Vector Short Answer Type Questions And Answers
Question 1. The angle subtended by the vector \(\vec{A}=\sqrt{3} \hat{i}-\hat{j}\) with the y-axis is
- \(60^{\circ}\)
- \(240^{\circ}\)
- \(120^{\circ}\)
- \(45^{\circ}\)
Answer: 3. \(120^{\circ}\)
The option 3 is correct
Question 2. Two non-collinear unit vector \(\hat{a}\) and \(\hat{b}\) are such that \(|\hat{a}+\hat{b}|=\sqrt{3}\). Find the angle between the two unit vectors.
Answer:
⇒ \(|\vec{a}+\vec{b}|=\sqrt{3}\)
or, \(a^2+b^2+2 a b \cos \theta=(\sqrt{3})^2=3\)
or, \(\cos \theta=\frac{3-a^2-b^2}{2 a b}=\frac{3-\left(1^2-1\right)}{2 \cdot 1 \cdot 1}=\frac{1}{2}=\cos 60^{\circ}\)
∴ \(\theta=60^{\circ}\)
Question 3. The ratio between the values of the cross product and the dot product of two vectors is \(\frac{1}{\sqrt{3}}\). The angle between them is
- 30
- 45
- 60
- 120
Answer:
For \(\vec{A}\) and \(\vec{B}\), \(\frac{\text { value of cross product }}{\text { value of dot product }}=\frac{A B \sin \theta}{A B \cos \theta}=\tan \theta\)
∴ \(\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ}\)
Question 4. \((2 \hat{i}+\hat{j}-\hat{k}) \mathrm{N}\) force is acting on a body of \(10 \mathrm{~kg}\) mass. If the body starts from rest, then after \(20 \mathrm{sec}\) what will be its velocity?
Answer:
Acceleration, \(\vec{a}=\frac{\text { force }}{\text { mass }}=\frac{2 \hat{i}+\hat{j}-\hat{k}}{10} \mathrm{~m} \cdot \mathrm{s}^{-2}=\text { constant }\)
So, the particle is moving with uniform acceleration.
After 20s, velocity, \(\vec{v}=\vec{u}+\vec{a} t=0+\frac{2 \hat{i}+\hat{j}-\hat{k}}{10} \cdot 20=(4 \hat{i}+2 \hat{j}-2 \hat{k}) \mathrm{m} \cdot \mathrm{s}^{-1}\)
Magnitude of the velocity, \(v=\sqrt{4^2+2^2+(-2)^2}=\sqrt{24}=2 \sqrt{6} \mathrm{~m} \cdot \mathrm{s}^{-1}\)
Question 5. If \(\vec{A}+\vec{B}+\vec{C}=0\) then show that \(\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A}\)
Answer:
⇒ \(\vec{B} \times \vec{C}=\vec{B} \times(-\vec{A}-\vec{B})=-\vec{B} \times \vec{A}-\vec{B} \times \vec{B}=+\vec{A} \times \vec{B}-0\)
= \(\vec{A} \times \vec{B}\)
Now, \(\vec{C} \times \vec{A}=(-\vec{A} \times-\vec{B}) \times \vec{A}=-\vec{A} \times \vec{A}-\vec{B} \times \vec{A}\)
= \(-0+\vec{A} \times \vec{B}=\vec{A} \times \vec{B}\)
∴ \(\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A}\)
WBBSE Class 11 Vector Short Answer Questions
Question 6. If \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\), then the angle between \(\vec{A}\) and \(\vec{B}\) is
- \(\pi\)
- \(\frac{\pi}{2}\)
- 0
- \(\frac{\pi}{4}\)
Answer:
⇒ \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\)
or, \(A B \cos \theta=A B \sin \theta\) or, \(\tan \theta=1\)
∴ \(\theta=\frac{\pi}{4}\)
The option 4 is correct
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Question 7. \(\vec{a} \times \vec{b}\) is not equal to \(\vec{b} \times \vec{a}\). Why?
Answer:
∴ \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\) both are equal in magnitude but opposite in direction.
