Law of Conservation of Momentum Derivation

The momentum of an object is the product of the velocity and mass of an object. It is a vector quantity. Conservation of momentum is a fundamental law of physics, which states that the total momentum of an isolated system is conserved. In other words, the total momentum of a system of objects remains constant during any interaction, if no external force acts on the system. The total momentum is the vector sum of individual momenta. Therefore, the component of the total momentum along any direction remains constant (whether the objects interact or not). Momentum remains conserved in any physical process.


Illustration in One - Dimension

Conservation of momentum can be explained through a one-dimensional collision of two objects. Two objects of masses \[m_{1}\] and \[m_{2}\] collide with each other while moving along a straight line with velocities \[u_{1}\] and \[u_{2}\] respectively.  After the collision, they acquire velocities \[v_{1}\] and \[v_{2}\] in the same direction.

Total momentum before collision \[p_{i} = m_{1}u_{1} + m_{2}u_{2}\]

Total momentum after collision \[p_{f} = m_{1}v_{1} + m_{2}v_{2}\]

If no other force acts on the system of the two objects, total momentum remains conserved. Therefore,

\[p_{i} = p_{f}\]


\[m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\]

(image will be uploaded soon)


Derivation of Conservation of Momentum

If no external force is exerted on the system of two colliding objects, the objects apply impulse on each other for a short interval of time at the point of contact. According to Newton’s third law of motion, the impulsive force applied by the first object on the second one is equal and opposite to the impulsive force applied by the second object on the first object.  


During the one-dimensional collision of two objects of masses \[m_{1}\] and \[m_{2}\], which have velocities \[u_{1}\] and \[u_{2}\] before collision and velocities \[v_{2}\] and \[v_{2}\] after the collision, the impulsive force on the first object is \[F_{21}\] (applied by the second object) and the impulsive force on the second object is \[F_{12}\] (applied by the first object). Applying Newton’s third law, these two impulsive forces are equal and opposite i.e.


\[F_{21} = -F_{12}\]


If the time of contact is t,  the impulse of the force \[F_{21}\] is equal to the change in momentum of the first object. 


\[F_{21}t = m_{1}v_{1} - m_{1}u_{1}\]


The impulse of force \[F_{12}\] is equal to the change in momentum of the second object.

\[F_{11}t = m_{2}v_{2} - m_{2}u_{2}\]


From \[F_{21} = -F_{12}\],


\[F_{21}t = -F_{12}t\]


\[m_{1}v_{1} - m_{1}u_{1}\] = \[-(m_{2}v_{2} - m_{2}u_{2})\]


\[m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\]


This relation suggests that momentum is conserved during the collision.


Conservation of Momentum Examples

  • Recoil of a Gun: If a bullet is fired from a gun, both the bullet and the gun are initially at rest i.e. the total momentum before firing is zero. The bullet acquires a forward momentum when it gets fired. According to the conservation of momentum, the gun receives a backward momentum. The bullet of mass m is fired with forward velocity v. The gun of mass M acquires a backward velocity u. Before firing, the total momentum is zero so that the total momentum after firing is also zero.

0 = mv + Mu

\[u = -\frac{m}{M} v\]

u is the recoil velocity of the gun. The mass of the bullet is much less than that of the gun i.e. m ≪ M. The backward velocity of the gun is very small,

u ≪ v

  • Rocket Propulsion: Rockets have a gas chamber at one end, from which gas is ejected with enormous velocity. Before the ejection, the total momentum is zero. Due to the ejection of gas, the rocket gains a recoil velocity and acceleration in the opposite direction. This is a consequence of the conservation of momentum

(image will be uploaded soon)

If a rocket of mass m ejects the propellent of small mass dm with an exhaust velocity \[v_{e}\] such that the residual rocket of mass m - dm acquires a velocity dv in the opposite direction, the momenta of the propellant and the residual rocket are equal in magnitude and opposite in direction.

\[v_{e}dm = -(m - dm)dv\]

Since both  dmand dvare small, the equation can be approximated as

\[dv = -v_{e} \frac{dm}{m}\]

If the mass of the rocket reduces from \[m_{0}\] to \[\overline{m}\] as its velocity increases from 0 to \[\overline{v}\], integrating the above equation

\[\int_{0}^{\overline{v}} dv = - v_{e} \int_{m_{0}}^{\overline{m}} \frac{dm}{m}\]

\[\overline{v} = v_{e} ln(\frac{m_{0}}{\overline{m}})\]


Solved Examples

I. A bullet of mass 6 g is fired with a speed 500 m/s from a gun of mass 4 kg. What would be the recoil velocity of the gun?


