JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Power Collisions
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Power Collisions
🎓 Class 11
Physics
CBSE
Theory
Ch 5 – Work, Energy and Power
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Power Collisions
Class 11 સ્તર, Work Energy Power માં, Advanced કઠિનતા સાથે લક્ષ્ય.
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Power Collisions
5.13 Power
Power is the time-rate at which work is done. The average power over a time interval \(\Delta t\) during which work \(W\) is done is:
\[P_{\text{avg}} = \frac{W}{\Delta t}\]
The instantaneous power is the limit of this ratio as \(\Delta t \to 0\):
\[P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}\]
The SI unit of power is the watt (W) = 1 J/s. Common alternative units include:
| Unit | Conversion | Typical Use |
|---|---|---|
| Watt (W) | 1 J/s | SI base, electronics, lighting |
| Kilowatt (kW) | 1000 W | Household appliances |
| Megawatt (MW) | 10⁶ W | Power plants |
| Horsepower (hp) | 746 W | Vehicles, motors |
| kWh (energy, not power) | 3.6 × 10⁶ J | Electricity bills |
kWh vs kW: Be careful — your electricity bill uses kWh, which is energy (power × time), not power. A 1 kW heater running for 1 hour consumes 1 kWh = 3.6 MJ.
5.14 Collisions
A collision is a brief interaction during which two bodies exert strong forces on each other. We classify collisions by what is conserved:
In ALL collisions (in absence of external forces):
- Total linear momentum is always conserved
- Total energy is always conserved (but some KE may convert to heat, sound, internal energy)
- Elastic collision: Total kinetic energy is conserved.
- Inelastic collision: Some KE is lost.
- Perfectly inelastic: Bodies stick together; maximum KE lost (consistent with momentum conservation).
5.14.1 Elastic Collisions in One Dimension
Let mass \(m_1\) move with velocity \(u_1\) and mass \(m_2\) be at rest (\(u_2 = 0\)). After elastic collision their velocities are \(v_1, v_2\). Apply momentum and KE conservation:
\[m_1 u_1 = m_1 v_1 + m_2 v_2\]
\[\tfrac{1}{2}m_1 u_1^2 = \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2\]
Solving these (a non-trivial algebra) gives:
\[v_1 = \frac{m_1 - m_2}{m_1 + m_2}\,u_1, \quad v_2 = \frac{2 m_1}{m_1 + m_2}\,u_1\]
| Mass relationship | v₁ (after) | v₂ (after) | Physical situation |
|---|---|---|---|
| m₁ = m₂ | 0 | u₁ | Equal masses: full velocity exchange |
| m₁ ≫ m₂ | ≈ u₁ | ≈ 2u₁ | Heavy hits light: light flies off fast |
| m₁ ≪ m₂ | ≈ −u₁ | ≈ 0 | Light hits heavy: light bounces back |
5.14.2 Perfectly Inelastic 1D Collisions
The two bodies stick together after collision. Only momentum is conserved:
\[m_1 u_1 = (m_1 + m_2) v\]
\[v = \frac{m_1 u_1}{m_1 + m_2}\]
Energy lost:
\[\Delta K = \tfrac{1}{2}m_1 u_1^2 - \tfrac{1}{2}(m_1+m_2)v^2 = \tfrac{1}{2}\frac{m_1 m_2}{m_1+m_2}u_1^2\]
5.14.3 Collisions in Two Dimensions
For 2D collisions, momentum conservation gives TWO equations (one for x, one for y). For elastic 2D collisions, KE conservation gives a third. With four unknowns (two final velocity components for each particle), one parameter (e.g., scattering angle) must be specified.
\[m_1 u_1 = m_1 v_1\cos\theta_1 + m_2 v_2\cos\theta_2 \;\;\;(\text{x-component})\]
\[0 = m_1 v_1\sin\theta_1 - m_2 v_2\sin\theta_2 \;\;\;(\text{y-component})\]
🎯 Interactive Simulation: 1D Elastic Collision
Choose masses and initial velocity of m₁. m₂ is initially at rest. The simulator shows post-collision velocities for an elastic collision.
Elastic outcome
v₁ = −1.43 m/s
v₂ = 2.86 m/s
Perfectly inelastic outcome
v = 1.43 m/s
Energy lost: 17.86 J
Example 5.10: Power of an Elevator
An elevator can carry up to 8 passengers each averaging 70 kg, total maximum cabin mass plus passengers = 800 kg. It must rise 10 m in 5 s at constant speed. What minimum motor power is needed? (g = 10 m/s², ignore friction)
Work to lift = mgh = 800 × 10 × 10 = 80 000 J.
Average power = W/t = 80 000/5 = 16 000 W = 16 kW ≈ 21.4 hp.
This is the minimum (ideal) — real motors need more to overcome friction and provide acceleration margin.
Average power = W/t = 80 000/5 = 16 000 W = 16 kW ≈ 21.4 hp.
This is the minimum (ideal) — real motors need more to overcome friction and provide acceleration margin.
Example 5.11: Equal-Mass Elastic Collision
A 5 kg ball moving at 6 m/s collides elastically with a stationary 5 kg ball. Find their velocities after the collision.
With m₁ = m₂, the formulae give v₁ = 0, v₂ = u₁:
\[v_1 = 0, \quad v_2 = 6\text{ m/s}\] Result: The first ball stops dead and the second moves off with the original velocity. This is the famous "Newton's cradle" effect — energy and momentum are passed from ball to ball.
\[v_1 = 0, \quad v_2 = 6\text{ m/s}\] Result: The first ball stops dead and the second moves off with the original velocity. This is the famous "Newton's cradle" effect — energy and momentum are passed from ball to ball.
