_i = mv + M(0) = mv$.
The final momentum of the system is $p_f = (M + m)v_f$.
Since $p_i = p_f$,we have:
$mv = (M + m)v_f$
Solving for $v_f$,we get:
$v_f = \frac{m}{M + m}v$

(B) Initial kinetic energy of the system: $K_i = \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 = \frac{1}{2} \times 2 \times (6)^2 + 0 = 36 \ J$.
By the law of conservation of linear momentum: $m u_1 + m u_2 = (m + m) v$.
$2 \times 6 + 2 \times 0 = (2 + 2) v \implies 12 = 4v \implies v = 3 \ m/s$.
Final kinetic energy of the system: $K_f = \frac{1}{2} (m + m) v^2 = \frac{1}{2} \times 4 \times (3)^2 = 2 \times 9 = 18 \ J$.
The heat evolved is equal to the loss in kinetic energy: $\Delta K = K_i - K_f = 36 \ J - 18 \ J = 18 \ J$.

(A) Let the mass of the first body be $m_1 = 5\,kg$ and its initial velocity be $u$. Let the mass of the second body be $m_2 = 2.5\,kg$ with initial velocity $u_2 = 0$.
By the law of conservation of linear momentum: $m_1 u + m_2 u_2 = (m_1 + m_2) v$,where $v$ is the final common velocity.
$5u + 0 = (5 + 2.5)v \Rightarrow 5u = 7.5v \Rightarrow v = \frac{5}{7.5}u = \frac{2}{3}u$.
The final kinetic energy of the combined mass is $K_f = \frac{1}{2}(m_1 + m_2)v^2 = 5\,J$.
Substituting the values: $\frac{1}{2} \times 7.5 \times (\frac{2}{3}u)^2 = 5$.
$3.75 \times \frac{4}{9}u^2 = 5 \Rightarrow \frac{15}{9}u^2 = 5 \Rightarrow \frac{5}{3}u^2 = 5 \Rightarrow u^2 = 3$.
The initial kinetic energy of the first body is $K_i = \frac{1}{2}m_1 u^2 = \frac{1}{2} \times 5 \times 3 = 7.5\,J$.

(A) Initial momentum of the system,$p_i = mv$.
Let $V$ be the velocity of the composite system after the collision.
Final momentum of the system,$p_f = (m+M)V$.
By the law of conservation of linear momentum,$p_i = p_f$,so $mv = (m+M)V$.
Therefore,the velocity of the composite block is $V = \frac{mv}{m+M}$.
The kinetic energy of the composite block is $K = \frac{1}{2}(m+M)V^2$.
Substituting the value of $V$,we get $K = \frac{1}{2}(m+M) \left( \frac{mv}{m+M} \right)^2$.
Simplifying this expression,$K = \frac{1}{2}(m+M) \frac{m^2v^2}{(m+M)^2} = \frac{1}{2}mv^2 \times \frac{m}{m+M}$.

(C) Given: $m_1 = m_2 = 0.25\, kg$,$u_1 = 3\, ms^{-1}$,$u_2 = -1\, ms^{-1}$.
Since the masses move towards each other,we take one velocity as positive and the other as negative.
This is a perfectly inelastic collision where the bodies stick together after the collision.
According to the law of conservation of linear momentum:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
Substituting the values:
$(0.25)(3) + (0.25)(-1) = (0.25 + 0.25) v$
$0.75 - 0.25 = 0.5 v$
$0.5 = 0.5 v$
$v = 1\, ms^{-1}$.

(B) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.
$m_1\vec{v}_1 + m_2\vec{v}_2 = (m_1 + m_2)\vec{v}_{final}$
Given $m_1 = 2 \, kg$,$\vec{v}_1 = (\hat{i} + \hat{j} + \hat{k}) \, m/s$ and $m_2 = 5 \, kg$,$\vec{v}_2 = (\hat{i} - 2\hat{j} + 3\hat{k}) \, m/s$.
Substituting the values:
$2(\hat{i} + \hat{j} + \hat{k}) + 5(\hat{i} - 2\hat{j} + 3\hat{k}) = (2 + 5)\vec{v}_{final}$
$(2\hat{i} + 2\hat{j} + 2\hat{k}) + (5\hat{i} - 10\hat{j} + 15\hat{k}) = 7\vec{v}_{final}$
$(7\hat{i} - 8\hat{j} + 17\hat{k}) = 7\vec{v}_{final}$
$\vec{v}_{final} = \frac{7}{7}\hat{i} - \frac{8}{7}\hat{j} + \frac{17}{7}\hat{k} = (\hat{i} - \frac{8}{7}\hat{j} + \frac{17}{7}\hat{k}) \, m/s$.

