Numerical Questions-1 - Chapter 2 Kinematics 11th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Questions
Question 1.
The position vector of the particle has length $1 \mathrm{~m}$ and makes $30^{\circ}$ with the $\mathrm{x}$-axis. What are the lengths of the $\mathrm{x}$ and $\mathrm{y}$ - components of the position vector?
Answer:
Given,
Length of position vector $=1 \mathrm{~m}$
Angle made with $\mathrm{x}$ axis $=30$
Solution:
Length of $\mathrm{X}$ component $(\mathrm{OB})=\mathrm{OA} \cos \theta$
$=1 \times \cos 30^{\circ}$
$=\frac{\sqrt{3}}{2}$ (or) $0.87 \mathrm{~m}$

Length of $Y$ component $(\mathrm{AB})=\mathrm{OA} \sin \theta=1 \mathrm{x} \sin 30^{\circ}=\frac{1}{2}=0.5 \mathrm{~m}$.

Question 2.
A particle has its position moved from $\vec{r}_1=3 \hat{i}+4 \hat{j}$ to $\mathrm{r} 2=\hat{i}+2 \hat{i}$. Calculate the displacement vector $(\Delta \overrightarrow{\mathrm{r}})$ and draw the $\vec{r}_1, \vec{r}_2$ and $\Delta \overrightarrow{\mathrm{r}}$ vector in a two dimensional Cartesian coordinate system.
Answer:
Given,
Position vectors $\vec{r}_1=3 \hat{i}+4 \hat{j}$
$\vec{r}_1=\hat{i}+2 \hat{j}$
Solution:
Displacement vector:
$
\begin{aligned}
& \Delta \mathrm{r}=\vec{r}_2-\vec{r}_1=(1-3) \hat{i}+(2-4) \hat{j} \\
& \Delta \mathrm{r}=-2 \hat{i}-2 \hat{j}=-2(\hat{i}+\hat{j})
\end{aligned}
$

Question 3.
Calculate the average velocity of the particle whose position vector changes from $\vec{r}_1=5 \hat{i}+6$ $\hat{j}$ to $\vec{r}_2=2 \hat{i}+3 \hat{j}$ in a time 5 seconds.
Answer:
Given.
Position vectors of a particle
$
\begin{aligned}
& \vec{r}_1=5 \hat{i}+6 \hat{j}, \\
& \vec{r}_2=2 \hat{i}+35 \hat{j} \\
& \text { time }(\mathrm{t})=5 \mathrm{~s}
\end{aligned}
$

Solution:
Average velocity:
$
\begin{aligned}
& \overrightarrow{\mathrm{V}}_{\mathrm{ave}}=\frac{\Delta r}{\Delta t}=\frac{\vec{r}_2-\vec{r}_1}{d t}=\frac{(2-5) \hat{i}+(3-6) \hat{j}}{5} \\
& \overrightarrow{\mathrm{V}}_{\mathrm{ave}}=\frac{-3 \hat{i}-3 \hat{j}}{5}=\frac{-3}{5}(\hat{i}+\hat{j})
\end{aligned}
$
Question 4.
Convert the vector $\overrightarrow{\mathrm{r}}=3 \hat{i}+2 \hat{j}$ into a unit vector.
Answer:
Given:
Position vector $\hat{r}=3 \hat{i}+2 \hat{j}$
Solution:
Unit vector $\hat{r}=\frac{\vec{r}}{|r|}$
$
\begin{aligned}
\text { Here } & =|r|=\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13} \\
\hat{r} & =\frac{\vec{r}}{|r|}=\frac{3 i+2 j}{\sqrt{13}}
\end{aligned}
$
Question 5.
What are the resultants of the vector product of two given vectors given by $\vec{A}=4 \hat{i}-2 \hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{B}}=5 \hat{i}+3 \hat{j}-4 \hat{k}$ ?
Answer:
Given,
Vectors $\vec{A}=4 \hat{i}-2 \hat{j}+\hat{k}$
$
\overrightarrow{\mathrm{B}}=5 \hat{i}+3 \hat{j}-4 \hat{k}
$
Solution:
$
\text { Vector product } \begin{aligned}
\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
4 & -2 & 1 \\
5 & 3 & -4
\end{array}\right| \\
& =\hat{i}(8-3)-\hat{j}(-16-5)+\hat{k}(12+10)=5 \hat{i}+21 \hat{j}+22 \hat{k}
\end{aligned}
$

