In the previous problem (5.3), the magnitude of the momentum transferred during the hit is (a) Zero (b) 0.75 k

NCERT Exemplar · Q4

Q. In the previous problem (5.3), the magnitude of the momentum transferred during the hit is

(a) Zero

(b)

0.75

kg m s

^{-1}

(c)

1.5

kg m s

^{-1}

(d)

14

kg m s

^{-1}

Kerala DhseMCQ· 1mImportance★★★★★est

51% · 39/77 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

The momentum transferred equals the magnitude of the ball's change in momentum, $|\Delta\vec{p}| = 1.5\ \text{kg m s}^{-1}$ , so option (C) is correct.

This question refers back to problem 5.3, where a cricket ball of mass $m = 150\ \text{g} = 0.15\ \text{kg}$ has initial velocity $\vec{u} = (3\hat{i} + 4\hat{j})\ \text{m s}^{-1}$ and, after being hit straight back by the bat, final velocity $\vec{v} = -(3\hat{i} + 4\hat{j})\ \text{m s}^{-1}$ — same speed, exactly reversed direction. The "momentum transferred" is the change in the ball's momentum, which by the impulse-momentum theorem equals the impulse delivered by the bat.

Change in momentum. …

Start your 14-day free trial to unlock the full solution →

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
$^{-1}$
›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 …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04194 marksMCQ
$g$
›Reveal solutionSolution

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

$\Delta p=mv-(-mv)=2mv=2(0.2\ \text{kg})(5\ \text{m s}^{-1})=2\ \text{kg m s}^{-1}.$ …
Start your 14-day free trial to see the full solution →
KEAM 2026Set eng-2026-04224 marksMCQ
$Q. If the area under the graph between the force on an object and time is 20 units, then the object experiences (A) an impulse of 10 units (B) a change of momentum of 10 units (C) an impulse of 20 units (D) a change of force by 20 units (E) a change of acceleration of 20 units$
›Reveal solutionSolution

The area under a force–time graph equals impulse (and hence the change in momentum), numerically $20$ .

Impulse $J=\int F\,dt=$ area under the $F$ – $t$ graph $=20$ units. …
Start your 14-day free trial to see the full solution →
KEAM 2026Set pha-2026-0420F4 marksMCQ
$^{-1}$
›Reveal solutionSolution

Taking rebound velocity as opposite in sign, $\Delta p = m(v_2-v_1)=0.2(18)=3.6$ Ns.

Impulse equals the change in momentum. Take the incoming direction as positive: $v_1 = +10\,\mathrm{m/s}$ , and after rebound the ball moves the other way, $v_2 = -8\,\mathrm{m/s}$ . …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04254 marksMCQ
$20$
›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).

Impulse = change in momentum:
$J = m v_f - m v_i = m(-v) - m(v) = -2mv.$
The magnitude is

$J = 2mv.$ …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04274 marksMCQ
$^{-1}$
›Reveal solutionSolution

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.

Take the initial direction as positive: $v_i=+20$ m/s, and after the strike the ball moves the opposite way at $30$ m/s, so $v_f=-30$ m/s. The change in momentum is

$\Delta p = m(v_f-v_i)=0.15\big(-30-20\big)=-7.5\ \text{kg m s}^{-1},$ …
Start your 14-day free trial to see the full solution →
KEAM 2025Set eng-2025-04284 marksMCQ
$50 \text{ ms}^{-1}$
›Reveal solutionSolution

On rebounding with equal speed, $\Delta p=m(2v)=500$ ; $F=\Delta p/\Delta t=500\times20=1\times10^4$ N.

The body rebounds with the same speed in the opposite direction, so the magnitude of the change in momentum is
$\Delta p=m\,|v_f-v_i|=5\times|50-(-50)|=5\times 100=500\ \text{kg\,ms}^{-1}.$
The average force is …
Start your 14-day free trial to see the full solution →
KEAM 2025Set pha-2025-0424A4 marksMCQ
$F$
›Reveal solutionSolution

Impulse $J=\int F\,dt$ , which is exactly the area under the $F$ vs $t$ curve, and equals the change in momentum. …

Start your 14-day free trial to see the full solution →
KEAM 2025Set pha-2025-0429F4 marksMCQ
$Q. A hockey player hits a ball with an impulse of 15 Ns. If time of hit is 0.2 s, the average force exerted by the player on the ball is (A) 75 N (B) 50 N (C) 15 N (D) 20 N (E) 25 N$
›Reveal solutionSolution

Impulse equals average force times contact time, so $F_{avg} = J/t = 15/0.2 = 75\,\text{N}$ .

Impulse is defined as $J = F_{avg}\,\Delta t$ . Rearranging for the average force: …
Start your 14-day free trial to see the full solution →
KEAM 2024Set eng-2024-06064 marksMCQ
$Q. When a cricketer catches a ball in 30 s, the force required is 2.5 N. The force required to catch that ball in 50 s is (A) 1.5 N (B) 1 N (C) 2.5 N (D) 3 N (E) 5 N$
›Reveal solutionSolution

Same momentum change over a longer time means a smaller force: $F=\Delta p/t$ 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 …
Start your 14-day free trial to see the full solution →
KEAM 2021Set eng-2021-P1-A14 marksMCQ
$Q. Area under the force-time graph gives the change in (A) velocity (B) acceleration (C) linear momentum (D) angular momentum (E) impulsive force$
›Reveal solutionSolution
Area under the force-time graph equals the change in linear momentum.

Concept and Intuition

Impulse is defined as $J=\int F\,dt$ , 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.

$F = \dfrac{dp}{dt}\Rightarrow \int F\,dt = \Delta p$ . …
Start your 14-day free trial to see the full solution →

←Previous questionQ3 — A cricket ball of mass 150 g has an initial velocity mathbfu = (3hati …→Next questionQ5 — Conservation of momentum in a collision between particles can be under…

Q.In the previous problem (5.3), the magnitude of the momentum transferred during the hit is

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: Magnitude of the Momentum Transferred (Impulse Magnitude)

Concept and Intuition

Step-by-Step Solution

Q.In the previous problem (5.3), the magnitude of the momentum transferred during the hit is

🎓Unlock everything free for 14 days

Unlock everything free for 14 days