Q.Derive the law of conservation of linear momentum from Newton's third law of motion. A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms^-1. What is the speed of the recoil of the gun? OR Derive an expression for maximum speed of a car on a banked road in circular motion. Find the angle through which a cyclist bends from the vertical, when he covers a circular path of 34.3 m circumference in sqrt(22) seconds.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Conservation of Momentum
Conservation of Momentum: From Push to Principle
Imagine you're standing on perfectly smooth ice, wearing skates. You're completely still. Now, you push a heavy medicine ball away from you. What happens? You roll backward. The harder you push the ball, the faster you roll back.
That's the core intuition: you can't push something away without being pushed back yourself. The push you give the ball is matched by an equal push on you, in the opposite direction. This isn't a special property of ice or skates — it's a fundamental rule of how forces work in the universe.
The Hidden Quantity That Never Changes
Physicists call the "amount of motion" an object has its momentum. For everyday speeds, momentum is simple:
p=mv
Where m is mass (how much stuff) and v is velocity (speed with direction). Momentum is a vector — it cares about which way you're going.
A truck creeping forward has huge momentum (big mass, small speed). A bullet zipping through air has moderate momentum (tiny mass, huge speed). A parked car has zero momentum (speed is zero).
Now here's the key: in any isolated system (no outside forces), total momentum stays the same. Always. Before, during, and after any interaction.
The Precise Statement
Law of Conservation of Momentum:
In a closed, isolated system (no external forces), the total vector momentum of the system remains constant over time.
Mathematically, for two objects that interact (collide, push apart, explode):
p1,initial+p2,initial=p1,final+p2,final
Or in terms of masses and velocities:
m1u1+m2u2=m1v1+m2v2
Where u means initial velocity and v means final velocity.
Why This Works: Newton's Third Law in Disguise
When you push the medicine ball, your hand exerts a force F on the ball. By Newton's Third Law, the ball exerts an equal and opposite force −F back on your hand. These forces act for the same time Δt.
Force times time equals impulse, which equals change in momentum:
FΔt=Δp
For you and the ball:
- Ball's momentum change: +FΔt (ball goes forward)
- Your momentum change: −FΔt (you go backward)
Add them: +FΔt+(−FΔt)=0
Total change is zero. Momentum is conserved because forces always come in equal-and-opposite pairs.
This is why a rocket works in the vacuum of space. It throws exhaust backward (one momentum change), and the rocket itself moves forward (equal opposite momentum change). No air needed — just Newton's Third Law and conservation of momentum.
What This Law Does NOT Mean
- It does NOT mean individual objects keep constant momentum. Only the total of all objects in the system stays constant. Individual momenta can change wildly.
- It does NOT apply if external forces act. If friction, gravity from outside, or a wall stops something, momentum is not conserved for that system. (You can expand the system to include the Earth or the wall, and then momentum is conserved again.)
- It does NOT require collisions to be elastic. Even in a messy, sticky, energy-losing collision, momentum is still perfectly conserved. Energy can be lost to heat or deformation, but momentum never disappears.
A Quick Example …
This part-D question offers a choice of two sub-questions; the primary option (momentum conservation and the gun-bullet recoil) is answered here. …
Newton's third law leads to conservation of linear momentum; applying it to the gun-bullet system gives a recoil speed of 2.5 m/s.
Part 1 — Derivation of conservation of linear momentum from Newton's third law:
Consider two bodies A and B that interact with each other (e.g., in a collision or explosion), with no external force acting on the system. Let FAB be the force exerted by B on A, and FBA be the force exerted by A on B. By Newton's third law:
FAB=−FBA
By Newton's second law, force equals the rate of change of momentum, so FAB=dtdpA and FBA=dtdpB. Substituting:
dtdpA=−dtdpB⇒dtdpA+dtdpB=0⇒dtd(pA+pB)=0
This means pA+pB=constant — the total linear momentum of the isolated two-body system does not change with time. This is the law of conservation of linear momentum: in the absence of an external force, the total momentum of a system remains constant.
Part 2 — Numerical (bullet and gun):
Given: mass of bullet m1=50 g=0.05 kg, mass of gun m2=10 kg, muzzle speed of bullet v1=500 m/s.
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- CBSE 2025Set ANNUAL1 markMCQQ.A bullet of mass m and velocity v is fired into a large block of wood of mass M which is at rest. If the bullet gets stuck in the block, the final velocity of the system is(a) (A) m/(m+M) × v(b) (B) M/(m+M) × v(c) (C) (m+M)/m × v(d) (D) (m+M)/M × v
›Reveal solutionSolution
[!TLDR]
(A) m/(m+M) × v
Why
Conservation of momentum: mv = (m+M)v' ⇒ v' = mv …
- CBSE 2023Set ANNUAL1 markMCQQ.A shell of mass 200 g is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, then the recoil speed of the gun is(1) 16 cms^-1(2) 8 cms^-1(3) 8 ms^-1(4) 16 ms^-1
›Reveal solutionSolution
The gun-shell system starts at rest, so by conservation of momentum the gun recoils with momentum equal and opposite to the shell's momentum.
Given:
Mass of shell, m1 = 200 g = 0.2 kg
Mass of gun, m2 = 100 kg
Muzzle speed of shell, v1 = 80 m/s
…
- CBSE 2023Set annual1 markQ.Gun recoils back when it is being fired is based on law of conservation of linear momentum. (True/False)
›Reveal solutionSolution
True. The recoil of a gun is a classic real-world demonstration of conservation of linear momentum.
Before firing, the gun and the bullet inside it are both at rest, so the total momentum of the (gun + bullet) system is zero.
When the gun is fired, the bullet is pushed forward and gains a forward momentum m(v). Since no external horizontal force acts on the (gun + bullet) system during firing (the explosive force is internal), the total momentum of the system must remain zero, exactly as it was before firing.