Question 6.
object at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Give,
Horizontal range $=4 \mathrm{H}_{\max }$
Solution:
$
\begin{aligned}
\text { Horizontal range } & =\frac{u^2 \sin 2 \theta}{g}=\frac{2 u^2 \sin \theta \cos \theta}{g} \\
\text { Maximum height } & =\frac{u^2 \sin ^2 \theta}{2 g} \\
\frac{2 u^2 \sin \theta \cos \theta}{g} & =\frac{4 u^2 \sin ^2 \theta}{2 g} \\
2 \cos \theta & =2 \sin \theta \\
\tan \theta & =1 \\
\therefore \quad \theta & =45^{\circ}
\end{aligned}
$

Question 7.
The following graphs represent velocity - time graph. Identity what kind of motion a particle undergoes in each graph.

Answer:
(a) At all the points, slope of the graph is constant.
$
\therefore \overrightarrow{\mathrm{a}}=\text { constant }
$
(b) No change in magnitude of velocity with respect to time
$
\therefore \overrightarrow{\mathrm{v}}=\text { constant }
$
(c) Slope of this graph is greater than graph (a) but constant $\therefore \overrightarrow{\mathrm{a}}=$ constant but greater than the graph (a)
(d) At each point slope of the curve increases. $\therefore \overrightarrow{\mathrm{a}}$ is a variable and object is in accelerated motion.
Question 8.
The following velocity - time graph represents a particle moving in the positive $\mathrm{x}$-direction. Analyse its motion from 0 to $7 \mathrm{~s}$. Calculate the displacement covered and distance travelled
by the particle from 0 to $2 \mathrm{~s}$.

Answer:
As per graph,
(a) From 0 to $1.5 \mathrm{~s}$ the particle moving in a opposite direction.
- From $1.5 \mathrm{~s}$ to $2 \mathrm{~s}$ the particle is moving with increasing velocity.
- From $2 \mathrm{~s}$ to $5 \mathrm{~s}$ velocity of the particle is constant of magnitude $1 \mathrm{~ms}^{-1}$
- From $5 \mathrm{~s}$ to $6 \mathrm{~s}$ velocity of the particle is decreasing.
- From $6 \mathrm{~s}$ to $7 \mathrm{~s}$ the particle is at rest.
(b) Distance covered by the particle - Area covered under (v -t) graph
$
=\frac{1}{2} \times 2 \times 1.5+\frac{1}{2} \times 1 \times 0.5=1.5 \mathrm{~m}+0.25 \mathrm{~m}=1.75 \mathrm{~m}
$
Displacement of the particle
$
\begin{aligned}
& =-\frac{1}{2} \times 2 \times 1.5+\frac{1}{2} \times 1 \times 0.5 \\
& =-1.5 \mathrm{~m} \text { to } 0.25 \mathrm{~m}=-1.25 \mathrm{~m}
\end{aligned}
$
Question 9.
A particle is projected at an angle of $\theta$ with respect to the horizontal direction. Match the following for the above motion.
(a) $v_x$ - decreases and increases
(b) $v_y$ - remains constant
(c) Acceleration - varies
(d) Position vector - remains downward
Answer:
(a) $v_x=$ remains constant
(b) $\mathrm{v}_{\mathrm{y}}=$ decreases and increases
(c) $\mathrm{a}=$ remains downward
(d) $r=$ varies

