Conservation of momentum

What is the law of conservation of momentum?

In plain words: add up the momenta of all the bodies in a system, as vectors. As long as no net external force acts on the system, this total stays the same, however the bodies push, pull, collide with or break away from one another. This is conservation of linear momentum, as opposed to angular momentum. “Linear” does not restrict the particles to straight paths: each vector component of total momentum is conserved when the corresponding external force component is zero.

Two words in the statement do the work:

• Isolated. The system has no net external force on it. Forces between the bodies inside the system (internal forces) are allowed; they can change each body’s momentum, but not the total.
• Total. Individual momenta do change. Only their vector sum stays fixed.

Recall that the momentum of a body of mass m moving with velocity $\vec v$ is $\vec p = m\vec v$ (Newton’s second law of motion defines it). Momentum is a vector, so every sum on this page is a vector sum: take one direction as positive and give each velocity its sign.

Deriving the law from Newton’s laws

Take two bodies A and B that interact, for example in a collision, for a short time Δt. Let their momenta be $\vec p_A$ and $\vec p_B$ before, and $\vec p\,'_A$ and $\vec p\,'_B$ after. Let $\vec F_{AB}$ be the force on A due to B, and $\vec F_{BA}$ the force on B due to A (their average values during the contact).

Step 1: the second law for each body. If no external force acts, the only force on A is $\vec F_{AB}$ and the only force on B is $\vec F_{BA}$. The impulse of each force equals the change in momentum of the body it acts on:

Step 2: the third law. By Newton’s third law of motion, the two forces are equal and opposite, and both act for the same contact time Δt:

Therefore

Step 3: rearrange.

Here body A has mass m₁ and body B mass m₂; $\vec u_1, \vec u_2$ are their velocities before the interaction and $\vec v_1, \vec v_2$ after it. NCERT stresses that this holds whether the collision is elastic or inelastic: kinetic energy may be lost to heat, sound or deformation, but momentum is not.

Why can kinetic energy change when momentum cannot? Equal and opposite internal forces give cancelling impulses, so they do not change total momentum. Energy transfer also depends on displacement and deformation: during an inelastic impact, the bodies’ bulk kinetic energy can become internal deformation, vibration and heat. Momentum conservation therefore does not guarantee conservation of their bulk kinetic energy.

The same argument works for any number of bodies, because internal forces always come in equal and opposite pairs:

Thus in an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged.

NCERT Physics Part I (Class 11), Section 4.7, page 57

In compact form: add the second law for every body in the system. The internal forces cancel in pairs, leaving

where $\vec P$ is the total momentum and $\vec F_{\text{ext}}$ the net external force. So

Two refinements are used again and again in JEE and NEET problems:

• One direction at a time. If the net external force has no component along some axis, the momentum along that axis is conserved even if it is not conserved along the others. For a shell that explodes in mid-air, gravity acts downward, so horizontal momentum is conserved.
• Short interactions. In a collision or an explosion, the internal forces are very large and act for a very short time. An external force such as gravity or friction changes the total momentum only a little in that time, so the momentum just before and just after the event can be taken as equal. But if an external force also becomes very large during the event, it cannot be ignored. A cannon on the ground that fires a shell at an angle is pushed hard into the ground, and the ground’s normal force is just as impulsive as the firing. Then conserve momentum only along the direction in which that force has no component: for a 500 kg cannon on smooth, level ground firing a 5 kg shell at 200 m s^⁻¹ relative to the ground at 60° above the horizontal, neglecting propellant-gas momentum, the cannon recoils horizontally at

Recoil of a gun

A gun and its bullet are at rest before firing, so their total momentum is zero. When the gun fires, the gun pushes the bullet forward and the bullet pushes the gun backward with an equal force for the same time. The total momentum stays zero:

Here m and $\vec v$ are the bullet’s mass and velocity, and M and $\vec V$ the gun’s, with both velocities measured in the same inertial frame. The model neglects gas momentum and external impulse during firing. The minus sign says the gun moves backward. Because M is much larger than m, the recoil speed is much smaller than the bullet’s speed.

The two momenta are equal in size, but the kinetic energies are not. For any body of mass m and speed v, the kinetic energy is $K = \tfrac{1}{2}mv^{2}$, which can be written as p^²/2m. With the same p for both,

so

So the light bullet carries away almost all the kinetic energy. Holding the gun firmly against the shoulder makes the gun and the shooter’s body recoil together. The same momentum is then shared by a much larger mass, so both the recoil speed and the recoil kinetic energy, p^²/2M, are much smaller.

