Question

State and prove low of conservation of momentum.

Answer 1

A principle in physics: the total linear momentum of a system of particles not acted upon by external forces is constant in magnitude and direction irrespective of any reactions among the parts of the system.

Consider two particles say A and B of mass m1 and m2 collide with each other and forces acting on these particles are only the ones they exert on each other.

Explanation:

Let $u_1$and $v_1$ be the initial and final velocities of particle A and similarly, $u_2$ and $v_2$ for particle B. Let the two particles be in contact for a time t.

So, the change in momentum of

$A=m_1(v_1-u_1)$

The change in momentum of

$B=m_2(v_2-u_2)$

Now, during the collision,

Let A impart an average force equal to $F_{BA}$on B and let B exert an average $F_{AB}$ on A.

We know that from third law of motion

$F_{BA}=-F_{AB}.....\left(I\right)$

Now newton’s second law

$F_{BA}=m_2\times a_2$

$F_{BA}=\frac{m_2(v_2-u_2)}{t}$

$F_{AB}=m_1\times a_1$

$F_{AB}=\frac{m_1(v_1-u_1)}{t}$

Now, putting the value of $F_{AB}$ and $F_{BA}$ in equation (I)

$\frac{m_2(v_2-u_2)}{t}=-\frac{m_1(v_1-u_1)}{t}$

$m_2{v}_2-m_2{u}_2=-m_1{v}_1+m_1{u}_1$

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2......\left(II\right)$

Now, $m_1{u}_1+m_2{u}_2$represents the total momentum of particles A and B before collision and $m_1{v}_1+m_2{v}_2$ represent the total momentum of particles after collision

Hence, the equation (II) which is known as the law of conservation of momentum