Collisions in one dimension: Class 11 practice
Momentum is conserved in every collision. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Momentum
- Kinetic energy
The idea behind collisions in one dimension
Momentum is conserved in every collision. In a perfectly elastic collision kinetic energy is also conserved; in a perfectly inelastic one the bodies stick together and KE is lost.
Step-by-step method
- Write momentum before and after.
- Elastic: also conserve KE (or use the velocity formulas).
- Inelastic: common final velocity; compare KE before and after.
Worked example 1 (Basic)
A 4 kg trolley at 4 m/s hits and sticks to a stationary 1 kg trolley. Find their common speed.
- 4 × 4 = 5v
- v = 3.2 m/s
- Answer: 3.2 m/s
Worked example 2 (Standard)
For the same sticky collision (1 kg at 9 m/s into 5 kg at rest), how much kinetic energy is lost?
- KE before = 40.5 J
- KE after = 6.75 J
- Lost = 33.75 J
- Answer: 33.75 J
Worked example 3 (Challenge)
A 2 kg ball at 5 m/s hits a stationary 5 kg ball elastically (head-on). Find both velocities after.
- v₁ = (m₁ − m₂)u/(m₁ + m₂) = -2.14 m/s
- v₂ = 2m₁u/(m₁ + m₂) = 2.86 m/s
- Answer: -2.14 m/s and 2.86 m/s
Common mistakes and misconceptions
- Conserving KE in sticky collisions
- Sign errors for opposite directions
Practice questions
- (Basic) A 4 kg trolley at 7 m/s hits and sticks to a stationary 5 kg trolley. Find their common speed.
- (Basic) A 3 kg trolley at 5 m/s hits and sticks to a stationary 5 kg trolley. Find their common speed.
- (Standard) For the same sticky collision (2 kg at 3 m/s into 3 kg at rest), how much kinetic energy is lost?
- (Standard) For the same sticky collision (2 kg at 9 m/s into 5 kg at rest), how much kinetic energy is lost?
- (Challenge) A 5 kg ball at 5 m/s hits a stationary 4 kg ball elastically (head-on). Find both velocities after.
- (Challenge) A 3 kg ball at 6 m/s hits a stationary 3 kg ball elastically (head-on). Find both velocities after.
Answer key with full working
- Q1: 4 × 7 = 9v → v = 3.11 m/s → Answer: 3.11 m/s
- Q2: 3 × 5 = 8v → v = 1.88 m/s → Answer: 1.88 m/s
- Q3: KE before = 9 J → KE after = 3.6 J → Lost = 5.4 J → Answer: 5.4 J
- Q4: KE before = 81 J → KE after = 23.14 J → Lost = 57.86 J → Answer: 57.86 J
- Q5: v₁ = (m₁ − m₂)u/(m₁ + m₂) = 0.56 m/s → v₂ = 2m₁u/(m₁ + m₂) = 5.56 m/s → Answer: 0.56 m/s and 5.56 m/s
- Q6: v₁ = (m₁ − m₂)u/(m₁ + m₂) = 0 m/s → v₂ = 2m₁u/(m₁ + m₂) = 6 m/s → Answer: 0 m/s and 6 m/s
Syllabus note
This skill is listed under "Work Energy and Power" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is collisions in one dimension in?
"Work Energy and Power", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.