The law in everyday life

• Rockets. A rocket pushes hot gas out backward at high speed, and gains an equal and opposite momentum forward. It needs nothing outside to push against, which is why it works in space.
• Jumping off a boat. When you jump forward from a small boat onto the bank, the boat moves backward. If the boat is light, you may fall short of the bank.
• Explosions. When a firecracker or a shell bursts, the fragments fly apart, but the vector sum of their momenta equals the momentum just before the burst.
• Collisions on the road. Cars that crash and lock together make a completely inelastic collision: they move off with the same total momentum they had together just before impact, even though much of their kinetic energy is lost.
• Skaters pushing apart. Two skaters at rest who push each other move off in opposite directions, the lighter one faster.

Solved examples

Common mistakes to avoid

Quick revision summary

• The total momentum of an isolated system, one with zero net external force, stays constant.
• The law follows from the second law (impulse equals change in momentum) and the third law (internal forces cancel in pairs).
• It holds for collisions, whether elastic or inelastic, for explosions and for recoil. Kinetic energy is conserved only in elastic collisions.
• If the external force has no component along an axis, the momentum along that axis is conserved.
• Recoil: a gun and bullet starting at rest get equal and opposite momenta; the light bullet takes almost all the kinetic energy.

Questions students ask

Is momentum conserved in an inelastic collision?

Yes, momentum is conserved in every collision of an isolated system, elastic or inelastic: the internal forces are equal and opposite and act for the same time, so the total momentum just after equals the total momentum just before. What an inelastic collision loses is kinetic energy, which goes into heat, sound and the deformation of the bodies. When the bodies stick together, they move off with v = (m₁u₁ + m₂u₂)/(m₁ + m₂).

What is the difference between conservation of momentum and conservation of energy?

Momentum is a vector, and the total momentum of a system stays constant only when the net external force on it is zero. Energy is a scalar, and the total energy of an isolated system, in all its forms, is always conserved, but kinetic energy alone is conserved only in elastic collisions. So in a collision you can use conservation of momentum whenever external forces are negligible during the impact, but conservation of kinetic energy only if the collision is elastic.

When is linear momentum not conserved?

Linear momentum is not conserved when a net external force acts on the system. A car braking on a level road loses momentum because friction from the road, an external force, acts on it; only the car and the Earth together keep their total momentum. Momentum can still be conserved along a direction in which the external force has no component, and very nearly during a short collision or explosion, when the external forces are small compared with the internal ones.

Why should a gun be held tightly against the shoulder while firing?

A gun is held tightly so that the gun and the shooter’s body recoil together as one large mass. The backward momentum, equal in size to the bullet’s forward momentum mv, is then shared by this much larger mass, so the recoil speed and the recoil kinetic energy, p^²/2M with M now the mass of gun and body together, are much smaller. A loosely held gun recoils fast on its own and hits the shoulder hard.

How does a rocket move in space when there is no air to push against?

A rocket does not push against the air. It throws hot exhaust gas backward at high speed, and by conservation of momentum the rocket gains an equal and opposite momentum forward. With external forces neglected, the rocket and all its exhaust form the isolated system whose total momentum is conserved; the rocket alone continually loses mass. The same mechanism works in empty space.

What is the SI unit of momentum?

The SI unit of momentum is the kilogram metre per second, kg m s^⁻¹, since p = mv. It is the same as the newton second, N s, the unit of impulse, because impulse equals the change in momentum. Momentum is a vector, pointing along the velocity.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A person of mass 60 kg standing on a boat of mass 40 kg at rest jumps off horizontally at 2 m s^⁻¹ relative to the water. With what speed does the boat move? Ignore the water’s resistance.

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   Total momentum stays zero:

   $$\begin{aligned} 60 \times 2 + 40\,V &= 0 \\ V &= -\frac{120}{40} \\ &= -3\ \text{m s}^{-1} \end{aligned}$$

   The boat moves back at 3 m s⁻¹.

2. Q2. A bullet of mass 50 g moving at 200 m s^⁻¹ hits a wooden block of mass 950 g at rest on a smooth surface and stays inside it. Find the velocity of the block just after the impact.

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   $$\begin{aligned} 0.05 \times 200 &= (0.05 + 0.95)\,v \\ v &= \frac{10}{1.0} = 10\ \text{m s}^{-1} \end{aligned}$$

   The block (with the bullet) moves at 10 m s⁻¹ in the bullet’s original direction.

3. Q3. A bomb at rest explodes into two pieces. Show that the pieces must move in opposite directions along the same line.

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   The total momentum is zero before the explosion, so after it

   $$m_1\vec v_1 + m_2\vec v_2 = 0$$

   $$\vec v_2 = -\frac{m_1}{m_2}\,\vec v_1$$ The velocity of the second piece is a negative multiple of the first’s, so the two pieces move along the same line in opposite directions.