Q.A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12m s−1. If the mass of the ball is 0.15kg, determine the impulse imparted to the ball. (Assume linear motion of the ball.)
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. …
The ball reverses direction at the same speed. Taking the initial direction (bowler to batsman) as positive, the initial velocity is vi=+12m/s and the final velocity after being hit back is vf=−12m/s.
The ball reverses direction at constant speed, so its velocity changes from +12m s−1 to −12m s−1; impulse equals the change in momentum, giving 3.6kg m s−1 in magnitude.
Why impulse is about change in momentum
Impulse measures the effect of a force acting over time. Newton's second law in its most general form tells us that the net force equals the rate of change of momentum:
F=dtdp
Integrating both sides over the collision time gives the impulse-momentum theorem:
J=∫Fdt=Δp=pfinal−pinitial
The beauty here is that we don't need to know the force profile or contact time — only the momentum before and after.
Step-by-step calculation
Set up a coordinate system.
Choose the direction from bowler to batsman as positive. The ball initially travels toward the batsman at vi=+12m s−1.
Find the initial momentum.
pi=mvi=0.15×12=1.8kg m s−1
Determine the final velocity.
The batsman hits the ball "straight back" at the same speed, so it now travels toward the bowler. In our coordinate system, vf=−12m s−1.
Calculate the final momentum.
pf=mvf=0.15×(−12)=−1.8kg m s−1
Compute the impulse.J=pf−pi=−1.8−1.8=−3.6kg m s−1 …