Question 8. Find a unit vector that is perpendicular to both \(\vec{A}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{B}=2 \hat{i}-2 \hat{j}+4 \hat{k}\).
Answer:
The unit vector perpendicular to both \(\vec{A}\) and \(\vec{B}\),
∴ \(\hat{n}= \pm \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\)
Now, \(\vec{A} \times \vec{B}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4\end{array}\right|=8 \hat{i}-8 \hat{j}-8 \hat{k}\)
∴ \(|\vec{A} \times \vec{B}|=8 \sqrt{3}\)
∴ \(\hat{n}= \pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})\)
Question 9.
- Determine the relation between the kinetic energy of a projectile at maximum height and at initial position for maximum range.
- What is a null vector?
Answer:
Let the mass of the projectile = m and the velocity of projection = v.
Kinetic energy at the time of projection, E = 1/2 mv².
For maximum horizontal range, the angle of projection, = 45
At highest point, kinetic energy, \(E^{\prime}=\frac{1}{2} m\left(v \cos 45^{\circ}\right)^2=\frac{1}{2} m v^2 \times \frac{1}{2}=\frac{E}{2}\)
A null vector is a vector having a magnitude zero and no fixed direction.
Question 10. Which quantity remains unchanged in the case of a projectile?
- Momentum
- Kinetic energy
- Vertical component of velocity
- Horizontal component of velocity
Answer:
There is no horizontal acceleration of a projectile.
The option 4 is correct.
Vector Addition and Subtraction Short Answers
Question 11. Determine the unit vector along the vector \(\vec{A}=\hat{i}+3 \hat{j}+4 \hat{k} \text {. }\)
Answer:
The unit vector, \(\hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{\hat{i}+3 \hat{j}+4 \hat{k}}{\sqrt{1^2+3^2+4^2}}=\frac{1}{\sqrt{26}} \hat{i}+\frac{3}{\sqrt{26}} \hat{j}+\frac{4}{\sqrt{26}} \hat{k}\)
Question 12. At what angle should the two forces \((\vec{A}+\vec{B})\) and \((\vec{A}-\vec{B})\) act so that their resultant be \(\sqrt{3 A^2+B^2} \text {. }\)
Answer:
If \(\vec{a}=(\vec{A}+\vec{B})\), then \(a^2=\vec{a} \cdot \vec{a}=(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=A^2+B^2+2 \vec{A} \cdot \vec{B}\)
If \(\vec{b}=(\vec{A}-\vec{B})\), similarly \(b^2=\vec{b} \cdot \vec{b}=A^2+B^2-2 \vec{A} \cdot \vec{B}\)
∴ \(\vec{a} \cdot \vec{b}=(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\)
= \(A^2-\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{B}-B^2=A^2-B^2\)
If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\alpha, \vec{a} \cdot \vec{b}=a b \cos \alpha\)
∴ \(a b \cos \alpha=A^2-B^2\)
If \(\vec{c}\) be the resultant of \(\vec{a}\) and \(\vec{b}\),
\(c^2=a^2+b^2+2 a b \cos \alpha or, \quad 3 A^2+B^2=\left(A^2+B^2+2 \vec{A} \cdot \vec{B}\right)\)+ \(\left(A^2+B^2-2 \vec{A} \cdot \vec{B}\right)+2\left(A^2-B^2\right)\)
or, \(3 A^2+B^2=4 A^2\) or, \(A^2-B^2=0\)
Hence, \(a b \cos \alpha=A^2-B^2=0\)
or, \(\cos \alpha=0\) or, \(\alpha=90^{\circ}\)
If the two forces act at an angle \(90^{\circ}\), their resultant will be \(\sqrt{3 A^2+B^2}\).
Question 13. If \(\vec{A}=0.4 \hat{i}+0.3 \hat{j}+c \hat{k}\) be a unit vector, then what is the value of c?
Answer:
∴ \(|\vec{A}|=\sqrt{(0.4)^2+(0.3)^2+c^2}=1\)
or, \(0.16+0.09+c^2=1\) or, \(c=\frac{\sqrt{3}}{2}\)
Question 14. Find the angle between the two vectors \(\vec{A}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{B}=2 \hat{i}+\hat{j}+3 \hat{k}\).
