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.)
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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 …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set eng-2026-04184 marksMCQ
Q.A batsman hits a cricket ball of mass 0.15 kg travelling at a speed of 54 kmph. The ball reverses its direction. The impulse imparted to the ball in kgms−1 is
(A) 4.5
(B) 45
(C) 810
(D) 8.1
(E) 16.2
›Reveal solutionSolution
The ball reverses, so the change in momentum is 2mv; with v=54kmph=15m/s this gives 4.5 kg m/s.
Convert 54kmph=54×185=15m/s. Reversal changes the velocity from +v to −v, so …
Q.A ball of 200 g mass moving with a speed of 5 ms−1 collides with a wall and bounces back with the same speed. If the force exerted on the wall is 1 N, then the ball is in contact with the wall for
(A) 2 s
(B) 1 s
(C) 0.5 s
(D) 1.5 s
(E) 0.75 s
›Reveal solutionSolution
The change in momentum on rebound is 2mv; contact time =Δp/F.
The ball reverses direction with the same speed, so the change in momentum has magnitude
Q.If the area under the graph between the force on an object and time is 20 units, then the object experiences
(A) an impulse of 10 units
(B) a change of momentum of 10 units
(C) an impulse of 20 units
(D) a change of force by 20 units
(E) a change of acceleration of 20 units
›Reveal solutionSolution
The area under a force–time graph equals impulse (and hence the change in momentum), numerically 20.
Impulse J=∫Fdt= area under the F–t graph =20 units. …
Q.0.2 kg ball strikes a wall with velocity 10 ms−1 and rebounds with 8 ms−1. The impulse delivered by the ball is
(A) 0.4 Ns
(B) 3.6 Ns
(C) 1.4 Ns
(D) 1.8 Ns
(E) 16.0 Ns
›Reveal solutionSolution
Taking rebound velocity as opposite in sign, Δp=m(v2−v1)=0.2(18)=3.6 Ns.
Impulse equals the change in momentum. Take the incoming direction as positive: v1=+10m/s, and after rebound the ball moves the other way, v2=−8m/s. …
Q.If a ball of mass 0.02 kg bowled by a bowler straight to a batsman is hit back with the same speed with an impulse of 2 Ns, then the speed of the ball bowled is
(A) 20 ms−1
(B) 80 ms−1
(C) 50 ms−1
(D) 60 ms−1
(E) 40 ms−1
›Reveal solutionSolution
The ball reverses direction with equal speed, so the impulse equals 2mv. Solving 2=2(0.02)v gives v=50 m s−1.
Take the incoming direction as positive. The ball arrives with velocity +v and leaves with velocity −v (same speed, reversed).
Q.A tennis ball of mass 150 g is moving at 20 ms−1. A racket strikes it, reversing its direction with a final speed of 30 ms−1. If the contact time is 0.02 s, then the magnitude of the force (in N) exerted by the racket is
(A) 1.5 N
(B) 3.75 N
(C) 15 N
(D) 150 N
(E) 375 N
›Reveal solutionSolution
The ball reverses direction, so the speed change in magnitude is 20+30=50 m/s. F=ΔtmΔv=0.020.15×50=375 N.
Take the initial direction as positive: vi=+20 m/s, and after the strike the ball moves the opposite way at 30 m/s, so vf=−30 m/s. The change in momentum is
Q.A body of mass 5 kg collides with a wall with a speed of 50 ms−1 and rebounds with the same speed. If the time of contact of the body with the wall is 201s the force exerted on the wall is
(A) 0.5×104N
(B) 2.5×104N
(C) 2×103N
(D) 1×104N
(E) 4×103N
›Reveal solutionSolution
On rebounding with equal speed, Δp=m(2v)=500; F=Δp/Δt=500×20=1×104 N.
The body rebounds with the same speed in the opposite direction, so the magnitude of the change in momentum is
Q.A hockey player hits a ball with an impulse of 15 Ns. If time of hit is 0.2 s, the average force exerted by the player on the ball is
(A) 75 N
(B) 50 N
(C) 15 N
(D) 20 N
(E) 25 N
›Reveal solutionSolution
Impulse equals average force times contact time, so Favg=J/t=15/0.2=75N.
Impulse is defined as J=FavgΔt. Rearranging for the average force: …
Q.When a cricketer catches a ball in 30 s, the force required is 2.5 N. The force required to catch that ball in 50 s is
(A) 1.5 N
(B) 1 N
(C) 2.5 N
(D) 3 N
(E) 5 N
›Reveal solutionSolution
Same momentum change over a longer time means a smaller force: F=Δp/t gives 1.5 N.
The ball's momentum change Δp is the same in both cases, and F=tΔp, so F∝t1. Therefore …
Q.Area under the force-time graph gives the change in
(A) velocity
(B) acceleration
(C) linear momentum
(D) angular momentum
(E) impulsive force
›Reveal solutionSolution
Area under the force-time graph equals the change in linear momentum.
Concept and Intuition
Impulse is defined as J=∫Fdt, the area under the force-time curve. The impulse-momentum theorem states this impulse equals the change in linear momentum Δp.