Newton's second law and momentum: Class 9 practice
Force equals the rate of change of momentum: F = (mv − mu)/t, which becomes F = ma for constant mass. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Equations of motion
The idea behind newton's second law and momentum
Force equals the rate of change of momentum: F = (mv − mu)/t, which becomes F = ma for constant mass. Momentum p = mv is measured in kg m/s.
Step-by-step method
- Convert mass to kg and speeds to m/s.
- Momentum change = m(v − u).
- F = change ÷ time, or F = ma.
Worked example 1 (Basic)
Find the force needed to give a 25 kg mass an acceleration of 3 m/s².
- F = ma = 25 × 3
- F = 75 N
- Answer: 75 N
Worked example 2 (Standard)
A 500 g ball's velocity changes from 0 m/s to 3 m/s in 2 s. Find the change in momentum and the force.
- m = 0.5 kg
- Δp = 0.5 × (3 − 0) = 1.50 kg m/s
- F = Δp ÷ t = 0.75 N
- Answer: Δp = 1.50 kg m/s, F = 0.75 N
Worked example 3 (Challenge)
A force of 6 N acts on a 3 kg trolley initially at rest for 4 s. Find its final velocity.
- a = F/m = 6 ÷ 3 = 2 m/s²
- v = 0 + 2 × 4 = 8 m/s
- Answer: 8 m/s
Common mistakes and misconceptions
- Using grams in F = ma
- Forgetting direction (sign) in momentum change
Practice questions
- (Basic) Find the force needed to give a 37 kg mass an acceleration of 5 m/s².
- (Basic) Find the force needed to give a 19 kg mass an acceleration of 6 m/s².
- (Standard) A 600 g ball's velocity changes from 1 m/s to 10 m/s in 2 s. Find the change in momentum and the force.
- (Standard) A 200 g ball's velocity changes from 0 m/s to 5 m/s in 4 s. Find the change in momentum and the force.
- (Challenge) A force of 56 N acts on a 7 kg trolley initially at rest for 2 s. Find its final velocity.
- (Challenge) A force of 768 N acts on a 48 kg trolley initially at rest for 4 s. Find its final velocity.
Answer key with full working
- Q1: F = ma = 37 × 5 → F = 185 N → Answer: 185 N
- Q2: F = ma = 19 × 6 → F = 114 N → Answer: 114 N
- Q3: m = 0.6 kg → Δp = 0.6 × (10 − 1) = 5.40 kg m/s → F = Δp ÷ t = 2.70 N → Answer: Δp = 5.40 kg m/s, F = 2.70 N
- Q4: m = 0.2 kg → Δp = 0.2 × (5 − 0) = 1.00 kg m/s → F = Δp ÷ t = 0.25 N → Answer: Δp = 1.00 kg m/s, F = 0.25 N
- Q5: a = F/m = 56 ÷ 7 = 8 m/s² → v = 0 + 8 × 2 = 16 m/s → Answer: 16 m/s
- Q6: a = F/m = 768 ÷ 48 = 16 m/s² → v = 0 + 16 × 4 = 64 m/s → Answer: 64 m/s
Syllabus note
This skill is listed under "Force and Laws of Motion" for Class 9 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 newton's second law and momentum in?
"Force and Laws of Motion", Class 9 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.