Answer:
⇒ \(\vec{A}= \hat{i}-2 \hat{j}+3 \hat{k}, \vec{B}=2 \hat{i}+\hat{j}+4 \hat{k}\)
⇒ \(\vec{A} \cdot \vec{B}=1 \times 2+(-2) \times 1+3 \times 4=12\)
⇒ \(|\vec{A}|=A=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}\)
⇒ \(|\vec{B}|=B=\sqrt{2^2+1^2+4^2}=\sqrt{21}\)
Now, \(\vec{A} \cdot \vec{B}=A B \cos \theta[\theta\) is the angle between the vectors \(\vec{A}\) and \(\vec{B}\).
or, \(\cos \theta=\frac{\vec{A} \cdot \vec{B}}{A B}=\frac{12}{7 \sqrt{6}}\)=\(\frac{2 \sqrt{6}}{7}\) or, \(\theta=\cos ^{-1}\left(\frac{2 \sqrt{6}}{7}\right)\)
Question 15. What are the quantities that remain constant during the motion of the particle?
Answer:
The horizontal component of the velocity of the particle and its downward acceleration (acceleration due to gravity) remain constant
Question 16. Consider three vectors \(\vec{A}=\hat{i}+\hat{j}-2 \hat{k}, \vec{B}=\hat{i}-\hat{j}+\hat{k}\) and \(\vec{C}=2 \hat{i}-3 \hat{j}+4 \hat{k}\). A vector \(\vec{X}\) of the form \(\alpha \vec{A}+\beta \vec{B}\)(α and β are numbers) is perpendicular to \(\vec{C}\). The ratio of α and β is
- 1: 1
- 2: 1
- -1: 1
- 3: 1
Answer:
∴ \((\alpha \vec{A}+\beta \vec{B}) \cdot \vec{C}=0\)
or, \(2(\alpha+\beta)-3(\alpha-\beta)+4(\beta-2 \alpha)=0\)
or, \(-9 \alpha+9 \beta=0 or, \alpha: \beta=1: 1\)
The option 1 is correct.
Real-Life Examples of Vector Applications
Question 17. A cricket ball thrown across a field is at heights h1 and h2 from the point of projection at times t1 and t2 respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is
- \(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\)
- \(\frac{h_1 t_2^2+h_2 t_2^2}{h_2 t_1+h_1 t_2}\)
- \(\frac{h_1 t_2^2+h_2 t_1^2}{h_1 t_2+h_2 t_1}\)
- \(\frac{h_1 t_1^2-h_2 t_2^2}{h_1 t_1-h_2 t_2}\)
Answer:
⇒ \(h_1=(u \sin \theta) t_1-\frac{1}{2} g t_1^2 ; h_2=(u \sin \theta) t_2-\frac{1}{2} g t_2^2\)
∴ \(\frac{h_1+\frac{1}{2} g t_1^2}{h_2+\frac{1}{2} g t_2^2}=\frac{t_1}{t_2} \quad \text { or, } h_1 t_2-h_2 t_1=\frac{g}{2}\left(t_1 t_2^2-t_1^2 t_2\right)\)
So, time of flight is given by T = \(\frac{2 u \sin \theta}{g}=\frac{2}{g}\left[\frac{h_1+\frac{1}{2} g t_1^2}{t_1}\right]=\frac{2}{t_1}\left[\frac{h_1}{g}+\frac{t_1^2}{2}\right]\)
= \(\frac{h_1}{t_1} \times\left(\frac{t_1 t_2^2-t_1^2 t_2}{h_1 t_2-h_2 t_1}\right)+t_1=\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\)
Question 18. Particle A moves along the x-axis with a uniform velocity of magnitude 10 m/s. Particle B moves with a uniform velocity of 20 m/s along a direction making an angle of 60° with the positive direction of the x-axis as shown The relative velocity of B with respect of that of A is
- 10 m/s along x-axis
- 10√3 m/s along the y-axis (perpendicular to the x-axis)
- 10√5 m/s along the bisection of the velocities of A and B
- 30 m/s along negative x-axis
Answer:
cos60° = \(\frac{1}{2}\) = \(\frac{10}{20}\)
So, the third side of the triangle will be parallel to the y-axis
Length of the third side = \(\sqrt{20^2-10^2}=10 \sqrt{3}\)
Therefore, the relative velocity of B with respect to that of A = 10√3 m/s along the y-axis
The option 2 is correct.