Question 10.
A water fountain on the ground sprinkles water all around it. If the speed of the water coming out of the fountain is $\mathrm{v}$, calculate the total area around the fountain that gets wet.
Answer:
Given,
Speed of water $=\mathrm{v}$
Solution:
Water comes from a fountain can be taken as projectile and the distance covered is maximum range of projectile i.e. $\theta=45^{\circ}$.
Range of the particle $\left(\mathrm{R}_{\max }\right)=\frac{v^2}{g} \sin 2 \theta=\frac{v^2}{g}$
here, $\mathrm{R}_{\max }$ is radius of the area covered.
$
\pi r^2=\pi \mathrm{R}_{\max }^2=\pi\left(\frac{v^2}{g}\right)^2=\frac{\pi v^4}{g^2}
$
Question 11.
The following table gives the range of a particle when thrown on different planets. All the particles are thrown at the same angle with the horizontal and with the same initial speed. Arrange the planets in ascending order according to their acceleration due to gravity, ( $\mathrm{g}$ value).

Answer:
Range $=\frac{v^2}{g} \sin 2 \theta \therefore \mathrm{g} \alpha \frac{1}{\text { range }}$
Ascending order of the planet with respect to their " $\mathrm{g}$ " is Mercury, Mars, Earth, Jupiter.
Question 12 .
The resultant of two vectors $A$ and $B$ is perpendicular to vector $A$ and its magnitude is equal to half of the magnitude of vector $B$. Then the angle between $A$ and $B$ is
(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $150^{\circ}$
(d) $120^{\circ}$
Answer:
Given:
Resultant of $\vec{A} \& \vec{B}$ is perpendicular to $\vec{A}$ and magnitude of resultant $(C)=\frac{1}{2} \vec{B}$ and $\alpha=$ $90^{\circ}$

Solution:
(i) Magnitude of resultant:
$
\begin{aligned}
\mathrm{C}^2 & =A^2+B^2+2 A B \cos \theta \\
\frac{B^2}{4} & =A^2+B^2+2 A B\left(\frac{A}{B}\right) \\
-\frac{3}{4} B^2 & =3 A^2 \Rightarrow B^2=4 A^2 \text { i.e. } B=2 A
\end{aligned}
$
(ii) direction of resultant:
$
\begin{aligned}
\tan \alpha & =\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \\
\infty & =\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta} \\
\mathrm{A}+\mathrm{B} \cos \theta & =0 \\
\cos \theta & =-\frac{\mathrm{A}}{\mathrm{B}}=-\frac{\mathrm{A}}{2 \mathrm{~A}}=-\frac{1}{2} \\
\cos \theta & =-\frac{1}{2} \quad \theta=\cos ^{-1}\left(-\frac{1}{2}\right) \\
\theta & =120^{\circ}
\end{aligned}
$
Question 13 .
Compare the components for the following vector equations
(a) $\mathrm{T} \hat{j}-\mathrm{mg} \hat{j}=\mathrm{ma} \hat{j}$
(b) $\overrightarrow{\mathrm{T}}+\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}$
(c) $\overrightarrow{\mathrm{T}}-\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}$
(d) $\mathrm{T} \hat{j}+\mathrm{mg} \hat{j}=\mathrm{ma} \hat{j}$
Answer:
Components of the vectors
(a) $\mathrm{T}-\mathrm{mg}=\mathrm{ma}$
(b) $\overline{\mathrm{T}}_x+\overline{\mathrm{F}}_x=\overline{\mathrm{A}}_x+\overline{\mathrm{B}}_x$ (or) $\overline{\mathrm{T}}_y+\overline{\mathrm{F}}_y=\overline{\mathrm{A}}_y+\overline{\mathrm{B}}_y$
(c) $\overline{\mathrm{T}}_x-\overline{\mathrm{F}}_x=\overline{\mathrm{A}}_x+\overline{\mathrm{B}}_x$ (or) $\overline{\mathrm{T}}_y-\overline{\mathrm{F}}_y=\overline{\mathrm{A}}_y+\overline{\mathrm{B}}_y$
(d) $\mathrm{T}+\mathrm{mg}=\mathrm{ma}$

