This MCQ module is based on: Second Law Momentum
TOPIC 15 OF 33
Second Law Momentum
🎓 Class 11
Physics
CBSE
Theory
Ch 4 – Laws of Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Second Law Momentum
4.5 Newton's Second Law of Motion
The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general situation when there is a net external force acting on the body. It relates the net external force to the acceleration of the body.
Momentum
Momentum of a body is defined as the product of its mass and velocity:
\[\vec{p} = m\vec{v}\]
Momentum is a vector quantity having the same direction as velocity. Its SI unit is kg·m/s. Some everyday observations show why momentum (not just speed or mass alone) is the important quantity:
- A light object (small mass) moving fast can have the same momentum as a heavy object moving slowly. A bullet (small m, huge v) can have momentum comparable to a large rolling stone.
- The greater the momentum, the greater the force needed to stop the body in a given time.
- The greater the time over which the change of momentum occurs, the smaller the average force required (this is why air bags work).
Statement of Newton's Second Law
Newton's Second Law: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.
\[\vec{F} = \frac{d\vec{p}}{dt}\]
If mass m is constant (which is true for most everyday situations):
\[\vec{F} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}\]
This gives the famous form:
\[\boxed{\vec{F} = m\vec{a}}\]
Important properties of Newton's Second Law:
- Vector law: F and a are vectors, in the same direction. Equivalent to three scalar equations: Fx = max, Fy = may, Fz = maz.
- Local relation: F at time t produces a at time t (instantaneous, not delayed).
- Defines force quantitatively: 1 newton (N) = the force that produces an acceleration of 1 m/s² on a mass of 1 kg.
- Reduces to first law: if F = 0, then a = 0, so v is constant (uniform motion or rest).
4.6 Impulse
We sometimes encounter examples where a large force acts for a very short time, producing a finite change of momentum. For example, when a ball hits a wall and bounces back, the force on the ball acts only during the brief contact (~0.01 s), but it changes the ball's momentum significantly.
The product of force and time is called impulse:
\[\vec{J} = \vec{F} \cdot \Delta t\]
From Newton's Second Law: \(F = \Delta p / \Delta t\), so:
\[\vec{J} = \vec{F}\,\Delta t = \Delta\vec{p} = m\vec{v_f} - m\vec{v_i}\]
Impulse-Momentum Theorem: The impulse of force on a body equals the change in its momentum.
\[\vec{J} = \Delta\vec{p}\]
🎯 Interactive: Force, Mass & Acceleration
Apply a force on a block. Observe how acceleration depends on mass.
Acceleration a = F/m = 2.00 m/s²
After 5 seconds: velocity = 10.0 m/s | distance = 25.0 m
Worked Examples
Worked Example 1 (NCERT Example 4.3): Stopping a Ball
A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
Given: m = 0.04 kg, u = 90 m/s, v = 0, s = 0.60 m.
Using \(v^2 = u^2 + 2as\): \[0 = 90^2 + 2 \times a \times 0.60\] \[a = -\frac{8100}{1.2} = -6750 \text{ m/s}^2\]
\[F = ma = 0.04 \times (-6750) = \boxed{-270 \text{ N}}\] The negative sign indicates the force is opposite to the bullet's direction (resistive). Magnitude = 270 N.
Using \(v^2 = u^2 + 2as\): \[0 = 90^2 + 2 \times a \times 0.60\] \[a = -\frac{8100}{1.2} = -6750 \text{ m/s}^2\]
\[F = ma = 0.04 \times (-6750) = \boxed{-270 \text{ N}}\] The negative sign indicates the force is opposite to the bullet's direction (resistive). Magnitude = 270 N.
Worked Example 2: Cricket Ball - Impulse Calculation
A cricket ball of mass 150 g moving at 15 m/s east is struck by a bat such that it returns at 25 m/s west. The contact time is 0.01 s. Find: (a) impulse, (b) average force on the ball.
Setup: Take east as +x. m = 0.15 kg, v_i = +15 m/s, v_f = −25 m/s.
(a) Impulse = change in momentum: \[J = m(v_f - v_i) = 0.15 \times (-25 - 15) = 0.15 \times (-40) = \boxed{-6 \text{ kg·m/s}}\] The negative sign means impulse is directed west (opposite to original motion).
(b) Average force: \[F_{avg} = \frac{J}{\Delta t} = \frac{-6}{0.01} = \boxed{-600 \text{ N}}\] Magnitude 600 N, directed west — explains why it stings the bat!
(a) Impulse = change in momentum: \[J = m(v_f - v_i) = 0.15 \times (-25 - 15) = 0.15 \times (-40) = \boxed{-6 \text{ kg·m/s}}\] The negative sign means impulse is directed west (opposite to original motion).