Question 19. The vectors \(\vec{A}\) and \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\).
- 0°
- 60°
- 90°
- 45°
Answer:
Here, \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\)
or, \(A^2+B^2+2 A B \cos \theta=A^2+B^2-2 A B \cos \theta\)
(θ is the angle between \(\vec{A}\) and \(\vec{B}\))
or, \(4 A B \cos \theta=0\) or, \(\cos \theta=0\) or, \(\theta=90^{\circ}\)
The option 3 is correct.
Question 20. Three vectors \(\vec{A}=a \hat{i}+\hat{j}+\hat{k}; \quad \vec{B}=\hat{i}+b \hat{j}+\hat{k}\) and \(\vec{C}=\hat{i}+\hat{j}+c \hat{k}\) are mutually perpendicular (\(\hat{i}, \hat{j} \text { and } \hat{k}\) are unit vectors along X Y and Z axis respectively). The respective values of a, b, and c are
- 0,0,0
- \(-\frac{1}{2}, \frac{-1}{2}, \frac{-1}{2}\)
- 1,-1,1
- \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\)
Answer:
As \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) are perpendicular to each other, \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{C}=\vec{C} \cdot \vec{A}=0\)
a + b + 1 = 1 + b + c = a + 1 + c = 0
a+ b + 1= 0……(1)
1 + b + c = 0…..(2)
a + 1 + c = 0………(3)
Solving equations (1), (2), and (3) we get a = b = c = -1/2
The option 4 is correct.
Step-by-Step Solutions to Vector Problems
Question 21. In a triangle ABC, the sides AB and AC are represented by the vectors \(3 \hat{i}+\hat{j}+\hat{k} \text { and } \hat{i}+2 \hat{j}+\hat{k}\) respectively. Calculate the angle ∠ABC.
- \(\cos ^{-1} \sqrt{\frac{5}{11}}\)
- \(\cos ^{-1} \sqrt{\frac{6}{11}}\)
- \(\left(90^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)\)
- \(\left(180^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)\)
Answer:
Given, \(\overrightarrow{A B}=(3 \hat{i}+\hat{j}+\hat{k}) and \overrightarrow{A C}=(\hat{i}+2 \hat{j}+\hat{k})\)
From the triangle law of vector, \(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\)
or, \(\overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B}=(\hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}+\hat{k})=-2 \hat{i}+\hat{j}\)
∠ABC is the angle between \(\overrightarrow{B A}\) and $\(\overrightarrow{B C}\).
∴ \(\overrightarrow{B A} \cdot \overrightarrow{B C}=|\overrightarrow{B A}||\overrightarrow{B C}| \cos \theta$\)
or, \((6-1)=\sqrt{3^2+1^2+1^2} \times \sqrt{(-2)^2+1^2} \times \cos \theta\)
or, \(\frac{5}{\sqrt{55}}=\cos \theta\) or, \(\theta=\cos ^{-1}\left(\sqrt{\frac{5}{11}}\right)\)
The option 1 is correct
Question 22. A projectile is fired from the surface of the earth with a velocity of 5 m · s-1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3m · s-1 at the same angle follows a trajectory that is identical to the trajectory of the projectile fired from the Earth. The value of the acceleration due to gravity on the planet is (in m · s-2) (given g = 9.8m· s-2)
- 3.5
- 5.9
- 16.3
- 110.8
Answer:
Since trajectories of both cases are identical, \(R_{\max }=R_{\max }^{\prime}\) (Here \(R_{\max }^{\prime}\) is the horizontal range in the second case)
∴ \(R_{\max }=\frac{u_1^2}{g} \sin 2 \theta \text { and } R_{\max }^{\prime}=\frac{u_2^2}{g^{\prime}} \sin 2 \theta\)
So, \(\frac{u_1^2}{g} \sin 2 \theta=\frac{u_2^2}{g^{\prime}} \sin 2 \theta\)
∴ \(g^{\prime}=\frac{u_2^2 g}{u_1^2}=\frac{3^2 \times 9.8}{5^2}=3.5\)
The option 1 is correct.
