Class 9 > Unit # 03: Dynamics >Law of Conservation of Momentum

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Unit No. 3 Dynamics - Law of Conservation of Momentum

Q. State and explain law of conservation of momentum.

Statement

"The total momentum of an isolated system of objects (which has no interactions with external agents) is constant."

Explanation & Mathematical Derivation

Consider two bodies $A$ and $B$ of masses $m_1$ and $m_2$ moving in the same straight line with initial velocities $u_1$ and $u_2$ respectively, such that $u_1$ is greater than $u_2$ ($u_1 > u_2$). Consequently, body $A$ collides with body $B$. After collision, let their velocities become $v_1$ and $v_2$ respectively.

Figure: Interaction of two bodies before and after collision in an isolated system

During the short time interval $t$ of collision, body $A$ exerts a force $F$ on body $B$, and according to Newton's Third Law, body $B$ exerts an equal and opposite force $-F$ on body $A$.

Momentum of the system before collision:

$\text{Total Momentum before collision} = m_1 u_1 + m_2 u_2$

Momentum of the system after collision:

$\text{Total Momentum after collision} = m_1 v_1 + m_2 v_2$

According to Newton's Second Law of Motion, the force exerted by body $A$ on $B$ is equal to the rate of change of momentum of body $B$:

$F = \frac{m_2 v_2 - m_2 u_2}{t}$

Similarly, the force exerted by body $B$ on $A$ is:

$-F = \frac{m_1 v_1 - m_1 u_1}{t}$

By Newton's Third Law of Motion, action and reaction are equal and opposite ($F = -F$):

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

Multiplying both sides by time $t$ and simplifying:

$m_2 v_2 - m_2 u_2 = -m_1 v_1 + m_1 u_1$

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Total Momentum Before Collision = Total Momentum After Collision

This proves the Law of Conservation of Momentum.

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