Vidaara.orgClass 9 · Physics
CodeVID-P9-02-SLM-01
Newton's Second Law & Momentum — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show full working with correct SI units in all numerical answers.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The formula for momentum is:
- A.$p = mv$
- B.$p = ma$
- C.$p = m/v$
- D.$p = v/m$
2.
The SI unit of momentum is:
- A.$\text{N}$
- B.$\text{kg m/s}$
- C.$\text{J}$
- D.$\text{m/s}$
3.
$1\ \text{N}$ is equal to:
- A.$1\ \text{kg m/s}$
- B.$1\ \text{kg m/s}^2$
- C.$1\ \text{kg/s}$
- D.$1\ \text{kg m}^2$
4.
For a fixed force, doubling the mass will:
- A.double acceleration
- B.halve acceleration
- C.not change acceleration
- D.make acceleration zero
5.
Newton's First Law is a special case of the Second Law when:
- A.$F = 0$
- B.$m = 0$
- C.$a$ is maximum
- D.$v = 0$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Find the momentum of a $3\ \text{kg}$ object moving at $6\ \text{m/s}$.
7.
State Newton's Second Law of Motion.
8.
Define one newton of force.
9.
A $10\ \text{N}$ force acts on a $2\ \text{kg}$ body. Find its acceleration.
Section C — Short Answer (3 marks)
3 × 3 = 9 marks
10.
A $1500\ \text{kg}$ car accelerates from $0$ to $20\ \text{m/s}$ in $10\ \text{s}$. Find the force.
11.
Why is it advised to bend the knees while landing after a jump?
12.
Derive $F = ma$ from Newton's Second Law.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
13.
A bullet of mass $0.05\ \text{kg}$ moving at $80\ \text{m/s}$ is brought to rest in $0.02\ \text{s}$. Find (a) the change in momentum and (b) the average force. Explain with one example how increasing contact time reduces force.
Answer Key
Section A — Multiple Choice Questions
- (A) $p = mv$
- (B) $\text{kg m/s}$
- (B) $1\ \text{kg m/s}^2$
- (B) halve acceleration
- (A) $F = 0$
Section B — Short Answer (2 marks)
- $p = mv = 3 \times 6 = 18\ \text{kg m/s}$.
- The rate of change of momentum is proportional to the applied force and acts in the direction of the force.
- The force that gives a $1\ \text{kg}$ mass an acceleration of $1\ \text{m/s}^2$.
- $a = F/m = 10/2 = 5\ \text{m/s}^2$.
Section C — Short Answer (3 marks)
- $a = (20-0)/10 = 2\ \text{m/s}^2$; $F = ma = 1500 \times 2 = 3000\ \text{N}$.
- Bending knees increases the stopping time, and since $F = \Delta p/\Delta t$, the impact force is reduced.
- $F \propto \Delta p/\Delta t = m(v-u)/t = ma$, taking the proportionality constant as 1.
Section D — Long Answer (5 marks)
- (a) $\Delta p = 0 - (0.05 \times 80) = -4\ \text{kg m/s}$. (b) $F = \Delta p/\Delta t = -4/0.02 = -200\ \text{N}$, i.e. $200\ \text{N}$ opposing motion. Increasing contact time (e.g. airbags) lowers the force because $F = \Delta p/\Delta t$.
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