Physics Mechanics - How to Solve: Conservation of Linear Momentum (Explosions, Recoil, Bullet-Block System) – IIT JEE Guide

How to Solve: Conservation of Linear Momentum (Explosions, Recoil, Bullet-Block System) – IIT JEE Guide

Introduction

Mastering conservation of linear momentum in explosions, recoil, and bullet-block systems unlocks 5-8 marks in IIT JEE (Main + Advanced)—enough to push you into the top 10%. Whether it’s a gun firing a bullet, a bomb splitting into fragments, or a block absorbing a bullet, this concept appears every year. Miss it, and you lose easy marks. Nail it, and you solve problems in under 2 minutes.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. Newton’s 3rd Law – Action-reaction forces are equal and opposite.
2. Definition of Linear Momentum (p = mv) – Momentum is mass × velocity.
3. Vector Nature of Momentum – Direction matters; assign signs (+/-) to velocities.

If any of these are unclear, stop and review them first.

KEY TERMS & FORMULAS

Key Terms

1. System – The objects we’re studying (e.g., gun + bullet, bomb + fragments).
2. External Force – A force acting on the system from outside (e.g., gravity, friction).
3. Internal Force – Forces between objects inside the system (e.g., explosion force, recoil).
4. Conservation of Momentum – If no external force acts, the total momentum before = total momentum after.

Formulas

1. Momentum of a Single Object [ p = mv ]
2. ( p ) = momentum (kg·m/s)
3. ( m ) = mass (kg)
4. ( v ) = velocity (m/s)

5. MEMORISE THIS

6. Conservation of Linear Momentum (1D) [ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 ]

7. ( u ) = initial velocity (before event)
8. ( v ) = final velocity (after event)

9. MEMORISE THIS (Given on JEE sheet, but know how to apply it)

10. Conservation of Momentum (2D – Vector Form) [ \vec{p}{\text{initial}} = \vec{p} ]}

11. Break into x and y components if needed.
12. MEMORISE THIS APPROACH

STEP-BY-STEP METHOD

Follow these exact steps for every momentum conservation problem.

Step 1: Define the System

• Circle all objects involved (e.g., gun + bullet, bomb + fragments).
• Check for external forces:
• If no external force (or negligible), momentum is conserved.
• If external force exists (e.g., friction, gravity), momentum is not conserved (unless the force is perpendicular to motion).

Step 2: Assign Directions & Signs

• Choose a positive direction (e.g., right = +, left = –).
• All velocities must have signs (+ or –).

Step 3: Write Initial & Final Momentum

• Initial Momentum (Before Event):
• Sum of momenta of all objects before the event (e.g., explosion, firing).
• If objects are at rest, initial momentum = 0.
• Final Momentum (After Event):
• Sum of momenta of all objects after the event.

Step 4: Apply Conservation of Momentum

[ \text{Initial Momentum} = \text{Final Momentum} ] - Plug in values with correct signs. - Solve for the unknown.

Step 5: Check Units & Reasonableness

• Ensure all masses are in kg, velocities in m/s.
• Does the answer make sense? (e.g., lighter object should move faster).

WORKED EXAMPLES

Example 1 – Basic (Gun Firing a Bullet)

Problem: A 2 kg gun fires a 0.01 kg bullet at 200 m/s. What is the recoil velocity of the gun?

Solution (Step-by-Step):

1. Define the System:
2. Objects: Gun + bullet.

3. No external force (friction neglected) → Momentum conserved.

4. Assign Directions:

5. Let bullet’s direction = +x.

6. Gun’s recoil = –x (opposite direction).

7. Initial Momentum:

8. Gun and bullet at rest → Initial momentum = 0.

9. Final Momentum:

10. Bullet: ( p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}} = 0.01 \times 200 = +2 \, \text{kg·m/s} )

11. Gun: ( p_{\text{gun}} = m_{\text{gun}} \times v_{\text{gun}} = 2 \times v_{\text{gun}} ) (negative, since recoil is opposite)

12. Apply Conservation: [ 0 = p_{\text{bullet}} + p_{\text{gun}} ] [ 0 = 2 + 2v_{\text{gun}} ] [ v_{\text{gun}} = -1 \, \text{m/s} ]

13. Negative sign means gun moves opposite to bullet.

14. Check:

15. Gun is 200× heavier → moves 200× slower (1 m/s vs 200 m/s). Makes sense!

What we did and why: - Used conservation of momentum because no external force acted. - Assigned signs to velocities to account for direction. - Solved for recoil velocity by equating initial and final momenta.

