AimMent
Challenging1 past question

Impulse, momentum, and conservation laws

Explore the principles of linear momentum, impulse, and their conservation in collisions and explosions.

Why this shows up in the exam

You must be able to solve problems involving momentum changes and conservation in both isolated and interacting systems.

How NEET tests this

Numerical · 2 QsApplication

Learn the idea

Linear momentum is a vector p=mv and any external impulse changes it by Δp. In an isolated explosion the internal forces cancel, so the vector sum of all fragment momenta stays zero.

🧠 Memory hook: Zero net push – three friends pulling a rope equally from different sides keep the rope still; the third pull must cancel the diagonal pull of the other two.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Momentum p equals mass times velocity (vector)
  • Impulse J equals change in momentum and also equals average force times time
  • In an isolated system total momentum is conserved (remains constant)
  • Momentum components add vectorially – use right‑angle triangle for perpendicular directions
  • For an explosion initial momentum is zero, so final momenta must cancel
  • Mass ratio determines each fragment’s mass: m_i = (given fraction)·total mass

How to approach it

  1. 1Write the masses of each fragment using the given ratio
  2. 2Assign velocity vectors to the known fragments (choose perpendicular axes)
  3. 3Apply conservation of momentum: sum of all momentum vectors = 0, resolve x and y components
  4. 4Find the magnitude of the unknown fragment’s momentum and divide by its mass to get its speed

Worked example — watch it click

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is:

  • A)v
  • B)√2v
  • ✅2√2v
  • D)3√2v

The concept behind this problem

The problem forces you to combine two equal perpendicular momenta vectorially and then use the zero‑total‑momentum condition to find the speed of the remaining fragment.

Step by step

  1. 1Mass ratio 2:2:1 means fragments have masses 2m/5, 2m/5, m/5.
  2. 2By momentum conservation, initial momentum = 0.
  3. 3Two equal fragments (each 2m/5) move perpendicular with speed v; their vector sum has magnitude (2m/5)v√2.
  4. 4Third fragment (m/5) must have momentum (2m/5)v√2 opposite, giving speed = [(2m/5)v√2]/(m/5) = 2√2v.

Watch out

Students often add the two equal momenta as 2 mv + 2 mv instead of taking the √2 vector sum, leading to a wrong speed.

Common slip-ups that cost marks

  • •Treating momentum as a scalar and adding magnitudes directly
  • •Using the wrong mass for the lighter fragment (forgetting the 1/5 factor)
  • •Assuming the third fragment moves along one of the given directions instead of opposite to the resultant

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 1NEET 2022

A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is:

More from Laws of Motion