Imagine you're catching a cricket ball. If you let your hands stay rigid, the ball stings and might bounce off. But if you give with the ball — pulling your hands back as you catch — the catch feels soft and the ball stops gently.
Same ball, same speed, same change in momentum. But the force you feel is completely different. Why?
The answer is time. When you pull your hands back, you increase the time over which the ball slows down. A longer time means a smaller force — even though the total "oomph" needed to stop the ball is the same. That "oomph" is called impulse.
Note
Impulse is not a mysterious new quantity. It's just force multiplied by the time it acts. If you push gently for a long time, or push hard for a short time, you can produce the same effect.
The Precise Statement
The Impulse-Momentum Theorem says:
The impulse delivered to an object equals the change in its momentum.
In symbols:
J=Δp
Where:
J is the impulse (a vector)
Δp is the change in momentum (also a vector)
And since impulse is force times time:
FavgΔt=mvf−mvi
J=FavgΔt=Δp
Breaking It Down Piece by Piece
Momentum (p) is mass times velocity: p=mv. It's a measure of how hard it is to stop a moving object. A truck moving slowly has large momentum; a bullet moving fast has large momentum too.
Impulse (J) is the product of the average force and the time interval over which it acts: J=FavgΔt.
The theorem connects them: the net impulse changes the momentum. If you apply a net force to an object for some time, its momentum changes by exactly that amount.
Watch out
A common mistake is to think impulse is just force. It's force × time. A huge force acting for a tiny time (like a bat hitting a ball) can produce the same impulse as a tiny force acting for a long time (like a gentle push).
Why This Matters: Real-World Examples
Catching a ball (soft vs. hard hands)
Hard hands: Δt is small → Favg is large (it hurts)
Soft hands: Δt is large → Favg is small (it's comfortable)
In both cases, Δp is the same (ball goes from moving to stopped)
Airbags in cars
Without airbag: your head hits the dashboard in ~0.01 s → huge force
With airbag: your head decelerates over ~0.1 s → force is 10 times smaller
Same change in momentum, but the airbag extends the time
A cricket bat hitting a ball
The bat is in contact with the ball for a few milliseconds
The force during that contact is enormous (hundreds of Newtons)
The impulse changes the ball's momentum from one direction to another
The Mathematical Derivation (Short)
Start from Newton's second law:
Fnet=ma=mdtdv
Multiply both sides by dt:
Fnetdt=mdv
Integrate over the time interval:
∫titfFnetdt=m∫vivfdv=mvf−mvi
The left side is the impulse (the area under the force-time graph). The right side is the change in momentum. …
Impulse-Momentum Theorem — the momentum transferred equals ∣Δp∣ for the ball.
From the referenced problem 5.3: a ball of mass m=0.15 kg has initial velocity u=(3i^+4j^) m/s and, after the hit, final velocity v=−(3i^+4j^) m/s. …
The momentum transferred equals the magnitude of the ball's change in momentum, ∣Δp∣=1.5kg m s−1, so option (C) is correct.
This question refers back to problem 5.3, where a cricket ball of mass m=150g=0.15kg has initial velocity u=(3i^+4j^)m s−1 and, after being hit straight back by the bat, final velocity v=−(3i^+4j^)m s−1 — same speed, exactly reversed direction. The "momentum transferred" is the change in the ball's momentum, which by the impulse-momentum theorem equals the impulse delivered by the bat.