Conservation Of Momentum

From Newton’s law, we know that the rate of change of the momentum of a particle over time is equal to the net force acting on the particle and is in the same direction as that force.

‘(\begin{array}{l}F_{net} = \frac{dP}{dt} \end{array})’

If there is no net external force, p cannot change, which clearly makes us understand that linear momentum can only be changed by a net external force.

$$F_{net} = 0; \frac{dP}{dt} = 0 \Rightarrow P~is~constant$$

The momentum is conserved, implying that there is no change in the momentum. i.e.

(\begin{array}{l}initial ;momentum = final; momentum\end{array})

In order to apply conservation of momentum, you must select a system in which the net external force is equal to zero.

Example: This is an example.

This is an example.

In the example given below, the two cars of masses (m_1) and (m_2) are moving with velocities (v_1) and (v_2) respectively before the collision. And their velocities change to (v_1^{’}) and (v_2^{’}) after collision. To apply the law of conservation of linear momentum, we cannot choose any one of the cars as the system. If so, then there is an external force on the car by another car. So we choose both the cars as our system of interest. This is why in all collisions, if both the colliding objects are considered as a system, then linear momentum is always conserved (irrespective of the type of collision).

Solved Problem

The recoil speed of the gun will be $$\frac{mv}{M}$$.

The system (gun + bullet) was initially at rest, so its initial momentum was zero. Since no external force acts on the system during their impact, the momentum of the system remains constant. Therefore, just after the impact, the momentum of the system (gun + bullet) will be zero.

‘(\begin{array}{l}M\vec{V}_b + M\vec{V}_g = 0\end{array})’

‘(\begin{array}{l}m\vec{V}_{bg} + (M+m)\vec{V}_g = 0\end{array})’

$\vec{V}g = -\frac{m\vec{V}{bg}}{M+m},$ where $\left|\vec{V}_{bg}\right| = velocity;of;bullet;relative;to;the;gun = v$

(\begin{array}{l}\vec{V}_g= - \frac{mv}{M+m}\end{array}) Types Of Collision



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