A cricket ball of mass 150 g has an initial velocity $\mathbf{u} = (3\hat{i} + 4\hat{j})$ m s$^{-1}$ and a final velocity $\mathbf{v} = -(3\hat{i} + 4\hat{j})$ m s$^{-1}$ after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s$^{1}$) (a) zero (b) $-(0.45\hat{i} + 0.6\hat{j})$ (c) $-(0.9\hat{i} + 1.2\hat{j})$ (d) $-5(\hat{i} + \hat{j})$.

A cricket ball of mass 150 g has an initial velocity mathbfu = (3hati + 4hatj) m s⁻¹ and a final velocity math

NCERT Exemplar · Q3

\mathbf{u} = (3\hat{i} + 4\hat{j})

(a) zero

(b)

-(0.45\hat{i} + 0.6\hat{j})

(c)

-(0.9\hat{i} + 1.2\hat{j})

(d)

-5(\hat{i} + \hat{j})

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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.

Important

The theorem works for any force, even if it varies wildly with time. The impulse is always the area under the $F$ - $t$ curve, and it always equals the change in momentum.

How to Use It in Problems

Identify the object whose momentum changes
Find initial and final velocities (and mass)
Compute $\Delta p$ = $m(v_f - v_i)$ (watch direction — use signs)
Set $\Delta p$ equal to $F_{\text{avg}} \Delta t$
Solve for the unknown (force, time, mass, or velocity)

Tip

If the force is not constant, use the average force. The impulse is still $F_{\text{avg}} \Delta t$ , and it still equals $\Delta p$ .

The Bottom Line

The Impulse-Momentum Theorem is not a new law — it's Newton's second law rewritten in a form that's often more useful. It tells you that to change an object's momentum, you need to apply a force for some time. The longer you apply it, the less force you need. That's why catching a ball with "give" feels easier, and why airbags save lives.

Final takeaway: Impulse = Force × Time = Change in Momentum.

"Impulse Momentum Theorem derivation" and "Impulse Momentum Theorem numerical problems" are two of the most common searches tied to this topic, and Impulse Momentum Theorem is a core, NCERT-aligned topic from the Laws of Motion portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Pairing this explanation with NCERT Physics textbook practice and previous years' questions is the surest way to lock the concept in before an exam.

The ball's velocity reverses direction completely, so the change in momentum is twice the initial momentum in the opposite direction: $\Delta \mathbf{p} = -(0.9\hat{i} + 1.2\hat{j})$ kg m s $^{-1}$ .

When a cricket ball is struck, its momentum changes. Momentum is a vector quantity $\mathbf{p} = m\mathbf{v}$ , and the change in momentum tells us about the impulse delivered by the bat. The key insight here is that the ball doesn't just stop—it reverses direction entirely, which means the momentum change is substantial.

The change in momentum is defined as:

\Delta \mathbf{p} = \mathbf{p}_{\text{final}} - \mathbf{p}_{\text{initial}} = m\mathbf{v} - m\mathbf{u} = m(\mathbf{v} - \mathbf{u})

This vector subtraction will account for both the magnitude and direction of the momentum change.

Watch out

A common mistake is to think that because the speeds are the same (5 m/s before and after), the momentum change is zero. But momentum is a vector—direction matters! The ball has completely reversed its velocity, so the momentum change is definitely non-zero.

Let me work through this systematically:

Convert the mass to SI units The mass is given as 150 g, which we need in kilograms:

m = 150 \text{ g} = 0.15 \text{ kg}

Identify the initial and final velocities

Initial velocity: $\mathbf{u} = (3\hat{i} + 4\hat{j})$ m/s

Final velocity: $\mathbf{v} = -(3\hat{i} + 4\hat{j})$ m/s

Notice that $\mathbf{v} = -\mathbf{u}$ . The ball has reversed direction completely.
Calculate the velocity change

\mathbf{v} - \mathbf{u} = -(3\hat{i} + 4\hat{j}) - (3\hat{i} + 4\hat{j})

= -3\hat{i} - 4\hat{j} - 3\hat{i} - 4\hat{j}

= -6\hat{i} - 8\hat{j} \text{ m/s}

Find the change in momentum Multiply the velocity change by the mass:

\Delta \mathbf{p} = m(\mathbf{v} - \mathbf{u}) = 0.15 \times (-6\hat{i} - 8\hat{j})

= -0.9\hat{i} - 1.2\hat{j} \text{ kg m/s}

This can be written as $-(0.9\hat{i} + 1.2\hat{j})$ kg m s $^{-1}$ .

Tip

When a ball bounces or reverses direction elastically (same speed, opposite direction), the momentum change is always $\Delta \mathbf{p} = -2m\mathbf{u}$ . Here: $-2 \times 0.15 \times (3\hat{i} + 4\hat{j}) = -(0.9\hat{i} + 1.2\hat{j})$ .

✓Final answer

-(0.9\hat{i} + 1.2\hat{j})

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Q.A cricket ball of mass 150 g has an initial velocity $\mathbf{u} = (3\hat{i} + 4\hat{j})$ m s $^{-1}$ and a final velocity $\mathbf{v} = -(3\hat{i} + 4\hat{j})$ m s $^{-1}$ after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s $^{1}$ )