Question 23. A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2s and (13m, 14m) at time t = 5s. Average velocity vector \(\left(\vec{V}_{\mathrm{av}}\right)\) from t = 0 and t = 5s.
- \(\frac{1}{5}(13 \hat{i}+14 \hat{j})\)
- \(\frac{7}{3}(\hat{i}+\hat{j})\)
- \(2(\hat{i}+\hat{j})\)
- \(\frac{11}{5}(\hat{i}+\hat{j})\)
Answer:
At t=0, \(\vec{r}_1=2 \hat{i}+3 \hat{j}\) and at \(t=5 \mathrm{~s}, \overrightarrow{r_3}=13 \hat{i}+14 \hat{j}\)
∴ \(V_{\mathrm{av}}=\frac{\left(x_2 \hat{i}-x_1 \hat{i}\right)+\left(y_2 \hat{j}-y_1 \hat{j}\right)}{5}\)
= \(\frac{(13-2) \hat{i}+(14-3) \hat{j}}{5}\)
= \(\frac{11 \hat{i}+11 \hat{j}}{5}=\frac{11}{5}(\hat{i}+\hat{j})\)
Option 4 is correct.
Question 24. A ship A is moving westwards with a speed of 10 km · h-1 and ship B, initially 100 km south of A, is moving northwards with a speed of 10 km · h-1. The time after which the distance between them becomes the shortest is
- 0 h
- 5 h
- 5√2 h
- 10√2 h
Answer:
Let the required time be t hours.
If we suppose the x-axis along the eastward direction and the y-axis along the northward direction, then the position of ship A after time t, \(\vec{r}_A=(-10 \hat{i}) t\)
The position of ship B after time t, \(\vec{r}_B=-100 \hat{j}+(10 \hat{j}) t\)
Therefore, the position of B with respect to A, \(\vec{r}_B-\vec{r}_A=(10 t) \hat{i}+(10 t-100) \hat{j}\)
So, \(\left|\vec{r}_B-\vec{r}_A\right|=\sqrt{(10 t)^2+(10 t-100)^2}\)
= \(\sqrt{100 t^2+100 t^2-2000 t+10000}\)
= \(10 \sqrt{2} \sqrt{t^2-10 t+50}\)
This distance becomes the shortest when \(\left(t^2-10 t+50\right)\) becomes minimum, i.e., \(\frac{d}{d t}\left(t^2-10 t+50\right)=0 \quad \text { or, } 2 t-10=0\)
∴ t = 5 hours
The option 2 is correct
Question 25. If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is
- 90°
- 45°
- 180°
- 0°
Answer: 1. 90°
The option 1 is correct.
Question 26. A ball of mass 1 kg is thrown vertically upwards and returns to the ground after 3 seconds. Another ball, thrown at 60° with vertical also stays in air for the same time before it touches the ground. The ratio of the two heights are
- 1:3
- 1:2
- 1:1
- 2:1
Answer:
Both of the balls stay in the air for the same time before touching the ground. Hence, the vertical components of the velocities of the balls along with the heights are equal. So, the ratio of the two heights is 1:1.
The option 3 is correct.
Vector Components and Resolution Questions
Question 27. The angle between \(\vec{A}-\vec{B} \text { and } \vec{A} \times \vec{B} \text { is }(\vec{A} \neq \vec{B})\)
- 60°
- 90°
- 120°
- 45°
Answer:
⇒ \(\vec{A}-\vec{B}\) lies on the same plane of \(\vec{A}\) and \(\vec{B}\). Again, the direction of vector \(\vec{A}\) x \(\vec{B}\) is along the perpendicular to \(\vec{A}\) – \(\vec{B}\).
