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CodeVID-P11-04-CH-01
Laws of Motion — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The SI unit of force is the:
- A.joule
- B.newton
- C.watt
- D.pascal
2.
Newton's second law in momentum form is:
- A.$\vec{F}=m\vec{v}$
- B.$\vec{F}=\frac{d\vec{p}}{dt}$
- C.$\vec{F}=\vec{p}\,t$
- D.$\vec{F}=\frac{\vec{p}}{t^2}$
3.
The SI unit of impulse is:
- A.$\text{N}$
- B.$\text{N}\cdot\text{s}$
- C.$\text{N/s}$
- D.$\text{J}$
4.
Kinetic friction is given by:
- A.$f_k=\mu_s N$
- B.$f_k=\mu_k N$
- C.$f_k\le\mu_k N$
- D.$f_k=\frac{N}{\mu_k}$
5.
The centripetal force is:
- A.$\frac{mv}{r}$
- B.$\frac{mv^2}{r}$
- C.$\frac{mr}{v^2}$
- D.$mvr$
6.
Ideal banking angle satisfies:
- A.$\tan\theta=\frac{rg}{v^2}$
- B.$\tan\theta=\frac{v^2}{rg}$
- C.$\cos\theta=\frac{v^2}{rg}$
- D.$\tan\theta=v^2 rg$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
A force of 15 N acts on a 3 kg body. Find its acceleration.
8.
State the impulse–momentum theorem.
9.
A 5 kg block on a floor has $\mu_k=0.2$. Find the kinetic friction. ($g=10$)
10.
Define the angle of repose.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
A 60 kg person stands in a lift accelerating upward at $2\ \text{m/s}^2$. Find the apparent weight. ($g=10$)
12.
A car of mass 800 kg turns a curve of radius 40 m at 10 m/s. Find the centripetal force.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Using Newton's laws, derive the law of conservation of linear momentum for two colliding bodies, and apply it to a 2 kg body at 5 m/s sticking to a 3 kg body at rest.
14.
Derive $\tan\theta=\frac{v^2}{rg}$ for the ideal banking of a road and find the banking angle for $r=90\ \text{m}$, $v=30\ \text{m/s}$. ($g=10$)
Answer Key
Section A — Multiple Choice Questions
- (B) newton
- (B) $\vec{F}=\frac{d\vec{p}}{dt}$
- (B) $\text{N}\cdot\text{s}$
- (B) $f_k=\mu_k N$
- (B) $\frac{mv^2}{r}$
- (B) $\tan\theta=\frac{v^2}{rg}$
Section B — Short Answer (2 marks)
- $a=\frac{15}{3}=5\ \text{m/s}^2$.
- Impulse equals the change in momentum: $\vec{J}=\vec{F}\Delta t=\Delta\vec{p}$.
- $f_k=\mu_k mg=0.2\times50=10\ \text{N}$.
- The incline angle at which a body just starts to slide; $\tan\theta=\mu_s$.
Section C — Short Answer (3 marks)
- $N=m(g+a)=60\times12=720\ \text{N}$.
- $F=\frac{mv^2}{r}=\frac{800\times100}{40}=2000\ \text{N}$.
Section D — Long Answer (5 marks)
- Equal-and-opposite internal forces give $\Delta p_1=-\Delta p_2$, so total $p$ is conserved; common velocity $=\frac{2\times5}{5}=2\ \text{m/s}$.
- $N\sin\theta=\frac{mv^2}{r}$ and $N\cos\theta=mg$ give $\tan\theta=\frac{v^2}{rg}=\frac{900}{900}=1$, so $\theta=45^{\circ}$.
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