(B) According to the law of conservation of linear momentum,the total momentum before collision equals the total momentum after collision.
$mv + (2m)(0) = (m + 2m)v'$
$mv = 3mv'$
$v' = \frac{v}{3}$
The initial kinetic energy is $KE_i = \frac{1}{2}mv^2$.
The final kinetic energy is $KE_f = \frac{1}{2}(m + 2m)(v')^2 = \frac{1}{2}(3m)(\frac{v}{3})^2 = \frac{1}{2}(3m)(\frac{v^2}{9}) = \frac{1}{6}mv^2$.
The loss in kinetic energy is $\Delta KE = KE_i - KE_f = \frac{1}{2}mv^2 - \frac{1}{6}mv^2 = \frac{3-1}{6}mv^2 = \frac{2}{6}mv^2 = \frac{1}{3}mv^2$.
The percentage loss in kinetic energy is $\frac{\Delta KE}{KE_i} \times 100 = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2} \times 100 = \frac{2}{3} \times 100 = \frac{200}{3}\,\%$.

(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.
Let the mass of the first body be $m_1 = 2\, kg$ and its velocity be $u_1 = 3\, m/s$.
Let the mass of the second body be $m_2 = 1\, kg$ and its velocity be $u_2 = -4\, m/s$ (since it is moving in the opposite direction).
Let the common velocity after the collision be $v$.
Using the conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$.
Substituting the values: $(2 \times 3) + (1 \times -4) = (2 + 1) v$.
$6 - 4 = 3v$.
$2 = 3v$.
$v = 2/3\, m/s$.

(B) Initial momentum of the system: $\vec{P}_i = m(u\hat{i}) + m\left(\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right) = \frac{3}{2}mu\hat{i} + \frac{1}{2}mu\hat{j}$.
Since the collision is completely inelastic,the particles stick together and move with a common velocity $\vec{v}_f$. By conservation of linear momentum: $\vec{P}_i = (m+m)\vec{v}_f = 2m\vec{v}_f$.
$\vec{v}_f = \frac{1}{2m} \left(\frac{3}{2}mu\hat{i} + \frac{1}{2}mu\hat{j}\right) = \frac{3}{4}u\hat{i} + \frac{1}{4}u\hat{j}$.
Initial kinetic energy: $K_i = \frac{1}{2}mu^2 + \frac{1}{2}m\left|\frac{u}{2}\hat{i} + \frac{u}{2}\hat{j}\right|^2 = \frac{1}{2}mu^2 + \frac{1}{2}m\left(\frac{u^2}{4} + \frac{u^2}{4}\right) = \frac{1}{2}mu^2 + \frac{1}{4}mu^2 = \frac{3}{4}mu^2$.
Final kinetic energy: $K_f = \frac{1}{2}(2m)|\vec{v}_f|^2 = m\left(\left(\frac{3}{4}u\right)^2 + \left(\frac{1}{4}u\right)^2\right) = m\left(\frac{9}{16}u^2 + \frac{1}{16}u^2\right) = \frac{10}{16}mu^2 = \frac{5}{8}mu^2$.
Energy lost: $\Delta K = K_i - K_f = \frac{3}{4}mu^2 - \frac{5}{8}mu^2 = \frac{6-5}{8}mu^2 = \frac{1}{8}mu^2$.