∴ The angle between \(\vec{A}\)–\(\vec{B}\) and \(\vec{A}\) x \(\vec{B}\) =90°
The option 2 is correct.
Question 28. The moment of the force, \(\vec{F}=4 \hat{i}+5 \hat{j}-6 \hat{k}\) at (2, 0, -3), about the point (2, -2, -2), is given by
- \(-7 \hat{i}-8 \hat{j}-4 \hat{k}\)
- \(-4 \hat{i}-\hat{j}-8 \hat{k}\)
- \(-8 \hat{i}-4 \hat{j}-7 \hat{k}\)
- \(-7 \hat{i}-4 \hat{j}-8 \hat{k}\)
Answer:
⇒ \(\vec{F}=4 \hat{i}+5 \hat{j}-6 \hat{k}\)
⇒ \(\vec{r}=(2 \hat{i}+0 \hat{j}-3 \hat{k})-(2 \hat{i}-2 \hat{j}-2 \hat{k})=0 \hat{i}+2 \hat{j}-\hat{k}\)
⇒ \(\vec{\tau}=\vec{r} \times \vec{F}=(2 \hat{j}-\hat{k}) \times(4 \hat{i}+5 \hat{j}-6 \hat{k})\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 2 & -1 \\
4 & 5 & -6
\end{array}\right|=-7 \hat{i}-4 \hat{j}-8 \hat{k}\)
The option 4 is correct.
Question 29. A body is projected horizontally from the top of a building of height h. The velocity of projection is u. Find
- The time it will take to reach the ground,
- The horizontal distance between the foot of the building and the ground where it will strike,
- The velocity with which the body will reach the ground.
Answer:
u = initial velocity at the point A
So, u = horizontal component of initial velocity,
0 = vertical component of the initial velocity.
The horizontal component, u = constant, since there is no acceleration in that direction. But the vertical motion is under a constant acceleration g, the acceler¬ation due to gravity.
Let B be the point where the body strikes the ground.
The corresponding vertical motion is through the distance AO = h.
1. For the vertical motion AO, h = \(0 \cdot t+\frac{1}{2} g t^2 \quad \text { or, } t=\sqrt{\frac{2 h}{g}}\)
As it corresponds to the actual of flight is, \(t=\sqrt{\frac{2 h}{g}}\)
2. Horizontal range, OB = uniform horizontal velocity x time of flight.
or, \(R=u \sqrt{\frac{2 h}{g}}\)
3. After a time t, the horizontal component of velocity = u, and the vertical component of velocity = 0 + gt = gt.
∴ The resultant velocity at B = \(\sqrt{u^2+(g t)^2}=\sqrt{u^2+g^2 \frac{2 h}{g}}=\sqrt{u^2+2 g h}\)
If it makes an angle θ with the horizontal, then \(\theta=\tan ^{-1}\left(\frac{g t}{u}\right)\)
Common Vector Questions for Class 11
Question 30. Explain why it is easier to pull a lawn mower than to push it.
Answer:
The force F is usually applied obliquely relative to the ground, during either a push or a pull of the lawn mower. In both cases, force F is resolved into two components, one is horizontal and the other is vertical.
The horizontal component H of the force F is responsible for the horizontal motion. The difference is, during the push the vertical component V increases the effective weight of the mower, whereas during the pull the effective weight decreases due to V. So it is easier to pull the mower than to push it.
Question 31. On a two-lane road, car A is traveling at a speed of 36 km · h-1. Two cars B and C approach car A from opposite directions with speeds of 54 km · h-1 each. At a certain instant, when both cars B and C are at a distance of 1 km from A, B decides to overtake car A before C does. What minimum acceleration is required of B to avert an accident?
Answer:
Velocity of car A, vA = 36km · h-1 = 10m · s-1
Let vB and vC be the velocities of cars B and C.
∴ vB = vC = 54 km · h-1 = 15 m · s-1
Velocity of car B relative to A, VBA = VB – VA = 15-10 = 5 m · s-1
Velocity of car C relative to A, VCA = vC + (vA) = VC+VA= 15 + 10 = 25 m · s-1
Also, the distance of B and C from A is 1 km = 100 m. Let t = time taken by car C to travel a distance of 1 km towards A.
