Conservation Of Linear Momentum

The second and the third laws of motion lead to an important result: the law of conservation of momentum. 

According to law of conservation of momentum, 

‘When the net external force on a system of objects is zero, the total momentum of the system remains constant’. 

Law of Conservation of Momentum : Proof

Let us consider two balls A and B having masses m_A and m_B, travelling in the same direction along a straight line at different velocities u_A and u_B, respectively. No other external unbalanced forces are acting on them. 

Let u_A > u_B and the two balls collide with each other. During collision which lasts for a very short time t, ball A exerts a force F_BA on ball B, and ball B exerts a force F_AB on ball A. Suppose v_A and v_B are the velocities of the two balls A and B after the collision, respectively.

Initial momentum of ball A = m_Au_A, and final momentum of ball A = m_Av_A

Force on A due to B, F_AB = rate of change of momentum of ball A

$F_{AB}=\frac{m_A{v}_A-m_A{u}_A}{t}=\frac{m_A(v_A-u_A)}{t}$ ...(1)

Initial momentum of ball B = m_Bu_B, and final momentum of ball B = m_Bv_B

Force on B due to A, F_BA = rate of change of momentum of ball B

$F_{BA}=\frac{m_B{v}_B-m_B{u}_B}{t}=\frac{m_B(v_B-u_B)}{t}$ ...(2)

According to Newton’s third law of motion, the force F_BA exerted by ball A on ball B (action) and the force F_AB exerted by ball B on ball A (reaction) must be equal and opposite to each other. That is,

F_BA = – F_AB

$\frac{m_B(v_B-u_B)}{t}=-\frac{m_A(v_A-u_A)}{t}$

m_B(v_B – u_B) = – m_A(v_A – u_A)

m_Bv_B – m_Bu_B = – m_Av_A + m_Au_A  

m_Bv_B + m_Av_A = m_Bu_B + m_Au_A ...(3)

Since, (m_Bv_B + m_Av_A) represents total final momentum and (m_Bu_B + m_Au_A) represents total initial momentum, from eq.(3), we can conclude that

Total final momentum = Total initial momentum

Thus, in an isolated system (a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. This fact is known as the law of conservation of momentum.

Alternate Method

If net external force on the system is zero then linear momentum of the system remains constant.

According to newtons II law.

$F_{ext}=\frac{dp}{dt}$

If $F_{ext}=0$

$\frac{dp}{dt}=0$

The preceding equation demonstrates that    

⇒  p=constant

$P_{Initial}=P_{Final}$

As a result when the force acting on the system is zero. The system’s total linear momentum is either conserved or constant. 

The total linear momentum for ‘n’ particles is represented as 

$\mathbf{p}={\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3+{\mathbf{p}}_4+\dots +{\mathbf{p}}_n$

$\Delta \mathbf{p}=\Delta {\mathbf{p}}_1+\Delta {\mathbf{p}}_2+\Delta {\mathbf{p}}_3+\dots +\Delta {\mathbf{p}}_n$

Now,

${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3+{\mathbf{p}}_4+\dots +{\mathbf{p}}_n=\text{constant}$

So, 

$\Delta {\mathbf{p}}_1+\Delta {\mathbf{p}}_2+\dots +\Delta {\mathbf{p}}_n=0$

for two particle system ${\mathbf{p}}_1+{\mathbf{p}}_2=\text{constant}$

So, $\Delta {\mathbf{p}}_1+\Delta {\mathbf{p}}_2=0$

$\Delta {\mathbf{p}}_1=-\Delta {\mathbf{p}}_2$

According to Newton’s second law, force F can be written as, 

$F=\frac{\Delta p}{t}$ where, Δp is the change in momentum of the object.

or Δp = F × t

This means, F × t is the change in momentum of the bullet and – F × t is the change in momentum of the gun. Since initially, both are at rest, the change in momentum equals the final momentum for each. Let p_b be the momentum of the bullet after firing and p_g is the recoil momentum of the gun, then, 

p_b = F × t ... (1)      

and  p_g = – F × t ... (2)

Adding eq.(1) and eq.(2), we get, p_b + p_g = 0. That is, the final momentum of the system (bullet plus gun) is zero. But, initial momentum of the system is also zero. This means initial momentum is equal to the final momentum i.e., total momentum is conserved.