(N/A) Consider a particle of mass $m_{1}$ moving with velocity $v_{1i}$ colliding with a stationary particle of mass $m_{2}$. After the collision,both particles move together in the direction of $v_{1i}$ with a common final velocity $v_{f}$.
From the law of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision:
$m_{1} v_{1i} + m_{2} v_{2i} = (m_{1} + m_{2}) v_{f}$
Since the second particle is initially at rest,$v_{2i} = 0$:
$m_{1} v_{1i} = (m_{1} + m_{2}) v_{f}$
$v_{f} = \frac{m_{1}}{m_{1} + m_{2}} v_{1i}$
Kinetic energy is not conserved in an inelastic collision. The loss in kinetic energy $\Delta K$ is given by:
$\Delta K = K_{i} - K_{f}$
$\Delta K = \frac{1}{2} m_{1} v_{1i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2}$
Substituting $v_{f}$:
$\Delta K = \frac{1}{2} m_{1} v_{1i}^{2} - \frac{1}{2} (m_{1} + m_{2}) \left( \frac{m_{1}}{m_{1} + m_{2}} \right)^{2} v_{1i}^{2}$
$\Delta K = \frac{1}{2} m_{1} v_{1i}^{2} \left( 1 - \frac{m_{1}}{m_{1} + m_{2}} \right)$
$\Delta K = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} v_{1i}^{2}$
Since $\Delta K > 0$,kinetic energy is dissipated during the collision.

(B) The total kinetic energy of a system of two bodies becomes zero after a collision if and only if the collision is perfectly inelastic and the total linear momentum of the system is zero.
If two bodies of masses $m_1$ and $m_2$ move with velocities $v_1$ and $v_2$ in opposite directions such that their total momentum $p = m_1v_1 + m_2v_2 = 0$,and they stick together after the collision,the final velocity $V$ of the combined mass will be $V = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = 0$.
Since the final velocity is zero,the final kinetic energy $K_f = \frac{1}{2}(m_1 + m_2)V^2 = 0$.

(B) Initial kinetic energy of the system is $K_i = 1/2 mv^2 + 1/2 (2m)(0)^2 = 1/2 mv^2$.
Since the collision is perfectly inelastic,the two bodies stick together after the collision and move with a common velocity $V$.
By the law of conservation of linear momentum: $mv + (2m)(0) = (m + 2m)V$.
$mv = 3mV$,which gives $V = v/3$.
The final kinetic energy of the system is $K_f = 1/2 (m + 2m)V^2 = 1/2 (3m)(v/3)^2 = 1/2 (3m)(v^2/9) = 1/6 mv^2$.
The loss in kinetic energy is $\Delta K = K_i - K_f = 1/2 mv^2 - 1/6 mv^2 = (3/6 - 1/6) mv^2 = 2/6 mv^2 = 1/3 mv^2$.

(B) Let the two bodies have mass $m$ and initial speed $v_0$. Let the angle between their initial velocity vectors be $2\theta$.
By symmetry,the combined mass $2m$ will move along the angle bisector of the initial velocity vectors with a final speed $v_f = v_0/2$.
Applying the law of conservation of linear momentum along the direction of the final velocity:
$m v_0 \cos \theta + m v_0 \cos \theta = (2m) v_f$
$2 m v_0 \cos \theta = 2 m (v_0/2)$
$2 m v_0 \cos \theta = m v_0$
$\cos \theta = 1/2$
$\theta = 60^\circ$
The angle between the initial velocities is $2\theta = 2 \times 60^\circ = 120^\circ$.

(A) Given: $m_1 = 5 \times 10^{3} \, kg$,$u_1 = 2 \, m/s$,$m_2 = 15 \times 10^{3} \, kg$,$u_2 = 0 \, m/s$.
Since the collision is perfectly inelastic $(e = 0)$,the bodies stick together.
The loss in kinetic energy $(\Delta K.E.)$ for a perfectly inelastic collision is given by:
$\Delta K.E. = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2$
Substituting the values:
$\Delta K.E. = \frac{1}{2} \times \frac{(5 \times 10^{3}) \times (15 \times 10^{3})}{5 \times 10^{3} + 15 \times 10^{3}} \times (2 - 0)^2$
$\Delta K.E. = \frac{1}{2} \times \frac{75 \times 10^{6}}{20 \times 10^{3}} \times 4$
$\Delta K.E. = \frac{1}{2} \times 3.75 \times 10^{3} \times 4$
$\Delta K.E. = 7.5 \times 10^{3} \, J = 7.5 \, kJ$.