∴ S = vCAt or, 1000 = 25 for, t = 40 s
Suppose a = minimum acceleration required for car B to avoid the accident.
With this acceleration, car B overtakes car A in 40 s.
So, \(S=v_{B A} t+\frac{1}{2} a t^2\) or, \(1000=5 \times 40+\frac{1}{2} a \times(40)^2\)
∴ a = \(\frac{(1000-200) \times 2}{(40)^2}=1 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Question 32. Find the condition for which \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}| ?\)
Answer:
⇒ \(|\vec{A}+\vec{B}|^2=|\vec{A}-\vec{B}|^2\)
or, \(A^2+B^2+2 A B \cos \theta=A^2+B^2-2 A B \cos \theta\)
or, \(4 A B \cos \theta=0\) or, \(\cos \theta=0=\cos 90^{\circ}\) or, \(\theta=90^{\circ}\)
∴ The condition is, \(\vec{A} \perp \vec{B}\)
Key Terms in Vectors Explained
Question 33. Two parallel rail tracks run north-south. Train A moves north at a speed of 54 km/h, and Train B moves south at a speed of 90 km/h. What is the
- The velocity of B with respect to A
- The velocity of ground with respect to B
- Velocity of a monkey, running on the roof of train A against its motion with a velocity of 18 km/h with respect to the train, as observed by a man standing on the ground.
Answer:
Let us consider the south-to-north direction to be positive.
Here, VA = +54 km/h = 15 m/s
vB = -90 km/h = -25 m/s
1. Velocity of B with respect to A, VBA = VB – VA = -25-15 = -40 m/s towards south
2. Velocity of ground with respect to B = 0-vB =0-(-25) =25 m/s towards north
3. Let the velocity of the monkey with respect to the ground be vm.
So, the velocity of monkey with respect to A = vm– vA =-18 km/h =-5 m/s
∴ vm = VA -5 = 15-5 =10 m/s towards north.
Question 34. Is it possible to have a constant rate of change of velocity when velocity changes both in magnitude and direction?
Answer:
Yes, for example—a body moving upwards or downwards where acceleration is constant while magnitude and direction change.
Question 35. What are two angles of projection of a projectile projected with a velocity 30 m/s, so that the horizontal range is 45m? Take g = 10 m/s².
Answer:
Horizontal range, R = \(\frac{u^2 \sin 2 \theta}{g}\)
or, \(\sin 2 \theta=R \cdot g / u^2\)
or, \(\sin 2 \theta=\frac{R \cdot g}{30^2}=\frac{45 \times 10}{30 \times 30}=\frac{1}{2}=\sin 30^{\circ}\)
∴ \(\theta=15^{\circ}\) and \(\left(90^{\circ}-15^{\circ}\right)=75^{\circ}\)
Question 36. If a projectile has a constant initial speed and angle of projection, find the relation between the changes in the horizontal range due to a change in acceleration due to gravity.
Answer:
R = \(\frac{u^2 \sin 2 \theta}{g}\)
⇒ \(R_1=\frac{u^2 \sin 2 \theta}{g_1} ; R_2=\frac{u^2 \sin 2 \theta}{g_2}\)
∴ \(\frac{R_1}{R_2}=\frac{g_2}{g_1}\)
Hence, the horizontal range is inversely proportional to the acceleration due to gravity.
Question 37. An old man walks 10 m due east from his house and then turns to his left at an angle of 60° east. He then walked 10 m in that direction fell down on the ground and got injured. His grandson observing him moves straight towards him from the initial position of his grandfather, helps him to stand, and takes him safely home,
- Should the boy follow the same path followed by the old man? If not, why?
- What are the values you suggest for the boy’s reply?
Answer:
Boy should take the shortest path, i.e., direct from A to C.
Caring, by holding and supporting his grandfa¬ther from his shoulders,
- Call nearby people for help,
- Gives first aid to his grandfather at home,
- Taking grandfather to hospital.