Kinematics-2D
Q1
Objective
Match the columns
Numerical
A particle is projected from ground with velocity u at angle $\theta$ from horizontal. Match the following two columns and mark the correct option from the codes given below.
| Column I | Column II | |||||||
| (A) | Average velocity between initial and final points | (p) | u sinθ | |||||
| (B) | Change in velocity between initial and final points | (q) | u cosθ | |||||
| (C) | Change in velocity between initial and peak points | (r) | zero | |||||
| (D) | Average velocity between initial and highest points | (s) | None | |||||
| Codes | ||||||||
| A | B | C | D | A | B | C | D | |
| (a) | p | s | r | q | (b) | p | r | q |
| (c) | q | s | p | s | (d) | r | p | q |
Correct Answer: C
Explanation:
C
Q2
Objective
Match the columns
Numerical
A particle is projected horizontally from a tower with velocity $10 \, ms^{-1}$ . Taking, $g = 10 \, ms^{-2}$ . Match the following two columns at time t = 1s and mark the correct option from the codes given below.
| Column I | Column II | |||||||
| (A) | Horizontal component of velocity | (p) | 5 SI units | |||||
| (B) | Vertical component of velocity | (q) | 10 SI unit | |||||
| (C) | Horizontal displacement | (r) | 15 SI unit | |||||
| (D) | Vertical displacement | (s) | 20 SI unit | |||||
| Codes | ||||||||
| A | B | C | D | A | B | C | D | |
| (a) | p | q | s | r | (b) | q | s | p |
| (c) | q | p | r | s | (d) | q | q | p |
Correct Answer: D
Explanation:
In horizontal projectile motion,
Horizontal component of velocity,
$u _ {x} = u = 1 0 \mathrm{ms} ^ {- 1}$
Vertical component of velocity,
$u _ {y} = g t = 1 0 \times 1 = 1 0 \mathrm{ms} ^ {- 1}$
Horizontal displacement
$= u \times t = 1 0 \times (1) = 1 0 \mathrm{m}$
Vertical displacement = $\frac{1}{2}gt^{2}=\frac{1}{2}\times10\times(1)^{2}=5m$
Horizontal component of velocity,
$u _ {x} = u = 1 0 \mathrm{ms} ^ {- 1}$
Vertical component of velocity,
$u _ {y} = g t = 1 0 \times 1 = 1 0 \mathrm{ms} ^ {- 1}$
Horizontal displacement
$= u \times t = 1 0 \times (1) = 1 0 \mathrm{m}$
Vertical displacement = $\frac{1}{2}gt^{2}=\frac{1}{2}\times10\times(1)^{2}=5m$