Q.A 100 kg gun fires a ball of 1kg horizontally from a cliff of height 500m. It falls on the ground at a distance of 400m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 m s−2)
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
Important
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.
Note
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.
The gun and ball form an isolated system in the horizontal direction. Initially both are at rest, so total horizontal momentum is zero. When the gun fires, momentum is conserved: the ball gains forward momentum and the gun recoils backward with equal magnitude.
Step 1: Find the time of flight from vertical motion.
The ball falls freely through height h=500 m under gravity g=10 m/s². Using h=21gt2:
t=g2h=102×500=10 s
Step 2: Find the horizontal velocity of the ball.
The ball travels horizontal distance x=400 m in time t=10 s, so:
The ball leaves at 40m/s horizontally (from its projectile motion); conserving momentum for the gun-ball system gives a gun recoil of 0.4m/s in the direction opposite to the ball.
The gun and ball start at rest, so the total horizontal momentum of the system is zero. The firing force is internal, so horizontal momentum is conserved: the forward momentum of the ball is balanced by the backward momentum of the gun.
Ball's horizontal velocity from projectile motion
Time of flight (vertical motion). The ball falls freely from height h=500m:
t=g2h=102×500=100=10s.
Horizontal (muzzle) velocity. It covers the range R=400m at constant horizontal speed:
Concept: Conservation of Momentum + Projectile Motion
The ball's horizontal launch speed is found from its projectile motion, then momentum conservation (initial momentum = 0) gives the gun's recoil velocity.
Step 1: Find the time of flight from the vertical fall
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set pha-2026-0419F4 marksMCQ
Q.When the total external force acting on a moving system is zero, the velocity of the center of mass
(A) remains constant
(B) becomes zero
(C) increases
(D) becomes infinity
(E) decreases
›Reveal solutionSolution
Fext=Macm; if Fext=0 then acm=0, so vcm stays constant (conservation of momentum). …
Q.A shell travelling along a parabolic path in the gravitational field of the earth undergoes explosion in mid air. The centre of mass of the fragments will move
(A) horizontally and then vertically down
(B) along the original parabolic path
(C) vertically down
(D) vertically up and then vertically down
(E) horizontally and then in the parabolic path
›Reveal solutionSolution
Explosion forces are internal and cancel in pairs; the only external force is gravity, so the centre of mass follows the same parabola it would have without the explosion.
During the explosion the fragments push on each other with internal forces that occur in equal-and-opposite pairs, so they contribute nothing to the motion of the centre of mass. The only external force on the system is gravity. …
Q.A block of mass $M$ moves with a velocity $v$ along a frictionless horizontal surface towards another block of mass $2M$ at rest. The velocity of the center of mass of the system of blocks is
(A) $\frac{v}{2}$
(B) $2v$
(C) $3v$
(D) $\frac{v}{3}$
(E) $\frac{v}{4}$
›Reveal solutionSolution
Centre-of-mass velocity = total momentum / total mass.
Q.An object at rest suddenly explodes into three parts of equal masses. Two of them move away at right angles to each other with equal speed of 10 m/s. The speed of the third part just after the explosion will be:
(A) 10 m/s
(B) 20 m/s
(C) 210 m/s
(D) 0
(E) 102 m/s
›Reveal solutionSolution
The third fragment moves at 102 m/s.
Concept and Intuition
Total momentum is conserved and starts at zero. Two equal-mass fragments fly off perpendicularly with equal speed; their resultant momentum is balanced by the third fragment's momentum.
Step-by-Step Solution
The two perpendicular momenta each have magnitude m(10); their resultant =m⋅102.
The third fragment (same mass m) must carry equal and opposite momentum: mv3=m⋅102.
Q.A man weighing 70 kg is riding on a cart of mass 30 kg which moves along a level floor at a speed of 3ms−1. If he runs on the cart so that his velocity relative to the cart is 4ms−1 in the direction opposite to the motion of the cart, the speed of centre of mass of the system is
(A) 0.3ms−1
(B) 0.5ms−1
(C) 0.2ms−1
(D) 0.1ms−1
(E) zero
›Reveal solutionSolution
With the man moving at -1 m/s and the cart at 3 m/s, the centre of mass moves at 0.2 m/s.
Concept and Intuition
The velocity of the centre of mass is the mass-weighted average of the parts' velocities. The man's velocity relative to the ground is his velocity relative to the cart added to the cart's velocity; running backward at 4 m/s relative to a cart moving forward at 3 m/s puts him at -1 m/s (1 m/s backward) in the ground frame.