The Law of Conservation of Linear Momentum: Explained with Examples

By Pankaj Kumar|Last Updated: July 12, 2026|

Key Takeaways

Linear momentum is conserved when the net external force on a system is zero.
Newton’s second law directly proves the conservation of linear momentum.
Gun recoil, collisions, and explosions all demonstrate this conservation law.
In elastic collisions between equal masses, the first object stops completely.
Center of mass position stays fixed when no external horizontal force acts.

The law of conservation of momentum is divided into two parts:

The law of conservation of linear momentum
The law of conservation of angular momentum

In this article, we will learn about the law of conservation of LINEAR momentum.

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The Law of Conservation of Linear Momentum

The law of conservation of linear momentum says that when the resultant external force acting on a system is zero, the system’s total momentum (vector sum) remains constant.

Examples of Linear Momentum Conservation

Explosions, the disintegration of nuclei, recoil of a gun, collisions, etc., can be explained using the law of conservation of linear momentum.

The recoil of the gun is one of the best daily life examples.

In this example, when a shot is fired from the gun, while the shot moves forwards, the gun moves backward. This motion of the gun is called recoil of the gun. When a gun of mass M fires a bullet of mass m with a muzzle velocity v, then the gun recoils with a velocity V given by $V = frac{{mv}}{M}$.

$F = frac{{dP}}{{dt}}$, If $F = 0$, then $P$ remains constant or conserved.

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Proof of the Linear Momentum Conservation Law

As we know, from Newton’s 2^nd law of motion, the rate of change of momentum of a body is directly proportional to the resultant (or) net external force acting on the body and taking place along the direction of the force.

Mathematically we can write, ${F_{net}} = frac{{dP}}{{dt}}$

If ${F_{net}} = 0$; this means, $frac{{dP}}{{dt}} = 0$.

Hence, the system’s total momentum will not change with respect to time. This will remain constant.

If $sum {F_{ext}} = 0,;,{left( {sum {P_{sy}}} right)_i} = {left( {sum {P_{sy}}} right)_f}$

If ${F_{net}} = frac{{dL}}{{dt}} = 0,;$ then L remains constant or conserved.

As we know, the angular momentum of a body is defined as,

$L = sum left( {{r_i} times {P_i}} right)$

Now, differentiating with respect to time,

$frac{{dL}}{{dt}} = frac{d}{{dt}}sum left( {{r_i} times {P_i}} right) = sumlimits_i^{} {left[ {frac{{d{r_i}}}{{dt}} times {P_i} + r times frac{{d{P_i}}}{{dt}}} right]} $

$frac{{dL}}{{dt}} = sumlimits_i^{} {left[ {{v_i} times m{v_i} + {r_i} times {F_i}} right]} $

Here, ${v_i} times m{v_i} = 0$, so,

$frac{{dL}}{{dt}} = sumlimits_i^{} {{r_i} times {F_i}} = {Gamma _{total}}$

Here, ${F_i}$ is the total force acting on the i^th particle. This includes any external force as well as the forces on the i^th particle by all the other particles. Taking summation of both sides, internal torque adds to zero.

$sumlimits_{}^{} {{Gamma ^{ext}}} = frac{{dL}}{{dt}}$

Here, ${Gamma ^{ext}}$ is the total torque due to all the external forces acting on the system.

If the torque of external forces about any particular point is equal to zero, then $frac{{dL}}{{dt}} = 0$. This means that the system’s total angular momentum about that particular point will remain constant.

${left( {sumlimits_{}^{} {{L_{sys,,0}}} } right)_i} = {left( {sumlimits_{}^{} {{L_{sys,,0}}} } right)_f}$ if $Gamma _0^{ext} = 0$.

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Conservation of Linear Momentum: Solved Problems

Problem 1: A Person Jumping on a Moving Car

A car of mass M is moving with uniform velocity v on a horizontal road when a person drops himself on it from above. Taking mass of person to be m; What will be the car’s velocity after the event?

Ans.: Consider the car plus the person as a system. In the horizontal direction, there is no external force since the momentum of that person is equal to zero.

Initially, the horizontal momentum of the system = MV.

Finally, the person sticks to the car’s roof, so they move with equal velocity V. Now, final horizontal momentum of system = (M + m)V.

Now, from the principle of conservation of angular momentum,

$Mv = (M + m)V$

$V = frac{{Mv}}{{M + m}}$

Problem 2: Force of Recoil of a Machine Gun

The hero of a stunt film fires bullets from a machine gun, each at a speed of 1km/s. If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period?