(b) Average force: \[F_{avg} = \frac{J}{\Delta t} = \frac{-6}{0.01} = \boxed{-600 \text{ N}}\] Magnitude 600 N, directed west — explains why it stings the bat!
Worked Example 3 (NCERT Example 4.5): Bowling Strike
A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upward with uniform speed of 10 m/s, (b) downward with uniform acceleration of 5 m/s², (c) upward with uniform acceleration of 5 m/s². What would the scale read in each case? Take g = 10 m/s².
The scale reads the normal force N (apparent weight).
(a) Uniform speed (a = 0): N − mg = 0 ⇒ \(N = mg = 70 \times 10 = \boxed{700 \text{ N}}\)
The reading is the same as when stationary.
(b) Downward acceleration (a = 5 m/s² down): mg − N = ma ⇒ N = m(g − a) = 70 × (10 − 5) = \(\boxed{350 \text{ N}}\)
The man feels lighter (apparent weight reduced).
(c) Upward acceleration (a = 5 m/s² up): N − mg = ma ⇒ N = m(g + a) = 70 × (10 + 5) = \(\boxed{1050 \text{ N}}\)
The man feels heavier (this is what happens at lift start-up going up, or stopping going down).
(a) Uniform speed (a = 0): N − mg = 0 ⇒ \(N = mg = 70 \times 10 = \boxed{700 \text{ N}}\)
The reading is the same as when stationary.
(b) Downward acceleration (a = 5 m/s² down): mg − N = ma ⇒ N = m(g − a) = 70 × (10 − 5) = \(\boxed{350 \text{ N}}\)
The man feels lighter (apparent weight reduced).
(c) Upward acceleration (a = 5 m/s² up): N − mg = ma ⇒ N = m(g + a) = 70 × (10 + 5) = \(\boxed{1050 \text{ N}}\)
The man feels heavier (this is what happens at lift start-up going up, or stopping going down).
Worked Example 4: Air Bag Physics
A car of mass 1000 kg moving at 20 m/s crashes into a wall. The crumple zone increases the stopping time from 0.01 s (no airbag, instant stop) to 0.5 s (with airbag). Compare the average forces in each case.
Change in momentum (same in both cases): \(\Delta p = m\Delta v = 1000 \times 20 = 20{,}000\) kg·m/s.
Without airbag (Δt = 0.01 s): \[F_1 = \frac{\Delta p}{\Delta t} = \frac{20000}{0.01} = \boxed{2 \times 10^6 \text{ N}}\]
With airbag (Δt = 0.5 s): \[F_2 = \frac{20000}{0.5} = \boxed{4 \times 10^4 \text{ N}}\]
Force ratio: F_1/F_2 = 50. The airbag reduces force by 50× by extending stopping time. This is why airbags save lives — same Δp, but smaller F because larger Δt.
Without airbag (Δt = 0.01 s): \[F_1 = \frac{\Delta p}{\Delta t} = \frac{20000}{0.01} = \boxed{2 \times 10^6 \text{ N}}\]
With airbag (Δt = 0.5 s): \[F_2 = \frac{20000}{0.5} = \boxed{4 \times 10^4 \text{ N}}\]
Force ratio: F_1/F_2 = 50. The airbag reduces force by 50× by extending stopping time. This is why airbags save lives — same Δp, but smaller F because larger Δt.
4.7 Conservation of Momentum
Newton's Second Law allows us to derive an extremely powerful principle. If the net external force on a system is zero, then \(d\vec{p}/dt = 0\), which means \(\vec{p}\) = constant. This is the law of conservation of momentum.
Law of Conservation of Linear Momentum: The total momentum of an isolated system of interacting particles is conserved.
\[\vec{p}_{initial} = \vec{p}_{final} \quad (\text{when } F_{ext} = 0)\]
Application: Recoil of a Gun
When a gun fires a bullet, the gun recoils backward. Why? Initially the gun + bullet system has zero momentum (at rest). Fired forward, the bullet gets momentum +mv. By conservation, the gun must get equal and opposite momentum:
\[m_{bullet} v_{bullet} + M_{gun} v_{gun} = 0\]
\[V_{gun} = -\frac{m_{bullet}}{M_{gun}} \times v_{bullet}\]
Since M >> m, the gun's recoil velocity is small but non-zero.
📐 Activity 4.2 — Egg Drop Challenge (Impulse Demo)
Setup: Drop a raw egg from a height of 1 m onto: (i) a stone slab, (ii) a thick foam pad, (iii) a cushion of crumpled newspapers.
Predict: Which surface will the egg survive?
Observation:
(i) On the stone — egg breaks instantly.
(ii) On the foam — egg survives.
(iii) On the newspapers — egg survives.
