Perfectly Inelastic Collision Questions in English

(C) In a perfectly inelastic collision,the two bodies stick together after the impact and move with a common final velocity.
Let the velocities of the two bodies after the collision be $v_1$ and $v_2$. Since they move together,$v_1 = v_2 = v$.
The relative velocity of the bodies after the impact is defined as the difference between their velocities:
$v_{\text{rel}} = v_1 - v_2$
Substituting $v_1 = v_2 = v$ into the equation:
$v_{\text{rel}} = v - v = 0$
Therefore,the relative velocity of the bodies after the impact is zero.

(A) Given: Mass $m_1 = 3 \,kg$, Velocity $v_1 = 8 \,ms^{-1}$.
Mass $m_2 = 1 \,kg$, Velocity $v_2 = -4 \,ms^{-1}$ (since it is moving in the opposite direction).
According to the law of conservation of linear momentum, the total momentum before collision equals the total momentum after collision.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$
Substituting the values:
$(3 \times 8) + (1 \times -4) = (3 + 1) \times v$
$24 - 4 = 4v$
$20 = 4v$
$v = 5 \,ms^{-1}$
Thus, the common velocity is $5 \,ms^{-1}$.

(B) When a bullet strikes a wooden block and gets embedded in it,the two bodies move together with a common velocity after the collision.
This type of collision,where the bodies stick together after impact,is defined as a perfectly inelastic collision.
In this process,the loss of kinetic energy is maximum.

(D) Mass of the bullet $= m$. Speed of the bullet $= v$.
According to the problem,the bullet gets embedded into the bag,so they move together with a common velocity $v_1$.
This is a case of a perfectly inelastic collision.
Initial kinetic energy of the bullet,$K_i = \frac{1}{2} m v^2$.
By the law of conservation of linear momentum: $m v = (M + m) v_1 \Rightarrow v_1 = \frac{m v}{M + m}$.
Final kinetic energy of the system,$K_f = \frac{1}{2} (M + m) v_1^2 = \frac{1}{2} (M + m) \left( \frac{m v}{M + m} \right)^2 = \frac{m^2 v^2}{2(M + m)}$.
Loss in kinetic energy $= K_i - K_f = \frac{1}{2} m v^2 - \frac{m^2 v^2}{2(M + m)}$.
$= \frac{1}{2} m v^2 \left( 1 - \frac{m}{M + m} \right) = \frac{1}{2} m v^2 \left( \frac{M + m - m}{M + m} \right) = \frac{m M v^2}{2(M + m)}$.

(D) In a perfectly elastic collision between two bodies of equal mass where one body is initially at rest, the bodies exchange their velocities.
Let the mass of the ball be $m$ and the mass of the bob be $m$.
Initial velocity of the ball $u_1 = 2 \,ms^{-1}$ and initial velocity of the bob $u_2 = 0 \,ms^{-1}$.
After the elastic collision, the ball comes to rest $(v_1 = 0)$ and the bob acquires the velocity of the ball $(v_2 = u_1 = 2 \,ms^{-1})$.
The kinetic energy of the bob immediately after the collision is converted into potential energy as it rises to a height $h$.
Using the conservation of energy: $\frac{1}{2}mv_2^2 = mgh$.
Substituting the values: $\frac{1}{2} \times (2)^2 = 10 \times h$.
$2 = 10h$.
$h = 0.2 \,m = 20 \,cm$.

(D) 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 $x$-direction be positive. The initial momentum of the system is:
$p_i = (4m)(v_1) + (2m)(-v_2) = 4mv_1 - 2mv_2$
After the collision,the two blocks move together as a single mass $(4m + 2m = 6m)$ with a final velocity $5v_2$ in the $x$-direction:
$p_f = (6m)(5v_2) = 30mv_2$
Equating the initial and final momentum:
$4mv_1 - 2mv_2 = 30mv_2$
$4mv_1 = 32mv_2$
$\frac{v_1}{v_2} = \frac{32}{4} = 8$