Example 5.12: Bullet Embedded in Block
A bullet of mass 0.020 kg moving at 600 m/s embeds itself in a 4 kg wooden block initially at rest on a frictionless table. Find (a) the common velocity after collision, (b) the kinetic energy lost.
(a) Perfectly inelastic — momentum conserved:
\[v = \frac{0.020 \times 600}{0.020 + 4.0} = \frac{12}{4.02} \approx 2.985\text{ m/s}\]
(b) Initial KE = ½(0.020)(600)² = 3600 J. Final KE = ½(4.02)(2.985)² ≈ 17.9 J.
ΔK = 3600 − 17.9 = 3582.1 J lost (≈ 99.5% of initial KE!) — converted to heat, sound, deformation. This is why bullets are dangerous: nearly all of the bullet's KE goes into damaging the target.
ΔK = 3600 − 17.9 = 3582.1 J lost (≈ 99.5% of initial KE!) — converted to heat, sound, deformation. This is why bullets are dangerous: nearly all of the bullet's KE goes into damaging the target.
⚡ Activity 5.4 — Newton's Cradle
L4 Analyse
Materials: Newton's cradle toy (5 identical metal balls hanging in a row), or 5 marbles in a straight track.
Procedure:
- Pull back one ball and release. Observe what happens.
- Pull back two balls and release them together.
- Pull back three balls.
Predict: If you pull 2 balls, will 1 ball pop out twice as fast, or 2 balls pop out at the original speed?
Observation: If 2 balls strike, 2 balls pop off the far end at the same speed (NOT 1 ball at twice the speed). Energy and momentum together pin down the answer.
Conclusion: Both momentum (mv) AND kinetic energy (½mv²) must balance. If 1 ball came off at 2u, momentum would be ok (2mu) but KE would be 2mu² — twice the input KE of 2 × ½mu² = mu². Only "n in, n out at original speed" satisfies BOTH laws simultaneously.
🎯 Competency-Based Questions
A railway shunting yard couples wagons on a level track. Wagon A (mass 8000 kg) moves east at 2 m/s and couples with stationary wagon B (mass 4000 kg). The collision is perfectly inelastic.
Q1. Find the common velocity after coupling.L3 Apply
Answer: (b). v = (8000 × 2 + 0)/(8000 + 4000) = 16000/12000 = 1.33 m/s east.
Q2. Calculate the kinetic energy lost in the coupling.L4 Analyse
Answer: K_i = ½(8000)(2)² = 16 000 J. K_f = ½(12 000)(1.33)² ≈ 10 667 J. Energy lost ≈ 5333 J. Equivalently, ½·m₁m₂/(m₁+m₂)·u² = ½·(8000·4000/12000)·4 = 5333 J.
Q3. A 1500 W vacuum cleaner runs for 2 hours. Compute the energy used in kWh.L3 Apply
Answer: Energy = P × t = 1.5 kW × 2 h = 3 kWh = 3 × 3.6 × 10⁶ J = 10.8 MJ.
Q4. State whether TRUE or FALSE: "Momentum is conserved only in elastic collisions." L5 Evaluate
Answer: FALSE. Momentum is conserved in ALL collisions (both elastic and inelastic) — provided no external force acts. Only KE distinguishes the two: KE is conserved only in elastic, lost in inelastic.
Q5. HOT: A neutron of mass m strikes a stationary proton of nearly equal mass head-on elastically. Calculate the fraction of KE transferred. Comment on its application in nuclear reactors. L6 Create
Answer: For m₁ = m₂, v₁ = 0, v₂ = u₁ — so 100% of KE is transferred. Application: In nuclear reactors, fast neutrons are slowed (moderated) by elastic collisions with light nuclei (protons in water, deuterium in heavy water). Equal mass gives maximum slowing per collision. Heavier moderators (graphite carbon-12) lose only ~28% per collision, requiring more collisions but with less neutron capture loss.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): In a perfectly inelastic collision, kinetic energy is lost.
Reason (R): In a perfectly inelastic collision, momentum is not conserved.
Answer: (C). Assertion is TRUE — KE is lost. Reason is FALSE — momentum IS conserved (we used it to find the common velocity).
Assertion (A): Power can be expressed as F·v.
Reason (R): Work done in time dt is F·v dt; dividing by dt gives instantaneous power.
Answer: (A). Both true and R explains A.
Assertion (A): When equal masses collide elastically in 1D with one initially at rest, they exchange velocities.
Reason (R): The formula v₁ = (m₁−m₂)/(m₁+m₂)·u₁ becomes 0 when m₁ = m₂.
Answer: (A). Both true and R explains A.
Frequently Asked Questions - Power Collisions
What is the main concept covered in Power Collisions?
In NCERT Class 11 Physics Chapter 5 (Work, Energy and Power), "Power Collisions" 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 Power Collisions useful in real-life applications?
Real-life applications of Power Collisions from NCERT Class 11 Physics Chapter 5 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 Power Collisions?
Key formulas in Power Collisions (NCERT Class 11 Physics Chapter 5 Work, Energy and Power) 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 5?
NCERT Class 11 Physics Chapter 5 (Work, Energy and Power) is structured so each part builds on the previous one. Power Collisions 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 Power Collisions?
CBSE board questions from Power Collisions 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 Power Collisions 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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JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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