(B) Initial momentum of the system: $P_i = m_b v_b + m_s v_s = 0.2 \, kg \times 10 \, ms^{-1} + 9.8 \, kg \times 0 = 2 \, kg \cdot ms^{-1}$.
Final momentum of the system: $P_f = (m_b + m_s) v = (0.2 + 9.8) \, kg \times v = 10 \, kg \times v$.
By the law of conservation of linear momentum: $P_i = P_f \implies 2 = 10v \implies v = 0.2 \, ms^{-1}$.
Initial kinetic energy: $K_i = \frac{1}{2} m_b v_b^2 = \frac{1}{2} \times 0.2 \times (10)^2 = 0.1 \times 100 = 10 \, J$.
Final kinetic energy: $K_f = \frac{1}{2} (m_b + m_s) v^2 = \frac{1}{2} \times 10 \times (0.2)^2 = 5 \times 0.04 = 0.2 \, J$.
Loss in kinetic energy: $\Delta K = K_i - K_f = 10 - 0.2 = 9.8 \, J$.

(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 $V$ be the final velocity of the combined system of mass $(m + m) = 2m$.
$m v + m(0) = (2m)V$
$m v = 2m V$
$V = v / 2$
Now,the kinetic energy of the system after the collision is given by:
$K_f = \frac{1}{2} (2m) V^2$
Substituting the value of $V$:
$K_f = \frac{1}{2} (2m) \left(\frac{v}{2}\right)^2$
$K_f = m \times \frac{v^2}{4} = \frac{m v^2}{4}$

(C) For a perfectly inelastic collision,the loss in kinetic energy $\Delta K$ is given by the formula:
$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$
Given:
$m_1 = 80 \,kg$,$u_1 = 2 \,ms^{-1}$
$m_2 = 20 \,kg$,$u_2 = 4 \,ms^{-1}$
Substituting the values:
$\Delta K = \frac{80 \times 20}{2(80 + 20)} (2 - 4)^2$
$\Delta K = \frac{1600}{2(100)} (-2)^2$
$\Delta K = \frac{1600}{200} \times 4$
$\Delta K = 8 \times 4 = 32 \,J$
Therefore,the loss of energy is $32 \,J$.

(B) According to the law of conservation of linear momentum,the total momentum before collision is equal to the total momentum after collision.
Let the direction towards the west be along the negative $x$-axis $(-\hat{i})$ and the direction towards the south be along the negative $y$-axis $(-\hat{j})$.
The initial momentum of the first particle is $\vec{p}_1 = m v(-\hat{i}) = -m v \hat{i}$.
The initial momentum of the second particle is $\vec{p}_2 = m v(-\hat{j}) = -m v \hat{j}$.
The total initial momentum is $\vec{P}_i = \vec{p}_1 + \vec{p}_2 = -m v \hat{i} - m v \hat{j}$.
After the collision,the particles stick together to form a single particle of mass $2m$ moving with velocity $\vec{V}$.
The final momentum is $\vec{P}_f = (2m) \vec{V}$.
By conservation of momentum,$\vec{P}_i = \vec{P}_f$,so $(2m) \vec{V} = -m v \hat{i} - m v \hat{j}$.
Dividing by $2m$,we get $\vec{V} = -\frac{v}{2} \hat{i} - \frac{v}{2} \hat{j}$.
The speed is the magnitude of the velocity vector: $|\vec{V}| = \sqrt{(-\frac{v}{2})^2 + (-\frac{v}{2})^2} = \sqrt{\frac{v^2}{4} + \frac{v^2}{4}} = \sqrt{\frac{v^2}{2}} = \frac{v}{\sqrt{2}}$.

(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 $m_1 = 10 \, kg$,$u_1 = 3 \, m/s$,$m_2 = 5 \, kg$,and $u_2 = 0 \, m/s$.
Let $V$ be the final velocity of the combined mass $(m_1 + m_2) = 15 \, kg$.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) V$
$10 \times 3 + 5 \times 0 = (10 + 5) V$
$30 = 15 V$
$V = 2 \, m/s$
Now,the kinetic energy $(KE)$ of the composite mass is given by:
$KE = \frac{1}{2} (m_1 + m_2) V^2$
$KE = \frac{1}{2} \times 15 \times (2)^2$
$KE = \frac{1}{2} \times 15 \times 4$
$KE = 30 \, J$.