Solution: The initial momenta of the bullet and the gun are zero such that the total initial momentum is zero. The bullet of mass m = 6g  is fired with forward velocity v = 500 m/s. The gun of mass M = 4kg acquires a backward velocity V. 


\[m = 6g = \frac{6}{1000}kg\]


According to the conservation of momentum formula,


0 = mv + MV


0 = \[\frac{6}{1000} kg (500 m/s) + (4 kg) V\]


V = -0.75 m/s


The recoil speed of the gun is 0.75 m/s. The negative sign implies that the recoil velocity is opposite to the velocity of the bullet.


II. An object of mass \[m_{1}\] is moving with speed \[v_{1}\] along the X - axis. Another object of mass \[m_{2}\] is moving with speed \[v_{2}\] long the Y - axis. Both the axes are perpendicular to each other. The objects collide and get stuck. What would be the velocity of the combined object?


Solution: The object of mass \[m_{1}\]moves with speed \[v_{1}\] along the X-axis. Its momentum is \[p_{1} = m_{1}v_{1}\], along the X - axis. 


The car of mass \[m_{2}\] is moving with speed \[v_{2}\] along the Y - axis. Its momentum is \[p_{2} = m_{2}v_{2}\] along the Y - axis. 


The final velocity of the combined object of mass (m + M) is u, along a direction that makes an angle θ with the X - axis. 


(image will be uploaded soon)


Collision in Two - Dimensions


Before the collision, the total momentum is \[p_{ix} = p_{1} = m_{1}v_{1}\], along X - axis and \[p_{iy} = p_{2} = m_{2}v_{2}\], along the Y - axis. After the collision, the total momentum is \[p_{fx} = (m + M)u cos\theta\], along X-axis and \[p_{fy} = (m + M)u sin\theta\].


Applying conservation of momentum,


\[p_{ix} = p_{fx}\]


\[m_{1}v_{1} = (m + M)u cos\theta\]    (1)


\[p_{iy} = p_{fy}\]


\[m_{2}v_{2} = (m + M)u sin\theta\]   (2)

         

Therefore, squaring and adding equations (1) and (2),


\[(m_{1}v_{1})^{2} + (m_{2}v_{2})^{2} = (m + M)^{2} u^{2} (cos\theta^{2} + sin\theta^{2})\]


\[u = \frac{\sqrt{m_{1}^{2}v_{1}^{2} + m_{2}^{2}v_{2}^{2}}}{m + M}\]


It is the speed of the combined object.


Dividing equation (2) by (1),


\[tan\theta = \frac{m_{2}v_{2}}{m_{1}v_{1}}\]


θ gives the direction of the velocity.


Did You Know?

  • Light consists of photons, which are massless particles. however, a photon has momentum, expressed in terms of its energy.

  • In quantum mechanics, conservation of momentum is an effect of translation or ‘shifting’ symmetry of space.

FAQ (Frequently Asked Questions)

1) What is Conservation of Momentum?

The total momentum of an isolated system remains constant if any external force does not act on the system. Consequently, if the total linear momentum of a system remains constant, the resultant force acting on the system is zero. Similarly, angular momentum is also conserved in the absence of external torque.

2) Give Examples of Conservation of Momentum.

All physical processes abide by the law of conservation of momentum. Some examples are,

  • Collision: The collision of objects follows the conservation of momentum and energy. 

  • Rocket motion: The momentum of the propellent gas gives the rocket an opposite momentum. This is a consequence of momentum conservation.

  • Ejection of a bullet from a gun: A gun experiences a recoil momentum due to the ejection of a bullet.