91Ó°ÊÓ

The given data can be listed below as,

• The mass of the rock 1 is$0.1 kg$.
• The velocity of the rock 1 is$5 m/s$.
• The mass of the rock 2 is$0.25 kg$.
• The velocity of the rock 2 is $7.5 m/s$.

The law of the conservation of momentum states that the total momentum for a body before and after the collision becomes equal.

The law of the center of mass states that if a rigid object is pushed then it will continue to move as a point mass.

The equation of the momentum gives the total momentum of the system and the equation of velocity gives the velocity of the system.

a) From the law of conservation of momentum, the equation of the total momentum of a system can be expressed as:

$p=m_1{v}_1+m_2{v}_2$

Here, p is the total momentum of the system, role="math" localid="1658129790340" $m_1{\text{andm}}_2$are the mass of the rock 1 and rock 2 respectively and are $v_1\text{and}{v}_2$the velocity of the rock 1 and rock 2 respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} p=(0.1\mathrm{kg})×(5\mathrm{m}/\mathrm{s})+(0.25\mathrm{kg})×(7.5\mathrm{m}/\mathrm{s}) \\ p=2.375kg.m/s \end{gathered}$$

Thus, the total momentum of the system consisting of both rocks and the string is $2.375 kg.m/s$.

b) From the law of center of mass, the equation of the velocity of the center of mass is expressed as:

$v=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

Substituting the values in the above equation, we get-

$$\begin{gathered} v=\frac{(0.1\mathrm{kg})×(5\mathrm{m}/\mathrm{s})+(0.25\mathrm{kg})×(7.5\mathrm{m}/\mathrm{s})}{(0.1\mathrm{kg})+(0.25\mathrm{kg})} \\ v=6.785 m/s \end{gathered}$$

Thus, the velocity of the center of mass of the system is $6.785 m/s.$