(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.
Let $m_1 = 10 \,g = 0.01 \,kg$ be the mass of the bullet and $v_1 = 300 \,m/s$ be its initial velocity.
Let $m_2 = 5 \,kg$ be the mass of the ice block and $v_2 = 0 \,m/s$ be its initial velocity.
After the collision,the bullet gets embedded in the block,so the total mass is $M = m_1 + m_2 = 0.01 \,kg + 5 \,kg = 5.01 \,kg$.
Let $V$ be the final velocity of the system.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) V$
$(0.01 \,kg)(300 \,m/s) + (5 \,kg)(0 \,m/s) = (5.01 \,kg) V$
$3 = 5.01 V$
$V = \frac{3}{5.01} \approx 0.5988 \,m/s \approx 0.6 \,m/s$.
To convert this into $cm/s$,we multiply by $100$:
$V = 0.6 \times 100 = 60 \,cm/s$.

(C) Let the velocity of the first particle be $\vec{v}_1 = v \hat{i}$ and the velocity of the second particle be $\vec{v}_2 = v \hat{j}$.
According to the law of conservation of linear momentum,the total initial momentum must equal the total final momentum:
$\vec{P}_i = \vec{P}_f$
$m \vec{v}_1 + m \vec{v}_2 = (2m) \vec{V}$
$m(v \hat{i}) + m(v \hat{j}) = 2m \vec{V}$
$m v (\hat{i} + \hat{j}) = 2m \vec{V}$
$\vec{V} = \frac{v}{2} (\hat{i} + \hat{j})$
The magnitude of the final 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 vector $(\hat{i} + \hat{j})$,which is North-East.
Thus,the new particle moves with a velocity of $\frac{v}{\sqrt{2}}$ in the North-East direction.

(C) Let the two balls have mass $m$ and velocities $\vec{v}_1 = 6 \hat{i} \, m/s$ and $\vec{v}_2 = 8 \hat{j} \, m/s$.
Since the collision is perfectly inelastic,the two balls stick together and move with a common velocity $\vec{V}$.
According to the law of conservation of linear momentum:
$\vec{P}_{initial} = \vec{P}_{final}$
$m \vec{v}_1 + m \vec{v}_2 = (m + m) \vec{V}$
$m(6 \hat{i} + 8 \hat{j}) = 2m \vec{V}$
$\vec{V} = \frac{6 \hat{i} + 8 \hat{j}}{2} = 3 \hat{i} + 4 \hat{j} \, m/s$
The magnitude of the combined velocity is:
$V = |\vec{V}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, m/s$.

(D) Given:
Mass of ball $A$,$m_A = 2 \, kg$
Velocity of ball $A$,$u_A = 5 \, m/s$
Mass of ball $B$,$m_B = 5 \, kg$
Velocity of ball $B$,$u_B = -2 \, m/s$ (since it is moving in the opposite direction)
In a perfectly inelastic collision,the two bodies stick together after the collision and move with a common velocity $v$.
According to the law of conservation of linear momentum:
$m_A u_A + m_B u_B = (m_A + m_B) v$
Substituting the values:
$(2 \, kg)(5 \, m/s) + (5 \, kg)(-2 \, m/s) = (2 \, kg + 5 \, kg) v$
$10 - 10 = 7v$
$0 = 7v$
$v = 0 \, m/s$
Thus,the velocity of the balls after the collision is $0 \, m/s$.

