Vidaara.orgClass 11 · Physics
CodeVID-P11-05-POC-01
Power & Collisions — 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 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
5 × 1 = 5 marks
1.
One horsepower is approximately:
- A.$100\ \text{W}$
- B.$746\ \text{W}$
- C.$1000\ \text{W}$
- D.$500\ \text{W}$
2.
Quantity conserved in every collision is:
- A.kinetic energy
- B.linear momentum
- C.speed
- D.height
3.
For a perfectly inelastic collision, $e$ equals:
- A.$1$
- B.$0$
- C.$0.5$
- D.$2$
4.
When equal masses collide elastically with one at rest, they:
- A.stick together
- B.exchange velocities
- C.both stop
- D.reverse and speed up
5.
The kilowatt-hour is a unit of:
- A.power
- B.force
- C.energy
- D.momentum
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
A machine does 5000 J of work in 10 s. Find its power.
7.
Define coefficient of restitution.
8.
A force of 200 N moves a body at 5 m/s along its direction. Find the power.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A 4 kg body at 5 m/s collides head-on with a 6 kg body at rest and they stick together. Find the common velocity.
10.
A ball dropped from 8 m rebounds to 2 m. Find the coefficient of restitution with the floor.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
For a 1D elastic collision between masses $m_1$ and $m_2$ (with $m_2$ initially at rest), derive the final velocities and show that equal masses exchange velocities.
Answer Key
Section A — Multiple Choice Questions
- (B) $746\ \text{W}$
- (B) linear momentum
- (B) $0$
- (B) exchange velocities
- (C) energy
Section B — Short Answer (2 marks)
- $P=\frac{W}{t}=\frac{5000}{10}=500\ \text{W}$.
- $e=\frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}=\frac{v_2-v_1}{u_1-u_2}$.
- $P=Fv=200\times 5=1000\ \text{W}$.
Section C — Short Answer (3 marks)
- $v=\frac{4\times 5+6\times 0}{4+6}=\frac{20}{10}=2\ \text{m/s}$.
- $e=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{2}{8}}=\sqrt{0.25}=0.5$.
Section D — Long Answer (5 marks)
- From momentum and KE conservation, $v_1=\frac{(m_1-m_2)u_1}{m_1+m_2}$ and $v_2=\frac{2m_1u_1}{m_1+m_2}$. Putting $m_1=m_2$ gives $v_1=0$ and $v_2=u_1$, so the velocities are exchanged.
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