A batsman deflects a ball by an angle of $45^{\circ}$ without changing its initial speed which is equal to $54\ \text{km/h}$. What is the impulse imparted to the ball? (Mass of the ball is $0.15\ \text{kg}$.)

A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54km/h. What

Exercises · 4.20

45^{\circ}

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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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Impulse equals the change in momentum. The speed is unchanged, so only the direction changes by $45^\circ$ ; treating the initial and final velocity vectors as two equal-magnitude sides of an isosceles triangle gives $|\Delta v| = 2v\sin(\theta/2)$ . With $v=15\ \text{m s}^{-1}$ and $\theta=45^\circ$ : $J \approx 1.72\ \text{N·s}$ .

Concept

By the impulse–momentum theorem, $\vec{J} = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i$ . Even though the speed is unchanged, momentum is a vector, so a change of direction gives a non-zero impulse. The initial velocity $\vec{v}_i$ and final velocity $\vec{v}_f$ have the same magnitude $v$ , with the angle between them equal to the $45^\circ$ deflection. Geometrically, $\vec{v}_i$ , $\vec{v}_f$ , and $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$ form an isosceles triangle with two sides of length $v$ and included angle $\theta$ .

Step 1 — Convert the speed

v = 54\ \text{km/h} = 54 \times \tfrac{5}{18} = 15\ \text{m s}^{-1}

Step 2 — Magnitude of the velocity change

For an isosceles triangle with two sides $v$ and included angle $\theta$ , the base (the magnitude of $\Delta \vec{v}$ ) is:

|\Delta \vec{v}| = \sqrt{v^2+v^2-2v^2\cos\theta} = v\sqrt{2(1-\cos\theta)} = 2v\sin\left(\frac{\theta}{2}\right)

With $\theta = 45^\circ$ :

$|\Delta \vec{v}| = 2(15)\sin(22.5^\circ) = 30 \times 0.3827 \approx 11.48\ \text{m s}^{-1}$ …

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Q.A batsman deflects a ball by an angle of $45^{\circ}$ without changing its initial speed which is equal to $54\ \text{km/h}$ . What is the impulse imparted to the ball? (Mass of the ball is $0.15\ \text{kg}$ .)

Concept understanding — Impulse Momentum Theorem

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

The Precise Statement

Breaking It Down Piece by Piece

Why This Matters: Real-World Examples

The Mathematical Derivation (Short)

Concept

Step 1 — Convert the speed

Step 2 — Magnitude of the velocity change

Concept: Impulse as a Vector Change in Momentum (Speed Unchanged, Direction Changed)

Q.A batsman deflects a ball by an angle of 45∘45^{\circ}45∘ without changing its initial speed which is equal to 54 km/h54\ \text{km/h}54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg0.15\ \text{kg}0.15 kg.)

Concept

Step 1 — Convert the speed

Step 2 — Magnitude of the velocity change

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Q.A batsman deflects a ball by an angle of $45^{\circ}$ without changing its initial speed which is equal to $54\ \text{km/h}$ . What is the impulse imparted to the ball? (Mass of the ball is $0.15\ \text{kg}$ .)

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