Vidaara.orgClass 9 · Physics
CodeVID-P9-04-ENG-01
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.
- Take $g=10\,\text{m/s}^2$ unless stated otherwise. Show all working for Sections B, C and D.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Energy is measured in
- A.newtons
- B.joules
- C.watts
- D.pascals
2.
Kinetic energy depends on the
- A.height
- B.square of speed
- C.colour
- D.shape
3.
Potential energy due to height is
- A.$\frac{1}{2}mv^2$
- B.$mgh$
- C.$Fs$
- D.$mv$
4.
Tripling the speed of a body makes its kinetic energy
- A.3 times
- B.6 times
- C.9 times
- D.unchanged
5.
A stretched rubber band stores
- A.kinetic energy
- B.potential energy
- C.heat only
- D.no energy
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Find the kinetic energy of a 3 kg body moving at 4 m/s.
7.
Find the potential energy of a 2 kg body at a height of 5 m.
8.
State the work-energy theorem.
9.
Why does kinetic energy increase rapidly with speed?
Section C — Short Answer (3 marks)
3 × 3 = 9 marks
10.
A 1000 kg car speeds up from 0 to 10 m/s. Find the change in its kinetic energy.
11.
A net force does 200 J of work on a 4 kg body at rest. Find its final speed.
12.
A 6 kg block is raised 4 m. Find the potential energy gained.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
13.
Derive an expression for the kinetic energy of a body of mass $m$ moving with speed $v$ using the work-energy idea, and use it to find the KE of a 0.5 kg ball moving at 8 m/s.
Answer Key
Section A — Multiple Choice Questions
- (B) joules
- (B) square of speed
- (B) $mgh$
- (C) 9 times
- (B) potential energy
Section B — Short Answer (2 marks)
- $E_k=\frac{1}{2}\times3\times4^2=24\,\text{J}$.
- $E_p=mgh=2\times10\times5=100\,\text{J}$.
- The work done by the net force on a body equals the change in its kinetic energy, $W=\Delta E_k$.
- Because $E_k=\frac{1}{2}mv^2$ depends on the square of the speed.
Section C — Short Answer (3 marks)
- $\Delta E_k=\frac{1}{2}\times1000\times10^2-0=50000\,\text{J}$.
- $200=\frac{1}{2}\times4\times v^2\Rightarrow v^2=100\Rightarrow v=10\,\text{m/s}$.
- $E_p=mgh=6\times10\times4=240\,\text{J}$.
Section D — Long Answer (5 marks)
- Work done to accelerate from rest to speed $v$ equals $E_k=\frac{1}{2}mv^2$. For the ball: $E_k=\frac{1}{2}\times0.5\times8^2=16\,\text{J}$.
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