Why? Same change in momentum in all cases (Δp = mv where v ≈ 4.4 m/s after falling 1 m). Stone stops the egg in ~0.001 s → huge force. Foam/papers stop the egg in ~0.05 s → small force. Force = Δp/Δt; larger Δt = smaller F = no breakage.
This is the same physics as: airbags, helmets, packaging foam, sand pits in long-jump landings, gymnastics mats.
🎯 Competency-Based Questions
Q1. A force of 20 N acts on a 4 kg body. Find its acceleration. L3 Apply
Answer: (a) 5 m/s². \(a = F/m = 20/4 = 5\) m/s².
Q2. A cricket player while catching a ball lowers his hands. Why? L4 Analyse
Answer: Lowering the hands increases the time interval Δt over which the ball decelerates. Since impulse Δp is fixed (ball stops, so Δp = mv), the force F = Δp/Δt decreases. So the impact on the hands is gentler — preventing pain or injury. Same physics as airbags and gymnasts' mats.
Q3. A 10-gram bullet is fired from a 5-kg gun with velocity 400 m/s. Calculate the recoil velocity of the gun. L3 Apply
Answer: By conservation of momentum: \(m_b v_b + M_g v_g = 0\). \(v_g = -\frac{m_b v_b}{M_g} = -\frac{0.01 \times 400}{5} = \boxed{-0.8 \text{ m/s}}\). The gun recoils at 0.8 m/s backward.
Q4. A boy of mass 50 kg jumps off a stationary boat of mass 200 kg with horizontal velocity 4 m/s. Find the velocity of the boat. L4 Analyse
Answer: Initial momentum = 0. Final momentum = m_boy v_boy + M_boat v_boat = 0.
\[V_{boat} = -\frac{50 \times 4}{200} = \boxed{-1 \text{ m/s}}\] The boat moves at 1 m/s in the OPPOSITE direction to the boy. (This is why jumping off a small boat is risky!)
\[V_{boat} = -\frac{50 \times 4}{200} = \boxed{-1 \text{ m/s}}\] The boat moves at 1 m/s in the OPPOSITE direction to the boy. (This is why jumping off a small boat is risky!)
Q5. HOT: Two ice skaters of masses 60 kg and 40 kg stand at rest, holding a long pole between them. They push apart along the pole. The 60-kg skater is observed to move at 2 m/s. Find: (a) velocity of the 40-kg skater, (b) which had a larger impulse from the pole? L6 Create
(a) Conservation of momentum: \(60 \times 2 + 40 \times v_{40} = 0\)
\[v_{40} = -\frac{120}{40} = \boxed{-3 \text{ m/s}}\]
The 40-kg skater moves at 3 m/s in the opposite direction.
(b) Impulse on each = change in momentum.
Impulse on 60-kg = 60 × 2 = 120 kg·m/s.
Impulse on 40-kg = 40 × 3 = 120 kg·m/s.
Same magnitude! By Newton's Third Law (preview), the forces are equal and opposite, applied for the same duration → equal impulses.
Insight: Lighter skater accelerates more, but both experience equal force/impulse.
(b) Impulse on each = change in momentum.
Impulse on 60-kg = 60 × 2 = 120 kg·m/s.
Impulse on 40-kg = 40 × 3 = 120 kg·m/s.
Same magnitude! By Newton's Third Law (preview), the forces are equal and opposite, applied for the same duration → equal impulses.
Insight: Lighter skater accelerates more, but both experience equal force/impulse.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: An athlete bends his knees while landing from a jump.
R: Bending the knees increases stopping time, reducing the force on the legs.
Answer: (A). Both true; R correctly explains A. F = Δp/Δt; larger Δt → smaller F.
A: A heavy truck and a small car moving with the same momentum experience the same force when stopped in the same time.
R: Force = Δp/Δt depends only on change in momentum and time, not on mass directly.
Answer: (A). Both true; R explains A. Same Δp and Δt → same F, regardless of how the momentum is distributed (large m × small v vs small m × large v).
A: Conservation of momentum is a consequence of Newton's Second Law for an isolated system.
R: If F_ext = 0, then dp/dt = 0, so p = constant.
Answer: (A). Both true; R is the direct mathematical proof of A.
Frequently Asked Questions - Second Law Momentum
What is the main concept covered in Second Law Momentum?
In NCERT Class 11 Physics Chapter 4 (Laws of Motion), "Second Law Momentum" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Second Law Momentum useful in real-life applications?
Real-life applications of Second Law Momentum from NCERT Class 11 Physics Chapter 4 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Second Law Momentum?
Key formulas in Second Law Momentum (NCERT Class 11 Physics Chapter 4 Laws of Motion) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 4?
NCERT Class 11 Physics Chapter 4 (Laws of Motion) is structured so each part builds on the previous one. Second Law Momentum connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Second Law Momentum?
CBSE board questions from Second Law Momentum typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Second Law Momentum lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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