Example 2 – Medium (Explosion into Two Fragments)

Problem: A 5 kg bomb at rest explodes into two fragments: - Fragment A: 2 kg, moves at 30 m/s at 30° above horizontal. - Fragment B: 3 kg, moves at unknown speed at θ below horizontal. Find the speed of Fragment B.

Solution (Step-by-Step):

1. Define the System:
2. Objects: Fragment A + Fragment B.

3. No external force → Momentum conserved.

4. Assign Directions:

5. Let horizontal = +x, vertical = +y.

6. Initial momentum = 0 (bomb at rest).

7. Break into Components:

8. Fragment A (2 kg, 30 m/s, 30°):
   • ( p_{Ax} = 2 \times 30 \cos 30° = 2 \times 30 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \, \text{kg·m/s} )
   • ( p_{Ay} = 2 \times 30 \sin 30° = 2 \times 30 \times 0.5 = 30 \, \text{kg·m/s} )

9. Fragment B (3 kg, v, θ):

   • ( p_{Bx} = 3v \cos θ )
   • ( p_{By} = -3v \sin θ ) (negative because it’s below horizontal)

10. Apply Conservation (x and y separately):

11. x-direction: [ 0 = p_{Ax} + p_{Bx} \implies 0 = 30\sqrt{3} + 3v \cos θ ] [ 3v \cos θ = -30\sqrt{3} \quad (1) ]

12. y-direction: [ 0 = p_{Ay} + p_{By} \implies 0 = 30 - 3v \sin θ ] [ 3v \sin θ = 30 \quad (2) ]

13. Solve for v:

14. Divide (2) by (1): [ \frac{3v \sin θ}{3v \cos θ} = \frac{30}{-30\sqrt{3}} \implies \tan θ = -\frac{1}{\sqrt{3}} ] [ θ = 150° \text{ or } -30° \quad (\text{We take } θ = 30° \text{ below horizontal}) ]

15. Plug ( θ = 30° ) into (2): [ 3v \sin 30° = 30 \implies 3v \times 0.5 = 30 \implies v = 20 \, \text{m/s} ]

16. Check:

17. Fragment B is heavier → moves slower (20 m/s vs 30 m/s). Makes sense!

What we did and why: - Used vector conservation (x and y components) because explosion was 2D. - Solved two equations to find speed and angle. - Verified that heavier fragment moves slower.

Example 3 – Exam-Style (Bullet-Block System with Friction)

Problem (JEE Advanced 2018-Style): A 0.01 kg bullet moving at 500 m/s hits a 2 kg block at rest on a rough surface (μ = 0.2). The bullet embeds in the block. How far does the block slide before stopping?

Solution (Step-by-Step):

1. Define the System:
2. Objects: Bullet + block.
3. External force = friction → Momentum not conserved during collision.

4. But: During collision (instantaneous), friction is negligible → Momentum conserved for collision only.

5. Step 1: Find Velocity After Collision (Momentum Conservation)

6. Initial momentum = ( 0.01 \times 500 + 2 \times 0 = 5 \, \text{kg·m/s} )
7. Final momentum = ( (0.01 + 2) \times v = 2.01v )

8. Apply conservation: [ 5 = 2.01v \implies v = \frac{5}{2.01} \approx 2.49 \, \text{m/s} ]

9. Step 2: Find Stopping Distance (Work-Energy Principle)

10. Friction force = ( μmg = 0.2 \times 2.01 \times 9.8 \approx 3.94 \, \text{N} )
11. Work done by friction = ( F \times d )
12. Initial KE = ( \frac{1}{2} \times 2.01 \times (2.49)^2 \approx 6.23 \, \text{J} )

13. Work-Energy Theorem: [ W = ΔKE \implies -3.94d = 0 - 6.23 ] [ d = \frac{6.23}{3.94} \approx 1.58 \, \text{m} ]

14. Check:

15. Bullet embeds → inelastic collision → KE not conserved, but momentum is (during collision).
16. Friction after collision slows the block.

What we did and why: - Split the problem into collision (momentum conserved) and post-collision (friction acts). - Used work-energy principle to find stopping distance. - Exam trap: If you applied momentum conservation after collision, you’d get the wrong answer.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is how you own momentum problems in JEE:

1. System first: Circle all objects. If no external force, momentum is conserved.
2. Signs matter: Assign + and – to velocities. Left = –, right = +.
3. Initial = Final: Write ( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 ). If at rest, initial momentum = 0.
4. 2D? Break into x and y. Solve each component separately.
5. Bullet-block? First collision (momentum conserved), then post-collision (work-energy).
6. Check units: kg, m/s. Does the answer make sense? Heavier object = slower speed.

You’ve got this. Now go solve those 5-8 marks like a pro."