A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its

NCERT Exemplar · Q18

Q. A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be

(a) 10.5 N

(b) 21 N

(c)

1.05 \times 10^4

(d)

2.1 \times 10^4

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Concept understanding — Impulse Momentum Theorem

The Intuition: Why Do We Need a "New" Idea?

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:

\vec{J} = \Delta \vec{p}

Where:

$\vec{J}$ is the impulse (a vector)
$\Delta \vec{p}$ is the change in momentum (also a vector)

And since impulse is force times time:

\vec{F}_{\text{avg}} \, \Delta t = m \vec{v}_f - m \vec{v}_i

\vec{J} = \vec{F}_{\text{avg}} \, \Delta t = \Delta \vec{p}

Breaking It Down Piece by Piece

Momentum ( $\vec{p}$ ) is mass times velocity: $\vec{p} = m\vec{v}$ . 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 ( $\vec{J}$ ) is the product of the average force and the time interval over which it acts: $\vec{J} = \vec{F}_{\text{avg}} \, \Delta t$ .

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: $\Delta t$ is small → $\vec{F}_{\text{avg}}$ is large (it hurts)
Soft hands: $\Delta t$ is large → $\vec{F}_{\text{avg}}$ is small (it's comfortable)
In both cases, $\Delta \vec{p}$ 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:

\vec{F}_{\text{net}} = m \vec{a} = m \frac{d\vec{v}}{dt}

Multiply both sides by $dt$ :

\vec{F}_{\text{net}} \, dt = m \, d\vec{v}

Integrate over the time interval:

\int_{t_i}^{t_f} \vec{F}_{\text{net}} \, dt = m \int_{\vec{v}_i}^{\vec{v}_f} d\vec{v} = m\vec{v}_f - m\vec{v}_i

The left side is the impulse (the area under the force-time graph). The right side is the change in momentum. …

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This problem requires applying the Impulse-Momentum Theorem to find the average force exerted during the collision. The ball's momentum changes significantly over a short contact time, and the force the batsman applies must counteract the force from the ball. The magnitude of this force is $\boxed{1.05 \times 10^4 \text{ N}}$ .

When a force acts on an object for a certain duration, it causes a change in the object's momentum. This relationship is precisely described by the Impulse-Momentum Theorem. This theorem is particularly useful in situations involving collisions, where large forces act over very short time intervals, leading to significant changes in momentum.

Here, a cricket ball undergoes a rapid change in its velocity (and thus momentum) upon hitting the bat. We are given the mass of the ball, its initial speed, the nature of the collision (elastic), and the contact time. We need to find the force the batsman applies to keep the bat steady. This force is equal in magnitude to the average force exerted by the ball on the bat during the collision.

\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p} = m(\vec{v}_f - \vec{v}_i)

Let's break down the solution step-by-step:

Convert Units: First, ensure all given quantities are in consistent SI units.
- Mass of the ball, $m = 150 \text{ g} = 0.150 \text{ kg}$ .
- Initial speed of the ball, $v_i = 126 \text{ km/h}$ . To convert this to meters per second (m/s), multiply by $\frac{1000 \text{ m}}{3600 \text{ s}}$ or simply $\frac{5}{18}$ :

v_i = 126 \times \frac{5}{18} \text{ m/s} = 7 \times 5 \text{ m/s} = 35 \text{ m/s}

*   Contact time, $\Delta t = 0.001 \text{ s} = 1 \times 10^{-3} \text{ s}$.

2. Determine Initial and Final Velocities:

Let's define the initial direction of the ball's motion as the positive direction.

* Initial velocity of the ball, $\vec{v}_i = +35 \text{ m/s}$ .

* The problem states the collision is "completely elastic" and the ball "moves straight back". When a ball undergoes a completely elastic collision with a much more massive, stationary object (like a firmly held bat), its speed after the collision remains the same, but its direction reverses.

* Therefore, the final velocity of the ball, $\vec{v}_f = -35 \text{ m/s}$ . The negative sign indicates the opposite direction of motion.

