Question

Prove the conservation of linear momentum for a head on collision of two bodies with the help of Newton's second law.

Answer 1

The law of conservation of linear momentum asserts that if no external force acts on an object or system of objects, the total momentum of the system is always conserved.

Consider a collision between two balls in which no energy is lost as a result of the impact.

In order to prove the law of conservation of momentum, let's take two bodies A and B having mass m₁ and m₂ respectively. Let the initial velocities of both bodies be u₁ and u₂, after the head-on collision let's assume that the final velocities of both bodies be v₁ and v₂ respectively. As we know that the momentum ‘p’ of a body having mass ‘m’ and velocity ‘v’ is;

$p=mv\left(1\right)$

We can write change in momentum by using equation (1);

$Changeinmomentum=Mass(finalvelocity-initialvelocity)\left(2\right)$

Change in momentum for body A will be;

$ChangeinmomentumofbodyA=m_1(v_1-u_1)$

Similarly, change in momentum for body B will be;

$ChangeinmomentumforbodyB=m_2(v_2-u_2)$

Now, we can claim that any action has an equal and opposite response, according to Newton's third law of motion. That is;

Force applied by B on A = Force applied by A on B

$F_{AB}=-F_{BA}\left(3\right)$

Here negative sign shows that both forces occur in opposite direction.

The force F applied to a body is equal to the rate of change of momentum p of the body with respect to time, according to Newton's second law of motion.

$$\begin{gathered} F=\frac{dp}{dt} \\ F=\frac{∆p}{dt}\left(4\right) \end{gathered}$$

By using (4), we have;

$$\begin{gathered} F_{AB}=\frac{ChangeinmomentumofbodyA}{Timetaken} \\ =\frac{m_1(v_1-u_1)}{t} \end{gathered}$$

Similarly,

$$\begin{gathered} F_{BA}=\frac{ChangeinmomentumofbodyB}{Timetaken} \\ =\frac{m_2(v_2-u_2)}{t} \end{gathered}$$

Now, by using equation (3), we get;

$$\begin{gathered} \Rightarrow \frac{m_1(v_1-u_1)}{t}=-\frac{m_2(v_2-u_2)}{t} \\ \Rightarrow m_1(v_1-u_1)=-m_2(v_2-u_2) \\ \Rightarrow m_1{v}_1-m_1{u}_1=-m_2{v}_2+m_2{u}_2 \\ \Rightarrow m_1{v}_1+m_2{v}_2=m_1{u}_1+m_2{u}_2 \end{gathered}$$

The total momentum of the system after collision equals the total momentum of the system before contact, according to the preceding equation. Hence, the law of conservation of momentum is proved.