Center of Mass and Collision

Given, mass of body, $m_1=5 \times 10^3 \mathrm{~kg}$

Velocity, $\quad v_1=2 \mathrm{~m} / \mathrm{s}$

Mass of another body, $m_2=15 \times 10^3 \mathrm{~kg}$

For perfectly inelastic collision $(e=0)$,

Loss in kinetic energy of system,

$\begin{aligned} \Delta E_K & =\frac{1}{2} \frac{m_1 m_2}{m_1+m_2} \times v_1^2 \\ & =\frac{1}{2} \times \frac{5 \times 10^3 \times 15 \times 10^3}{5 \times 10^3+15 \times 10^3} \times 2^2 \\ & =7.5 \times 10^3 \mathrm{~J}=7.5 \mathrm{~kJ} \end{aligned}$

In elastic collision, total energy, kinetic energy and momentum remain conserved, therefore no loss in energy occurs in elastic collision. Hence, both assertion and reason are true but reason is not the correct explanation of assertion.

By Newton second law of motion,

Net external force applied on a system is equal to rate of change momentum.

i.e., $\mathbf{F}_{\mathrm{ext}}=\frac{d \mathbf{p}}{d t}$

If $\mathbf{F}_{\text {ext }}=0$

$\frac{d \mathbf{p}}{d t}=0 \Rightarrow \mathbf{p}=\text { constant. }$

i.e., $F_{\text {ext }}=0$, then momentum of the system remains conserved.

The internal interaction between particles of a body do not contribute to change in total momentum of body.

Hence, both the assertion and reason are false.

From conservation of momentum,

$\begin{aligned} & M v+m \times 0=M v_1+m v_2 \\ \Rightarrow & \quad M\left(v-v_1\right)=m v_2 \quad \text{.... (i)} \end{aligned}$

Again, from the conservation of kinetic energy (as collision is of elastic nature)

$\begin{aligned} & & \frac{1}{2} M v^2+\frac{1}{2} m \times 0 & =\frac{1}{2} M v_1^2+\frac{1}{2} m v_2^2 \\ \Rightarrow & & M\left(v^2-v_1^2\right) & =m v_2^2 \quad \text{..... (ii)} \end{aligned}$

On solving Eqs. (i) and (ii), we get

$\begin{gathered} \frac{M\left(v-v_1\right)}{M\left(v+v_1\right)\left(v-v_1\right)}=\frac{m v_2}{m v_2^2} \\ v_2=v+v_1 \quad \text{..... (iii)} \end{gathered}$

Now, solving Eqs. (i) and (iii), we get

$v_1=\frac{(M-m) v}{(M+m)} \quad \text { and } \quad v_2=\frac{2 M v}{(M+m)}$

As $\quad M \gg>m$

So, $\quad v_1=v \Rightarrow v_2=v+v=2 v$

To find the velocity of the center of mass of the system, we use the formula for the velocity of the center of mass (CM) of several bodies given by:

$ v_{\text{CM}} = \frac{m_1 v_1 + m_2 v_2 + m_3 v_3}{m_1 + m_2 + m_3} $

Given:

Masses: $ m_1 = 5 \, \text{kg}, \, m_2 = 4 \, \text{kg}, \, m_3 = 2 \, \text{kg} $

Velocities: $ v_1 = 5 \, \text{m/s}, \, v_2 = 4 \, \text{m/s}, \, v_3 = 2 \, \text{m/s} $

Substitute these values into the formula:

$ v_{\text{CM}} = \frac{(5 \, \text{kg} \times 5 \, \text{m/s}) + (4 \, \text{kg} \times 4 \, \text{m/s}) + (2 \, \text{kg} \times 2 \, \text{m/s})}{5 \, \text{kg} + 4 \, \text{kg} + 2 \, \text{kg}} $

Calculate the numerator:

$ 5 \, \text{kg} \times 5 \, \text{m/s} = 25 \, \text{kg} \cdot \text{m/s} $

$ 4 \, \text{kg} \times 4 \, \text{m/s} = 16 \, \text{kg} \cdot \text{m/s} $

$ 2 \, \text{kg} \times 2 \, \text{m/s} = 4 \, \text{kg} \cdot \text{m/s} $

Thus, the total momentum is:

$ 25 \, \text{kg} \cdot \text{m/s} + 16 \, \text{kg} \cdot \text{m/s} + 4 \, \text{kg} \cdot \text{m/s} = 45 \, \text{kg} \cdot \text{m/s} $

The total mass is:

$ 5 \, \text{kg} + 4 \, \text{kg} + 2 \, \text{kg} = 11 \, \text{kg} $

Now, calculate $ v_{\text{CM}} $:

$ v_{\text{CM}} = \frac{45 \, \text{kg} \cdot \text{m/s}}{11 \, \text{kg}} = 4.09 \, \text{m/s} $

Thus, the magnitude of the velocity of the center of mass is approximately 4.1 m/s. Choice not matching entirely the given options, however, closest is Option B 4 m/s following rounding approximations commonly in physics problem settings.

If it is completely inelastic collision, then

$\begin{aligned} m_1 v_1+m_2 v_2 & =m_1 v+m_2 v \\\\ v & =\frac{m_1 v_1+m_2 v_2}{m_1+m_2} \\\\ \therefore \mathrm{KE} & =\frac{m_1^2}{2 m_1}+\frac{p_2^2}{2 m_2} \end{aligned}$

As, $p_1$ and $p_2$ both simultaneously cannot be zero, therefore total KE cannot be lost.

$\begin{aligned} \text {Loss of } \mathrm{KE} & =\frac{1}{2} m_1 u_1^2-\frac{m_1^2 u_1^2}{2\left(m_1+m_2\right)} \\ & =\frac{m_1 m_2 u_1^2}{2\left(m_1+m_2\right)} \end{aligned}$

$\begin{aligned} \text{Given,} \quad m_1 & =4 \mathrm{~kg} \\ u_1 & =12 \mathrm{~m} / \mathrm{s} \\ m_2 & =6 \mathrm{~kg} \end{aligned}$

$\begin{aligned} u_2 & =0 \\ \therefore \quad \Delta \mathrm{KE} & =\frac{1}{2} \frac{4 \times 6}{(4+6)}(12)^2 \\ & =\frac{1}{2} \times \frac{24}{10} \times(12)^2=\frac{12}{10} \times 144 \\ & =172.8 \mathrm{~J} \end{aligned}$