Calculate the Change in Momentum of the Ball: The change in momentum ( $\Delta \vec{p}$ ) is the final momentum minus the initial momentum:

\Delta \vec{p} = m\vec{v}_f - m\vec{v}_i = m(\vec{v}_f - \vec{v}_i)

Substitute the values:

\Delta \vec{p} = 0.150 \text{ kg} \times (-35 \text{ m/s} - (+35 \text{ m/s}))

\Delta \vec{p} = 0.150 \text{ kg} \times (-70 \text{ m/s})

\Delta \vec{p} = -10.5 \text{ kg m/s}

The negative sign indicates that the change in momentum is in the direction opposite to the ball's initial motion. This is the direction of the force exerted *on the ball* by the bat.

4. Apply the Impulse-Momentum Theorem to find the Force on the Ball:

The average force ( $\vec{F}_{bat\_on\_ball}$ ) exerted by the bat on the ball is given by:

\vec{F}_{bat\_on\_ball} = \frac{\Delta \vec{p}}{\Delta t}

\vec{F}_{bat\_on\_ball} = \frac{-10.5 \text{ kg m/s}}{0.001 \text{ s}}

\vec{F}_{bat\_on\_ball} = -10500 \text{ N}

$$ \vec{F}_{bat\_on\_ball} = -1.05 \times 10^4 \text{ N} $$ …

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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
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›Reveal solutionSolution

The ball reverses, so the change in momentum is $2mv$ ; with $v=54\ \text{kmph}=15\ \text{m/s}$ this gives $4.5$ kg m/s.

Convert $54\ \text{kmph}=54\times\tfrac{5}{18}=15\ \text{m/s}$ . Reversal changes the velocity from $+v$ to $-v$ , so …
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The change in momentum on rebound is $2mv$ ; contact time $=\Delta p/F$ .

The ball reverses direction with the same speed, so the change in momentum has magnitude

$\Delta p=mv-(-mv)=2mv=2(0.2\ \text{kg})(5\ \text{m s}^{-1})=2\ \text{kg m s}^{-1}.$ …
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$Q. If the area under the graph between the force on an object and time is 20 units, then the object experiences (A) an impulse of 10 units (B) a change of momentum of 10 units (C) an impulse of 20 units (D) a change of force by 20 units (E) a change of acceleration of 20 units$
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The area under a force–time graph equals impulse (and hence the change in momentum), numerically $20$ .

Impulse $J=\int F\,dt=$ area under the $F$ – $t$ graph $=20$ units. …
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Taking rebound velocity as opposite in sign, $\Delta p = m(v_2-v_1)=0.2(18)=3.6$ Ns.

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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).

Impulse = change in momentum:
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$J = 2mv.$ …
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The ball reverses direction, so the speed change in magnitude is $20+30=50$ m/s. $F=\dfrac{m\Delta v}{\Delta t}=\dfrac{0.15\times50}{0.02}=375$ N.

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$\Delta p = m(v_f-v_i)=0.15\big(-30-20\big)=-7.5\ \text{kg m s}^{-1},$ …
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On rebounding with equal speed, $\Delta p=m(2v)=500$ ; $F=\Delta p/\Delta t=500\times20=1\times10^4$ N.

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$\Delta p=m\,|v_f-v_i|=5\times|50-(-50)|=5\times 100=500\ \text{kg\,ms}^{-1}.$
The average force is …
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Impulse $J=\int F\,dt$ , which is exactly the area under the $F$ vs $t$ curve, and equals the change in momentum. …

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$Q. A hockey player hits a ball with an impulse of 15 Ns. If time of hit is 0.2 s, the average force exerted by the player on the ball is (A) 75 N (B) 50 N (C) 15 N (D) 20 N (E) 25 N$
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Impulse equals average force times contact time, so $F_{avg} = J/t = 15/0.2 = 75\,\text{N}$ .

Impulse is defined as $J = F_{avg}\,\Delta t$ . Rearranging for the average force: …
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Same momentum change over a longer time means a smaller force: $F=\Delta p/t$ gives $1.5$ N.

The ball's momentum change $\Delta p$ is the same in both cases, and $F=\dfrac{\Delta p}{t}$ , so $F\propto\dfrac1t$ . Therefore …
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Area under the force-time graph equals the change in linear momentum.

Concept and Intuition

Impulse is defined as $J=\int F\,dt$ , the area under the force-time curve. The impulse-momentum theorem states this impulse equals the change in linear momentum $\Delta p$ .

Step-by-Step Solution

$J = \int F\,dt =$ area under the $F$ - $t$ graph.

$F = \dfrac{dp}{dt}\Rightarrow \int F\,dt = \Delta p$ . …
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