(A) Initial kinetic energy of the system:
$K_i = \frac{1}{2} m (2v)^2 + \frac{1}{2} (2m) v^2 = 2mv^2 + mv^2 = 3mv^2$
Initial momentum of the system:
$\vec{p}_i = m(2v)\hat{i} + (2m)v\hat{j} = 2mv\hat{i} + 2mv\hat{j}$
Since the collision is perfectly inelastic,the particles stick together and move with a common velocity $\vec{V}$.
By the law of conservation of linear momentum:
$\vec{p}_f = \vec{p}_i$
$(m + 2m)\vec{V} = 2mv\hat{i} + 2mv\hat{j}$
$3m\vec{V} = 2mv(\hat{i} + \hat{j})$
$\vec{V} = \frac{2v}{3}(\hat{i} + \hat{j})$
Final kinetic energy of the system:
$K_f = \frac{1}{2} (m + 2m) |\vec{V}|^2 = \frac{1}{2} (3m) \left( \sqrt{(\frac{2v}{3})^2 + (\frac{2v}{3})^2} \right)^2$
$K_f = \frac{3m}{2} \left( \frac{4v^2}{9} + \frac{4v^2}{9} \right) = \frac{3m}{2} \left( \frac{8v^2}{9} \right) = \frac{4}{3} mv^2$
Loss in energy:
$\Delta K = K_i - K_f = 3mv^2 - \frac{4}{3} mv^2 = \frac{5}{3} mv^2$
Percentage loss in energy:
$\text{Percentage Loss} = \frac{\Delta K}{K_i} \times 100 = \frac{\frac{5}{3} mv^2}{3mv^2} \times 100 = \frac{5}{9} \times 100 \approx 55.55\% \approx 56\%$

(C) According to the law of conservation of linear momentum,the total initial momentum of the system must be equal to the total final momentum.
Initial momentum of the system: $P_i = m \cdot v + 2m \cdot 0 = mv$
After the collision,the two particles stick together,forming a combined mass of $(m + 2m) = 3m$.
Let the final velocity of the combined mass be $v'$.
Final momentum of the system: $P_f = (3m) \cdot v'$
Equating initial and final momentum: $mv = 3m \cdot v'$
Solving for $v'$: $v' = \frac{mv}{3m} = \frac{v}{3}$

(A) Let the mass of both bodies $A$ and $B$ be $m$.
Before the collision,body $A$ has velocity $v_1$ and body $B$ is at rest $(v_B = 0)$.
After the collision,since it is a completely inelastic collision,both bodies stick together and move with a common velocity $v_2$.
According to the law of conservation of linear momentum:
Initial momentum = Final momentum
$m v_1 + m(0) = (m + m) v_2$
$m v_1 = 2m v_2$
Dividing both sides by $m v_2$,we get:
$\frac{v_1}{v_2} = \frac{2}{1}$
Therefore,the ratio $v_1: v_2$ is $2: 1$.

(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 $m_1 = 2 \ kg$,$v_1 = 3 \ m/s$,$m_2 = 1 \ kg$,and $v_2 = -4 \ m/s$ (since it is moving in the opposite direction).
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$
Substituting the values:
$2(3) + 1(-4) = (2 + 1) v$
$6 - 4 = 3v$
$2 = 3v$
$v = \frac{2}{3} \ m/s$
Thus,the common velocity is $\frac{2}{3} \ m/s$.

(D) Initial kinetic energy of the system is $K_i = \frac{1}{2} m u^2$.
By the law of conservation of linear momentum,the final velocity $v$ of the combined mass $(m + 4m = 5m)$ is given by:
$m u + (4m)(0) = (5m) v$
$m u = 5m v$
$v = \frac{u}{5}$
Final kinetic energy of the system is $K_f = \frac{1}{2} (5m) v^2 = \frac{1}{2} (5m) \left(\frac{u}{5}\right)^2 = \frac{1}{2} (5m) \frac{u^2}{25} = \frac{1}{10} m u^2$.
The loss in kinetic energy is $\Delta K = K_i - K_f = \frac{1}{2} m u^2 - \frac{1}{10} m u^2 = \frac{5-1}{10} m u^2 = \frac{4}{10} m u^2 = \frac{2}{5} m u^2$.
The percentage loss in kinetic energy is $\frac{\Delta K}{K_i} \times 100 = \frac{\frac{2}{5} m u^2}{\frac{1}{2} m u^2} \times 100 = \frac{4}{5} \times 100 = 80 \%$.

(D) This is a case of a perfectly inelastic collision.
In any collision where external forces are absent,the total linear momentum of the system is conserved.
However,in an inelastic collision,some kinetic energy is converted into other forms of energy (such as heat or deformation energy) due to the internal forces acting during the impact.
Therefore,only the linear momentum is conserved,while the kinetic energy is not.