(B) Given,mass of the first body $m_1 = 3 \ kg$ and its velocity $v_1 = (2 \hat{i}+3 \hat{j}+3 \hat{k}) \ m/s$.
Mass of the second body $m_2 = 4 \ kg$ and its velocity $v_2 = (3 \hat{i}+2 \hat{j}-3 \hat{k}) \ m/s$.
According to the law of conservation of linear momentum,the total momentum before collision is equal to the total momentum after collision.
$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$
Substituting the values:
$3(2 \hat{i}+3 \hat{j}+3 \hat{k}) + 4(3 \hat{i}+2 \hat{j}-3 \hat{k}) = (3 + 4) v$
$(6 \hat{i}+9 \hat{j}+9 \hat{k}) + (12 \hat{i}+8 \hat{j}-12 \hat{k}) = 7 v$
$(6+12) \hat{i} + (9+8) \hat{j} + (9-12) \hat{k} = 7 v$
$18 \hat{i} + 17 \hat{j} - 3 \hat{k} = 7 v$
$v = \frac{1}{7}(18 \hat{i}+17 \hat{j}-3 \hat{k}) \ m/s$.

(A) Initial kinetic energy of the bullet is $K_i = \frac{1}{2} m v_0^2$.
After the collision,the bullet gets lodged in the block,and they move together with a common velocity $v$. By the law of conservation of linear momentum:
$m v_0 = (m + M) v \implies v = \frac{m v_0}{m + M}$.
The kinetic energy of the combined system immediately after the collision is $K_f = \frac{1}{2} (m + M) v^2$.
Substituting the value of $v$:
$K_f = \frac{1}{2} (m + M) \left( \frac{m v_0}{m + M} \right)^2 = \frac{1}{2} (m + M) \frac{m^2 v_0^2}{(m + M)^2} = \frac{1}{2} \frac{m^2 v_0^2}{m + M}$.
The energy lost in the collision is $\Delta K = K_i - K_f$:
$\Delta K = \frac{1}{2} m v_0^2 - \frac{1}{2} \frac{m^2 v_0^2}{m + M} = \frac{1}{2} m v_0^2 \left( 1 - \frac{m}{m + M} \right) = \frac{1}{2} m v_0^2 \left( \frac{m + M - m}{m + M} \right) = \frac{1}{2} m v_0^2 \left( \frac{M}{m + M} \right)$.

(A) The total kinetic energy of the two blocks before the collision is $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Since the collision is perfectly inelastic (the blocks melt),the maximum loss in kinetic energy is equal to the total initial kinetic energy,which is $mv^2$.
This energy is used to raise the temperature of the ice from $-8^{\circ}C$ to $0^{\circ}C$ and then to melt it.
The energy required for one block is $Q = ms\Delta\theta + mL$.
For two blocks,the total energy required is $2(ms\Delta\theta + mL)$.
Equating the energy loss to the heat required: $mv^2 = 2m(s\Delta\theta + L)$.
Canceling $m$ from both sides: $v^2 = 2(s\Delta\theta + L)$.
Given: $s = 2100 \ Jkg^{-1}K^{-1}$,$\Delta\theta = 8^{\circ}C$,$L = 3.36 \times 10^5 \ Jkg^{-1}$.
$v^2 = 2(2100 \times 8 + 3.36 \times 10^5) = 2(16800 + 336000) = 2(352800) = 705600$.
$v = \sqrt{705600} = 840 \ ms^{-1}$.

(D) Mass of bullet $= m$,Initial speed of bullet $= v$.
Mass of block $= M$,Initial speed of block $= 0$.
Let the common velocity of the system after collision be $V$.
According to the law of conservation of linear momentum:
$m v + M(0) = (m + M)V$
$V = \frac{m v}{m + M}$
Heat generated is equal to the loss in kinetic energy $(KE)$.
$\Delta KE = KE_{initial} - KE_{final}$
$\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) V^{2}$
Substituting the value of $V$:
$\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) \left( \frac{m v}{m + M} \right)^{2}$
$\Delta KE = \frac{1}{2} m v^{2} - \frac{1}{2} (m + M) \frac{m^{2} v^{2}}{(m + M)^{2}}$
$\Delta KE = \frac{1}{2} m v^{2} \left( 1 - \frac{m}{m + M} \right)$
$\Delta KE = \frac{1}{2} m v^{2} \left( \frac{m + M - m}{m + M} \right)$
$\Delta KE = \frac{1}{2} \frac{m M v^{2}}{m + M}$