A woman throws an object of mass 500 g with a speed of 25 m s¹. (a) What is the impulse imparted to the object

NCERT Exemplar · Q24

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(a) What is the impulse imparted to the object?

(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?

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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 is the change in momentum. For part (a), it equals the object's momentum since it starts from rest. For part (b), the vector nature of momentum makes the magnitude of change larger than either individual momentum, because direction reverses.

The impulse imparted is $\boxed{12.5\ \text{kg m/s}}$ .
The change in momentum has magnitude $\boxed{18.75\ \text{kg m/s}}$ , opposite to the initial direction of motion.

Momentum ( $p$ ) is a vector quantity defined as the product of an object's mass ( $m$ ) and its velocity ( $v$ ):

p = mv

Impulse ( $J$ ) is also a vector quantity and equals the change in momentum ( $\Delta p$ ) of an object:

J = \Delta p = p_{final} - p_{initial}

Given:

Mass of the object, $m = 500\ \text{g} = 0.500\ \text{kg}$
Initial speed when thrown, $u = 25\ \text{m/s}$

(a) What is the impulse imparted to the object?

This asks for the impulse given to the object by the woman to bring it from rest to a speed of $25\ \text{m/s}$ .

Initial and final states: $v_i = 0\ \text{m/s}$ (at rest before the throw); $v_f = 25\ \text{m/s}$ (its direction taken as positive).
Initial momentum: $p_i = m v_i = 0.500 \times 0 = 0\ \text{kg m/s}$ .
Final momentum: $p_f = m v_f = 0.500 \times 25 = 12.5\ \text{kg m/s}$ .
Impulse imparted: $J = p_f - p_i = 12.5 - 0 = 12.5\ \text{kg m/s}$ , in the direction the object is thrown.

(b) Change in momentum on rebounding

Coordinate system: take the direction of motion toward the wall as positive. Velocity just before hitting the wall: $v_1 = +25\ \text{m/s}$ .
Momentum just before impact: $p_1 = m v_1 = 0.500 \times 25 = +12.5\ \text{kg m/s}$ .
Velocity after rebound: half the original speed, direction reversed: $v_2 = -\tfrac{1}{2}(25) = -12.5\ \text{m/s}$ . …

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Q.A woman throws an object of mass 500 g with a speed of 25 m s $^{1}$ .

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)

(a) What is the impulse imparted to the object?

(b) Change in momentum on rebounding

Concept: Impulse-Momentum Theorem

Q.A woman throws an object of mass 500 g with a speed of 25 m s1^{1}1.

(a) What is the impulse imparted to the object?

(b) Change in momentum on rebounding

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Q.A woman throws an object of mass 500 g with a speed of 25 m s $^{1}$ .

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