Vidaara.orgClass 11 · Physics
CodeVID-P11-04-MOM-01
Momentum, Impulse & Conservation — 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.
Linear momentum is defined as:
- A.$\frac{m}{v}$
- B.$m v$
- C.$\frac{v}{m}$
- D.$m v^2$
2.
Impulse is equal to:
- A.$F t^2$
- B.$F\Delta t$
- C.$\frac{F}{t}$
- D.$m a$
3.
The SI unit of impulse is:
- A.$\text{N}$
- B.$\text{N}\cdot\text{s}$
- C.$\text{N/s}$
- D.$\text{J}$
4.
In a perfectly inelastic collision, the colliding bodies:
- A.bounce back
- B.stick and move together
- C.stop instantly
- D.exchange velocities
5.
Conservation of momentum holds when the external force on the system is:
- A.maximum
- B.zero
- C.constant non-zero
- D.increasing
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Find the momentum of a 5 kg body moving at 8 m/s.
7.
State the impulse–momentum theorem.
8.
Why are vehicles fitted with crumple zones?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
A gun of mass 5 kg fires a 0.01 kg bullet at 500 m/s. Find the recoil velocity.
10.
A 2 kg body moving at 5 m/s sticks to a 3 kg body at rest. Find the common velocity.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Using Newton's second and third laws, derive the law of conservation of linear momentum for two colliding bodies, and apply it to find the velocity of a 1 kg body if a 2 kg body at 3 m/s strikes it (at rest) and the 2 kg body continues at 1 m/s.
Answer Key
Section A — Multiple Choice Questions
- (B) $m v$
- (B) $F\Delta t$
- (B) $\text{N}\cdot\text{s}$
- (B) stick and move together
- (B) zero
Section B — Short Answer (2 marks)
- $p=mv=40\ \text{kg}\cdot\text{m/s}$.
- Impulse equals change in momentum: $\vec{J}=\vec{F}\Delta t=\Delta\vec{p}$.
- They increase collision time $\Delta t$, reducing the force on occupants.
Section C — Short Answer (3 marks)
- $v_g=-\frac{0.01\times500}{5}=-1\ \text{m/s}$ (recoil 1 m/s backward).
- $v=\frac{2\times5+3\times0}{5}=2\ \text{m/s}$.
Section D — Long Answer (5 marks)
- Equal-and-opposite internal forces give $\Delta p_1=-\Delta p_2$, so total $p$ is conserved; $2(3)+1(0)=2(1)+1 v_2\Rightarrow v_2=4\ \text{m/s}$.
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