…
- CBSE 2023Set ANNUAL1 markMCQQ.Rocket projection is based on which principle:(a) nuclear fussion(b) nuclear fission(c) conservation of momentum(d) none of these
›Reveal solutionSolution
Rocket propulsion is based on the conservation of linear momentum.
A rocket burns fuel and expels hot gases downward/backward with large momentum. Since no external force acts (ideally) on the rocket-gas system, total momentum is conserved: the backward momentum of the ejected gases i …
- CBSE 2022Set TERM11 markMCQQ.A bullet of mass 40 gm is fired from a gun of mass 8 kg with a velocity of 800 m/s. Calculate the recoil velocity of the gun.(1) 1 m/s(2) -1 m/s(3) 2 m/s(4) -4 m/s
›Reveal solutionSolution
Conservation of momentum for the gun-bullet system (initially at rest) requires the gun's recoil momentum to exactly cancel the bullet's forward momentum: m_gun v_gun = -m_bullet v_bullet, giving v_gun = -4 m/s.
Before firing, the gun+bullet system is at rest, so its total momentum is zero.
By conservation of momentum, the total momentum immediately after firing must also be zero:
m_bullet v_bullet + M_gun v_gun = 0
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- CBSE 2021Set ANN1 markQ.Momentum is a _______ quantity. (vector/scalar)
›Reveal solutionSolution
Momentum p = mv is a vector because it is the product of a scalar (mass) and a vector (velocity); it therefore has both magnitude and direction, the direction being the same as that of the velocity.
Momentum is defined as p = mv, where m (mass) is a positive scalar and v (velocity) is a vector. Multiplying a vector by a positive scalar only scales its magnitude — it never removes its directionality. Hence momentum inherits a direction, alwa …
- CBSE 2021Set hz1 markQ.On what principle does a rocket work?
›Reveal solutionSolution
Rocket propulsion is a direct application of the law of conservation of linear momentum.
Initially the rocket (with its fuel) is at rest, so the total momentum of the rocket-plus-fuel system is zero. As fuel burns, gas is ejected backward through the nozzle at very high velocity. Since no external horizontal force acts on the rocket-gas system, its total momentum must remain conserved.
If a small mass dm of gas is ejected backward with velocity v (relative to the rocket), it carries backward momentum. For the total momentum of the system to stay constant (zero, if starting from rest), the rocket must gain an equal and opposite forward momentum. This gives rise to a forward force called thrust:
Thrust F = v (dm/dt)
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- CBSE 2020Set ANNUAL1 markMCQQ.In the motion of a rocket, the physical quantity conserved is:(a) Angular momentum(b) Linear momentum(c) Force(d) Work
›Reveal solutionSolution
A rocket has no external horizontal force acting on the rocket-plus-exhaust system, so total linear momentum is conserved as fuel is ejected.
In rocket propulsion, hot gases are ejected backward at high speed. Since the rocket and the ejected gas together form an isolated system (ignoring gravity/air resistance for the basic principle), the total linear momentum of the system remains constant.
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- CBSE 2020Set ANNUAL1 markMCQQ.A man fires a bullet of mass 0.2 Kg with a speed of 5 m/s. The gun is of 1Kg mass. By what velocity does the gun recoil backwards?(a) 0.01 m/s(b) 0.1 m/s(c) 1 m/s(d) 10 m/s
›Reveal solutionSolution
By conservation of momentum (initial total momentum = 0, since both gun and bullet start at rest), the gun recoils with a velocity that makes its momentum exactly cancel the bullet's forward momentum, giving 1 m/s.
Before firing, both the gun and bullet are at rest, so the total momentum of the system is zero. Since the force of firing is entirely internal to the gun-bullet system, total momentum is conserved:
m_bullet v_bullet + m_gun (-v_gun) = 0 …
- CBSE 2019Set ANNUAL1 markMCQQ.Match the columns. Column A item: Momentum. Pick its correct dimensional formula from Column B.(a) [M L^2 T^-3](b) [M^0 L T^-2](c) [M L^2 T^-2](d) [M L T^-2](e) [M L T^-1](f) [M^0 L T^0]
›Reveal solutionSolution
Momentum = mass x velocity = M x (L T^-1) = M L T^-1.
Linear momentum is defined as p = mv, the product of mass and velocity. Mass has dimension [M] and velocity has dimension [L T^-1], so momentum has dimensional formul …
- CBSE 2018Set hz1 markQ.Rocket propulsion is based on law of conservation of linear momentum. (True/False)
›Reveal solutionSolution
The statement is True: a rocket's forward thrust arises purely from conservation of linear momentum as it ejects mass (exhaust gases) backward.
A rocket carries its own fuel and oxidizer, and propels itself by burning fuel and ejecting the hot exhaust gases backward at very high speed. Before ignition, the total momentum of the rocket + fuel system is zero (or constant, if already moving). As gas is ejected backward with momentum, by the law of conservation of linear momentum (no external force needed for this internal process), the rocket itself must gain an equal and opposite momentum forward. This is exactly analogous to the recoil of a gun. …
- CBSE 2018Set annual1 markMCQQ.The working of a rocket is based on the principle of: (A) Elasticity (B) Kepler's law (C) Newton's law of gravitation (D) Conservation of momentum
›Reveal solutionSolution
Rocket propulsion is a direct consequence of conservation of linear momentum: gases are expelled backward, so the rocket is pushed forward.
A rocket works by burning fuel and ejecting the resulting hot gases at very high speed out of its rear nozzle. Before ignition, the rocket + fuel system is at rest (or moving with some velocity), with a certain total momentum. As fuel burns, exhaust gases of small mass but very high backward velocity are continuously ejected.
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