Ans.: The momentum of each bullet,

$P = (0.05,kg),(1000,m{rm{/s)}} = 50,{rm{kg – m/s}}$

The gun is imparted this much momentum by each bullet fired. So, the rate of change of momentum of the gun (total)

$ = (50,k{rm{g – m/s) times }}frac{{({rm{Number of bullets}})}}{{Delta t}}$

$ = (50,k{rm{g – m/s) times }}frac{{20}}{4} = 250,N$

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Problem 3: Two Men Walking on a Boat

Mr. John (50 kg) and Mr. Alex (60 kg) are sitting at the two extremes of a 4m long boat (40kg) still in the water. They come to the middle of the boat to discuss a mechanical problem. Neglecting friction with water, how far does the boat move on the water during the process?

Ans.: (Mr. John + Mr. Alex + Boat) as a system.

Here velocity of the center of the system along the x-axis equals zero. The sum of all external forces acting on the system along the x-axis is equal to zero; hence, the momentum (total) of the system along the x-axis must be conserved.

Since, ${({V_{cm}})_x} = 0quad Rightarrow quad frac{d}{{dt}}{X_{cm}} = 0$

$ Rightarrow quad {({X_{cm}})_i} = {({X_{cm}})_f}$ …(1)

Diagram showing two men walking on a boat on water

Let the boat moves along the –ve x-axis by a distance $ell $ as they approach the middle of the boat.

Now, from (1),

$frac{{50 times d + 40(d + 2) + 60(4 + d)}}{{(60 + 40 + 50)}} = frac{{(50 + 40 + 60)(2)}}{{(60 + 40 + 50)}}$

$ Rightarrow quad 50d + 40d + 80 + 240 + 60d = 150 times 2$

$ Rightarrow quad 150d + 320 = 300quad Rightarrow quad 150d = – 20$

$ Rightarrow quad d = – frac{2}{{15}}m = – frac{2}{{15}} times 100,cm$

Hence, boat will move along +ve x-direction by a distance of $frac{2}{{15}}m$ or $ approx 13,cm$.

Problem 4: Head-on Collision of Two Perfectly Elastic Balls

A marble going at a speed of 2 m/s hits another marble of equal mass at rest. If the collision is perfectly elastic, then find the velocity of the first marble after the collision.

Ans.: Assuming both marbles as a system,

Diagram showing head-on collision of two elastic marbles

Let speeds of marbles become ${v_1}’$ and ${v_2}’$ after the collision.

Here $left( {sumlimits_{}^{} {{F_{ext}}} } right)$ along x-axis equals to zero.

${left( {{P_{sys}}} right)_i} = {left( {{P_{sys}}} right)_f}$

$ Rightarrow quad {m_A}{v_1} + {m_B}{v_2} = {m_A}{v_1}’ + {m_B}{v_2}’$

Since ${m_A} = {m_B}$ and ${v_2} = 0$,

$ Rightarrow quad {v_1} + 0 = {v_1}’ + {v_2}’$

$ Rightarrow quad 2,{rm{m/s}},hat i = {v_1}’ + {v_2}’$ …(1)

Also, in this case, we can write, the velocity of approach = velocity of separation:

${v_1} – {v_2} = {v_2}’ – {v_1}’$

$ Rightarrow quad 2,{rm{m/s}},hat i – 0 = {v_2}’ – {v_1}’$ …(2)

Subtracting (2) from (1),

${v_1}’ + {v_2}’ – ({v_2}’ – {v_1}’) = 0$

$2{v_1}’ = 0quad Rightarrow quad {v_1}’ = 0$

So, the final velocity of the first marble will be zero.

Problem 5: Falling off a Vertical Pen on a Frictionless Surface

A uniform pen is kept vertically on a smooth horizontal surface at a point O. If it is rotated slightly and released, it falls on the horizontal surface. The lower end will remain:

(a) at point O

(b) at a distance, less than $frac{ell }{2}$ from O

(d) at a distance larger than $frac{ell }{2}$ from O

Ans.: Since the given pen is slightly rotated, the velocity of the center of mass along the x-axis is equal to zero. Hence, position of centre of mass of the pen $({X_{cm}})$ will be constant.

So, ${({X_{cm}})_i} = {({X_{cm}})_f}$.

Diagram showing a vertical pen falling on a frictionless horizontal surface

Hence, the lower end will be at a distance $frac{ell }{2}$ left with respect to the point O when the pen becomes horizontal on the surface.

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