Question 14.
Calculate the area of the triangle for which two of its sides are given by the vectors $\mathrm{A}=5 \hat{i}-3$ $\hat{j}, \mathrm{~B}=4 \hat{i}+6 \hat{j}$
Solution:
Area of the triangle $=\frac{1}{2}|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{A}}|$
$
\begin{aligned}
\overline{\mathrm{A}} \times \overline{\mathrm{B}} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
5 & -5 & 0 \\
4 & 6 & 0
\end{array}\right|=\hat{i}(0)+\hat{j}(0)+\hat{k}(30+12)=42 \hat{k} \\
|\overline{\mathrm{A}} \times \overline{\mathrm{B}}| & =\sqrt{42^2}=42
\end{aligned}
$
$\therefore \quad$ Area of the triangle $=\frac{1}{2}|\overline{\mathrm{A}} \times \overline{\mathrm{B}}|=\frac{1}{2} \times 42=21 \mathrm{~m}^2$
Question 15.
If Earth completes one revolution in 24 hours, what is the angular displacement made by Earth in one hour? Express your answer in both radian and degree.
Answer:
Given,
time period of earth $=24$ hours
Solution:
Earth covers $360^{\circ}$ in 24 hours
$\therefore$ Angular displacement $\mathrm{m} 1$ hour $=\frac{360^{\circ}}{24}=15^{\circ}$ (or) $\frac{\pi}{12}$
Angular displacement in radian $=\frac{15^{-}}{57.295^{\circ}}=0.262 \mathrm{rad}$

Question 16.
An object is thrown with initial speed $5 \mathrm{~ms}^{-1}$ with an angle of projection $30^{\circ}$. What is the height and range reached by the particle?
Answer:
Given,
Initial speed $(\mathrm{u})=5 \mathrm{~ms}^{-1}$
Angle of projection $\theta=30^{\circ}$
Solution:
Max height reached $\left(\mathrm{H}_{\max }\right)=\frac{u^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{25 \times\left(\frac{1}{2}\right)^2}{2 \times 9.8}=0.318 \mathrm{~m}$
Range $(\mathrm{R})=\frac{u^2 \sin 2 \theta}{g}=\frac{25 \times \sin 60^{\circ}}{9.8}=2.21 \mathrm{~m}$
Question 17.
A foot - ball player hits the ball with speed $20 \mathrm{~ms}^{-1}$ with angle $30^{\circ}$ with respect to horizontal
direction as shown in the figure. The goal post is at a distance of $40 \mathrm{~m}$ from him. Find out whether ball reaches the goal post.

Answer:
Given:
Initial speed $(\mathrm{u})=20 \mathrm{~ms}^{-1}$
Angle of projection $(\theta)=30^{\circ}$
The distance of the goal post $=40 \mathrm{~m}$
Solution:
Range of the projectile
$
\mathrm{R}=\frac{u^2 \sin 2 \theta}{g}=\frac{400 \times \sin 60^{\circ}}{9.8}=\frac{400 \times \sqrt{3} / 2}{9.8}=35.35 \mathrm{~m}
$
The distance of goal post is $40 \mathrm{~m}$. But the range of the ball is $35.35 \mathrm{~m}$ only. So ball will not reach the goal post.
Question 18.
If an object is thrown horizontally with an initial speed $10 \mathrm{~ms}^{-1}$ from the top of a building of height $100 \mathrm{~m}$. What is the horizontal distance covered by the particle?
Answer:
Given,
Initial speed $=10 \mathrm{~ms}^{-1}$
Height of the building $(\mathrm{h})=100 \mathrm{~m}$
Range $=$ ?
Solution:
Range of the object $=\mathrm{R}=u \sqrt{\frac{2 h}{g}}=10 \sqrt{\frac{200}{9.8}}=45.1 \mathrm{~m}$
$\mathrm{R}=45 \mathrm{~m}$.
Question 19.
An object is executing uniform circular motion with an angular speed of $\frac{\pi}{12}$ radian per second. At $\mathrm{t}=0$ the object starts at angle $\theta=0$. What is the angular displacement of the particle after $4 \mathrm{~s}$ ?
Answer:

Given:
Angular speed $=\frac{\pi}{12} \mathrm{rad} / \mathrm{sec}$
Solution:
Angular speed $=\frac{\text { Angulardisplacement }}{\text { timetaken }}$
Angular displacement $=\frac{\pi}{12} \times 4=\frac{\pi}{12}=60^{\circ}$
Question 20.
Consider the $\mathrm{x}$-axis as representing east, the $\mathrm{v}$-axis as north and $\mathrm{z}$-axis as vertically upwards. Give the vector representing each of the following points.
(a) $5 \mathrm{~m}$ north east and $2 \mathrm{~m}$ up
(b) $4 \mathrm{~m}$ south east and $3 \mathrm{~m}$ up
(c) $2 \mathrm{~m}$ north west and $4 \mathrm{~m}$ up
Answer:
Given,
Solution:
(a) Length along $X$ - axis $=5 \cos 45^{\circ}=\frac{5}{\sqrt{2}} \mathrm{~m}$
Length along $Y$ - axis $=5 \sin 45^{\circ}=\frac{5}{\sqrt{2}} \mathrm{~m}$
Length along $\mathrm{Z}-$ Axis $=2 \mathrm{~m}$
In vector rotation $=\frac{5}{\sqrt{2}} \hat{i}+\frac{5}{\sqrt{2}} \hat{j}+2 \hat{k}=\frac{5(\hat{i}+\hat{j})}{\sqrt{2}}+2 \hat{k}$
North

(b) Length along $X=4 \cos 45^{\circ}=\frac{4}{\sqrt{2}} \mathrm{~m}$
Length along $\mathrm{Y}=4 \sin 45^{\circ}=\frac{4}{\sqrt{2}} \mathrm{~m}$
Length along Z-axis $=3 \mathrm{~m}$
In vector rotation $=\frac{4}{\sqrt{2}} \hat{i}-\frac{4}{\sqrt{2}} \hat{i}+3 \mathrm{k}=4(\hat{i}-\hat{j}) \sqrt{2}+3 \hat{k}$
(c) Length along $\mathrm{X}=-2 \cos 45^{\circ}=\sqrt{2}+3 \hat{k}=\frac{2}{\sqrt{2}} m=\sqrt{2} m$
Length along $\mathrm{Y}=2 \sin 45^{\circ}=\frac{2}{\sqrt{2}} m=\sqrt{2} m$
length along $Z=4 \mathrm{~m}$
$\therefore$ In vector rotation $=-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+4 \hat{k}$

Question 21.
The Moon is orbiting the Earth approximately once in 27 days, what is the angle transformed by the Moon per day?
Answer:
Given,
period of moon $=27$ days
Solution:
i.e. in 27 days moon covers $360^{\circ}$
In one day angle traversed by moon $=\frac{360^{\circ}}{2 H}=13.3^{\circ}$
Question 22.
An object of mass $\mathrm{m}$ has angular acceleration $\mathrm{a}=0.2 \mathrm{rad} \mathrm{s} 2$. What is the angular displacement covered by the object after 3 second? (Assume that the object started with angle zero with zero angular velocity).
Answer:
Given,
Angular acceleration $=\alpha=0.2 \mathrm{rad} \mathrm{s}^{-2}$
Time $=3 \mathrm{~s}$
Initial velocity $=0$
Solution:
$
\begin{aligned}
& \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\
& \theta=\frac{1}{2} \times 0.2 \times 3^2=0.9 \mathrm{rad} \text { (or) } 51^{\circ} 54
\end{aligned}
$