$(A)$ According to the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
Given $m_1 = 2 \,kg$, $u_1 = 10 \,ms^{-1}$, $m_2 = 3 \,kg$, and $u_2 = 0 \,ms^{-1}$.
$2 \times 10 + 3 \times 0 = (2 + 3)v$
$20 = 5v \implies v = 4 \,ms^{-1}$.
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 = $K_i - K_f = 100 \,J - 40 \,J = 60 \,J$.

(A) 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 the mass of each block be $m$.
The initial velocity of the first block is $3 \vec{V}$ and the second block is $0$.
After the collision,the two blocks stick together and move with a common velocity $V_{c}$.
Applying the conservation of momentum:
$m(3 \vec{V}) + m(0) = (m + m) V_{c}$
$3 m \vec{V} = 2 m V_{c}$
$V_{c} = \frac{3 m \vec{V}}{2 m} = \frac{3}{2} \vec{V}$

(D) Initial mass $m_1 = m$,initial velocity $u_1 = 3 \ m/s$.
Initial mass $m_2 = 2m$,initial velocity $u_2 = 0 \ m/s$.
According to the law of conservation of linear momentum,the total initial momentum equals the total final momentum.
$P_i = m_1 u_1 + m_2 u_2 = m(3) + 2m(0) = 3m$.
After collision,the bodies coalesce,so the final mass is $M = m_1 + m_2 = m + 2m = 3m$.
Let the final velocity be $V$.
$P_f = (m_1 + m_2)V = 3mV$.
Equating $P_i = P_f$:
$3m = 3mV$.
$V = 1 \ m/s$.

(A) According to the law of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.
Initial momentum of mass $M$: $\vec{p}_1 = M V \hat{i}$.
Initial momentum of mass $2M$: $\vec{p}_2 = 2M (3V \hat{j}) = 6MV \hat{j}$.
Total initial momentum: $\vec{p}_{initial} = M V \hat{i} + 6MV \hat{j}$.
After the collision,the masses stick together to form a single body of mass $3M$ moving with velocity $\vec{v}_{final}$.
Total final momentum: $\vec{p}_{final} = (M + 2M) \vec{v}_{final} = 3M \vec{v}_{final}$.
Equating the two: $3M \vec{v}_{final} = M V \hat{i} + 6MV \hat{j}$.
Dividing by $3M$: $\vec{v}_{final} = \frac{V}{3} \hat{i} + 2V \hat{j}$.

(B) Let the initial velocity of the body of mass $m$ be $v$. The initial kinetic energy is $K_i = \frac{1}{2}mv^2$.
By the law of conservation of linear momentum: $mv + (2m)(0) = (m + 2m)v'$,where $v'$ is the final common velocity.
$mv = 3mv' \implies v' = v/3$.
The final kinetic energy is $K_f = \frac{1}{2}(m + 2m)(v')^2 = \frac{1}{2}(3m)(v/3)^2 = \frac{1}{2}(3m)(v^2/9) = \frac{1}{6}mv^2$.
The loss in kinetic energy is $\Delta K = K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{6}mv^2 = \frac{3-1}{6}mv^2 = \frac{2}{6}mv^2 = \frac{1}{3}mv^2$.
The fraction of kinetic energy lost is $\frac{\Delta K}{K_i} = \frac{\frac{1}{3}mv^2}{\frac{1}{2}mv^2} = \frac{1}{3} \times 2 = 2/3$.

(A) For ball $A$: $m_A = 1 \,kg$, $u_A = 4 \,ms^{-1}$. For ball $B$: $m_B = 3 \,kg$, $u_B = 0 \,ms^{-1}$.
Total initial momentum $p_i = m_A u_A + m_B u_B = (1 \times 4) + (3 \times 0) = 4 \,kg \cdot ms^{-1}$.
After collision, both bodies stick together and move with a common velocity $v$.
Total final momentum $p_f = (m_A + m_B)v = (1 + 3)v = 4v$.
By the law of conservation of linear momentum, $p_i = p_f$, so $4 = 4v$, which gives $v = 1 \,ms^{-1}$.
The force exerted on ball $B$ is the rate of change of momentum of ball $B$:
$F_B = \frac{\Delta p_B}{\Delta t} = \frac{m_B(v - u_B)}{\Delta t} = \frac{3(1 - 0)}{0.1} = \frac{3}{0.1} = 30 \,N$.